<TEI xmlns="http://www.tei-c.org/ns/1.0">
   <teiHeader>
      <fileDesc>
         <titleStmt>
            <title>The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed</title>
            <title>Elements. English</title>
            <author>Euclid.</author>
         </titleStmt>
         <editionStmt>
            <edition>
               <date>1570</date>
            </edition>
         </editionStmt>
         <extent>Approx. 3245 KB of XML-encoded text transcribed from 494 1-bit group-IV TIFF page images.</extent>
         <publicationStmt>
            <publisher>Text Creation Partnership,</publisher>
            <pubPlace>Ann Arbor, MI ; Oxford (UK) :</pubPlace>
            <date when="2003-01">2003-01 (EEBO-TCP Phase 1).</date>
            <idno type="DLPS">A00429</idno>
            <idno type="STC">STC 10560</idno>
            <idno type="STC">ESTC S106699</idno>
            <idno type="EEBO-CITATION">99842412</idno>
            <idno type="PROQUEST">99842412</idno>
            <idno type="VID">7062</idno>
            <availability>
               <p>This keyboarded and encoded edition of the
	       work described above is co-owned by the institutions
	       providing financial support to the Early English Books
	       Online Text Creation Partnership. This Phase I text is
	       available for reuse, according to the terms of <ref target="https://creativecommons.org/publicdomain/zero/1.0/">Creative
	       Commons 0 1.0 Universal</ref>. The text can be copied,
	       modified, distributed and performed, even for
	       commercial purposes, all without asking permission.</p>
            </availability>
         </publicationStmt>
         <seriesStmt>
            <title>Early English books online.</title>
         </seriesStmt>
         <notesStmt>
            <note>(EEBO-TCP ; phase 1, no. A00429)</note>
            <note>Transcribed from: (Early English Books Online ; image set 7062)</note>
            <note>Images scanned from microfilm: (Early English books, 1475-1640 ; 342:05)</note>
         </notesStmt>
         <sourceDesc>
            <biblFull>
               <titleStmt>
                  <title>The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed</title>
                  <title>Elements. English</title>
                  <author>Euclid.</author>
                  <author>Dee, John, 1527-1608.</author>
                  <author>Candale, François de Foix, comte de, 1502-1594.</author>
                  <author>Billingsley, Henry, Sir, d. 1606.</author>
               </titleStmt>
               <extent>[28], 203, 205-464, [1] leaves, folded table : ill.  </extent>
               <publicationStmt>
                  <publisher>By Iohn Daye,</publisher>
                  <pubPlace>Imprinted at London :</pubPlace>
                  <date>[1570 (3 Feb.]]</date>
               </publicationStmt>
               <notesStmt>
                  <note>With a sixteenth book added by François de Foix, comte de Candale.</note>
                  <note>Colophon reads: At London printed by Iohn Daye, dvvelling ouer Aldersgate beneath Saint Martins. These bookes are to be solde at his shop vnder the gate. 1570.</note>
                  <note>The folded table, dated 25 Feb. 1570 and showing the topics covered in Dee's preface, gives the date of the book's publication as 3 Feb.</note>
                  <note>Book 11 has onlays intended to be pasted at one edge over the illustrations. These were printed together on six bifolia. Variant: these bifolia are still intact and bound with book.</note>
                  <note>The last leaf is blank.</note>
                  <note>Reproduction of the original in the Henry E. Huntington Library and Art Gallery.</note>
               </notesStmt>
            </biblFull>
         </sourceDesc>
      </fileDesc>
      <encodingDesc>
         <projectDesc>
            <p>Created by converting TCP files to TEI P5 using tcp2tei.xsl,
      TEI @ Oxford.
      </p>
         </projectDesc>
         <editorialDecl>
            <p>EEBO-TCP is a partnership between the Universities of Michigan and Oxford and the publisher ProQuest to create accurately transcribed and encoded texts based on the image sets published by ProQuest via their Early English Books Online (EEBO) database (http://eebo.chadwyck.com). The general aim of EEBO-TCP is to encode one copy (usually the first edition) of every monographic English-language title published between 1473 and 1700 available in EEBO.</p>
            <p>EEBO-TCP aimed to produce large quantities of textual data within the usual project restraints of time and funding, and therefore chose to create diplomatic transcriptions (as opposed to critical editions) with light-touch, mainly structural encoding based on the Text Encoding Initiative (http://www.tei-c.org).</p>
            <p>The EEBO-TCP project was divided into two phases. The 25,363 texts created during Phase 1 of the project have been released into the public domain as of 1 January 2015. Anyone can now take and use these texts for their own purposes, but we respectfully request that due credit and attribution is given to their original source.</p>
            <p>Users should be aware of the process of creating the TCP texts, and therefore of any assumptions that can be made about the data.</p>
            <p>Text selection was based on the New Cambridge Bibliography of English Literature (NCBEL). If an author (or for an anonymous work, the title) appears in NCBEL, then their works are eligible for inclusion. Selection was intended to range over a wide variety of subject areas, to reflect the true nature of the print record of the period. In general, first editions of a works in English were prioritized, although there are a number of works in other languages, notably Latin and Welsh, included and sometimes a second or later edition of a work was chosen if there was a compelling reason to do so.</p>
            <p>Image sets were sent to external keying companies for transcription and basic encoding. Quality assurance was then carried out by editorial teams in Oxford and Michigan. 5% (or 5 pages, whichever is the greater) of each text was proofread for accuracy and those which did not meet QA standards were returned to the keyers to be redone. After proofreading, the encoding was enhanced and/or corrected and characters marked as illegible were corrected where possible up to a limit of 100 instances per text. Any remaining illegibles were encoded as &lt;gap&gt;s. Understanding these processes should make clear that, while the overall quality of TCP data is very good, some errors will remain and some readable characters will be marked as illegible. Users should bear in mind that in all likelihood such instances will never have been looked at by a TCP editor.</p>
            <p>The texts were encoded and linked to page images in accordance with level 4 of the TEI in Libraries guidelines.</p>
            <p>Copies of the texts have been issued variously as SGML (TCP schema; ASCII text with mnemonic sdata character entities); displayable XML (TCP schema; characters represented either as UTF-8 Unicode or text strings within braces); or lossless XML (TEI P5, characters represented either as UTF-8 Unicode or TEI g elements).</p>
            <p>Keying and markup guidelines are available at the <ref target="http://www.textcreationpartnership.org/docs/.">Text Creation Partnership web site</ref>.</p>
         </editorialDecl>
         <listPrefixDef>
            <prefixDef ident="tcp"
                       matchPattern="([0-9\-]+):([0-9IVX]+)"
                       replacementPattern="http://eebo.chadwyck.com/downloadtiff?vid=$1&amp;page=$2"/>
            <prefixDef ident="char"
                       matchPattern="(.+)"
                       replacementPattern="https://raw.githubusercontent.com/textcreationpartnership/Texts/master/tcpchars.xml#$1"/>
         </listPrefixDef>
      </encodingDesc>
      <profileDesc>
         <langUsage>
            <language ident="eng">eng</language>
         </langUsage>
         <textClass>
            <keywords scheme="http://authorities.loc.gov/">
               <term>Geometry --  Early works to 1800.</term>
            </keywords>
         </textClass>
      </profileDesc>
      <revisionDesc>
         <change>
            <date>2000-00</date>
            <label>TCP</label>Assigned for keying and markup</change>
         <change>
            <date>2002-01</date>
            <label>SPi Global</label>Keyed and coded from ProQuest page images</change>
         <change>
            <date>2002-03</date>
            <label>TCP Staff (Michigan)</label>Sampled and proofread</change>
         <change>
            <date>2002-03</date>
            <label>Mona Logarbo</label>Text and markup reviewed and edited</change>
         <change>
            <date>2002-04</date>
            <label>pfs</label>Batch review (QC) and XML conversion</change>
      </revisionDesc>
   </teiHeader>
   <text xml:lang="unk">
      <front>
         <div type="title_page">
            <pb facs="tcp:7062:1" rendition="simple:additions"/>
            <pb facs="tcp:7062:1" rendition="simple:additions"/>
            <figure>
               <head>VIRESCIT VULNERE VERITAS</head>
            </figure>
            <p>THE ELEMENTS OF GEOMETRIE of the most aunci<g ref="char:EOLhyphen"/>ent Philosopher EVCLIDE of Megara.</p>
            <p>Faithfully (now first) tran<g ref="char:EOLhyphen"/>slated into the Englishe toung, by <hi>H. Billingsley,</hi> Citizen of London. Whereunto are annexed certaine Scholics, Annotations, and Inuenti<g ref="char:EOLhyphen"/>ons, of the best Mathematici<g ref="char:EOLhyphen"/>ans, both of time past, and i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> this our age.</p>
            <p>With a very fruitfull Praeface made by <hi>M. I. Dee,</hi> specifying the chiefe Mathematicall Scie<g ref="char:cmbAbbrStroke">̄</g>ces, what they are, and wherunto commodious<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> where, also, are disclosed certaine new Secrets. Mathematicall and Mech<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>icall, vntill these our daies greatly missed:</p>
            <p>Imprinted at London by <hi>Iohn Daye.</hi>
            </p>
         </div>
         <div type="translator_to_the_reader">
            <pb facs="tcp:7062:2"/>
            <pb facs="tcp:7062:2"/>
            <head>The Translator to the Reader.</head>
            <p>THere is (gentle Reader) nothing (the word of God onely set apart) which so much beautifieth and a<g ref="char:EOLhyphen"/>dorneth the soule and minde of ma<g ref="char:cmbAbbrStroke">̄</g>, as doth the knowledge of good artes and sciences: as the know<g ref="char:EOLhyphen"/>ledge of naturall and morall Phi<g ref="char:EOLhyphen"/>losophie. The one setteth before our eyes, the creatures of God, both in the heauens aboue, and in the earth beneath: in which as in a glasse, we beholde the exceding maiestie and wisedome of God, in adorning and beautifying them as we see: in geuing vn<g ref="char:EOLhyphen"/>to them such wonderfull and manifolde proprieties, and naturall workinges, and that so diuersly and in such varietie: farther in maintaining and conseruing them continually, whereby to praise and adore him, as by S. Paule we are taught. The other tea<g ref="char:EOLhyphen"/>cheth vs rules and preceptes of vertue, how, in common life a<g ref="char:EOLhyphen"/>mongest men, we ought to walke vprightly: what dueties per<g ref="char:EOLhyphen"/>taine to our selues, what pertaine to the gouernment or good or<g ref="char:EOLhyphen"/>der both of an housholde, and also of a citie or common wealth. The reading likewise of histories, conduceth not a litle, to the ad<g ref="char:EOLhyphen"/>orning of the soule &amp; minde of man, a studie of all men co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>ded: by it are seene and knowen the artes and doinges of infinite wise men gone before vs. In histories are contained infinite ex<g ref="char:EOLhyphen"/>amples of heroicall vertues to be of vs followed, and horrible ex<g ref="char:EOLhyphen"/>amples of vices to be of vs eschewed. Many other artes also there are which beautifie the minde of man: but of all other none do more garnishe &amp; beautifie it, then those artes which are cal<g ref="char:EOLhyphen"/>led Mathematicall. Vnto the knowledge of which no man can attaine, without the perfecte knowledge and instruction of the principles, groundes, and Elementes of Geometrie. But per<g ref="char:EOLhyphen"/>fectly
<pb facs="tcp:7062:3"/>
to be instructed in them, requireth diligent studie and rea<g ref="char:EOLhyphen"/>ding of olde auncient authors. Amongest which, none for a be<g ref="char:EOLhyphen"/>ginner is to be preferred before the most auncient Philosopher <hi>Euclide</hi> of <hi>Megara.</hi> For of all others he hath in a true me<g ref="char:EOLhyphen"/>thode and iuste order, gathered together whatsoeuer any before him had of these Elementes written: inuenting also and adding many thinges of his owne: wherby he hath in due forme accom<g ref="char:EOLhyphen"/>plished the arte: first geuing definitions, principles, &amp; groundes, wherof he deduceth his Propositions or conclusions, in such won<g ref="char:EOLhyphen"/>derfull wise, that that which goeth before, is of necessitie requi<g ref="char:EOLhyphen"/>red to the proufe of that which followeth. So that without the diligent studie of <hi>Euclides</hi> Elementes, it is impossible to attaine vnto the perfecte knowledge of Geometrie, and consequently of any of the other Mathematicall sciences. Wherefore conside<g ref="char:EOLhyphen"/>ring the want &amp; lacke of such good authors hitherto in our Eng<g ref="char:EOLhyphen"/>lishe tounge, lamenting also the negligence, and lacke of zeale to their countrey in those of our nation, to whom God hath geuen both knowledge, &amp; also abilitie to translate into our tounge, and to publishe abroad such good authors, and bookes (the chiefe in<g ref="char:EOLhyphen"/>strumentes of all learninges): seing moreouer that many good wittes both of gentlemen and of others of all degrees, much de<g ref="char:EOLhyphen"/>sirous and studious of these artes, and seeking for them as much as they can, sparing no paines, and yet frustrate of their intent, by no meanes attaining to that which they seeke: I haue for their sakes, with some charge &amp; great trauaile, faithfully tran<g ref="char:EOLhyphen"/>slated into our vulgare tou<g ref="char:cmbAbbrStroke">̄</g>ge, &amp; set abroad in Print, this booke of <hi>Euclide.</hi> Whereunto I haue added easie and plaine decla<g ref="char:EOLhyphen"/>rations and examples by figures, of the definitions. In which booke also ye shall in due place finde manifolde additions, Scho<g ref="char:EOLhyphen"/>lies, Annotations, and Inuentions: which I haue gathered out of many of the most famous &amp; chiefe Mathematicies, both of old time, and in our age: as by diligent reading it in course, ye shall
<pb facs="tcp:7062:3"/>
well perceaue. The fruite and gaine which I require for these my paines and trauaile, shall be nothing els, but onely that thou gentle reader, will gratefully accept the same: and that thou mayest thereby receaue some profite: and moreouer to excite and stirre vp others learned, to do the like, &amp; to take paines in that behalfe. By meanes wherof, our Englishe tounge shall no lesse be enriched with good Authors, then are other straunge tounges: as the Dutch, French, Italian, and Spanishe: in which are red all good authors in a manner, found amongest the Grekes or Latines. Which is the chiefest cause, that amongest the<g ref="char:cmbAbbrStroke">̄</g> do flo<g ref="char:EOLhyphen"/>rishe so many cunning and skilfull men, in the inuentions of straunge and wonderfull thinges, as in these our daies we see there do. Which fruite and gaine if I attaine vnto, it shall encourage me hereafter, in such like sort to translate, and set abroad some other good authors, both pertaining to religion (as partly I haue already done) and also pertaining to the Mathe<g ref="char:EOLhyphen"/>maticall Artes. Thus gentle reader farewell.</p>
         </div>
         <div type="preface">
            <pb facs="tcp:7062:4"/>
            <pb facs="tcp:7062:4"/>
            <head>TO THE VNFAINED LOVERS of truthe, and constant Studentes of Noble Sciences, <hi>IOHN DEE</hi> of London, hartily wisheth grace from heauen, and most prospe<g ref="char:EOLhyphen"/>rous successe in all their honest attemptes and exercises.</head>
            <p>DIuine <hi>Plato,</hi> the great Master of many worthy Philosophers, and the constant auoucher, and pithy perswader of <hi>Vnum, Bo<g ref="char:EOLhyphen"/>num,</hi> and <hi>Ens:</hi> in his Schole and Academie, sundry times (besides his ordinary Scholers) was visited of a certaine kinde of men, allured by the noble fame of <hi>Plato,</hi> and the great commendation of hys profound and profitable doctrine. But when such Hearers, after long harkening to him, perceaued, that the drift of his discourses issued out, to conclude, this <hi>Vnum, Bo<g ref="char:EOLhyphen"/>num,</hi> and <hi>Ens,</hi> to be Spirituall, Infi<g ref="char:EOLhyphen"/>nite, AEternall, Omnipotent, &amp;c. Nothyng beyng alledged or expressed, How, worldly goods: how, worldly digni<g ref="char:EOLhyphen"/>tie: how, health, Stre<g ref="char:cmbAbbrStroke">̄</g>gth or Iustines of body: nor yet the meanes, how a merue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>lous sensible and bodyly blysse and felicitie hereafter, might be atteyned: Straightway, the fantasies of those hearers, were dampt: their opinion of <hi>Plato,</hi> was clene chaun<g ref="char:EOLhyphen"/>ged: yea his doctrine was by them despised: and his schole, no more of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hem visi<g ref="char:EOLhyphen"/>ted. <q>Which thing, his Scholer, <hi>Aristotle,</hi> narrowly co<g ref="char:cmbAbbrStroke">̄</g>sidering, founde the cause ther<g ref="char:EOLhyphen"/>of, to be, For that they had no forwarnyng and information, in generall, whereto his doctrine tended.</q> For, so, might they haue had occasion, either to haue forborne his schole hauntyng: (if they, then, had misliked his Scope and purpose) or con<g ref="char:EOLhyphen"/>stantly to haue continued therin: to their full satisfaction: if such finall scope &amp; intent, had ben to their desire.
<q>Wherfore, <hi>Aristotle,</hi> euer, after that, vsed in brief, to forewarne his owne Scholers and hearers, both of what matter, and also to what ende, he tooke in hand to speake, or teach. While I consider the diuerse trades of these two excellent Philosophers (and am most sure, both, that <hi>Plato</hi> right well, o<g ref="char:EOLhyphen"/>therwise could teach:</q>
and that <hi>Aristotle</hi> mought boldely, with his hearers, haue dealt in like sorte as <hi>Plato</hi> did) I am in no little pang of perplexitie: Bycause, that, which I mislike, is most easy for me to performe (and to haue <hi>Plato</hi> for my exa<g ref="char:cmbAbbrStroke">̄</g>ple.) And that, which I know to be most commendable: and (in this first bringyng, into common handling, the <hi>Artes Mathematicall</hi>) to be most necessary: is full of great difficultie and sundry daungers. Yet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> neither do I think it me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e, for so straunge mat<g ref="char:EOLhyphen"/>ter (as now is ment to be published) and to so straunge an audience, to be bluntly, at first, put forth, without a peculiar Preface: Nor (Imitatyng <hi>Aristotle</hi>) well can I hope, that accordyng to the amplenes and dignitie of the <hi>State Mathematicall,</hi> I am able, either playnly to prescribe the materiall boundes: or precisely to expresse the chief purposes, and most wonderfull applications therof. And though I am sure, that such as did shrinke from <hi>Plato</hi> his s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hole, after they had perc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>iued his fi<g ref="char:EOLhyphen"/>nall
<pb facs="tcp:7062:5"/>
conclusion, would in these thinges haue ben his most diligent hearers) so infi<g ref="char:EOLhyphen"/>nitely mought their desires, in fine and at length, by our <hi>Artes Mathematicall</hi> be sa<g ref="char:EOLhyphen"/>tisfied) yet, by this my Praeface &amp; forewarnyng, Aswell all such, may (to their great behofe) the soner, hither be allured: as also the <hi>Pythagoricall,</hi> and <hi>Platonicall</hi> perfect scholer, and the constant profound Philosopher, with more ease and spede, may (like the Bee,) gather, hereby, both wax and hony.</p>
            <p>
               <q>Wherfore, seyng I finde great occasion (for the causes alleged, and farder, in re<g ref="char:EOLhyphen"/>spect of my <hi>Art Mathematike generall</hi>) to vse a certaine forewarnyng and Praeface, whose content shalbe, that mighty, most plesaunt, and frutefull <hi>Mathematicall Tree,</hi> with his chief armes and second (grifted) braunches:<note place="margin">
                     <hi>The intent of this Preface.</hi>
                  </note> Both, what euery one is, and also, what commodity, in generall, is to be looked for, aswell of griff as stocke: And forasmuch as this enterprise is so great, that, to this our tyme, it neuer was (to my knowledge) by any achieued: And also it is most hard, in these our drery dayes, to such rare and straunge Artes, to wyn due and common credit:</q>
Neuertheles, if, for my sincere endeuour to satisfie your honest expectation, you will but lend me your tha<g ref="char:cmbAbbrStroke">̄</g>kefull mynde a while: and, to such matter as, for this time, my penne (with spede) is hable to deliuer, apply your eye or eare attentifely: perchaunce, at once, and for the first salutyng, this Preface you will finde a lesson long enough. And ei<g ref="char:EOLhyphen"/>ther you will, for a second (by this) be made much the apter: or shortly become, well hable your selues, of the lyons claw, to coniecture his royall symmetrie, and farder propertie. Now then, gentle, my frendes, and countrey men, Turne your eyes, and bend your myndes to that doctrine, which for our present purpose, my simple talent is hable to yeld you.</p>
            <p>All thinges which are, &amp; haue beyng, are found vnder a triple diuersitie generall.</p>
            <p>For, either, they are demed Supernaturall, Naturall, or, of a third being. Thinges Supernaturall, are immateriall, simple, indiuisible, incorruptible, &amp; vnchangeable. Things Naturall, are materiall, compounded, diuisible, corruptible, and chaungea<g ref="char:EOLhyphen"/>ble. Thinges Supernaturall, are, of the minde onely, comprehended: Things Natu<g ref="char:EOLhyphen"/>rall, of the sense exterior, ar hable to be perceiued. In thinges Naturall, probabilitie and coniecture hath place: But in things Supernaturall, chief demo<g ref="char:cmbAbbrStroke">̄</g>stration, &amp; most sure Science is to be had. By which properties &amp; comparasons of these two, more easily may be described, the state, condition, nature and property of those thinges, which, we before termed of a third being: which, by a peculier name also, are called <hi>Thynges Mathematicall.</hi> For, these, beyng (in a maner) middle, betwene thinges su<g ref="char:EOLhyphen"/>pernaturall and naturall: are not so absolute and excellent, as thinges supernatural: Nor yet so base and grosse, as things naturall: But are thinges immateriall: and ne<g ref="char:EOLhyphen"/>uerthelesse, by materiall things hable somewhat to be signified. And though their particular Images, by Art, are agg<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>egable and diuisible: yet the generall <hi>Formes,</hi> notwithstandyng, are constant, vnchaungeable, vntra<g ref="char:cmbAbbrStroke">̄</g>sformable, and incorruptible. Neither of the sense, can they, at any tyme, be perceiued or iudged. Nor yet for all that, in the royall mynde of man, first conceiued. But, surmountyng the imperfectio<g ref="char:cmbAbbrStroke">̄</g> of coniecture, weenyng and opinion: and commyng short of high intellectuall co<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>ceptio<g ref="char:cmbAbbrStroke">̄</g>, are the Mercurial fruite of <hi>Dianoeticall</hi> discourse, in perfect imagination sub<g ref="char:EOLhyphen"/>sistyng. A meruaylous newtralitie haue these thinges <hi>Mathematicall.</hi> and also a straunge participatio<g ref="char:cmbAbbrStroke">̄</g> betwene thinges supernaturall, immortall, intellectual, simple and indiuisible: and thynges naturall, mortall, sensible, compounded and diuisible. Probabilitie and sensible profe, may well serue in thinges naturall: and is commen<g ref="char:EOLhyphen"/>dable: In Mathematicall reasoninges, a probable Argument, is nothyng regarded: nor yet the testimony of sense, any whit credited: But onely a perfect demonstra<g ref="char:EOLhyphen"/>tion, of truthes certaine, necessary, and inuincible: vniuersally and necessaryly con<g ref="char:EOLhyphen"/>cluded:
<pb facs="tcp:7062:5"/>
is allowed as sufficient for an Argument exactly and purely Mathematical."</p>
            <p>Of <hi>Mathematicall</hi> thinges, are two principall kinde<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> namely, <hi>Number,</hi>
               <note place="margin">
                  <hi>Number.</hi>
               </note> and <hi>Mag<g ref="char:EOLhyphen"/>nitude. Number,</hi> we define, to be, a certayne Mathematicall Su<g ref="char:cmbAbbrStroke">̄</g>me, of <hi>Vnits.</hi>
               <note place="margin">
                  <hi>Note the worde, Vnit, to expresse the Greke Mona<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, &amp; not Vnitie: as we hau<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> all, commonly, till now, vsed.</hi>
               </note> And, an <hi>Vnit,</hi> is that thing Mathematicall, Indiuisible, by participation of some likenes of whose property, any thing, which is in de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>de, or is counted One, may resonably be called One. We account an <hi>Vnit,</hi> a thing <hi>Mathematicall,</hi> though it be no <hi>Number,</hi> and also indiuisible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> because, of it, materially, Number doth consist<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> which, princi<g ref="char:EOLhyphen"/>pally, is a thing <hi>Mathematicall. Magnitude</hi> is a thing <hi>Mathematicall,</hi>
               <note place="margin">
                  <hi>Magnitude.</hi>
               </note> by participation of some likenes of whose nature, any thing is iudged long, broade, or thicke. A thicke <hi>Magnitude</hi> we call a <hi>Solide,</hi> or a <hi>Body.</hi> What <hi>Magnitude</hi> so euer, is Solide or" Thicke, is also broade, &amp; long. A broade magnitude, we call a <hi>Superficies</hi> or a Plaine. Euery playne magnitude, hath also length. A long magnitude, we terme a <hi>Line.</hi> A <hi>Line</hi> is neither thicke nor broade, but onely long: Euery certayne Line, hath two endes: The endes of a line, are <hi>Pointes</hi> called.<note place="margin">
                  <hi>A point.</hi>
               </note> A <hi>Point,</hi> is a thing <hi>Mathematicall,</hi> indi<g ref="char:EOLhyphen"/>uisible, which may haue a certayne determined situation. If a Poynt moue from a" determined situation, the way wherein it moued, is also a <hi>Line</hi>: mathematically produced. whereupon, of the auncient Mathematiciens, a <hi>Line</hi> is called the race or course of a <hi>Point.</hi>
               <note place="margin">
                  <hi>A Line.</hi>
               </note> A Poynt we define, by the name of a thing Mathematicall: though it be no Magnitude, and indiuisible: because it is the propre ende, and bound of a <hi>Line</hi>: which is a true <hi>Magnitude.</hi>
               <note place="margin">
                  <hi>Magnitude.</hi>
               </note> And <hi>Magnitude</hi> we may define to be that thing <hi>Mathematicall,</hi> which is diuisible for euer, in partes diuisible, long, broade or thicke. Therefore though a Poynt be no <hi>Magnitude,</hi> yet <hi>Terminatiuely</hi> we rec<g ref="char:EOLhyphen"/>ken it a thing <hi>Mathematicall</hi> (as I sayd) by reason it is properly the end, and bound of a line.</p>
            <p>Neither <hi>Number,</hi> nor <hi>Magnitude,</hi> haue any Materialitie. First, we will consider of <hi>Number,</hi> and of the Science <hi>Mathe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icall,</hi> to it appropriate, called <hi>Arithmetike</hi>: and afterward of <hi>Magnitude,</hi> and his Science, called <hi>Geometrie.</hi> But that name con<g ref="char:EOLhyphen"/>tenteth me not: whereof a word or two hereafter shall be sayd. How Immateriall and free from all matter, <hi>Number</hi> is, who doth not perceaue? ye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, who doth not wonderfully wo<g ref="char:cmbAbbrStroke">̄</g>der at it: For, neither pure <hi>Element,</hi> nor <hi>Aristoteles, Quinta Essentia,</hi> is hable to serue for Number, as his propre matter. Nor yet the puritie and simple<g ref="char:EOLhyphen"/>nes of Substance Spirituall or Angelicall, will be found propre enough thereto. And therefore the great &amp; godly Philosopher <hi>Anitius Boetius,</hi> sayd: <hi>Omnia quaecun<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> a primaua rerum na<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ra constructa sunt, Numerorum videntur ratione formata. Hoc enim fuit principale in animo Conditoris Exemplar.</hi> That is: <hi>All thinges (which from the very first originall be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ng of thinges, haue bene framed and made) do appeare to be Formed by the reason of Numbers. For this was the principall example or patterne in the minde of the Creator.</hi> O comfor<g ref="char:EOLhyphen"/>table alluremen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, O rauishing perswasion, to deale with a Science, whose Subiect, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> so Auncient, so pure, so excellent, so surmounting all creatures, so vsed of the Al<g ref="char:EOLhyphen"/>mighty and incomprehensible wisdome of the Creator, in the distinct creation of all creatures<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> in all their distinct partes, properties, natures, and vertues, by order, and most absolute number, brought, from <hi>Nothing,</hi> to the <hi>Formalitie</hi> of their being and state. By <hi>Numbers</hi> propertie therefore, of vs, by all possible meanes, (to the per<g ref="char:EOLhyphen"/>fection of the Science) learned, we may both winde and draw our selues into the inward and deepe search and v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>w, of all creatures distinct vertues, natures, proper<g ref="char:EOLhyphen"/>ties, and <hi>Former</hi>: And also, farder, arise, clime, ascend, and mount vp (with Specula<g ref="char:EOLhyphen"/>tiue winges) in spirit, to behold in the Glas of Creation, the <hi>Forme</hi> of <hi>Formes,</hi> the <hi>Exemplar Number</hi> of all thinges <hi>Numerable:</hi> both visible and inuisible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> mortall and
<pb facs="tcp:7062:6"/>
immortall, Corporall and Spirituall. Part of this profound and diuine Science, had <hi>Ioachim</hi> the Prophesier atteyned vnto: by <hi>Numbers Formall, Naturall,</hi> and <hi>Rationall,</hi> forseyng, concludyng, and forshewyng great particular euents, long before their comming. His bookes yet remainyng, hereof, are good profe: And the noble Earle of <hi>Mirandula,</hi> (besides that,) a sufficient witnesse: that <hi>Ioachim, in his prophesies, proce<g ref="char:EOLhyphen"/>ded by no other way, then by Numbers Formall.</hi> And this Earle hym selfe, in Rome, <note place="margin">
                  <hi>Ano.</hi> 1488.</note> set vp 900. Conclusions, in all kinde of Sciences, openly to be disputed of: and among the rest, in his Conclusions <hi>Mathematicall,</hi> (in the eleuenth Conclusion) hath in Latin, this English sentence. <hi>By Numbers, a way is had, to the searchyng out, aud vnder<g ref="char:EOLhyphen"/>standyng of euery thyng, hable to be knowen. For the verifying of which Conclusion, I pro<g ref="char:EOLhyphen"/>mise to aunswere to the 74. Quaestions, vnder written, by the way of Numbers.</hi> Which Co<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>clusions, I omit here to rehearse: aswell auoidyng superfluous prolixitie: as, by<g ref="char:EOLhyphen"/>cause <hi>Ioannes Picus, workes,</hi> are commonly had. But, in any case, I would wish that those Conclusions were red diligently, and perceiued of such, as are earnest Ob<g ref="char:EOLhyphen"/>seruers and Considerers of the constant law of nu<g ref="char:cmbAbbrStroke">̄</g>bers: which is planted in thyngs Naturall and Supernaturall: and is prescribed to all Creatures, inuiolably to be kept. For, so, besides many other thinges, in those Conclusions to be marked, it would apeare, how sincerely, &amp; within my boundes, I disclose the wonderfull my<g ref="char:EOLhyphen"/>steries, by numbers, to be atteyned vnto.</p>
            <p>Of my former wordes, easy it is to be gathered, that <hi>Number</hi> hath a treble state: One, in the Creator: an other in euery Creature (in respect of his complete consti<g ref="char:EOLhyphen"/>tution:) and the third, in Spirituall and Angelicall Myndes, and in the Soule of ma<g ref="char:cmbAbbrStroke">̄</g>. In the first and third state, <hi>Number,</hi> is termed <hi>Number Numbryng.</hi> But in all Crea<g ref="char:EOLhyphen"/>tures, otherwise, <hi>Number,</hi> is termed <hi>Nu<g ref="char:cmbAbbrStroke">̄</g>ber Numbred.</hi> And in our Soule, Nu<g ref="char:cmbAbbrStroke">̄</g>ber bea<g ref="char:EOLhyphen"/>reth such a swaye, and hath such an affinitie therwith: that some of the old <hi>Philoso<g ref="char:EOLhyphen"/>phers</hi> taught, <hi>Mans Soule, to be a Number mouyng it selfe.</hi> And in dede, in vs, though it be a very Accident: yet such an Accident it is, that before all Creatures it had per<g ref="char:EOLhyphen"/>fect beyng, in the Creator, Sempiternally. <hi>Number Numbryng</hi> therfore, is the discre<g ref="char:EOLhyphen"/>tion discerning, and distincting of thinges. But in God the Creator, This discre<g ref="char:EOLhyphen"/>tion, in the beginnyng, produced orderly and distinctly all thinges. For his <hi>Num<g ref="char:EOLhyphen"/>bryng,</hi> then, was his Creatyng of all thinges. And his Continuall <hi>Numbryng,</hi> of all thinges, is the Conseruation of them in being: And, where and when he will lacke an <hi>Vnit:</hi> there and then, that particular thyng shalbe <hi>Discreated.</hi> Here I stay. But our Seuerallyng, distinctyng, and <hi>Numbryng,</hi> createth nothyng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> but of Multitude con<g ref="char:EOLhyphen"/>sidered, maketh certaine and distinct determination. And albeit these thynges be waighty and truthes of great importance, yet (by the infinite goodnes of the Al<g ref="char:EOLhyphen"/>mighty <hi>Ternarie,</hi>) Artificiall Methods and easy wayes are made, by which the ze<g ref="char:EOLhyphen"/>lous Philosopher, may wyn nere this Riuerish <hi>Ida</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> this Mountayne of Contempla<g ref="char:EOLhyphen"/>tion: and more then Contemplation. And also, though <hi>Number,</hi> be a thyng so Im<g ref="char:EOLhyphen"/>materiall, so diuine, and aeternall: yet by degrees, by litle and litle, stretchyng forth, and applying some likenes of it, as first, to thinges Spirituall: and then, bryngyng it lower, to thynges sensibly perceiued: as of a momentanye sounde iterated: then to the least thynges that may be seen, numerable: And at length, (most grossely,) to a multitude of any corporall thynges seen, or felt: and so, of these grosse and sensible thynges, we are trayned to learne a certaine Image or likenes of numbers: and to vse Arte in them to our pleasure and proffit. So grosse is our conuersation, and dull is our apprehension: while mortall Sense, in vs, ruleth the common wealth of our litle world. Hereby we say, Three Lyons, are three: or a <hi>Ternarie.</hi> Three Egles, are three, or a <hi>Ternarie.</hi> Which<note place="margin">☞</note> 
               <hi>Ternaries,</hi> are eche, the <hi>Vnion, knot,</hi> and <hi>Vniformitie,</hi> of three discrete and distinct <hi>Vnits.</hi> That is, we may in eche <hi>Ternarie,</hi> thrise, seuerally pointe, and shew a part, <hi>One, One,</hi> and <hi>One.</hi> Where, in Numbryng, we say One, two,
<pb facs="tcp:7062:6"/>
Three. But how farre, these visible Ones, do differre from our Indiuisible Vnits (in pure <hi>Arithmetike,</hi> principally considered) no man is ignorant. Yet from these grosse and materiall thynges, may we be led vpward, by degrees, so, informyng our rude Imagination, toward the coceiuyng of <hi>Numbers,</hi> absolutely (Not supposing, nor admixtyng any thyng created, Corporall or Spirituall, to support, conteyne, or represent those <hi>Numbers</hi> imagined:) that at length, we may be hable, to finde the number of our owne name, gloriously exemplified and registred in the booke of the <hi>Trinitie</hi> most blessed and aeternall.</p>
            <p>But farder vnderstand, that vulgar Practisers, haue Numbers, otherwise, in sun<g ref="char:EOLhyphen"/>dry Considerations: and extend their name farder, then to Numbers, whose least part is an <hi>Vnit.</hi> For the common Logist, Reckenmaster, or Arithmeticien, in hys v<g ref="char:EOLhyphen"/>sing of Numbers: of an Vnit, imagineth lesse partes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and calleth them <hi>Fractions.</hi> As of an <hi>Vnit,</hi> he maketh an halfe, and thus noteth it, ½ and so of other, (infinitely di<g ref="char:EOLhyphen"/>uerse) partes of an <hi>Vnit,</hi> Yea and farder, hath, <hi>Fractions of Fractions. &amp;c.</hi> And, foras<g ref="char:EOLhyphen"/>much, as, <hi>Addition, Substraction, Multiplication, Diuision</hi> and <hi>Extraction of Rotes,</hi> are the chief, and sufficient partes of <hi>Arithmetike:</hi>
               <note place="margin">
                  <hi>Arithmetike.</hi>
               </note> which is, the <hi>Science that demonstra<g ref="char:EOLhyphen"/>teth the properties, of Numbers, and all operatio<g ref="char:cmbAbbrStroke">̄</g>s, in numbers to be performed.</hi>
               <q>How often, therfore,<note place="margin">
                     <hi>Note.</hi>
                  </note> these fiue sundry sortes of Operations, do, for the most part, of their exe<g ref="char:EOLhyphen"/>cution, differre from the fiue operations of like generall property and name, in our Whole numbers practisable, So often, (for a more distinct doctrine) we, vulgarly account and name it, an other kynde of <hi>Arithmetike.</hi>
               </q>
And by this reason: the Con<g ref="char:EOLhyphen"/>sideration,<milestone type="tcpmilestone" unit="unspecified" n="1"/> doctrine, and working, in whole numbers onely: where, of an <hi>Vnit,</hi> is no lesse part to be allowed: is named (as it were) an <hi>Ari<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hmetike</hi> by it selfe. And so of the <hi>Arithmetike of Fractions.</hi> In lyke sorte, the necessary, wonderfull and Secret doc<g ref="char:EOLhyphen"/>trine of Proportion, and proportionalytie hath purchased vnto it selfe a peculier <milestone type="tcpmilestone" unit="unspecified" n="2"/> maner of handlyng and workyng: and so may seme an other forme of <hi>Arithmetike.</hi> Moreouer, the <hi>Astronomers,</hi> for spede and more commodious calculation, haue de<g ref="char:EOLhyphen"/>uised <milestone type="tcpmilestone" unit="unspecified" n="3"/> a peculier maner of orderyng nu<g ref="char:cmbAbbrStroke">̄</g>bers, about theyr circular motions, by Sexa<g ref="char:EOLhyphen"/>genes, and Sexagesmes. By Signes, Degrees and Minutes &amp;c. which commonly is called the <hi>Arithmetike</hi> of <hi>Astronomical</hi> or <hi>Phisicall Fractions.</hi> That, haue I briefly no<g ref="char:EOLhyphen"/>ted, by the name of <hi>Arithmetike Circular.</hi> By cause it is also vsed in circles, not <hi>Astro<g ref="char:EOLhyphen"/>nomicall. &amp;c.</hi> Practise hath led <hi>Numbers</hi> farder, and hath framed them, to take vpon <milestone type="tcpmilestone" unit="unspecified" n="4"/> them, the shew of <hi>Magnitudes</hi> propertie: Which is <hi>Incommensurabilitie</hi> and <hi>Irratio<g ref="char:EOLhyphen"/>nalitie.</hi> (For in pu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e <hi>Arithmetike,</hi> an <hi>Vnit,</hi> is the common Measure of all Numbers.) And, here, Nu<g ref="char:cmbAbbrStroke">̄</g>bers are become, as Lynes, Playnes and Solides: some tymes <hi>Ratio<g ref="char:EOLhyphen"/>nall,</hi> some tymes <hi>Irrationall</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> And haue propre and peculier characters, (as √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>.√<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. and so of other. Which is to signifie <hi>Ro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e Square, Rote Cubik: and so forth:</hi>) &amp; propre and peculier fashions in the fiue principall partes: Wherfore the practiser, estemeth this, a diuerse <hi>Arithmetike</hi> from the other. Practise bryngeth in, here, diuerse com<g ref="char:EOLhyphen"/>poundyng of Numbers: as some tyme, two, three, foure (or more) <hi>Radicall</hi> nu<g ref="char:cmbAbbrStroke">̄</g>bers, diuersly knit, by signes, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> More &amp; Lesse: as thus √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 12 + √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 15. Or <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hus √<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 19 + √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 12 - √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> &amp;c. And some tyme with whole numbers, or fractions of whole Number, amo<g ref="char:cmbAbbrStroke">̄</g>g them as 20 + √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> + 33 - √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 10, √<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 44 + 12 - +√<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>9. And so, infinitely, may hap the varie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ie. After this: Both the one and the other hath fractions incident: and so is this <hi>Arithmetike</hi> greately enlarged, by diuerse ex<g ref="char:EOLhyphen"/>hibityng and vse of Compositions and mixtynges. Consider how<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> I (beyng desi<g ref="char:EOLhyphen"/>rous to deliuer the student from error and Cauillation) do giue to this <hi>Practise,</hi> the name of the <hi>Arithmetike of Radicall numbers:</hi> Not of <hi>Irrationall</hi> or <hi>Surd Numbers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> which other while, are Rationall: though they haue the Signe of a Rote before
<pb facs="tcp:7062:7"/>
them, which, <hi>Arithmetike</hi> of whole Numbers most vsuall, would say they had no such Roote: and so account them <hi>Surd Numbers:</hi> which generally spoke<g ref="char:cmbAbbrStroke">̄</g>, is vntrue: as <hi>Euclides</hi> tenth booke may teach you. Therfore to call them, generally, <hi>Radicall Numbers,</hi> (by reason of the signe √. prefixed,) is a sure way: and a sufficient generall distinction from all other ordryng and vsing of Numbers: And yet (beside all this) Consider: the infinite desire of knowledge, and incredible power of mans Search and Capacitye: how, they, ioyntly haue waded farder (by mixtyng of spe<g ref="char:EOLhyphen"/>culation and practise) and haue found out, and atteyned to the very chief perfec<g ref="char:EOLhyphen"/>tion (almost) of <hi>Numbers</hi> Practicall vse. Which thing, is well to be perceiued in that great Arithmeticall Arte of <hi>AEquation:</hi> commonly called the <hi>Rule of Coss.</hi> or <hi>Alge<g ref="char:EOLhyphen"/>bra.</hi> The Latines termed it, <hi>Regulam Rei &amp; Census,</hi> that is, the <hi>Rule of the thyng and his value.</hi> With an apt name: comprehendyng the first and last pointes of the worke. And the vulgar names, both in Italian, Frenche and Spanish, depend (in namyng it,) vpon the signification of the Latin word, <hi>Res: A thing:</hi> vnleast they vse the name of <hi>Algebra.</hi> And therin (commonly) is a dubble error. The one, of them, which thinke it to be of <hi>Geber</hi> his inuentyng: the other of such as call it <hi>Algebra.</hi> For, first, though <hi>Geber</hi> for his great skill in Numbers, Geometry, Astronomy, and other maruailous Artes, mought haue semed hable to haue first deuised the sayd Rule: and also the name carryeth with it a very nere likenes of <hi>Geber</hi> his name: yet true it is, that a <hi>Greke</hi> Philosopher and Mathematicien, named <hi>Diophantus,</hi> before <hi>Geber</hi> his tyme, wrote 13. bookes therof (of which, six are yet extant: and I had them to <note n="*" place="margin">
                  <hi>Anno.</hi> 1550.</note> vse, of the famous Mathematicien, and my great frende, <hi>Petrus Mon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>au<g ref="char:EOLhyphen"/>reus:</hi>) And secondly, the very name, is <hi>Algiebar,</hi> and not <hi>Algebra:</hi> as by the Arabien <hi>Auicen,</hi> may be proued: who hath these precise wordes in Latine, by <hi>Andreas Alpa<g ref="char:EOLhyphen"/>gus</hi> (most perfect in the Arabik tung) so translated. <hi>Scientia faciendi Algiebar &amp; Almachabel.</hi> i. <hi>Scientia inueniendi numerum ignotum, per additionem Numeri, &amp; diuisio<g ref="char:EOLhyphen"/>nem &amp; aequationem.</hi> Which is to say: <hi>The Science of workyng Algiebar and Al<g ref="char:EOLhyphen"/>machabel,</hi> that is, the <hi>Science of findyng an vnknowen number, by Addyng of a Number, &amp; Diuision &amp; aequation.</hi> Here haue you the name: and also the prin<g ref="char:EOLhyphen"/>cipall partes of the Rule, touched. To name it, <hi>The rule, or Art of AEquation,</hi> doth sig<g ref="char:EOLhyphen"/>ni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ie the middle part and the State of the Rule. This Rule, hath his peculier Cha<g ref="char:EOLhyphen"/>racters: <milestone type="tcpmilestone" unit="unspecified" n="5"/> and the principal partes of <hi>Arithmetike,</hi> to it appertayning, do differe from the other <hi>Arithmeticall operations.</hi> This <hi>Arithmetike, hath Nu<g ref="char:cmbAbbrStroke">̄</g>bers</hi> Simple, Co<g ref="char:cmbAbbrStroke">̄</g>pound, Mixt: and Fractions, accordingly. This Rule, and <hi>Arithmetike</hi> of <hi>Algiebar,</hi> is so pro<g ref="char:EOLhyphen"/>found, so generall and so (in maner) conteyneth the whole power of Numbers Application practicall: that mans witt, can deale with nothyng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> more proffitable a<g ref="char:EOLhyphen"/>bout numbers: nor match, with a thyng, more mete for the diuine force of the Soule, (in humane Studies, affaires, or exercises) to be tryed in. Perchaunce you looked for, (long ere now,) to haue had some particular profe, or euident testimo<g ref="char:EOLhyphen"/>ny of the vse, proffit and Commodity of Arithmetike vulgar, in the Common lyfe and trade of men. Therto, then, I will now frame my selfe: But herein great care I haue, least length of sundry profes, might make you deme, that either I did mis<g ref="char:EOLhyphen"/>doute your zelous mynde to vertues schole: or els mistrust your hable witts, by some, to gesse much more. A profe then, foure, fiue, or six, such, will I bryng, as any reasonable man, therwith may be persuaded, to loue &amp; honor, yea learne and exercise the excellent Science of <hi>Arithmetike.</hi>
            </p>
            <p>And first: who, nerer at hand, can be a better witnesse of the frute receiued by <hi>Arithmetike,</hi> then all kynde of Marchants? Though not all, alike, either nede it, or vse it. How could they forbeare the vse and helpe of the Rule, called the Golden
<pb facs="tcp:7062:7"/>
Rule? Simple and Compounde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> both fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ward and backward? How might they misse <hi>Arithmeticall</hi> helpe in the Rules of Felowshyp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> either without tyme, or with tyme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and betwene the Marchant &amp; his <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>actor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> The Rul<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> of Ba<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>tering in wares onely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> or part in wares, and part in money, would they gladly want? Our Mar<g ref="char:EOLhyphen"/>chant venturers, and Trauaylers ouer Sea, how could they order their doynges iustly and without losse, vnleast certa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ne and generall Rules for Exchau<g ref="char:cmbAbbrStroke">̄</g>ge of mo<g ref="char:EOLhyphen"/>ney, and Rechaunge, were, for their vse, deuised? The Rule of Alligation, in how sundry cases, doth it conclude for them, such precise verities, as neither by naturall witt, nor other experience, they, were hable, els, to know? And (with the Mar<g ref="char:EOLhyphen"/>chant then to make an end) how ample &amp; wonderfull is the Rule of False positi<g ref="char:EOLhyphen"/>ons? especially as it is now, by two excellent Mathematiciens (of my familer ac<g ref="char:EOLhyphen"/>quayntance in their life time) enlarged? I meane <hi>Gemma Frisius,</hi> and <hi>Simon Iacob.</hi> Who can either in brief conclude, the generall and Capitall Rules? or who can I<g ref="char:EOLhyphen"/>magine the Myriades of sundry Cases, and particular examples, in Act and earnest, continually wrought, tried and concluded by the forenamed Rules, onely? How sundry other <hi>Arithmeticall practises,</hi> are commonly in Marchantes handes, and knowledge: They them selues, can, at large, testifie.</p>
            <p>The Mintmaster, and Goldsmith, in their Mixture of Metals, either of diuerse kindes, or diuerse values: how are they, or may they, exactly be directed, and mer<g ref="char:EOLhyphen"/>uailously pleasured, if <hi>Arithmetike</hi> be their guide? And the honorable Phisicia<g ref="char:cmbAbbrStroke">̄</g>s, will gladly confesse them selues, much beholding to the Science of <hi>Arithmetike,</hi> and that sundry wayes: But chiefly in their Art of Graduation, and compounde Medicines. And though <hi>Galenus, Auer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ois, Arnoldus, Lullus,</hi> and other haue pu<g ref="char:EOLhyphen"/>blished their positions, aswell in the quantities of the Degrees aboue Tempera<g ref="char:EOLhyphen"/>ment, as in the Rules, concluding the new <hi>Forme</hi> resulting: yet a more precise, commodious, and easy <hi>Method,</hi> is extant:<note place="margin">
                  <hi>R. B.</hi>
               </note> by a Countreyman of ours (aboue 200 yeares ago) inuented. And forasmuch as I am vncertaine, who hath the same: or when that litle Latin treatise, (as the Author writ it,) shall come to be Printed: (Both to declare the desire I haue to pleasure my Countrey, wherin I may: and al<g ref="char:EOLhyphen"/>so, for very good profe of Numbers vse, in this most subtile and frutefull, Philoso<g ref="char:EOLhyphen"/>phicall Conclusion,) I entend in the meane while, most briefly, and with my far<g ref="char:EOLhyphen"/>der helpe, to communicate the pith therof vnto you.</p>
            <p>First describe a circle: whose diameter let be an inch. Diuide the Circumfe<g ref="char:EOLhyphen"/>rence into foure equall partes. Fro<g ref="char:cmbAbbrStroke">̄</g> the Center, by those 4. sections, extend 4. right lines: eche of 4. inches and a halfe long: or of as many as you liste, aboue 4. with<g ref="char:EOLhyphen"/>out the circumference of the circle: So that they shall be of 4. inches long (at the lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>t) without the Circle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Make good euident markes, at euery inches end. If you list, you may subdiuide the inches againe into 10. or 12. smaller partes, equall. At the endes of the lines, write the names of the 4. principall elementall Qualities. <hi>Hote</hi> and <hi>Cold<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,</hi> one against the other. And likewise <hi>Moyst</hi> and <hi>Dry,</hi> one against the other. And in the Circle write <hi>Temperate.</hi> Which <hi>Temperature</hi> hath a good La<g ref="char:EOLhyphen"/>titude: as appeareth by the Complexion of man. And therefore we haue allow<g ref="char:EOLhyphen"/>ed vnto it, the foresayd Circle: and not a point Mathematicall or Physicall.</p>
            <p>Now, when you haue two thinges Miscible, whose degrees are <note n="*" place="margin">
                  <hi>Take some part of Lullus counsayle in his booke de Q. Essentia.</hi>
               </note> truely knowen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Of necessitie, either they are of one Quantitie and waight, or of diuerse. If they be of one Quantitie and waight: whether their formes, be Contrary Qua<g ref="char:EOLhyphen"/>lities, or of one kinde (but of diuerse intentions and degrees) or a <hi>Temperate,</hi> and a Contrary, <hi>The form<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> resulting of their Mixture, is in the Middle betwene the degrees of
<pb facs="tcp:7062:8"/>
the formes mixt.</hi> As for example, let <hi>A,</hi> be <hi>Moist</hi> in the first degree: and <hi>B, Dry</hi> in the third degree. Adde 1. and 3. that maketh 4: the halfe or middle of 4. is 2. This 2. is the middle, equally distant from <hi>A</hi> and <hi>B</hi> (for the<note n="*" place="margin">
                  <hi>No<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> 
               <hi>Temperament</hi> is coun<g ref="char:EOLhyphen"/>ted none. And for it, you must put a Ciphre, if at any time, it be in mixture).
<figure/>
Counting then from <hi>B,</hi> 2. degrees, toward <hi>A:</hi> you finde it to be <hi>Dry</hi> in the first degree: So is the <hi>Forme resulting</hi> of the Mixture of <hi>A,</hi> and <hi>B,</hi> in our example, I will geue you an other example. Suppose, you haue two thinges, as <hi>C,</hi> and <hi>D</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and of <hi>C,</hi> the Heate to be in the 4. degree: and of <hi>D,</hi> the Colde, to be remisse, euen vnto the <hi>Temperament.</hi> Now, for <hi>C,</hi> you take 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and for <hi>D,</hi> you take a Ciphre: which, added vnto 4, yeldeth onely 4. The middle, or halfe, whereof, is 2. Wherefore the <hi>Forme resulting</hi> of <hi>C,</hi> and <hi>D,</hi> is Hote in the second degree: for 2. degrees, accoun<g ref="char:EOLhyphen"/>ted from <hi>C,</hi> toward <hi>D,</hi> ende iuste in the 2. degree of heate. Of the third ma<g ref="char:EOLhyphen"/>ner, I will geue also an example: which let be this: I haue a liquid Medicine whose Qualitie of heate is in the 4. degree exalted:<note place="margin">
                  <hi>Note.</hi>
               </note> as was <hi>C,</hi> in the example foregoing: and an other liquid Medicine I haue: whose Qualitie, is heate, in the first degree. Of eche of these, I mixt a like quantitie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Subtract here, the lesse fro<g ref="char:cmbAbbrStroke">̄</g> the more: and the residue diuide into two equall partes: whereof, the one part, either added to the lesse, or subtracted from the higher degree, doth produce the degree of the
<pb facs="tcp:7062:8"/>
Forme resulting, by this mixture of <hi>C,</hi> and <hi>E.</hi> As, if from 4. ye abate 1. there resteth 3, the halfe of 3, is 1 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: Adde to 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> this 1 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: you haue 2 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. Or subtract from 4. this 1 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: you haue likewise 2 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> remayning. Which declareth, the <hi>Forme resul<g ref="char:EOLhyphen"/>ting,</hi> to be <hi>Heate,</hi> in the middle of the third degree.</p>
            <p>
               <q>But if the Quantities of two thinges Commixt,<note place="margin">
                     <hi>The Se<g ref="char:EOLhyphen"/>cond Rule.</hi>
                  </note> be diuerse, and the Intensi<g ref="char:EOLhyphen"/>ons (of their Formes Miscible) be in diuerse degree<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and heigthes. (Whether those Formes be of one kinde, or of Contrary kindes, or of a Temperate and a Contrary, <hi>What proportion <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lesse quantiti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the greater, the same shall be of the difference, which is betwene the degree of the Forme resulting, and the degree of the greater quantitie of the thing miscible, to the difference, which is betwene the same degree of the Forme resulting, and the degree of the lesse quantiti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> As for example. Let two pound of Liquor be geuen, hote in the 4. degree: &amp; one pound of Liqour be geuen, hote in the third degree, I would gladly know the Forme resulting, in the Mixture of these two Liquors.</q>
Set downe your nu<g ref="char:cmbAbbrStroke">̄</g>bers in order, thus.
<figure/>
Now by the rule of Algiebar, haue I deuised a very <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>aste, briefe, and generall maner of working in this case. Let vs first, suppose that <hi>Middle Forme resulting,</hi> to be 1<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> at that Rule teacheth. And because (by our Rule, here geuen) as the waight of 1. is to 2: So is the difference betwene 4. (the degree of the greater quantitie) and 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> to the difference betwene 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> (the degree of the thing, in lesse qua<g ref="char:cmbAbbrStroke">̄</g>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>tie. And with all, 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, being alwayes in a cer<g ref="char:EOLhyphen"/>taine middell, betwene the two heigthes or degrees). For the first difference, I set—1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: and for the second, I set 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>—3. And, now againe, I say, as 1. is to 2. so is 4—1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> to 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>—3. Wherfore, of these foure proportionall numbers, the first and the fourth Multiplied, one by the other, do make as much, as the second and the third Multiplied the one by <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> other? Let these Multiplications be made accor<g ref="char:EOLhyphen"/>dingly. And of the first and the fourth, we haue 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>—3. and of the second &amp; the third, 8—<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>. Wherfore, our AEquation is betwene <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>—3: and 8—2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. Which may be reduced, according to the Arte of Algiebar as, here, adding 3. to eche part, ge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>eth the AEquation, thus, 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>=11—2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. And yet againe, contracting, or Redu<g ref="char:EOLhyphen"/>cing it: Adde to eche part, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>? Then haue you 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> equall to 11: thus represen<g ref="char:EOLhyphen"/>ted 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>=11. Wherefore, diuiding 11 by 3: the Quotient is 3 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: the <hi>Valew</hi> of our 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, <hi>Coss,</hi> or <hi>Thing,</hi> first supposed. And that is the heigth, or Intension of the <hi>Forme resulting:</hi> which is, <hi>Heate,</hi> in two thirdes of the fourth degree: And here I set the shew of the worke in conclusion, thus. The proufe hereof is easie by subtracting 3. from 3 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> resteth
<figure/>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. Subtracte the same heigth of the Forme resulting, (which is 3 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>) fro<g ref="char:cmbAbbrStroke">̄</g> 4: then resteth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: You see, that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> is double to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>: as 2. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. is double to 1. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. So should it be: by the rule here geuen. Note. As you ad<g ref="char:EOLhyphen"/>ded to eche part of the AEquation, 3: so if ye first added to eche part 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, it would stand, 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>—3=8. And now adding to eche par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> you haue (as afore) 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>=11.</p>
            <p>And though I, here, speake onely of two thyngs Miscible: and most common<g ref="char:EOLhyphen"/>ly, mo then three, foure, fiue or six, (&amp;c.) are to be Mixed: (and in one Compound
<pb facs="tcp:7062:9"/>
to be reduced<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> &amp; the Forme resultyng of the same, to serue the turne) yet these Ru<g ref="char:EOLhyphen"/>les are sufficient: duely repeated and iterated.<note place="margin">
                  <hi>Note.</hi>
               </note> In procedyng first, with any two: and then, with the Fonne Resulting, and an other: &amp; so forth: For, the last worke, con<g ref="char:EOLhyphen"/>cludeth the Forme resultyng of them all: I nede nothing to speake, of the Mixture (here supposed) what it is. Common Philosophie hath defined it, saying, <hi>Mixtio est miscibilium, alteratorum, per minima coniunctorum, Vnio.</hi> Euery word in the de<g ref="char:EOLhyphen"/>finition, is of great importance. I nede not also spend any time, to shew, how, the other manner of distributing of degrees, doth agree to these Rules. Neither nede I of the farder vse belonging to the Crosse of Graduation (before described) in this place declare, vnto such as are capable of that, which I haue all ready sayd. Neither yet with examples specifie the Manifold varie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ies, by the foresayd two gene<g ref="char:EOLhyphen"/>rall Rules, to be ordered. The witty and Studious, here, haue sufficient: And they which are not hable to atteine to this, without liuely teaching, and more in parti<g ref="char:EOLhyphen"/>cular: would haue larger discoursing, then is mete in this place to be dealt w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>thall: And other (perchaunce) with a proude snuffe will disdaine this litl<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> and would be vnthankefull for much more. I, therfore conclude: and wish such as haue modest and earnest Philosophicall mindes, to laude God highly for this: and to Meruayle, that the profoundest and subtilest point, concerning <hi>Mixture of Formes and Quali<g ref="char:EOLhyphen"/>ties Naturall,</hi> is so Match<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and maryed with the most simple, easie, and short way of the noble Rule of <hi>Algiebar.</hi> Who can remaine, therfore vnpersuaded, to loue, a<g ref="char:EOLhyphen"/>low, and honor the excellent Science of <hi>Arithmetike?</hi> For, here, you may perceiue that the litle finger of <hi>Arithmetike,</hi> is of more might and contriuing then a hun<g ref="char:EOLhyphen"/>derd thousand mens wittes, of the middle sorte, are hable to perfourme, or truely to conclude, with out helpe thereof.</p>
            <p>Now will we f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>rder, by the wise and valiant Capitaine, be certified, what helpe he hath, by the Rules of <hi>Arithmetike:</hi> in one of the Artes to him appertaining:
<q>And of the Grekes named <gap reason="foreign">
                     <desc>〈 in non-Latin alphabet 〉</desc>
                  </gap>.<note place="margin">
                     <gap reason="foreign">
                        <desc>〈 in non-Latin alphabet 〉</desc>
                     </gap>.</note> That is, the Skill of O<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dring Souldiers in Battell ray after the best maner to all purposes.</q>
This Art so much dependeth vppon Numbers vse, and the Mathematicals, that <hi>AElianus</hi> (the best writer therof,) in his worke, tö the <hi>Emperour Hadrianus,</hi> by his perfection, in the Mathematicals, (beyng greater, then other before him had,) thinketh his booke to passe all other the excellent workes, written of that Art, vnto his dayes. For, of it, had written <hi>AEneas: Gyneas</hi> of <hi>Thessaly: Pyrrhus Epirota:</hi> and <hi>Alexander</hi> his sonne: <hi>Clearchus: Pausanias: Euangelus: Polybi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s,</hi> familier frende to <hi>Scipio: Eupolemus: Iphicrates, Possidonius:</hi> and very many other worthy Capitaines, Philosophers and Princes of Immortall fame and me<g ref="char:EOLhyphen"/>mory: Whose fayrest floure of their garland (in this feat) was <hi>Arithmetike:</hi> and a litle perceiuerance, in <hi>Geometricall</hi> Figures. But in many other cases doth <hi>Arith<g ref="char:EOLhyphen"/>metike</hi> stand the Capitaine in great stede. As in proportionyng of vittayles, for the Army, either remaining at a stay: or suddenly to be encreased with a certaine number of Souldiers: and for a certain tyme. Or by good Art to diminish his com<g ref="char:EOLhyphen"/>pany, to make the victuals, longer to serue the remanent, &amp; for a certaine determi<g ref="char:EOLhyphen"/>ned tyme: if nede so require. And so in sundry his other accountes, Recke<g ref="char:EOLhyphen"/>ninges, Measurynges, and proportionynges, the wise, expert, and Circumspect <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>a<g ref="char:EOLhyphen"/>pitaine will affirme the Science of <hi>Arithmetike,</hi> to be one of his chief Counsaylors, directers and aiders. Which thing (by good meanes) was euident to the Noble, the Couragious,<note place="margin">☞</note> the loyall, and Curteous <hi>Iohn,</hi> late Earle of Warwicke. Who was a yong Gentleman, throughly knowne to very few. Albeit his lusty valiant<g ref="char:EOLhyphen"/>nes, force, and Skill in Chiualrous feates and exerci<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>es: his humblenes, and frende<g ref="char:EOLhyphen"/>lynes to all men, were thinges, openly, of the world perceiued. But what rotes (otherwise,) vertue had fastened in his brest, what Rules of godly and honorable
<pb facs="tcp:7062:9"/>
life he had framed to him selfe: what vices, (in some then liuing) notable, he tooke great care to eschew: what manly vertues, in other noble men, (florishing before his eyes,) he Sythingly aspired after: what prowesses he purposed and ment to a<g ref="char:EOLhyphen"/>chieue: with what feats and Artes, he began to furnish and fraught him selfe, for the better seruice of his Kyng and Countrey, both in peace &amp; warre. These (I say) his Heroicall Meditations, forecas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>inges and determinations, no twayne, (I thinke) beside my selfe, can so perfectly, and truely report. And therfore, in Con<g ref="char:EOLhyphen"/>science, I count it my part, for the honor, preferment, &amp; procuring of vertue (thus, briefly) to haue put his Name, in the Register of <hi>Fa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> Immortall.</hi>
            </p>
            <p>To our purpose. This <hi>Iohn,</hi> by one of his actes (besides many other: both in En<g ref="char:EOLhyphen"/>gland and Fraunce, by me, in him noted.) did dislose his harty loue to vertuous Sciences: and his noble intent, to excell in Martiall prowesse: When he, with hum<g ref="char:EOLhyphen"/>ble request, and instant Solliciting: got the best Rules (either in time past by Greke or Romaine, or in our time vsed: and new Stratagemes therin de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ised) for ordring of all Companies, summes and Number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> of me<g ref="char:cmbAbbrStroke">̄</g>, (Many or few) with one kinde of weapon, or mo, appointed: with Artillery, or without: on horsebacke, or on fote: to giue, or take onset: to seem many, being few<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> to s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>em few, being many. To marche in battaile or Iornay: with many such feate<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, to Foughten f<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>ld, Skarmoush, or Ambushe appartaining: And of all these, liuely desi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nementes (most curiously) to be in velame parchement described: with Notes &amp; peculier markes, as the Arte requireth: and all these Rules<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and descriptions Arithmeticall,<note place="margin">This noble Earle, dyed Anno. 1554. skarse of 24. yeares of age<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hauing no is<g ref="char:EOLhyphen"/>sue by his wife: Daugh<g ref="char:EOLhyphen"/>ter to the Duke of So<g ref="char:EOLhyphen"/>merset.</note> inclosed in a riche Case of Gold, he vsed to weare about his necke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> as his Iuell most precious, and Counsaylour most trusty. Thus, <hi>Arithmetik<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,</hi> of him was shryned in gold: Of <hi>Numbers</hi> frute, he had good hope. Now, Numbers therfore innumerable, in <hi>Numbers</hi> prayse, his sh<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>ne shall finde.</p>
            <p>What nede I, (for farder <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>rofe to you) of the Scholemasters of Iustice, to require testimony: how nedefull, how frutefull, how skillfull a thing <hi>Arithmetike</hi> is? I meane<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> the Lawyers of all sortes. Vndoubtedly, the Ciuilians, can meruaylous<g ref="char:EOLhyphen"/>ly declare: how, neither the Auncient Romaine lawes, without good knowledge of <hi>Numbers art,</hi> can be perceiued: Nor (Iustice in infinite Cases) without due pro<g ref="char:EOLhyphen"/>portion, (narrowly considered,) is hable to be executed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> How Iustly, &amp; with great knowledge of Arte, did <hi>Pap<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>anus</hi> institute a law of partition, and allowance, be<g ref="char:EOLhyphen"/>twene man and wife a<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap> a <gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>orce: But how <hi>Acc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rsius, Bald<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, Bartolus, Iason, Alex<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nder,</hi> and finally <hi>Alciat<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
               </hi> (bei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>g o<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>rwise, notably well learned) do iumble, gesse, and erre, from the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>quity<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>rt <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nd I<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>ent of the lawmaker: <hi>Arithmetike</hi> can detect, and conuince: and clerely, make the truth to shine. Good <hi>Bartolus,</hi> tyred in the examining &amp; proportioning of the matter: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nd with <hi>Accursius</hi> Glosse, much cum<g ref="char:EOLhyphen"/>bred: burst out, and sayd: <hi>Nulla est i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> libr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> glossa difficili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r: Cui<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mputatio<g ref="char:EOLhyphen"/>nem nec Scholas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ici <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ec Doct<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>res intellig<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>t. &amp;c.</hi> That is<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <hi>In the whole booke, there is no <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>losse harder then thi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> Whose accoump<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or reckenyng, neither the Scho<g ref="char:EOLhyphen"/>lers, nor the Doctours vnderstand. &amp;c.</hi> What can they say of <hi>I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lianus</hi> law, <hi>Si ita S<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ript<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>m. &amp;c.</hi> Of the Testators will iustly performing, betwene the wife, Sonne and daughter? How can they perceiue the <gap reason="illegible" resp="#PDCC" extent="5 letters">
                  <desc>•••••</desc>
               </gap>tie of <hi>Aphricanus, Arithmeticall</hi> Reckening, where he treateth of <hi>Lex Falcid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> How <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>n they deliuer him, from his Reprouers: and their maintainers: as <hi>I<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>es, Accursius Hypolitus</hi> and <hi>Alciatus?</hi> How <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ustly and artificially, was <hi>African<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s</hi> reckening made? Proportionating to the Sommes bequeathed, the Contributions of eche part? Namely, for the hundred presently receiued, 17 1/7. And for the hundred, receiued after ten monethes, 12 6/7: which make the 30: which were to be co<g ref="char:cmbAbbrStroke">̄</g>tributed by the legatari<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s to the heire.
<pb facs="tcp:7062:10"/>
For, what proportion, 100 hath to 75: the same hath 17 1/7 to 12 6/7: Which is Ses<g ref="char:EOLhyphen"/>quitertia: that is, as 4, to 3. which make 7. Wonderfull many places, in the Ciuile law, require an expert <hi>Arithmeticien,</hi> to vnderstand the deepe Iudgeme<g ref="char:cmbAbbrStroke">̄</g>t, &amp; Iust de<g ref="char:EOLhyphen"/>terminatio<g ref="char:cmbAbbrStroke">̄</g> of the Auncient Romaine Lawmakers. But much more expert ought he to be, who should be hable, to decide with aequitie, the infinite varietie of Cases, which do, or may happen, vnder euery one of those lawes and ordinances Ciuile. Hereby, easely, ye may now coniecture: that in the Canon law: and in the lawes of the Realme (which with vs, beare the chief Authoritie), Iustice and e<g ref="char:EOLhyphen"/>quity might be greately preferred, and skilfully executed, through due skill of A<g ref="char:EOLhyphen"/>rithmetike, and proportions appertainyng. The worthy Philosophers, and pru<g ref="char:EOLhyphen"/>dent lawmakers (who haue written many bookes <hi>De Republica:</hi> How the best state of Common wealthes might be procured and mainteined,) haue very well deter<g ref="char:EOLhyphen"/>mined of Iustice: (which, not onely, is the Base and foundacion of Common weales:
<q>but also the totall perfection of all our workes, words, and thoughtes:) de<g ref="char:EOLhyphen"/>fining it, to be that vertue,<note place="margin">
                     <hi>Iustice.</hi>
                  </note> by which, to euery one, is rendred, that to him appertai<g ref="char:EOLhyphen"/>neth. God challengeth this at our handes, to be honored as God:</q>
to beloued, as a father: to be feared as a Lord &amp; master. Our neighbours proportio<g ref="char:cmbAbbrStroke">̄</g>, is also pres<g ref="char:EOLhyphen"/>cribed of the Almighty lawmaker: which is, to do to other, euen as we would be done vnto. These proportions, are in Iustice necessary<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> in duety, commendable: and of Common wealthes, the life, strength, stay and florishing. <hi>Aristotle</hi> in his <hi>Ethikes</hi> (to fatch the sede of Iustice, and light of direction, to vse and execute the sam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>) was fayne to fly to the perfection, and power of Numbers: for proportions Arithmeticall and Geometricall. <hi>Plato</hi> in his booke called <hi>Epinomis</hi> (which boke, is the Threasury of all his doctrine) where, his purpose is, to seke a Science, which, when a man had it, perfectly: he might seme, and so be, in dede, <hi>Wise.</hi> He, br<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>efly, of other Sciences discoursing, findeth them, not hable to bring it to passe: But of the Science of Numbers, he sayth. <hi>Illa, qua numerum mortalium generi d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>n, id profecto ef<g ref="char:EOLhyphen"/>ficiet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Deum antem aliquem, magis quam fortunam, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d sa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tem nostram, hoc m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nus nobis arbitror contulisse. &amp;c. Nam ipsum <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>onorum omnium Authorem, cur non maximi boni, Prudentiae dico, causam arbitramur? That Science, verely, which hath taught man<g ref="char:EOLhyphen"/>kynde number, shall be able to bryng it to passe. And, I thinke, a certaine God, rather then fortune, to haue giuen vs this gift, for our blisse. For, why should we not Iudge him, who is the Author of all good things, to be also the cause of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> greatest good thyng, namely, Wisedome?</hi> There, at length, he proueth <hi>Wisedome</hi> to be atteyned, by good Skill of <hi>N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mbers.</hi> With which great Testimony, and the manifold profes, and reasons, before expressed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> you may be sufficiently and fully persuaded: of the perfect Science of <hi>Arithmetike,</hi> to make this accounte: That of all Sciences,<note place="margin">☞</note> next to <hi>Theologie,</hi> it is most diuine, most pure, most ample and generall, most profounde, most subtile, most commodious and most necessary. Whose next Sister, is the Absolute Science of <hi>Magnitudes:</hi> of which (by the Direction and aide of him, whose <hi>Magnitude</hi> is Infinite, and of vs Incomprehensible) I now en<g ref="char:EOLhyphen"/>tend, so to write, that both with the <hi>Multitude,</hi> and also with the <hi>Magnitude</hi> of Meruaylous and frutefull verities, you (my frendes and Countreymen) may be stird vp, and awaked, to behold what certaine Artes and Sciences, (to our vn<g ref="char:EOLhyphen"/>speakable behofe) our heauenly father, hath for vs prepared, and reuealed, by sun<g ref="char:EOLhyphen"/>dry <hi>Philosophers</hi> and <hi>Mathematiciens.</hi>
            </p>
            <p>BOth, <hi>Number</hi> and <hi>Magnitude,</hi> haue a certaine Originall sede, (as it were) of an incredible property: and of man, neuer hable, Fully, to be declared. Of <hi>Number,</hi> an Vnit, and of <hi>Magnitude,</hi> a Poynte, doo seeme to be much like Origi<g ref="char:EOLhyphen"/>nall
<pb facs="tcp:7062:10"/>
               <gap reason="illegible" resp="#PDCC" extent="1 page">
                  <desc>〈1 page〉</desc>
               </gap>
               <pb facs="tcp:7062:11"/>
pretence of iust content, and me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>sure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> lan<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>es and groundes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>re<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap>, <gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>
               <g ref="char:EOLhyphen"/>quietnes, murder, and warre did (full oft) ensue: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ill<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> by Gods <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> and <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> in<g ref="char:EOLhyphen"/>dustrie, The perfect Science of Lines, Plaines, and Solides (lik<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> diuine <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap>,) gaue vnto euery man, his owne. The people then, by this art pleasured, and great<g ref="char:EOLhyphen"/>ly relieued, in their landes iust measuring: &amp; other Philosophers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> writing Rules for land measuring<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> betwene them both, thus, con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>rmed the name of <hi>G<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tria,</hi> that is, (according to the very etimologie of the word) Land measuring. Wherin, the p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>o<g ref="char:EOLhyphen"/>ple knew no farder, of Magnitudes vse, but in Plai<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>: and the Philosophers, of the<g ref="char:cmbAbbrStroke">̄</g>, had no feethearers, or Scholers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>der to disclose vnto, then of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>lat, plaine <hi>Geome<g ref="char:EOLhyphen"/>trie.</hi> And though, these Philosophers, knew of farder vse, and best vnderstode the etymologye of the worde, yet this name <hi>G<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>etria,</hi> was of them applyed generally to all sortes of Magnitudes: vnleast, otherwhile, of <hi>Plato,</hi> and <hi>Pythagoras<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> When they would precisely declare their owne doctrine. Then, was<note n="*" place="margin">
                  <hi>Plato.</hi> 7. <hi>de Rep.</hi>
               </note> 
               <hi>Geometria,</hi> with them, <hi>Studium quod circa planum versatur.</hi> But, well you may perceiue by <hi>Euclides Elementes,</hi> that more ample is our Science, then to measure Plaines: and nothyng lesse therin is tought (of purpose) then how to measure Land.
<q>An other name, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<g ref="char:EOLhyphen"/>fore, must nedes be had, for our Mathematicall Science of Magnitudes: which re<g ref="char:EOLhyphen"/>gardeth neither clod, no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> turff<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> neither hill, nor dale<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> neither <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>th nor heauen: but is absolute <hi>Megeth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>logia:</hi> not creping on ground, and dissellng the eye, with pole perche, rod or lyne:<note place="margin">☞</note> but liftyng the hart aboue the heauens, by <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> lines, and immortall beames<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> meteth with the reflexions, of the light incomprehensible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and so procureth Ioye, and perfection vnspeakable.</q>
Of which true vse of our <hi>Meg<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>thica,</hi> or <hi>Megethologia, Diuine Plato</hi> seemed to haue good taste, and iudgement<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and (by the name of <hi>Geometrie</hi>) so noted it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and warned his Scholers therof: as, in hys seuenth <hi>Dialog,</hi> of the Common wealth, may <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>uidently be sene. Where (in La<g ref="char:EOLhyphen"/>tin) thus it is: right well translated: <hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> quid Geometri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> loquuntur, qui in ipsa vers<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>.</hi> In English, <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <hi>Verely</hi> (sayth <hi>Plat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi>) <hi>whosoeuer haue, (but euen very litle) tasted of Geometrie, will not denye vnto vs, this: but that this Science, is of an other condicion, quite contrary to that, which they that are exercised in it, do speake of it.</hi> And there it followeth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> of our <hi>Geometrie, Quod q<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ritur cognoscendi illius gratia, quod se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>per est, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> &amp; Interit. Geometria, eius quod est semper, Cognit<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> est. <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> Veritatem, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nimum: at<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> ita, ad Philosophandum <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> qua, nunc, contra quam decet, ad inf<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> &amp;c. <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> praecipiendum est, vt qui praclarissimam hanc <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> spernant. Nam &amp; quoe pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter ipsius propositu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> sunt. &amp;c.</hi> It must nedes be confessed (saith <hi>Plat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi>) <hi>That [<gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>] i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> learned, for the knowyng of that, which is <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>: and not of that, which, in tyme, both is bred and is brought to an ende. &amp;c. Geometrie is the knowledge of that which is euer<g ref="char:EOLhyphen"/>lastyng. It will lif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vp therfore (O Gentle Syr) our mynde to the Veritie: and by that meanes, it will prepare the Thought, to the Philosophicall loue of wisdome: that we may turne or con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ert, toward heauenly thinges [<gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>] which now, otherwise then b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>comme<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> vs, we cast <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> or inferior things &amp;c. Chiefly, therfore, Comma<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>dement must be giuen, that such as do <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nhabit this most honorable Citie, by no meanes, despise Geometrie. For euen those thinges [done by <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>] which, in manner, seame to be, beside the purpose of Geometrie: are of</hi>
               <pb facs="tcp:7062:11"/>
               <gap reason="illegible" resp="#PDCC" extent="1 page">
                  <desc>〈1 page〉</desc>
               </gap>
               <pb facs="tcp:7062:12"/>
               <gap reason="illegible" resp="#PDCC" extent="1 page">
                  <desc>〈1 page〉</desc>
               </gap>
               <pb facs="tcp:7062:12"/>
by vntrue measuring and surueying of Land or Woods, any way. And, this I am sure: that the Value of the difference, betwene the truth and such Surueyes, would haue bene hable to haue fou<g ref="char:cmbAbbrStroke">̄</g>d (foreuer) in eche of our two Vniuersities, an excel<g ref="char:EOLhyphen"/>lent Mathematicall Reader: to eche, allowing (yearly) a hundred Markes of lawfull money of this realme: which, in dede, would seme requisit, here, to be had (though by other wayes prouided for) as well, as the famou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>ersit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e of Paris, hath two Mathematicall Readers: and eche, two hundreth French Crownes yearly, of the French Kinges magnificent liberalitie onely. Now, againe, to our purpose retur<g ref="char:EOLhyphen"/>ning: Moreouer, of the former knowledge Geometricall, are growen the Skills of <hi>Geographie, Chorographie, Hydrographie,</hi> and <hi>S<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>rithmetrie.</hi>
            </p>
            <p>
               <q>Geographie teacheth wayes, by which, in su<g ref="char:cmbAbbrStroke">̄</g>dry formes, (as <hi>Spharike, Plaine</hi> or other), the Situation of Cities, Townes, Villages, Fortes, Castells, Mountaines, Woods, Hauens, Riuers, Crekes, &amp; such other things, vpo<g ref="char:cmbAbbrStroke">̄</g> the outface of the earth<g ref="char:EOLhyphen"/>ly Globe (either in the whole, or in some principall me<g ref="char:cmbAbbrStroke">̄</g>ber and portion therof co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned) may be described and designed, in co<g ref="char:cmbAbbrStroke">̄</g>mensurations Analogicall to Nature and veritie: and most aptly to our vew, may be represented. Of this Arte how great pleasure, and how manifolde commodities do come vnto vs, daily and hourely: of most men, is perceaued.</q>
While, some, to beautifie their Halls, Parlers, Chambers, Galeries, Studies, or Libraries with: other some for thinges past, as battels fought, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>arthquakes, heauenly fyringes, &amp; such occurentes, in histories mentioned: therby liuely, as it were, to vewe the place, the region adioyning, the distance from vs: and such other circumstances. Some other, presently to vewe the large dominion of the Turke: the wide Empire of the Moschouite: and the litle morsell of ground, where Christendome (by profession) is certainly knowen. Litle, I say, in respecte of the rest. &amp;c. Some, either for their owne iorneyes directing into farre landes: or to vnderstand of other m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> trauaile<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. To conclude, some, for one purpose: and some, for another, liketh, loueth, ge<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>th, and vse<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>, Mappes, Chartes, &amp; Geo<g ref="char:EOLhyphen"/>graphicall Globes. Of whose vse, to speake su<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>iciently, would require a booke peculi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>r.</p>
            <p>Chorographie <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap>
            </p>
            <p>
               <q>Hydrographie, deliuereth to our knowledge, on Globe or in Plaine, the perfect Analogicall description of the Ocean Sea coastes, through the whole world <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or in the chiefe and principall partes thereof: with the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>les and chiefe
<pb facs="tcp:7062:13"/>
particular places of daungers, conteyned within the boundes, and Sea coastes de<g ref="char:EOLhyphen"/>scribed: as, of Quicksandes, Bankes, Pittes, Rockes, Races, Countertides, Whorle<g ref="char:EOLhyphen"/>pooles. &amp;c.</q>
This, dealeth with the Element of the water chiefly: as <hi>Geographie</hi> did principally take the Element of the Earthes description (with his apperte<g ref="char:EOLhyphen"/>nances) to taske. And besides thys, <hi>Hydrographie,</hi> requireth a particular Register of certaine Landmarkes (where markes may be had) from the sea, well ha<g ref="char:EOLhyphen"/>ble to be skried, in what point of the Seacumpase they appeare, and what apparent forme, Situation, and bignes they haue, in respecte of any daungerous place in the sea, or nere vnto it, assigned: And in all Coastes, what Mone, maketh full Sea: and what way, the Tides and Ebbes, come and go, the <hi>Hydrographer</hi> ought to recorde. The Soundinges likewise: and the Chanels wayes: their number, and depthes or<g ref="char:EOLhyphen"/>dinarily, at ebbe and flud, ought the <hi>Hydrographer,</hi> by obseruation and diligence of <hi>Measuring,</hi> to haue certainly knowen. And many other pointes, are belonging to perfecte <hi>Hydrographie,</hi> and for to make a <hi>Rutter,</hi> by: of which, I nede not here speake: as of the describing, in any place, vpon Globe or Plaine, the 32. pointes of the Compase, truely: (wherof, scarsly foure, in England, haue right knowledge: bycause, the lines therof, are no straight lines, nor Circles.) Of making due pro<g ref="char:EOLhyphen"/>iection of a Sphere in plaine. Of the Variacion of the Compas, from true Northe: And such like matters (of great importance, all) I leaue to speake of, in this place: bycause, I may seame (al ready) to haue enlarged the boundes, and duety of an Hy<g ref="char:EOLhyphen"/>dographer, much more, then any man (to this day) hath noted, or prescribed. Yet am I well hable to proue, all these thinges, to appertaine, and also to be proper to the Hydrographer. The chief vse and ende of this Art, is the Art of Nauigation; but it hath other diuerse vses: euen by them to be enioyed, that neuer lacke sight of land.</p>
            <p>Stratarithmetrie, is the Skill, (appertainyng to the warre,) by which a man can set in figure, analogicall to any <hi>Geometricall</hi> figure appointed, any certaine number or summe of men: of such a figure capable: (by reason of the vsuall spaces betwene Souldiers allowed: and for that, of men, can be made no Fractions. Yet, neuertheles, he can order the giuen summe of men, for the greatest such figure, that of them, ca<g ref="char:cmbAbbrStroke">̄</g> be ordred) and certifie, of the ouerplus: (if any be) and of the next certaine summe, which, with the ouerplus, will admit a figure exactly proportionall to the figure assigned. By which Skill, also, of any army or company of men: (the figure &amp; sides of whose orderly standing, or array, is knowen) he is able to expresse the iust number of men, within that figure conteined: or (orderly) able to be con<g ref="char:EOLhyphen"/>teined. <note n="*" place="margin">
                  <hi>Note.</hi>
               </note> And this figure, and sides therof, he is hable to know: either beyng by, and at hand: or a farre of.
<q>Thus farre, stretcheth the description and property of <hi>Stratarithmetrie:</hi> sufficient for this tyme and place. It differreth from the Feate <hi>Tacticall, De acic<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us instruendis</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> bycause,<note place="margin">
                     <hi>The diffe<g ref="char:EOLhyphen"/>rence be<g ref="char:EOLhyphen"/>twene Stra<g ref="char:EOLhyphen"/>taruhme<g ref="char:EOLhyphen"/>trie and Tacticie.</hi>
                  </note> there, is necessary the wisedome and fore<g ref="char:EOLhyphen"/>sight, to what purpose he so ordreth the men: and Skillfull hability, also, for any occasion, or purpose, to deuise and vse the aptest and most necessary order, array and figure of his Company and Summe of men. By figure, I meane: as, either of a <hi>Perfect Square, Triangle, Circle, Ouale, long square,</hi> (of the Grekes it is called <hi>Etero<g ref="char:EOLhyphen"/>mekes</hi>) <hi>Rhombe, Rhomboïd, Lunular, Ryng, Serpentine,</hi> and such other Geometricall figures:</q>
Which, in warres, haue ben, and are to be vsed: for commodiousnes, ne<g ref="char:EOLhyphen"/>cessity, and auauntage &amp;c. And no small skill ought he to haue, that should make true report, or nere the truth, of the numbers and Summes, of footemen or horse<g ref="char:EOLhyphen"/>men, in the Enemyes ordring. A farre of, to make an estimate, betwene nere termes of More and Lesse, is not a thyng very rife, among those that gladly would
<pb facs="tcp:7062:13"/>
do it. Great pollicy may be vsed of the Capitaines,<note place="margin">
                  <hi>I.D. F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>end<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> you will finde it hard, to perform<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> my descripti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>his F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ate. But by Ch<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>graphie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> you may helpe your self<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> some <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hat: wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Figures knowne (in Sid<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd Angles) are not Regular: And where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Resolu<g ref="char:EOLhyphen"/>tion into Triangles can s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. &amp;c. And yet you will finde it strange to deale thus gener<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>lly with A<g ref="char:EOLhyphen"/>rithmeticall figures: and, that for Bat<g ref="char:EOLhyphen"/>tayle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ay. Their co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>tent<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> differ so much from like Geometr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>call Figur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s.</hi>
               </note> (at tymes fete, and in places conuenient) as to vse Figures, which make greatest shew, of so many as he hath: and vsing the aduauntage of the three kindes of vsuall spaces: (betwene footemen or horsemen) to take the largest: or when he would seme to haue few, (beyng ma<g ref="char:EOLhyphen"/>ny:) contrary wise, in Figure, and space. The Herald, Purseuant, Sergeant Royall, Capitaine, or who soeuer is carefull to come nere the truth herein<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, besides the Iudgement of his expert eye, his skill of Ordering <hi>Tacticall,</hi> the helpe of his Geo<g ref="char:EOLhyphen"/>metricall instrument: Ring, or Staffe Astronomicall: (commodiously framed for cariage and vse.) He may wonderfully helpe him selfe, by perspectiue Glasses. In which, (I trust) our posterity will proue more skillfull and expert, and to greater purposes, then in these dayes, can (almost) be credited to be possible.</p>
            <p>Thus haue I lightly passed ouer the Artificiall Feates, chiefly dependyng vpon vulgar <hi>Geometrie:</hi> &amp; commonly and generally reckened vnder the name of <hi>Geome<g ref="char:EOLhyphen"/>trie.</hi> But there are other (very many) <hi>Methodicall Artes,</hi> which, declyning from the purity, simplicitie, and Immateriality, of our Principall Science of <hi>Magnitudes:</hi> do yet neuertheles vse the great ayde, direction, and Method of the sayd principall Science, and haue propre names, and distinct: both from the Science of <hi>Geometrie,</hi> (from which they are deriued) and one from the other. As Per<g ref="char:EOLhyphen"/>spectiue, Astronomie, Musike, Cosmographie, Astrologie, Statike, Anthropographie, Trochilike, Helicosophie, Pneumatithmie, Me<g ref="char:EOLhyphen"/>nadrie, Hypogeiodie, Hydragogie, Horometrie, Zographie, Archi<g ref="char:EOLhyphen"/>tecture, Nauigation, Thaumaturgike and Archemastrie. I thinke it necessary, orderly, of these to giue some peculier descriptions: and withall, to touch some of their commodious vses, and so to make this Preface, to be a little swete, pleasant Nosegaye for you: to comfort your Spirites, beyng almost out of courage, and in despayre, (through brutish brute) Weenyng that <hi>Geometrie,</hi> had but serued for buildyng of an house, or a curious bridge, or the roufe of Westmin<g ref="char:EOLhyphen"/>ster hall, or some witty pretty deuise, or engyn, appropriate to a Carpenter, or a Ioyner &amp;c. That the thing is farre otherwise, then the world, (commonly) to this day, hath demed, by worde and worke, good profe wilbe made.</p>
            <p>Among these Artes, by good reason, Perspectiue ought to be had, ere of <hi>Astronomicall Apparences,</hi> perfect knowledge can be atteyned. And bycause of the prerogatiue of <hi>Light,</hi> beyng the first of <hi>Gods Creatures:</hi> and the eye, the light of our body, and his Sense most mighty, and his organ most Artificiall and <hi>Geome<g ref="char:EOLhyphen"/>tricall:</hi> At <hi>Perspectiue,</hi> we will begyn therfore. Perspectiue, is an Art Mathe<g ref="char:EOLhyphen"/>maticall, which demonstrateth the maner, and properties, of all Ra<g ref="char:EOLhyphen"/>diations Direct, Broken, and Reflected. This Description, or Notation, is brief<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> but it reacheth so farre, as the world is wyde. It concerneth all Creatures, all Actions, and passions, by Emanation of beames perfourmed. Beames, or na<g ref="char:EOLhyphen"/>turall lines, (here) I meane, not of light onely, or of colour (though they, to eye, giue shew, witnes, and profe, wherby to ground the Arte vpon) but also of other <hi>Formes,</hi> both <hi>Substantiall,</hi> and <hi>Accidentall,</hi> the certaine and determined actiue Ra<g ref="char:EOLhyphen"/>diall em<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nations. By this Art (omitting to speake of the highest pointes) we may vse our eyes, and the light, with greater pleasure: and perfecter Iudgement: both of thing<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, in l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ght seen; &amp; of other: which by like order of Lightes Radiations, worke and produce their effectes. We may be ashamed to be ignorant of the cause, why so sundry wayes our eye is deceiued, and abused: as, while the eye weeneth a rou<g ref="char:cmbAbbrStroke">̄</g>d Globe or Sphere (beyng farre of) to be a flat and plaine Circle, and so likewise iud<g ref="char:EOLhyphen"/>geth
<pb facs="tcp:7062:14"/>
a plaine Square, to be rou<g ref="char:cmbAbbrStroke">̄</g>d: supposeth walles parallels, to approche, a farre of: rofe and floure parallels, the one to bend downward, the other to rise vpward, at a little distance from you. Againe, of thinges being in like swiftnes of mouing, to thinke the nerer, to moue faster: and the farder, much slower. Nay, of two thinges, wherof the one (incomparably) doth moue swifter then the other, to deme the slower to moue very swift, &amp; the other to stand: what an error is this, of our eye? Of the Raynbow, both of his Colours, of the order of the colours, of the bignes of it, the place and heith of it, (&amp;c) to know the causes demonstratiue, is it not pleasant, is it not necessary? of two or three Sonnes appearing: of Blasing Sterres: and such like thinges: by naturall causes, brought to passe, (and yet neuertheles, of farder matter, Significatiue) is it not commodious for man to know the very true cause, &amp; occasion Naturall? Yea, rather, is it not, greatly, against the Souerainty of Mans nature, to be so ouershot and abused, with thinges (at hand) before his eyes? as with a Pecockes tayle, and a Doues necke: or a whole ore, in water, hol<g ref="char:EOLhyphen"/>den, to seme broken. Thynges, farre of, to seeme nere: and nere, to seme farre of. Small thinges, to seme great: and great, to seme small. One man, to seme an Army. Or a man to be curstly affrayed of his owne shad<g ref="char:EOLhyphen"/>dow. Yea, so much, to feare, that, if you, being (alone) nere a certaine glasse, and proffer, with dagger or sword, to foyne at the glasse, you shall suddenly be moued to giue backe (in maner) by reason of an Image,<note place="margin">
                  <hi>A marueilous Glasse. ☞</hi>
               </note> appearing in the ayre, betwene you &amp; the glasse, with like hand, sword or dagger, &amp; with like quicknes, foyning at your very eye, likewise as you do at the Glasse. Straunge, this is, to heare of: but more meruailous to behold, then these my wordes can signifie. And neuerthe<g ref="char:EOLhyphen"/>lesse by demonstration Opticall, the order and cause therof, is certified: euen so, as the effect is consequent. Yea, thus much more, dare I take vpon me, toward the sa<g ref="char:EOLhyphen"/>tisfying of the noble courrage, that longeth ardently for the wisedome of Causes Naturall: as to let him vnderstand, that, in London, he may wish his owne eyes, haue profe of that, which I haue sayd herein. A Gentleman, (which, for his good seruice,<note place="margin">
                  <hi>S.W.P.</hi>
               </note> done to his Countrey, is famous and honorable: and for skill in the Ma<g ref="char:EOLhyphen"/>thematicall Sciences, and Languages, is the Od man of this kind. &amp;c.) euen he, is hable: and (I am sure) will, very willingly, let the Glasse, and profe be sene: and so I (here) request him: for the encrease of wisedome, in the honorable: and for the stopping of the mouthes malicious: and repressing the arrogancy of the ignorant. Ye may easily gesse, what I meane. This Art of <hi>Perspectiue,</hi> is of that excellency, and may be led, to the certifying, and executing of such thinges, as no man would easily beleue: without Actuall profe perceiued. I speake nothing of <hi>Naturall Phi<g ref="char:EOLhyphen"/>losophie,</hi> which, without <hi>Perspectiue,</hi> can not be fully vnderstanded, nor perfectly at<g ref="char:EOLhyphen"/>teined vnto. Nor, of <hi>Astronomie</hi>: which, without <hi>Perspectiue,</hi> can not well be groun<g ref="char:EOLhyphen"/>ded: Nor <hi>Astrologie,</hi> naturally Verified, and auouched. That part hereof, which dealeth with Glasses (which name, Glasse, is a generall name, in this Arte, for any thing, from which, a Beame reboundeth) is called <hi>Catoptrike</hi> and hath so many v<g ref="char:EOLhyphen"/>ses, both merueilous, and proffitable: that, both, it would hold me to long, to no<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> therin the principall conclusions, all ready knowne: And also (perchaunce) some thinges, might lacke due credite with you: And I, therby, to leese my labor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and you, to slip into light Iudgement<note place="margin">☞</note>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Before you haue learned sufficiently the powre of Nature and Arte.</p>
            <p>NOw, to procede: Astronomie, is an Arte Mathematicall, which demonstrateth the distance, magnitudes, and all naturall motions, apparences, and passions propre to the Planets and fixed St<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>rtes: for
<pb facs="tcp:7062:14"/>
any time past, present and to come: in respect of a certaine Horizon, or without respect of any Horizon. By this Arte we are certified of the di<g ref="char:EOLhyphen"/>stance of the Starry Skye, and of eche <hi>Planete</hi> from the Centre of the Earth: and of the greatnes of any Fixed starre sene, or <hi>Planete,</hi> in respect of the Earthes greatnes. As, we are sure (by this Arte) that the Solidity, Massines and Body of the <hi>Sonne,</hi> conteineth the quantitie of the whole Earth and Sea, a hundred thre score and two times, lesse by 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> one eight parte of the earth. But the Body of the whole earthly globe and Sea, is bigger then the body of the Mone, three and forty times lesse by 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> of the Mone. Wherfore the <hi>Sonne</hi> is bigger then the <hi>Mone,</hi> 7000 times, lesse, by 59 <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>/164 that is, precisely 6940 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>5/<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> bigger then the <hi>Mone.</hi> And yet the vnskillfull man, would iudge them a like bigge. Wherfore, of Necessity, the one is much farder from vs, then the other. The <hi>Sonne</hi>; when he is fardest from the earth (which, now, in our age, is, when he is in the 8. degree, of Cancer) is, 1179 Semidiameters of the Earth, distante. And the <hi>Mone</hi> when she is fardest from the earth, is 68 Semidiameters of the earth and 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> The nerest, that the <hi>Mone</hi> com<g ref="char:EOLhyphen"/>meth to the earth, is Semidiameters 52 ¼ The distance of the Starry Skye is, fro<g ref="char:cmbAbbrStroke">̄</g> vs, in Semidiameters of the earth 20081 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Twenty thousand fourescore, one, and almost a halfe. Subtract from this, the <hi>Mones</hi> nerest distance, from the Earth: and therof remaineth Semidiameters of the earth 20029 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Twenty thousand nine and twenty and a quarter.<note place="margin">
                  <hi>Note.</hi>
               </note> So thicke is the heauenly Palace, that the <hi>Pla<g ref="char:EOLhyphen"/>netes</hi> haue all their exercise in, and most meruailously perfourme the Commau<g ref="char:cmbAbbrStroke">̄</g>de<g ref="char:EOLhyphen"/>ment and Charge to them giuen by the omnipotent Maiestie of the king of kings. This is that, which in <hi>Genesis</hi> is called <hi>Ha Rakia.</hi> Consider it well. The Semidia<g ref="char:EOLhyphen"/>meter of the earth; coteineth of our common miles 3436 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> three thousand, foure hundred thirty six and foure eleuenth partes of one myle: Such as the whole earth and Sea, round about, is 21600. One and twenty thousand six hundred of our myles. Allowyng for euery degree of the greatest circle, thre score myles. Now if you way well with your selfe but this litle parcell of f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ute <hi>Astronomicall,</hi> as con<g ref="char:EOLhyphen"/>cerning the bignesse, Dist<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nces of <hi>Sonne, Mone, Sterry Sky,</hi> and the huge massines of <hi>Ha Rakia,</hi> will you not finde your Consciences moued, with the kingly Prophet, to sing the confession of Gods Glory, and say, <hi>The Heauens declare the glo<g ref="char:EOLhyphen"/>ry of God, and the Firmament [Ha-Rakia] sheweth forth the workes of his handes.</hi> And so forth, for those fiue first staues, of that kingly Psalme. Well, well, It is time for some to lay hold on wisedome, and to Iudge truly of thinges: and not so to ex<g ref="char:EOLhyphen"/>pound the Holy word, all by Allegories: as to Neglect the wisedome, powre and Goodnes of God, in, and by his Creatures, and Creation to be seen and learned. By parables and Analogies of whose natures and properties, the course of the Ho<g ref="char:EOLhyphen"/>ly Scripture, also, declareth to vs very many, Mysteries. The whole Frame of Gods Creatures; (which is the whole world,) is to vs, a bright glasse: from which, by re<g ref="char:EOLhyphen"/>flexion, reboundeth to our knowledge and perceiuerance, Beames, and Radiati<g ref="char:EOLhyphen"/>ons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> representing the Image of his Infinite goodnes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Omnipote<g ref="char:cmbAbbrStroke">̄</g>cy, and wisedome. And we therby, are taught and persuaded to Glorifie our Creator as God: and be thankefull therfore. Could the Heathenistes finde these vses, of these most pure, beawtifull, and Mighty Corporall Creatures: and shall we after that the true <hi>Sonne</hi> of right wisenesse is risen aboue the <hi>Horizon,</hi> of our tempor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ll <hi>Hemispharie,</hi> and hath so abundantly streamed into our hartes, the direct beames of his goodnes, mercy, and grace: Whose heat All Creatures feele: Spirituall and Corpor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ll<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Visible and
<pb facs="tcp:7062:15"/>
Inuisible: Shall we (I say) looke vpon the <hi>Heauen, Sterres,</hi> and <hi>Planets,</hi> as an Oxe and an Asse doth: no furder carefull or inquisitiue, what they are: why were they Cre<g ref="char:EOLhyphen"/>ated, How do they execute that they were Created for? Seing, All Creatures, were for our sake created: and both we, and they, Created, chiefly to glorifie the Al<g ref="char:EOLhyphen"/>mighty Creator: and that, by all meanes, to vs possible. <hi>Nolite ignorare</hi> (saith <hi>Plat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in Epinomis</hi>) <hi>Astronomiam, Sapientissimu<g ref="char:cmbAbbrStroke">̄</g> quiddam esse. Be ye not ignorant, Astro<g ref="char:EOLhyphen"/>nomie to be a thyng of excellent wisedome. Astronomie,</hi> was to vs, from the be<g ref="char:EOLhyphen"/>ginning commended, and in maner commaunded by God him selfe. In asmuch as he made the <hi>Sonne, Mone,</hi> and <hi>Sterres,</hi> to be to vs, for <hi>Signes,</hi> and knowledge of Sea<g ref="char:EOLhyphen"/>sons, and for Distinctions of Dayes, and yeares. Many wordes nede not. But I wish, euery man should way this word, <hi>Signes.</hi> And besides that, conferre it also with the tenth Chapter of <hi>Hieremie.</hi> And though Some thinke, that there, they haue found a rod: Yet Modest Reason, will be indifferent Iudge, who ought to be beaten therwith, in respect of our purpose. Leaui<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>g that: I pray you vnderstand this: that without great diligence of Obseruation, examination and Calculation, their periods and courses (wherby <hi>Distinction</hi> of Seasons, yeares, and New Mones might precisely be knowne) could not exactely be certified. Which thing to per<g ref="char:EOLhyphen"/>forme, is that <hi>Art,</hi> which we here haue Defined to be <hi>Astronomie.</hi> Wherby, we may haue the distinct Course of Times, dayes, yeares, and Ages: aswell for Consi<g ref="char:EOLhyphen"/>deratio<g ref="char:cmbAbbrStroke">̄</g> of Sacred Prophesies, accomplished in due time, foretold: as for high My<g ref="char:EOLhyphen"/>sticall Solemnities holding: And for all other humaine affaires, Conditions, and couenantes, vpon certaine time, betwene man and man<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> with many other great vses: Wherin, (verely), would be great incertainty, Confusion, vntruth, and bru<g ref="char:EOLhyphen"/>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ish Barbarousnes: without the wonderfull diligence and skill of this Arte: conti<g ref="char:EOLhyphen"/>nually learning, and determining Times, and periodes of Time, by the Record of the heauenly booke, wherin all times are written: and to be read with an <hi>Astron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>micall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>taffe,</hi> in stede of a fest<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e.</p>
            <p>Musike, of Motion, hath his Originall cause: Therfore, after the motions most swift, and most Slow, which are in the Firmament, of Nature performed: and vnder the <hi>Astronomers Consideration:</hi> now I will Speake of an other kinde of <hi>Motion,</hi> producing sound, audible, and of Man numerable. <hi>Musike</hi> I call here that <hi>Science,</hi> which of the Grekes is called <hi>Harmonice.</hi> Not medling with the Controuersie be<g ref="char:EOLhyphen"/>twene the auncient <hi>Harmonistes,</hi> and <hi>Ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>onistes.</hi> Musike is a Mathematicall Science, which teacheth; by sense and reason, perfectly to iudge, and order the diuersities of soundes, hye and low. <hi>Astronomie</hi> and <hi>Musike</hi> are Sisters; saith <hi>Plato.</hi> As, for <hi>Astronomie,</hi> the eyes: So, for <hi>Harmonious Motion,</hi> the cares were made. But as <hi>Astronomie</hi> hath a more diuine Contemplation, and co<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>modity, then mortall eye can perceiue: So, is <hi>Musike</hi> to be considered, that the <note place="margin">1.</note> Minde may be pref<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ed, before the eare. And from audible sound, we ought to ascende, to the examination: which numbers are <hi>Harmonious,</hi> and which not. And why, either, the one are: or the other are not. I could at large, in the heauenly <note place="margin">2.</note> motions and distances, describe a meruallous Harmonie, of <hi>Pythagoras.</hi> Harpe <milestone type="tcpmilestone" unit="unspecified" n="4"/> with <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ight stringes. Also, somwhat might be sayd of <hi>Mercurius</hi> 
               <note place="margin">3.</note> two Harpes, eche of foure Stringes Elementall. And very straunge matter, might be alledged of the <hi>Harmo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e,</hi> to our <note place="margin">5.</note> Spirituall part appropriate. As in <hi>Pt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lom<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>us</hi> third boke, in the fourth and sixth Chapters may appeare. <note place="margin">6.</note> And what is the cause of the apt bonde, and frendly felowship, of the Intellectuall and Mentall part of vs, with our grosse &amp; corruptible body: but a certaine Meane, and <hi>Harmonious Spiritualitie, with
<pb facs="tcp:7062:15"/>
both participatyng, &amp; of both (i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a maner) <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>? In the</hi> 
               <note place="margin">7.</note> 
               <hi>T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne of Mans voyce, and also</hi> 
               <note place="margin">8.</note> 
               <hi>the sound of Instrument,</hi> what might be sayd, of <hi>Harmonie:</hi> No common Musicien would lightly beleue. But of the sundry Mixture (as I may terme it) and concurse,<note place="margin">
                  <hi>I.D. Read in A<g ref="char:EOLhyphen"/>ristotle his</hi> 8. <hi>booke of Politikes: the</hi> 5, 6, <hi>and</hi> 7. <hi>chapters. Where you shall haue some occasion farder to thinke of Mu<g ref="char:EOLunhyphen"/>sike, than commonly is thought.</hi>
               </note> diuerse collation, and Application of these <hi>Harmonies:</hi> as of thre, foure, fiue, or mo: Maruailous haue the effectes ben: and yet may be founde, and produced the like: with some proportionall consideration for our time, and being: in respect of the State, of the thinges then: in which, and by which, the wondrous effectes were wrought. <hi>Democritus</hi> and <hi>Theophrastus</hi> affirmed, that, by <hi>Musike,</hi> griefes and di<g ref="char:EOLhyphen"/>seases of the Minde, and body might be cured, or inferred. And we finde in Re<g ref="char:EOLhyphen"/>corde, that <hi>Terpander, Arion, Ismenias, Orpheus, Amphion, Dauid, Pythagoras, Empedo<g ref="char:EOLhyphen"/>cles, Asclepiades</hi> and <hi>Timotheus,</hi> by <hi>Harmonicall</hi> Consona<g ref="char:cmbAbbrStroke">̄</g>cy, haue done, and brought to pas, thinges, more then meruailous, to here of. Of them then, making no far<g ref="char:EOLhyphen"/>der discourse, in this place: Sure I am, that Common <hi>Musike,</hi> commonly vsed, is found to the <hi>Musiciens</hi> and Hearers, to be so Commodious and pleasant, That if I would say and dispute, but thus much: That it were to be otherwise vsed, then it is, I should finde more repreeuers, then I could finde priuy, or skilfull of my mea<g ref="char:EOLhyphen"/>ning. In thinges therfore euident, and better knowen, then I can expresse: and so allowed and liked of, (as I would wish, some other thinges, had the like hap) I will spare to enlarge my lines any farder, but consequently follow my purpose.</p>
            <p>Of Cosmographie, I appointed briefly in this place, to geue you some intelligence. Cosmographie, is the whole and perfect description of the heauenly, and also elementall parte of the world, and their ho<g ref="char:EOLhyphen"/>mologall application, and mutuall collation necessarie. This Art, requireth <hi>Astronomie, Geographie, Hydrographie</hi> and <hi>Musike.</hi> Therfore, it is no small Arte, nor so simple, as in common practise, it is (slightly) considered. This matcheth Heauen, and the Earth, in one frame, and aptly applieth parts Correspo<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>dent: So, as, the Heauenly Globe, may (in practise) be duely described vpon the Geographicall, and Hydrographicall Globe. And there, for vs to consider an <hi>AEquinoctiall Circle, an Ecliptike line, Colures, Poles, Sterr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s</hi> in their true Longitudes, Latitudes, Declinations, and Verticalitie: also Climes, and Parallels: and by an <hi>Ho<g ref="char:EOLhyphen"/>rizon</hi> annexed, and reuolution of the earthly Globe (as the Heauen, is<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> by the <hi>Pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>mouan<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,</hi> caried about in 24<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> aequall Houres) to learne the Risinges and Settinges of Sterres (of <hi>Virgill</hi> in his <hi>Georgikes</hi>: of <hi>Hes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>od</hi>: of <hi>Hippocrates</hi> in his <hi>Medicinall Sphare,</hi> to Perdic<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> King of the Macedonians: of <hi>Diocles,</hi> to King <hi>Antigonus,</hi> and of other fa<g ref="char:EOLhyphen"/>mous <hi>Philosophers</hi> prescribed) a thing necessary, for due manuring of the earth, for <hi>Nauigation,</hi> for the Alteration of mans body<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> being, whole, Sicke, wounded, or bru<g ref="char:EOLhyphen"/>sed. By the Reuolution, also, or mouing of the Globe Cosmographicall, the Rising and Setting of the Sonne: the Lengthes, of dayes and nightes: the Houres and times (both night and day) are knowne: with very many other pleasant and necessary vses: Wherof, some are knowne: but better remaine, for such to know and vse who of a sparke of true fire, can make a wonderfull bonfire, by applying of due matter, duely.<note place="margin">☜</note>
            </p>
            <p>Of Astrologie, here I make an Ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e, seuerall from <hi>Astronomie</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> not by new deuise, but by good reason and authoritie: for, Astrologie, is an Arte Mathematicall, which reasonably demonstrateth the operations and effectes, of the naturall beames, of light, and secrete influence: of the Sterres and Planets: in euery element and elementall body:
<pb facs="tcp:7062:16"/>
at all times, in any Horizon assigned. This Arte is furnished with ma<g ref="char:EOLhyphen"/>ny other great Artes and experiences: As with perfecte <hi>Perspectiue, Astronomie, Cosmographie, Naturall Philosophie</hi> of the 4. Elementes, the Arte of Graduation, and some good vndersta<g ref="char:cmbAbbrStroke">̄</g>ding in <hi>Musike:</hi> and yet moreouer, with an other great Arte, hereafter following, though I, here, set this before, for some considerations me mouing. Sufficient (you see) is the stuffe, to make this rare and secrete Arte, of: and hard enough to frame to the Conclusion Syllogisticall. Yet both the mani<g ref="char:EOLhyphen"/>folde and continuall trauailes of the most auncient and wise Philosophers, for the atteyning of this Arte: and by examples of effectes, to confirme the same: hath left vnto vs sufficient proufe and witnesse: and we, also, daily may perceaue, That mans body, and all other Elementall bodies, are altered, disposed, ordred, pleasu<g ref="char:EOLhyphen"/>red, and displeasured, by the Influentiall working of the <hi>Sunne, Mone,</hi> and the other Starres and Planets. And therfore, sayth <hi>Aristotle,</hi> in the first of his <hi>Meteorologicall</hi> bookes, in the second Chapter: <hi>Est autem necessariò Mundus iste, supernis lationibus ferè continuus. Vt, inde, vis eius vniuersa regatur. Ea siquidem Causa prima putanda omnibus est, vnde motus principium existit.</hi> That is: <hi>This [Elementall] World is of necessitie, almost, next adioyning, to the heauenly motions: That, from thence, all his vertue or force may be gouerned. For, that is to be thought the first Cause vnto all: from which, the beginning of motion, is.</hi> And againe, in the tenth Chapter. <hi>Op<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rtet igitur &amp; horum principia sumamus, &amp; causas omnium similiter. Principium igitur vy mouens, praecipuum<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> &amp; omnium primum, Circulus ille est, in quo manifeste Solis latio, &amp;c.</hi> And so forth. His <hi>Meteorologicall</hi> bookes, are full of argu<g ref="char:EOLhyphen"/>mentes, and effectuall demonstrations, of the vertue, operation, and power of the heauenly bodies, in and vpon the fower Elementes, and other bodies, of them (either perfectly, or vnperfectly) composed. And in his second booke, <hi>De Genera<g ref="char:EOLhyphen"/>tione &amp; Corruptione,</hi> in the tenth Chapter. <hi>Quocirca &amp; prima lati<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, Or<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>us &amp; Interi<g ref="char:EOLhyphen"/>tus causa non est: Sed obliqui Circuli latio: ea nam<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> &amp; continua est, &amp; duobus motibus fit:</hi> In Englishe, thus. <hi>Wherefore the vppermost motion, is not the cause of Gene<g ref="char:EOLhyphen"/>ration and Corruption, but the motion of the Zodiake: for, that, both, is con<g ref="char:EOLhyphen"/>tinuall, and is caused of two mouinges.</hi> And in his second booke, and second Chapter of hys <hi>Physikes. Homo nam<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> generat hominem, at<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> Sol. For Man</hi> (sayth he) <hi>and the Sonne, are cause of mans generation.</hi> Authorities may be brought, very many: both of 1000. 2000. yea and 3000. yeares Antiquitie: of great <hi>Philo<g ref="char:EOLhyphen"/>sophers, Expert, Wise,</hi> and godly men, for that Conclusion: which, daily and houre<g ref="char:EOLhyphen"/>ly, we men, may discerne and perceaue by sense and reason: All beastes do feele, and simply shew, by their actions and passions, outward and inward; All Plants, Herbes, Trees, Flowers, and Fruites. And finally, the Elementes, and all thinges of the Elementes composed, do geue Testimonie (as <hi>Aristotle</hi> sayd) that theyr <hi>Whole Dispositions, vertues, and naturall motions, depend of the Actiuitie of the heauenly motions and Influences. Whereby, beside the specificall order and forme, due to euery seede: and beside the Nature, propre to the Indiuiduall Ma<g ref="char:EOLhyphen"/>trix, of the thing produced: What shall be the heauenly Impression, the perfect and circumspecte Astrologien hath to Conclude.</hi> Not onely (by <hi>Apotelesmes</hi>) <gap reason="foreign">
                  <desc>〈 in non-Latin alphabet 〉</desc>
               </gap> but by Naturall and Mathematicall demonstration <gap reason="foreign">
                  <desc>〈 in non-Latin alphabet 〉</desc>
               </gap>. Whereunto, what Sciences are requis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>te (without exception) I partly haue here warned: And in my <hi>Brop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>
               </hi> besides other matter there disclosed) I haue Mathematically furni<g ref="char:EOLhyphen"/>shed vp the whole Method: To this our age, not so carefully handled by any, that
<pb facs="tcp:7062:16"/>
euer I saw, or heard of. I was, (for <note n="*" place="margin">
                  <hi>Anno.</hi> 1548 <hi>and</hi> 1549. <hi>in Louayn.</hi>
               </note> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>1. yeares ago) by certaine earnest disputati<g ref="char:EOLhyphen"/>ons, of the Learned <hi>G<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ardus M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>,</hi> and <hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> Goga<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a,</hi> (and other,) therto so prouoked: and (by my constant and inuincible zeale to the veritie) in obseruations of Heauenly Influencies (to the Min<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>te of time,) than, so diligent: And chiefly by the Supernaturall influence, from the Starre of Iacob, so directed; That any Modest and Sober Student, carefully and diligently sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ing for the Truth, will both finde &amp; co<g ref="char:cmbAbbrStroke">̄</g>fesse, therin, to be the Veritie, of these my wordes: And also become a Reaso<g ref="char:EOLhyphen"/>nable Reformer, of three Sortes of people: about these Influentiall Operations, greatly erring from the truth. Wherof, the one, is Light Beleuers, the other,<note place="margin">
                  <hi>Note.</hi>
               </note> Light Despisers, and the third Light Practisers. The first, &amp; most co<g ref="char:cmbAbbrStroke">̄</g>mon Sort, thinke the Heauen and Sterres, to be answerable to any their doutes or de<g ref="char:EOLhyphen"/>sires:<milestone type="tcpmilestone" unit="unspecified" n="1"/> which is not so: and, in dede, they, to much, ouer reache. The Second sorte thinke no Influentiall vertue (fro<g ref="char:cmbAbbrStroke">̄</g> the heauenly bodies) to beare any Sway in Ge<g ref="char:EOLhyphen"/>neration <milestone type="tcpmilestone" unit="unspecified" n="2"/> and Corruption, in this Elementall world. And to the <hi>Sunne, Mone</hi> and Sterres (being so many, so pure, so bright, so wonderfull bigge, so farre in distance, so manifold in their motions, so constant in their periodes. &amp;c.) they assigne a sleight, simple office or two, and so allow vnto the<g ref="char:cmbAbbrStroke">̄</g> (according to their capacities) as much vertue, and power Influentiall, as to the Signe of the <hi>Sunne, Mone,</hi> and seuen Sterres, hanged vp (for Signes) in London, for distinction of houses, &amp; such grosse helpes, in our wordly affaires: And they vnderstand not (or will not vnderstand) of the other workinges, and vertues of the Heauenly <hi>Sunne, Mone,</hi> and <hi>Sterres:</hi> not so much, as the Mariner, or Husband man<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> no, not so much as the <hi>Elephant</hi> doth, as the <hi>Cynocephalus,</hi> as the Por<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>entine do<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>h: nor will allow these perfect, and incor<g ref="char:EOLhyphen"/>ruptible mighty bodies, so much vertuall Radiation, &amp; Force, as they see in a litle peece of a <hi>Magnes stone:</hi> which, at great distance, sheweth his operation. And per<g ref="char:EOLhyphen"/>chaunce they thinke, the Sea &amp; Riuers (as the Thames) to be some quicke thing, and s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> to ebbe, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>d slow, run in and out, of them selues, at <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> owne fantasies. God helpe God helpe. Surely, these men come to short: and either are to dull: or willfully blind: or, perhaps, to malicious. The third man, is the common and vulgare <hi>Astrologien,</hi> or Practiser: who, being not duely, artificially, and perfectly <milestone type="tcpmilestone" unit="unspecified" n="3"/> furnished: yet, either for vaine glory, or gayne: or like a simple dolt, &amp; blinde Bay<g ref="char:EOLhyphen"/>ard<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> both in matter and maner, erreth: to the discredit of the <hi>Wary,</hi> and modest <hi>A<g ref="char:EOLhyphen"/>strologien:</hi> and to the robbing of those most noble corporall Creatures, of their Na<g ref="char:EOLhyphen"/>turall Vertue: being most mighty: most beneficiall to all elementall Generation, Corr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ion and the appa<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>nances<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and most Harmonious in thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Monarchie: For which thinges, being <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nowen, and modestly vsed: we might highly <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nd conti<g ref="char:EOLhyphen"/>nually glorifie God, with the princely Prophet saying. <hi>The Heauens declare the Glorie of God: who made the Heaue<g ref="char:cmbAbbrStroke">̄</g>s in his wisedome: who made the Sonne, for to haue dominion of the day: the Mone and Sterres to haue dominion of the nyght: whereby, Day to day <gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>reth tal<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> and night, to night declareth know<g ref="char:EOLhyphen"/>ledge. Prayse him, all ye St<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, and Light. Amen.</hi>
            </p>
            <p>IN order, now foloweth, of Statike, somewhat to say, what we meane by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hat name<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and what <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> doth, on <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> Art, depend. Statike, is an Arte Mathematicall, which demonstra<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>th the causes of heauynes, and lightnes of all thynges: and of motions and properties, to hea<g ref="char:EOLhyphen"/>uynes and lightnes belonging<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> And for asmuch as by the Bilanx, or Ba<g ref="char:EOLhyphen"/>lance (as the chief sensible Instrument,) Experience of these demonstrations may
<pb facs="tcp:7062:17"/>
be had: we call this Art, <hi>Statike:</hi> that is, <hi>the Experimentes of the Balance.</hi>
               <q>Oh; that men wist, what proffit, (all maner of wayes) by this Arte might grow, to the hable exa<g ref="char:EOLhyphen"/>miner, and diligent practiser. Thou onely, knowest all thinges precisely (O God) who hast made weight and Balance, thy Iudgement: who hast created all thinges in <hi>Number, Waight, and Measure</hi>: and hast wayed the mountaines and hils in a Ba<g ref="char:EOLhyphen"/>lance: who hast peysed in thy hand, both Heauen and earth. We therfore war<g ref="char:EOLhyphen"/>ned by the Sacred word, to Consider thy Creatures: and by that consideration, to wynne a glyms (as it were,) or shaddow of perceiuerance, that thy wisedome, might, and goodnes is infinite, and vnspeakable, in thy Creatures declared: And being farder aduertised, by thy mercifull goodnes, that, three principall wayes, were, of the, vsed in Creation of all thy Creatures, namely, <hi>Number, Waight</hi> and <hi>Measure,</hi> And for as much as, of <hi>Number</hi> and <hi>Measure,</hi> the two Artes (auncient, fa<g ref="char:EOLhyphen"/>mous, and to humaine vses most necessary,) are, all ready, sufficiently knowen and extant: This third key, we beseche thee (through thy accustomed goodnes,) that it may come to the nedefull and sufficient knowledge, of such thy Seruauntes, as in thy workemanship, would gladly finde, thy true occasions (purposely of the vsed) whereby we should glorifie thy name, and shew forth (to the weaklinges in faith) thy wondrous wisedome and Goodnes.</q>
Amen.</p>
            <p>Meruaile nothing at this pang (godly frend, you Gentle and zelous Student.) An other day, perchaunce, you will perceiue, what occasion moued me. Here, as now, I will giue you some ground, and withall some shew, of certaine commodi<g ref="char:EOLhyphen"/>ties, by this Arte arising. And bycause this Arte is rare, my wordes and practises might be to darke: vnleast you had some light, holden before the matter: and that, best will be, in giuing you, out of <hi>Archimedes</hi> demonstrations, a few principal Con<g ref="char:EOLhyphen"/>clusions, as foloweth.</p>
            <list>
               <label>1.</label>
               <item>The Superficies of euery Liquor, by it selfe consistyng, and in quyet, is Sphaericall: the centre whereof, is the same, which is the centre of the Earth.</item>
               <label>2.</label>
               <item>If Solide Magnitudes, being of the same bignes, or qua<g ref="char:cmbAbbrStroke">̄</g>titie, that any Liquor is, and hauyng also the same Waight: be let downe in<g ref="char:EOLhyphen"/>to the same Liquor, they will settle downeward, so, that no parte of them, shall be aboue the Superficies of the Liquor: and yet neuer<g ref="char:EOLhyphen"/>theles, they will not sinke vtterly downe, or drowne.</item>
               <label>3.</label>
               <item>If any Solide Magnitude beyng Lighter then a Liquor, be let downe into the same Liquor, it will settle downe, so farre into the same Liquor, that so great a quantitie of that Liquor, as is the parte of the Solid Magnitude, settled dow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e into the same Liquor <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is in Waight, aequall, to the waight of the whole Solid Magni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ude.</item>
               <label>4.</label>
               <item>Any Solide Magnitude, Lighter then a Liquor, forced downe
<pb facs="tcp:7062:17"/>
into the same Liquor, will moue vpward, with so great a power, by how much, the Liquor hauyng aequall quantitie to the whole Magnitude, is heauyer then the same Magnitude.</item>
               <label>5.</label>
               <item>Any Solid Magnitude, heauyer then a Liquor, beyng let downe into the same Liquor, will sinke downe vtterly: And wilbe in that Liquor, Lighter by so much, as is the waight or heauynes of the Liquor, hauing bygnes or quantitie, aequall to the Solid Magnitude.</item>
               <label>6.</label>
               <item>If any Solide Magnitude, Lighter then a Liquor,<note place="margin">
                     <hi>I.D. The Cutting of a Sphare according to any proportion as<g ref="char:EOLhyphen"/>signed may by this proposition be done Mechanically by tempering L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>quor to a certayne waight in respect of the waight of the Sphare <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> Swy<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>ng.</hi>
                  </note> be let downe into the same Liquor, the waight of the same Magnitude, will be, to the Waight of the Liquor. (Which is aequall in quantitie to the whole Magnitude,) in that proportion, that the parte, of the Mag<g ref="char:EOLhyphen"/>nitude settled downe, is to the whole Magnitude.</item>
            </list>
            <p>BY these verities, great Errors may be reformed, in Opinion of the Naturall Motion of thinges, Light and Heauy. Which errors, are in Naturall Philosophie (almost) of all me<g ref="char:cmbAbbrStroke">̄</g> allowed: to much trusting to Authority: and false Suppositions. As, Of any two bodyes, the heauyer, to moue downward faster then the lighter. This error, is not first by me, Noted:<note place="margin">
                  <hi>A common error<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> noted.</hi>
               </note> but by one <hi>Iohn Baptist de Be<g ref="char:EOLhyphen"/>nedictis.</hi> The chief of his propositions, is this: which seemeth a Paradox.</p>
            <p>If there be two bodyes of one forme, and of one kynde, aequall in quantitie or vnaequall, they will moue by aequall space,<note place="margin">
                  <hi>A paradox.</hi>
               </note> in aequall tyme: So that both theyr mouynges be in ayre, or both in water: or in any one Middle.</p>
            <p>Hereupon, in the feate of Gunnyng, certaine good discourses (otherwise) may receiue great amendement, and furderance.<note place="margin">
                  <hi>N. T.</hi>
               </note> In the entended purpose, also, allowing somwhat to the imperfection of Nature:<note place="margin">
                  <hi>The wonder<g ref="char:EOLhyphen"/>full vse of these Proposi<g ref="char:EOLhyphen"/>tions.</hi>
               </note> not aunswerable to the preci<g ref="char:EOLhyphen"/>senes of demonstration. Moreouer, by the foresaid propositions (wisely vsed.) The Ayre, the water, the Earth, the Fire, may be nerely, knowen, how light or hea<g ref="char:EOLhyphen"/>uy they are (Naturally) in their <gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap>gned partes: or in the whole. And then, to thinges Elementall, turning your practise: you may deale for the proportion of the Elementes, in the thinges Compounded. Then, to the proportions of the Hu<g ref="char:EOLhyphen"/>mours in Man: their waightes: and the waight of his bones, and flesh. &amp;c. Than, by waight, to haue consideration of the Force of man, any maner of way: in whole or in part. Then, may you, of Ships water drawing, diuersly, in the Sea and in fresh water, haue pleasant consideration: and of waying vp of any thing, sonken in Sea or in fresh water &amp;c. And (to lift vp your head a loft:) by waight, you may, as precisely, as by any instrument els, measure the Diameters of <hi>Sonne</hi> and <hi>Mone. &amp;c.</hi> Frende, I pray you, way these thinges, with the iust Balance of Reason. And you will finde Meruailes vpon Meruailes: And esteme one Drop of Truth (yea in Naturall Philosophie) more worth, then whole Libraries of Opinions, vndemon<g ref="char:EOLhyphen"/>strated: or not aunswering to Natures Law, and your experience. Leauing these
<pb facs="tcp:7062:18"/>
thinges, thus: I will giue you two or three, light practises, to great purpose<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and so finish my Annotation <hi>Staticall.</hi> In Mathematicall matters, by the Mechaniciens ayde, we will behold, here, the Commodity of waight. Make a Cube, of any one Vniforme:<note place="margin">
                  <hi>The practise Staticall, to know the pro<g ref="char:EOLhyphen"/>portion, be<g ref="char:EOLhyphen"/>twene the Cube, and the Sphare.</hi>
               </note> and through like heauy stuffe: of the same Stuffe, make a Sphaere or Globe, precisely, of a Diameter aequall to the Radicall side of the Cube. Your stuffe, may be wood, Copper, Tinne, Lead, Siluer. &amp;c. (being, as I sayd, of like na<g ref="char:EOLhyphen"/>ture, condition, and like waight throughout.) And you may, by Say Balance, haue prepared a great number of the smallest waightes: which, by those Balance can be discerned or tryed: and so, haue proceded to make you a perfect Pyle, com<g ref="char:EOLhyphen"/>pany &amp; Number of waightes: to the waight of six, eight, or twelue pound waight: most diligently tryed, all. And of euery one, the Content knowen, in your least waight, that is wayable. [They that can not haue these waightes of precisenes: may, by Sand, Vniforme, and well dusted, make them a number of waightes, some<g ref="char:EOLhyphen"/>what nere precisenes: by halfing euer the Sand: they shall, at length, come to a least common waight. Therein, I leaue the farder matter, to their discretion, whom nede shall pinche.] The <hi>Venetians</hi> consideration of waight, may seme precise enough:<note place="margin">I. D.</note> by eight descentes progressionall, <note n="*" place="margin">
                  <hi>For, so, haue you.</hi> 256. <hi>partes of a Graine.</hi>
               </note> halfing, from a grayne. Your Cube, Sphaere, apt Balance, and conuenient waightes, being ready: fall to worke. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>. First, way your Cube. Note the Number of the waight. Way, after that, your Sphaere. Note likewise, the Nu<g ref="char:cmbAbbrStroke">̄</g>ber of the waight. If you now find the waight of your Cube, to be to the waight of the Sphaere, as 21. is to 11: Then you see, how the Mechani<g ref="char:EOLhyphen"/>cien and <hi>Experimenter,</hi> without Geometrie and Demonstration, are (as nerely in effect) tought the proportion of the Cube to the Sphere: as I haue demonstrated it, in the end of the twelfth boke of <hi>Euclide.</hi> Often, try with the same Cube and Sphaere. Then, chaunge, your Sphaere and Cube, to an other matter: or to an other bignes: till you haue made a perfect vniuersall Experience of it. Possible it is, that you shall wynne to nerer termes, in the proportion.</p>
            <p>When you haue found this one certaine Drop of Naturall veritie, procede on, to Inferre, and duely to make assay, of matter depending. As, bycause it is well de<g ref="char:EOLhyphen"/>monstrated, that a Cylinder, whose heith, and Diameter of his base, is aequall to the Diameter of the Sphaere, is Sesquialter to the same Sphaere (that is, as 3. to 2:) To the number of the waight of the Sphaere, adde halfe so much, as it is: and so haue you the number of the waight of that Cylinder. Which is also Compre<g ref="char:EOLhyphen"/>hended of our former Cube: So, that the base of that Cylinder, is a Circle descri<g ref="char:EOLhyphen"/>bed in the Square, which is the base of our Cube. But the Cube and the Cy<g ref="char:EOLhyphen"/>linder, being both of one heith, haue their Bases in the same proportion, in the which, they are, one to an other, in their Massines or Soliditie. But, before, we haue two numbers, expressing their Massines, Solidities, and Quantities, by waight: wherfore, we haue <note n="*" place="margin">
                  <hi>The proportion of the Square to the Circle ins<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ribed.</hi>
               </note> the proportion of the Square, to the Circle, inscribed in the same Square. And so are we fallen into the knowledge sensible, and Expe<g ref="char:EOLhyphen"/>rimentall of <hi>Archimedes</hi> great Secret: of him, by great trauaile of minde, sought and found. Wherfore, to any Circle giuen, you can giue a Square aequall: <note n="*" place="margin">
                  <hi>The Squa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ing of the Cir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>le. M<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ha<g ref="char:EOLhyphen"/>ni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ally.</hi>
               </note> as I haue taught, in my Annotation, vpon the first proposition of the twelfth boke, And likewise, to any Square giuen, you may giue a Circle aequall: <note n="*" place="margin">
                  <hi>To any Squir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>.</hi>
               </note>If you describe a Circle, which shall be in that proportion, to your Circle inscribed, as the Square is to the same Circle: This, you may do, by my Annotations, vpon the second pro<g ref="char:EOLhyphen"/>position of the twelfth boke of <hi>Euclide,</hi> in my third Probleme there. Your dili<g ref="char:EOLhyphen"/>gence may come to a proportion, of the Square to the Circle inscribed, nerer the truth, then is the proportion of 14. to 11. And consider, that you may begyn at the Circle and Square, and so come to conclude of the Sphaere, &amp; the Cube, what
<pb facs="tcp:7062:18"/>
their proportion is: as now, you came from the Sph<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ere to the Circle. For, of Sil<g ref="char:EOLhyphen"/>uer, or Gold, or Latton Lamyns or plates (thorough one hole drawe<g ref="char:cmbAbbrStroke">̄</g>, as the maner is) if you make a Square figure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> &amp; way it: and then, describing theron, the Circle in<g ref="char:EOLunhyphen"/>scribed: &amp; cut of, &amp; file away, precisely (to the Circle) the ouerplus of the Square: you shall then, waying your Circle, see, whether the waight of the Square, be to your Circle, as 14. to 11. As I haue Noted, in the beginning of <hi>Euclides</hi> twelfth boke. &amp;c. after this resort to my last proposition, vpon the last of the twelfth. And there, helpe your selfe, to the end. And, here, Note this,<note place="margin">
                  <hi>Note Squaring of the Circle without know<g ref="char:EOLhyphen"/>ledge of the proportion be<g ref="char:EOLhyphen"/>twene Cir<g ref="char:EOLhyphen"/>cumference and Diame<g ref="char:EOLhyphen"/>ter.</hi>
               </note> by the way. That we may Square the Circle, without hauing knowledge of the proportion, of the Cir<g ref="char:EOLhyphen"/>cumference to the Diameter: as you haue here perceiued. And otherwayes also, I can demonstrate it. So that, many haue cumberd them selues superfluously, by trauailing in that point first, which was not of necessitie, first: and also very in<g ref="char:EOLhyphen"/>tricate. And easily, you may, (and that diuersly) come to the knowledge of the Circumference: the Circles Quantitie, being first knowen. Which thing, I leaue to your consideration: making hast to despatch an other Magistrall Probleme: and to bring it, nerer to your knowledge, and readier dealing with, then the world (be<g ref="char:EOLhyphen"/>fore this day,) had it for you, that I can tell of. And that is, <hi>A Mechanicall Dubblyng of the Cube: &amp;c.</hi>
               <note place="margin">
                  <hi>To Dubble the Cube re<g ref="char:EOLhyphen"/>dily: by Art Mechanicall: depending vp<g ref="char:EOLhyphen"/>pon Demon<g ref="char:EOLhyphen"/>stration Ma<g ref="char:EOLhyphen"/>thematicall.</hi>
               </note> Which may, thus, be done: Make of Copper plates, or Tyn plates, a foursquare vpright Pyramis, or a Cone: perfectly fashioned in the holow, within. Wherin, let great diligence be vsed, to ap<g ref="char:EOLhyphen"/>proche (as nere as may be) to the Mathematicall perfection of those figures. At their bases, let them be all open: euery where, els, most close, and iust to. From the vertex, to the Circumference of the base of the Cone: &amp; to the sides of the base of the Pyramis: Let 4. straight lines be drawen, in the inside of the Cone and Pyramis:<note place="margin">
                  <hi>I. D. The 4. sides of this Pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> must be 4. Isosceles Triangles <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lik<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>quall.</hi>
               </note> makyng at their fall, on the perimeters of the bases, equall angles on both sides them selues, with the sayd perimeters. These 4. lines (in the Pyra<g ref="char:EOLhyphen"/>mis: and as many, in the Cone) diuide: one, in 12. aequall partes: and an other, in 24. an other, in 60, and an other, in 100. (reckenyng vp from the vertex.) Or vse other numbers of diuision,<note place="margin">
                  <hi>I. D.</hi>
               </note> as experience shall reach you<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Then,<note n="*" place="margin">
                  <hi>In all work<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> with this Pyramis or Cone. Let their Situation<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be in all Pointe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and Condi<g ref="char:EOLhyphen"/>tions, a like, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> all one: while you are about <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ne worke. Els you will <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>.</hi>
               </note> set your Cone or Pyramis, with the vertex downward, perpendicularly, in respect of the Base. (Though it be otherwayes, it hindreth nothyng.) So let the<g ref="char:cmbAbbrStroke">̄</g> most stedily be stayed. Now, if there be a Cube, which you wold haue Dubbled. Make you a prety Cube of Copper, Siluer, Lead, Tynne, Wood, Stone, or Bone. Or els make a hollow Cube, or Cubi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> coffen, of Copper, Siluer, Tynne, or Wood &amp;c. These, you may so proportio<g ref="char:cmbAbbrStroke">̄</g> in respect of your Pyramis or Cone, that the Pyramis or Cone, will be hable to conteine the waight of them, in wa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>, 3. or 4. times: at the least: what stuff so euer they be made of<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Let not your Solid angle, at the vertex, be to sharpe: but that the water may come with ease, to the very vertex, of your hollow Cone or Pyramis. Put one of your Solid Cubes in a Balance apt:<note place="margin">
                  <hi>I. D.</hi>
               </note> take the waight therof ex<g ref="char:EOLhyphen"/>actly in water. Powre that water, (without losse) into the hollow Pyramis or Cone, quietly. Marke in your lines, what numbers the water Cutteth: Take the waight of the same Cube againe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> in the same kinde of water, which you had be<g ref="char:EOLhyphen"/>fore: put that<note n="*" place="margin">
                  <hi>Consider well whan you must put your wate<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> togyther: and whan, you must emp<g ref="char:EOLhyphen"/>ty you<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> first water<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> out of your Pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or Cone. El<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> you will <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>.</hi>
               </note> also, into the Pyramis or Cone, where you did put the first. Marke now againe, in what number or place of the lines, the water Cutteth them. Two
<pb facs="tcp:7062:19"/>
wayes you may conclude your purpose: it is to wete, either by numbers or lines. By numbers: as, if you diuide the side of your Fundamentall Cube into so many aequall partes, as it is capable of, conueniently, with your ease, and pre<g ref="char:EOLhyphen"/>cisenes of the diuision. For, as the number of your first and lesse line (in your hollow Pyramis or Cone,) is to the second or greater (both being counted from the vertex) so shall the number of the side of your Fundamentall Cube, be to the nu<g ref="char:cmbAbbrStroke">̄</g>ber belonging to the Radicall side, of the Cube, dubble to your Fun<g ref="char:EOLhyphen"/>damentall Cube: Which being multiplied Cubik wise, will sone shew it selfe, whe<g ref="char:EOLhyphen"/>ther it be dubble or no, to the Cubik number of your Fundamentall Cube. By lines, thus: As your lesse and first line, (in your hollow Pyramis or Cone,) is to the second or greater, so let the Radical side of your Fundame<g ref="char:cmbAbbrStroke">̄</g>tall Cube, be to a fourth proportionall line, by the 12. proposition, of the sixth boke of <hi>Euclide.</hi> Which fourth line, shall be the Rote Cubik, or Radicall side of the Cube, dubble to your Fundamentall Cube: which is the thing we desired. For this, may I (with ioy) say, EYPHKA, EYPHKA, EYPHKA: thanking the holy and glorious Trinity: hauing greater cause therto, then <note n="*" place="margin">
                  <p>
                     <hi>Vitruuius. Lib.</hi> 9. <hi>Cap.</hi> 3.</p>
                  <p>☞ <hi>God b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> than<g ref="char:EOLhyphen"/>ked <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>or this Inuention<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp; the frui<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n<g ref="char:EOLhyphen"/>suing.</hi>
                  </p>
               </note> 
               <hi>Archimedes</hi> had (for finding the fraude vsed in the Kinges Crowne, of Gold): as all men may easily Iudge: by the diuersitie of the frute following of the one, and the other. Where I spake before, of a hollow Cu<g ref="char:EOLhyphen"/>bik Coffen: the like vse, is of it: and without waight. Thus. Fill it with water, preci<g ref="char:EOLhyphen"/>sely full, and poure that water into your Pyramis or Cone. And here note the lines cutting in your Pyramis or Cone. Againe, fill your coffen, like as you did before. Put that Water, also, to the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Marke the second cutting of your lines. Now, as you proceded before, so must you here procede. <note n="*" place="margin">
                  <hi>Note.</hi>
               </note> And if the Cube, which you should Double, be neuer so great: you haue, thus, the proportion (in small) be<g ref="char:EOLhyphen"/>twene your two litle Cubes: And then, the side, of that great Cube (to be doubled) being the third, will haue the fourth, found, to it proportionall: by the 12. of the sixth of Eu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>lide.</p>
            <p>
               <note place="margin">
                  <hi>Note, as con<g ref="char:EOLhyphen"/>cerning the Sphaericall Super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies of the water.</hi>
               </note>Note, that all this while, I forget not my first Proposition Staticall, here rehear<g ref="char:EOLhyphen"/>sed: that, the Supersicies of the water, is Sphaericall. Wherein, vse your discretion: to the first line, adding a small heare breadth, more: and to the second, halfe a heare breadth more, to his length. For, you will easily perceaue, that the difference can be no greater, in any Pyramis or Cone, of you to be handled. Which you shall thus trye. <hi>For <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inding the swelling of the water aboue leuell.</hi>
               <q>Square the Semidiame<g ref="char:EOLhyphen"/>ter,<note place="margin">☞</note> from the Centre of the earth, to your first Waters Superficies. Square then, halfe the Subtendent of that watry Superficies (which Subtendent must haue the equall partes of his measure, all one, with those of the Semidiameter of the earth to your watry Superficies): Subtracte this square, from the first: Of the residue, take the Rote Square. That Ro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e, Subtracte from your first Semidiameter of the earth to your watry Superficies: that, which remaineth, is the heith of the water, in the middle, aboue the leuell.</q>
Which, you will finde, to be a thing insensible. And though it were greatly sensible,<note n="*" place="margin">
                  <hi>Note.</hi>
               </note> yet, by helpe of my sixt Theoreme vpon the last Proposition of Euclides twelfth booke, noted: you may reduce all, to a true Leuell. But, farther diligence, of you is to be vsed, against accidentall causes of the waters swelling: as by hauing (somwhat) with a moy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>t Sponge, before, made moyst your hollow Pyramis or Cone, will preuent an accidentall cause of Swelling, &amp;c. Experience will teach you abundantly: with great ease, pleasure, and co<g ref="char:cmbAbbrStroke">̄</g>moditie.</p>
            <p>
               <note place="margin">
                  <hi>Note this A<g ref="char:EOLhyphen"/>bridge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ent of Dubbling <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> Cube. <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</hi>
               </note>Thus, may you Double the Cube Mechanically, Treble it, and so forth, in any proportion. Now will I Abridge your paine, cost, and Care herein. Without all preparing of your Fundamentall Cubes: you may (alike) worke this Conclusion. For, that, was rather a kinde of Experimentall demo<g ref="char:cmbAbbrStroke">̄</g>stration, then the shortest way:
<pb facs="tcp:7062:19"/>
and all, vpon one Mathematicall Demonstration depending.
<q>Take water (as much as conueniently will serue your turne: as I warned before of your Funda<g ref="char:EOLhyphen"/>mentall Cubes bignes) Way it precisely. Put that water, into your Pyramis or Cone. Of the same kinde of water, then take againe, the same waight you had before: put that likewise into the Pyramis or Cone. For, in eche time, your mar<g ref="char:EOLhyphen"/>king of the lines, how the Water doth cut them, shall geue you the proportion be<g ref="char:EOLhyphen"/>twen the Radicall sides, of any two Cubes, wherof the one is Double to the other: working as before I haue taught you:<note n="*" place="margin">
                     <hi>Note. * ☜</hi>
                  </note>
               </q>
sauing that for you Fundamentall Cube his Radicall side: here, you may take a right line, at pleasure.</p>
            <p>Yet farther proceding with our droppe of Naturall truth: you may (now) geue Cubes, one to the other, in any proportio<g ref="char:cmbAbbrStroke">̄</g> geue<g ref="char:cmbAbbrStroke">̄</g>:<note place="margin">
                  <hi>To giue Cubes one to the o<g ref="char:EOLhyphen"/>ther in any proportion, Rationall or Irrationall.</hi>
               </note> Rationall or Ir<g ref="char:EOLhyphen"/>rationall: on this maner. Make a hollow Parallelipipedon of Copper or Tinne: with one Base wa<g ref="char:cmbAbbrStroke">̄</g>ting, or open: as in our Cubike Coffen. Fro<g ref="char:cmbAbbrStroke">̄</g> the bottome of that Parallelipipedon, raise vp, many perpendiculars, in euery of his fower sides. Now if any proportion be assigned you, in right lines:
<q>Cut one of your perpendiculars (or a line equall to it, or lesse then it) likewise: by the 10. of the sixth of Euclide. And those two partes, set in two sundry lines of those perpendiculars (or you may set them both, in one line) making their beginninges, to be, at the base: and so their lengthes to extend vpward. Now, set your hollow Parallelipipedon, vpright, perpendicularly, steadie. Poure in water, handsomly, to the heith of your shorter line. Poure that water, into the hollow Pyramis or Cone. Marke the place of the rising. Settle your hollow Parallelipipedon againe. Poure water into it: vnto the heith of the second line, exactly. Poure that water<note n="*" place="margin">
                     <hi>Empty<g ref="char:EOLhyphen"/>ing the first.</hi>
                  </note> duely into the hollow Pyramis or Cone: Marke now againe, where the water cutteth the same line which you marked before. For, there, as the first marked line, is to the se<g ref="char:EOLhyphen"/>cond: So shall the two Radicall sides be, one to the other, of any two Cubes: which, in their Soliditie, shall haue the same proportion, which was at the first as<g ref="char:EOLhyphen"/>signed: were it Rationall or Irrationall.</q>
            </p>
            <p>Thus, in sundry waies you may furnishe your selfe with such straunge and pro<g ref="char:EOLhyphen"/>fitable matter: which, long hath bene wished for. And though it be Naturally done and Mechanically: yet hath it a good Demonstration Mathematicall.<note place="margin">
                  <hi>The demonstrations of this Dubbling of the Cube, and o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the rest.</hi>
               </note> Which is this<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Alwaies, you haue two Like Pyramids: or two Like Cones, in the proporti<g ref="char:EOLhyphen"/>ons assigned: and like Pyramids or Cones, are in proportion, one to the other, in the proportion of their Homologall sides (or lines) tripled. Wherefore, if to the first, and second lines, found in your hollow Pyramis or Cone, you ioyne a third and a fourth, in continuall proportion: that fourth line, shall be to the first, as the greater Pyramis or Cone, is to the lesse: by the 33<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> of the eleuenth of Euclide. If Pyramis to Pyramis,<note place="margin">
                  <hi>I.D.</hi>
               </note> or Cone to Cone, be double, then shall <note n="*" place="margin">
                  <hi>Here <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> of the water.</hi>
               </note> Line to Line, be also double, &amp;c. But, as our first line, is to the second, so is the Radicall side of our Fundamentall Cube, to the Radicall side of the Cube to be made, or to be dou<g ref="char:EOLhyphen"/>bled: and therefore, to those twaine also, a third and a fourth line, in continuall proportion, ioyned: will geue the fourth line in that proportion to the first, as our fourth Pyramidall, or Conike line, was to his first: but that was double, or tre<g ref="char:EOLhyphen"/>ble, &amp;c. as the Pyramids or Cones were, one to an other (as we haue proued) ther<g ref="char:EOLhyphen"/>fore, this fourth, shalbe also double or treble to the first, as the Pyramids or Cones were one to an other: But our made Cube, is described of the second in proporti<g ref="char:EOLhyphen"/>on, of the fower proportionall lines: therfore<note n="*" place="margin">
                  <hi>By the 33. of the e<g ref="char:EOLhyphen"/>leuenth books of Euclide.</hi>
               </note> as the fourth line, is to the first, so is that Cube, to the first Cube: and we haue proued the fourth line, to be to the first, as the Pyramis or Cone, is to the Pyramis o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Cone: Wherefore the Cube is
<pb facs="tcp:7062:20"/>
to the Cube,<note place="margin">
                  <hi>I.D.</hi>
               </note> as Pyramis is to Pyramis, or Cone is to Cone. But we <note n="*" place="margin">
                  <hi>And your diligence in pra<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ise, can <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o (in waight of wa<g ref="char:EOLhyphen"/>te<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>forme it: Therefore. now, you ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> able to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eue good reason of your whole doing.</hi>
               </note> Suppose Py<g ref="char:EOLhyphen"/>ramis to Pyramis, or Cone to Cone, to be double or treble. &amp;c. Therfore Cube, is to Cube, double, or treble, &amp;c. Which was to be demonstrated. And of the Paralle<g ref="char:EOLhyphen"/>lipipedo<g ref="char:cmbAbbrStroke">̄</g>, it is euide<g ref="char:cmbAbbrStroke">̄</g>t, that the water Solide Parallelipipedons, are one to the other, as their heithes are, seing they haue one base. Wherfore the Pyramids or Cones, made of those water Parallelipipedons, are one to the other, as the lines are (one to the other) betwene which, our proportion was assigned. But the Cubes made of lines, after the proportio<g ref="char:cmbAbbrStroke">̄</g> of the Pyramidal or Conik <hi>homologall</hi> lines, are one to the other, as the Pyramides or Cones are, one to the other (as we before did proue) therfore, the Cubes made, shalbe one to the other, as the lines assigned, are one to the other: Which was to be demonstrated. Note.<note n="*" place="margin">
                  <hi>Note this Corollary.</hi>
               </note>This, my Demonstratio<g ref="char:cmbAbbrStroke">̄</g> is more generall, then onely in Square Pyramis or Cone: Consider well. Thus, haue I, both Mathematically and Mechanically, ben very long in wordes: yet (I trust) no<g ref="char:EOLhyphen"/>thing tedious to them, who, to these thinges, are well a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>fected. And verily I am forced (auoiding prolixitie) to omit sundry such things, easie to be practised: which to the Mathematicien, would be a great Threasure: and to the Mechanicien, no small gaine. <note n="*" place="margin">
                  <hi>The great Commodities following of these new In<g ref="char:EOLhyphen"/>uentions.</hi>
               </note> Now may you, Betwene two lines giuen, finde two middle proportionals, in Continuall proportion: by the hollow Paralleli<g ref="char:EOLhyphen"/>pipedon, and the hollow Pyramis, or Cone. Now, any Parallelipipedon rectangle being giuen: thre right lines may be found, proportionall in any propor<g ref="char:EOLhyphen"/>tion assigned, of which, shal be produced a Parallelipipedon, aequall to the Paralle<g ref="char:EOLhyphen"/>lipipedon giuen. Hereof, I noted somwhat, vpon the 36. proposition, of the 11. boke of <hi>Euclide.</hi> Now, all those thinges, which <hi>Vitruuius</hi> in his Architecture, specified hable to be done, by dubbling of the Cube<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Or, by finding of two middle propor<g ref="char:EOLhyphen"/>tionall lines, betwene two lines giuen, may easely be performed. Now, that Pro<g ref="char:EOLhyphen"/>bleme, which I noted vnto you, in the end of my Addition, vpon the 34. of the 11. boke of <hi>Euclide,</hi> is proued possible. Now may any regular body, be Transformed into an other, &amp;c. Now, any regular body: any Sphere, yea any Mixt Solid: and (that more is) Irregular Solides, may be made (in any proportio<g ref="char:cmbAbbrStroke">̄</g> assigned) like vnto the body, first giuen. Thus, of a <hi>Manneken,</hi> (as the <hi>Dutch</hi> Painters terme it) in the same <hi>Symmetrie,</hi> may a Giant be made: and that, with any gesture, by the Manne<g ref="char:EOLhyphen"/>ken vsed: and contrarywise. Now, may you, of any Mould, or Modell of a Ship, make one, of the same Mould (in any assigned proportion) bigger or lesser. Now, may you, of any<note n="*" place="margin">☞</note> Gunne, or little peece of ordinau<g ref="char:cmbAbbrStroke">̄</g>ce, make an other, with the same <hi>Symm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> (in all pointes) as great, and as little, as you will. Marke that: and thinke on it. Infinitely, may you apply this, so long sought for, and now so easily concluded: and withall, so willingly and frankly communi<g ref="char:EOLhyphen"/>cated to such, as faithfully deale with vertuous studies. Thus, can the Mathematicall minde,<note place="margin">
                  <hi>Such is the Fruite of the Mathemati<g ref="char:EOLhyphen"/>call Sciences and Artes.</hi>
               </note> deale Speculatiuely in his own Arte: and by good meanes, Mount aboue the cloudes and sterres<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> And thirdly, he can, by order, Descend, to frame Naturall thinges, to wonderfull vses: and when he list, retire home into his owne Centre: and there, prepare more Meanes, to Ascend or Descend by: and, all, to the glory of God, and our honest delectation in earth.</p>
            <p>Although, the Printer, hath looked for this Praeface, a day or two, yet could I not bring my pen from the paper, before I had giuen you comfortable warning, and brief instructions, of some of the Commodities, by <hi>Statike,</hi> hable to be reaped: In the rest, I will therfore, be as brief, as it is possible: and with all, describing them, somwhat accordingly. And that, you shall perceiue, by this, which in order com<g ref="char:EOLhyphen"/>meth
<pb facs="tcp:7062:20"/>
next. For, wheras, it <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s so ample and wonderful, that, an whole yeare long, one might finde fruitfull matter therin, to speake of: and also in pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ctise, is a Threa<g ref="char:EOLhyphen"/>sure endeles: yet will I glanse ouer it, with wordes very few.</p>
            <p>THis do I call Anthropographie. Which is an Art restored, and of my preferment to your Seruice. I pray you, thinke of it, as of one of the chief pointes, of Humane knowledge. Although it be, but now, first Co<g ref="char:cmbAbbrStroke">̄</g>firmed, with this new name: yet the matter, hath from the beginning, ben in consideration of all perfect Philosophers. Anthropographie, is the description of the Num<g ref="char:EOLhyphen"/>ber, Measure, Waight, figure, Situation, and colour of euery diuerse thing, conteyned in the perfect body of MAN: with certain know<g ref="char:EOLhyphen"/>ledge of the Symmetrie, figure, waight, Characterization, and due locall motion, of any parcell of the sayd body, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>signed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and of Nu<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>bers, to the sayd parcell appertainyng. This, is the one part of the Desini<g ref="char:EOLhyphen"/>tion, mete for this place: Sufficient to notifie, the particularitie, and excellency of the Arte: and why it is, here, ascribed to the Mathematicals. Yf the description of the heauenly part of the world, had <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>eculier Art, called <hi>Astronomie:</hi> If the de<g ref="char:EOLhyphen"/>scription of the earthly Globe, hath h<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>culier arte, called <hi>Geographie.</hi> If the Mat<g ref="char:EOLhyphen"/>ching of both, hath his peculier Arte, called <hi>Cosmographie:</hi> Which is the Descriptio<g ref="char:cmbAbbrStroke">̄</g> of the whole, and vniuersall frame of the world: Why should not the description of him, who is the Lesse world: and, fro<g ref="char:cmbAbbrStroke">̄</g> the beginning, called <hi>Microcosmus</hi> (that is. <hi>The Lesse World.</hi>
               <note place="margin">
                  <hi>MAN is the Lesse World.</hi>
               </note>) And for whose sake, and seruice, all bodily creatures els, were created: Who, also, participateth with Spirites, and Angels: and is made to the I<g ref="char:EOLhyphen"/>mage and similitude of <hi>God</hi>: haue his peculier Art? and be called the <hi>Arte of Artes:</hi> rather, then, either to want a name, or to haue to base and impropre a name? You must of sundry professions, borow or challenge home, peculier partes hereof: and farder procede: as, God, Nature, Reason and Experience shall informe you. The Anatomistes will restore to you, some part: The Physiognomistes, some: The Chy<g ref="char:EOLhyphen"/>romantistes some. The Metaposcopistes, some: The excellent, <hi>Albert Durer,</hi> a good part: the Arte of Perspectiue, will somwhat, for the Eye, helpe forward: <hi>Pythagoras, Hipocrates, Plato, Galenus, Meletius,</hi> &amp; many other (in certaine thinges) will be Con<g ref="char:EOLhyphen"/>tributaries. And farder, the Heauen, the Earth, and all other Creatures, will eche shew, and offer their Harmonious seruice, to fill vp, that, which wanteth hereof: and with your own Experience, concluding: you may Methodically register the whole, for the posteritie: Whereby, good profe will be had, of our Harmonious, and Microcosmicall constitution.<note n="*" place="margin">☜ <hi>Microcos<g ref="char:EOLhyphen"/>mus.</hi>
               </note> The outward Image, and vew hereof: to the Art of <hi>Zographie</hi> and Painting, to Sculpture, and Architecture: (for Church, House, Fort, or Ship) is most necessary and profitable: for that, it is the chiefe base and foundation of them. Looke in <note n="*" place="margin">
                  <hi>Lib.</hi> 3. <hi>Cap.</hi> 1.</note> 
               <hi>Vitruuius,</hi> whether I deale sincerely for your behoufe, or no. Looke in <hi>Albertus Durerus, De Symmetria humani Corporis.</hi> Looke in the 27. and 28. Chapters, of the second booke, <hi>De occulta Philosophia.</hi> Consi<g ref="char:EOLhyphen"/>der the <hi>Arke</hi> of <hi>Noe.</hi> And by that, wade farther. Remember the <hi>Delphicall Oracle NOSCE TEIPSVM (Knowe thy selfe)</hi> so long agoe pronounced: of so many a Philosopher repeated: and of the <hi>Wisest</hi> attempted: And then, you will perceaue, how long agoe, you haue bene called to the Schole, where this Arte might be learned. Well. I am nothing affrayde, of the disdayne of some such, as thinke Sciences and Artes, to be but Seuen. Perhaps, those Such may, with igno<g ref="char:EOLhyphen"/>rance, and shame enough, come short of them Seuen also: and yet neuerthelesse
<pb facs="tcp:7062:21"/>
they can not prescribe a certaine number of Artes: and in eche, certaine vnpassable boundes, to God, Nature, and mans Industrie. New Artes, dayly rise vp: and there was no such order taken,<note place="margin">☞</note> that, All Artes, should in one age, or in one land, or of one man, be made knowen to the world. Let vs embrace the giftes of God, and wayes to wisedome, in this time of grace, from aboue, continually bestowed on them, who thankefully will receiue them: <hi>Et bonis Omnia Cooperabuntur in bonum.</hi>
            </p>
            <p>Trochilike, is that Art Mathematicall, which demonstrateth the properties of all Circular motions, Simple and Compounde. And bycause the frute hereof, vulgarly receiued, is in Wheles, it hath the name of <hi>Trochilike:</hi> as a man would say, <hi>Whele Art.</hi> By this art, a Whele may be geuen which shall moue ones about, in any tyme assigned. Two Wheles may be giuen, whose turnynges about in one and the same tyme, (or equall tymes), shall haue, one to the other, any proportion appointed. By Wheles, may a straight line be described: Likewise, a Spirall line in plaine, Conicall Section lines, and other Irre<g ref="char:EOLhyphen"/>gular lines, at pleasure, may be drawen. These, and such like, are principall Con<g ref="char:EOLhyphen"/>clusions of this Arte: and helpe forward many pleasant and profitable Mechani<g ref="char:EOLhyphen"/>call workes: As Milles,<note place="margin">
                  <hi>Saw Milles.</hi>
               </note> to Saw great and very long Deale bordes, no man being by. Such haue I seene in Germany: and in the Citie of Prage: in the kingdome of Bohemia: Coyning Milles, Hand Milles for Corne grinding: And all maner of Milles, and Whele worke: By Winde, Smoke, Water, Waight, Spring, Man or Beast, moued. Take in your hand, <hi>Agricola <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ere Metallica:</hi> and then shall you (in all Mines) perceaue, how great nede is, of Whele worke. By Wheles, straunge workes and incredible, are done as will, in other Artes hereafter, appeare. A won<g ref="char:EOLhyphen"/>derfull example of farther possibilitie, and present commoditie, was sene in my time, in a certaine Instrument: which by the Inuenter and Artificer (before) was solde for xx. Talentes of Golde: and then had (by misfortune) receaued some iniu<g ref="char:EOLhyphen"/>rie and hurt: And one <hi>Ianellus</hi> of <hi>Cremona</hi> did mend the same, and presented it vn<g ref="char:EOLhyphen"/>to the Emperour <hi>Charles</hi> the fifth. <hi>Hieronymus Cardanus,</hi> can be my witnesse, that therein, was one Whele, which moued, and that, in such rate, that, in 7000. yeares onely, his owne periode should be finished. A thing almost incredible: But how farre, I keepe me within my boundes: very many men (yet aliue) can tell.</p>
            <p>Helicosophie, is nere Sister to <hi>Trochilike:</hi> and is, An Arte Mathema<g ref="char:EOLhyphen"/>ticall, which demonstrateth the designing of all Spirall lines in Plaine, on Cylinder, Cone, Sphaere, Conoid, and Sphaeroid, and their properties appertayning. The vse hereof, in <hi>Architecture,</hi> and di<g ref="char:EOLhyphen"/>uerse Instrumentes and Engines, is most necessary. For, in many thinges, the Skrue worketh the feate, which, els, could not be performed. By helpe hereof, it is <note n="*" place="margin">
                  <hi>Atheneus Lib.</hi> 5. <hi>cap.</hi> 8.</note> recorded, that, where all the power of the Citie of Syracusa, was not hable to moue a certaine Ship (being on ground) mightie <hi>Archimedes,</hi> setting to, his Skruish Engine, caused <hi>Hiero</hi> the king, by him self, at ease, to remoue her, as he would. Wherat, the King wondring:<note place="margin">
                  <hi>Proclus. Pag.</hi> 18.</note> 
               <gap reason="foreign">
                  <desc>〈 in non-Latin alphabet 〉</desc>
               </gap>. <hi>From this day, forward</hi> (said the King) <hi>Credit ought to be giuen to Archimedes, what so<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uer he sayth.</hi>
            </p>
            <p>Pneumatithmie demonstrateth by close hollow Geometri<g ref="char:EOLhyphen"/>call Figures, (regular and irregular) the straunge properties (in mo<g ref="char:EOLhyphen"/>tion or stay) of the Water, Ayre, Smoke, and Fire, in theyr co<g ref="char:cmbAbbrStroke">̄</g>tinuitie,
<pb facs="tcp:7062:21"/>
and as they are ioyned to the Elementes next them. This Arte, to the Naturall Philosopher, is very proffitable: to proue, that <hi>Vacuum,</hi> or <hi>Emptines</hi> is not in the world. And that, all Nature, abhorreth it so much: that, contrary to ordi<g ref="char:EOLhyphen"/>nary law, the Elementes will moue or stand. As, Water to ascend: rather then be<g ref="char:EOLhyphen"/>twene him and Ayre, Spac<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> or place should be left, more then (naturally) that qua<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>titie of Ayre requireth, or can fill. Againe, Water to hang, and not descend: rather then by descending, to leaue Emptines at his backe. The like, is of Fire and Ayre: they will descend: when, either, their Co<g ref="char:cmbAbbrStroke">̄</g>tinuitie should be dissolued: or their next Element forced from them. And as they will not be extended, to discontinuitie: So, will they not, nor yet of mans force, can be prest or pent, in space, not sufficient and aunswerable to their bodily substance. Great force and violence will they vse, to enioy their naturall right and libertie.<note place="margin">
                  <hi>To go to the bottom of the Sea without daunger.</hi>
               </note> Hereupon, two or three men together, by keping Ayre vnder a great Cauldron, and forcyng the same downe, orderly, may without harme descend to the Sea bottome: and continue there a tyme &amp;c. Where, Note, how the thicker Element (as the Water) giueth place to the thynner (as, is the ayre:) and receiueth violence of the thinner, in maner. &amp;c. Pumps and all maner of Bellowes, haue their ground of this Art: and many other straunge de<g ref="char:EOLhyphen"/>uises. As <hi>Hydraulica,</hi> Organes goyng by water. &amp;c. Of this Feat, (called common<g ref="char:EOLhyphen"/>ly <hi>Pneumatica,</hi>) goodly workes are extant, both in Greke, and Latin. With old and learned Schole men, it is called <hi>Scientia de pleno &amp; vacuo.</hi>
            </p>
            <p>Menadrie, is an Arte Mathematicall, which demonstrateth, how, aboue Natures vertue and power simple: Vertue and force may be multiplied: and so, to direct, to lift, to pull to, and to put o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> cast fro, any multiplied or simple, determined Vertue, Waight or Force: naturally, not, so, directible or moueable. Very much is this Art furdred by other Artes as, in some pointes, by <hi>Perspectiue:</hi> in some, by <hi>Statike:</hi> in some, by <hi>Trochilike:</hi> and in other, by <hi>Helicosophie:</hi> and <hi>Pneumatithmie.</hi> By this Art, all Granes, Gybbettes, &amp; Ingines to lift vp, or to force any thing, any maner way, are ordred: and the certaine cause of their force<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> is knowne: As, the force which one man hath with the Duche waghen Racke: therwith, to set vp agayne, a mighty waghen laden, being ouerthrowne. The force of the Crossebow Racke, is certain<g ref="char:EOLhyphen"/>ly, here, demonstrated. The reason, why one ma<g ref="char:cmbAbbrStroke">̄</g>, doth with a leauer, lift that, which Sixe men, with their handes onely, could no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, so easily do. By this Arte, in out common Cranes in London, where powre is to Crane vp, the waight of 2000. pound: by two Wheles more (by good order added) Arte concludeth, that there may be Craned vp 200000. pound waight &amp;c. So well knew <hi>Archimedes</hi> this Arte: that he alone, with his deuises and engynes, (twise or thrise) spoyled and discomfi<g ref="char:EOLhyphen"/>ted the whole<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Army and Hoste of the Romaines, besieging <hi>Syracusa, Marcus Mar<g ref="char:EOLhyphen"/>cellas the Consul,</hi>
               <note place="margin">
                  <p>Plut<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>bus in Mar<g ref="char:EOLhyphen"/>co M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rcello.</p>
                  <p>
                     <hi>Sy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>asius in Episto<g ref="char:EOLhyphen"/>lis.</hi>
                  </p>
                  <p>
                     <hi>Polybius.</hi>
                  </p>
                  <p>
                     <hi>Plinius.</hi>
                  </p>
                  <p>
                     <hi>Quint<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>lianus.</hi>
                  </p>
                  <p>
                     <hi>T. Liuius.</hi>
                  </p>
               </note> being their Generall Capitaine. Such huge Stones, so many, with such force, and so farre, did he with his <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ngynes hayle among them, out of the Citie. And by Sea likewise: though their Ships might come to the walls of <hi>Syra<g ref="char:EOLhyphen"/>cusa,</hi> yet hee vtterly confounded the Romaine N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>uye. What with his mighty Stones hurlyng: what with Pikes of<note n="*" place="margin">
                  <hi>Athena<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s.</hi>
               </note> 18 fote long, made like sliaftes: which he for<g ref="char:EOLhyphen"/>ced almost a quarter of a myle. What, with his catchyng hold of their Shyps, and hoysing them vp aboue the Water, and suddenly letting them fall into the Sea a<g ref="char:EOLhyphen"/>gaine: what with his<note n="*" place="margin">
                  <hi>Gale<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us. Anthemius.</hi>
               </note> Burning Glasses<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> by which he fired their other Shippes a far-of: what, with his other pollicies, deuises, and engines, he so manfully acquit him selfe: that all the Force, courage, and pollicie of the Romaines (for a great season)
<pb facs="tcp:7062:22"/>
could nothing preuaile, for the winning of Syracusa. Wherupon, the Romanes named <hi>Archimedes, Briareus,</hi> and <hi>Centimanus. Zonaras</hi> maketh mention of one <hi>Pro<g ref="char:EOLhyphen"/>clus,</hi> who so well had perceiued <hi>Archimedes</hi> Arte of <hi>Menadrie</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and had so well in<g ref="char:EOLhyphen"/>uented of his owne, that with his Burning Glasses,<note place="margin">
                  <hi>Burning Glasses.</hi>
               </note> being placed vpon the walles of Bysance, he multiplied so the heate of the Sunne, and directed the beames of the same against his enemies Nauie with such force, and so sodeinly (like lighte<g ref="char:EOLhyphen"/>ning) that he burned and destroyed both man and ship. And <hi>Dion</hi> specifieth of <hi>Priscus,</hi> a <hi>Geometricien</hi> in Bysance, who inuented and vsed sondry Engins, of Force multiplied: Which was cause, that the <hi>Emperour Senerus</hi> pardoned him, his life, af<g ref="char:EOLhyphen"/>ter he had wonne Bysance: Bycause he honored the Arte, wytt, and rare industrie of <hi>Priscus.</hi> But nothing inferior to the inuention of these engines of Force, was the inuention of Gunnes.<note place="margin">
                  <hi>Gunnes.</hi>
               </note> Which, from an English man, had the occasion and order of first inuenting: though in an other land, and by other men, it was first executed. And they that should see the record, where the occasion and order generall, of <q>Gunning, is first discoursed of, would thinke: that, small thinges, flight, and co<g ref="char:cmbAbbrStroke">̄</g>mon: comming to wise mens consideration, and industrious mens handling, may grow to be of force incredible.</q>
            </p>
            <p>Hypogeiodie, is an Arte Mathematicall, demonstratyng, how, vnder the Sphaericall Superficies of the earth, at any depth, to any perpendicular line assigned (whose distance from the perpendicular of the entrance: and the Azimuth, likewise, in respect of the said en<g ref="char:EOLhyphen"/>trance, is knowen) certaine way may be praescribed and gone: And how, any way aboue the Superficies of the earth designed, may vn<g ref="char:EOLhyphen"/>der earth, at any depth limited, be kept: goyng alwayes, perpendi<g ref="char:EOLhyphen"/>cularly, vnder the way, on earth designed: And, contrarywise, Any way, (straight or croked,) vnder the earth, beyng giuen: vppon the vtface, or Superficies of the earth, to Lyne out the same: So, as, from the Centre of the earth, perpendiculars drawen to the Sphaericall Superficies of the earth, shall precisely fall in the Correspondent pointes of those two wayes. This, with all other Cases and cir<g ref="char:EOLhyphen"/>cumstances herein, and appertenances, this Arte demonstrateth. This Arte, is very ample in varietie of Conclusions<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and very profitable sundry wayes to the Common Wealth. The occasion of my Inuenting this Arte, was at the request of two Gentlemen, who had a certaine worke (of gaine) vnder ground: and their groundes did ioyne ouer the worke: and by reason of the crokednes, diuers depthes, and heithes of the way vnder ground, they were in doubt, and at controuersie, vnder whose ground, as then, the worke was: The name onely (be<g ref="char:EOLhyphen"/>fore this) was of me published, <hi>De Itin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Subterranco:</hi> The rest, be at Gods will. For Pioners, Miners, Diggers for Mettalls, Stone, Cole, and for secrete passage<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> vnder ground, betwene place and place (as this land hath diuerse) and for other purposes, any man may easily perceaue, both the great fruite of this Arte, and also in this Arte, the great aide of Geometrie.</p>
            <p>Hydragogie, demonstrateth the possible leading of Water, by Natures lawe, and by artificiall helpe, from any head (being a Spring, standing, or running Water) to any other place assigned.
<pb facs="tcp:7062:22"/>
Long, hath this Arte bene in vse: and much thereof written: and very marueilous workes therein, performed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> as may ye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> appeare, in Italy: by the Ruynes remaining of the Aqueductes. In other places, of Riuers leading through the Maine land, Nauigable many a Mile. And in other places, of the marueilous forcinges of Wa<g ref="char:EOLhyphen"/>ter to Ascend which all, declare the great skill, to be required of him, who should in this Arte be perfecte, for all occasions of waters possible leading. To speake of the allowance of the Fall, for euery hundred foote: or of the Ventills (if the wa<g ref="char:EOLhyphen"/>ters labour be farre, and great) I neede not: Seing, at hand (about vs) many expert men can sufficiently testi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ie, in effecte, the order: though the Demonstration of the Necessiti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> thereof, they know not: Not yet, if they should be led, vp and downe, and about Mountaines, from the head of the Spring: and then, a place be<g ref="char:EOLhyphen"/>ing assigned: and of them, to be demaunded, how low or high, that last place is, in respecte of the head, from which (so crokedly, and vp and downe) they be come: Perhaps, they would not, or could not, very redily, or nerely assoyle that question. <hi>Geometrie</hi> therefore, is necessary to <hi>Hydragogie.</hi> Of the sundry wayes to force wa<g ref="char:EOLhyphen"/>ter to ascend, eyther by <hi>Tympane, Kettell mills, Skrue, Ctesibike,</hi> or such like: in <hi>Vi<g ref="char:EOLhyphen"/>truuius, Agricola,</hi> (and other,) fully, the maner may appeare. And so, thereby, also be most <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>uident, how the Artes, of <hi>Pneumatithmie, Helicosophie, Statik<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, Trochilike,</hi> and <hi>Menadrie,</hi> come to the furniture of this, in Speculation, and to the Commo<g ref="char:EOLhyphen"/>ditie of the Common Wealth, in practise.</p>
            <p>Horometrie, is an Arte Mathematicall, which demo<g ref="char:cmbAbbrStroke">̄</g>strateth, how, at all times appointed, the precise vsuall denominatio<g ref="char:cmbAbbrStroke">̄</g> of time, may be knowen, for any place assigned. These wordes, are smoth and plaine eas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e Englishe, but the reach of their meaning, is farther, then you woulde lightly imagine. Some part of this Arte, was called in olde time, <hi>Gnomonice:</hi> and of late, <hi>Ho<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ologiographia:</hi>) and in Englishe, may be termed, <hi>Dialling.</hi> Auncient is the vse, and more auncient, is the Inuention. The vse, doth well appeare to haue bene (at the least) aboue two thousand and three hundred yeare agoe: in <note place="margin">4. <hi>Reg.</hi> 20.</note> King <hi>Acha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> Diall, then, by the Sunne, shewing the distinction of time. By Sunne, Mone, and Sterres, this Dialling may be performed, and the precise Time of day or night knowen. But the demonstratiue delineation of these Dialls, of all sortes, requireth good skill, both of <hi>Astronomie,</hi> and <hi>Geometrie</hi> Elementall, Sphaericall, Phae<g ref="char:EOLhyphen"/>nomenall, and Conikall. Then, to vse the groundes of the Arte, for any regular Superficies, in any place offred: and (in any possible apt position therof) th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ron, to describe (all maner of wayes) how, vsuall howers, may be (by the <hi>Sunnes</hi> sha<g ref="char:EOLhyphen"/>dow) truely determined: will be found no sleight Painters worke. So to Paint, and prescribe the Sunnes Motion, to the breadth of a heare. In this Feate (in my youth) I Inuented a way, How in any Horizontall, Murall, or AEquino<g ref="char:EOLhyphen"/>ctiall Diall, &amp;c. At all howers (the Sunne shining) the Signe and De<g ref="char:EOLhyphen"/>gree ascendent, may be knowen. Which is a thing very necessary for the Rising of those fixed Sterres: whose Operation in the Ayre, is of great might, euidently. I speake no further, of the vse hereof. But forasmuch as, Mans affaires require knowledge of Times &amp; Momentes, when, neither Sunne, Mone, or Sterre, can be sene: Therefore, by Industrie Mechanicall, was inuented, first, how, by Wa<g ref="char:EOLhyphen"/>ter, running orderly, the Time and howers might be knowen: whereof, the famous <hi>Ctesibius,</hi> was Inuentor: a man, of <hi>Vitruuius,</hi> to the Skie (iustly) extolled. Then, after that, by Sand running, were howers measured: Then, by <hi>Trochilike</hi> with waight: And of late time, by <hi>Trochilike</hi> with Spring: without waight. All these,
<pb facs="tcp:7062:23"/>
by Sunne or Sterres direction (in certaine time) require ouersight and reformati<g ref="char:EOLhyphen"/>on, according to the heauenly AEquinoctiall Motion: besides the inaequalitie of their owne Operation. There remayneth (without parabolicall meaning herein) among the Philosophers,<note place="margin">
                  <hi>A perpetuall Motion.</hi>
               </note> a more excellent, more commodious, and more maruei<g ref="char:EOLhyphen"/>lous way, then all these: of hauing the motion of the Primouant (or first <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>quino<g ref="char:EOLhyphen"/>ctiall motion,) by Nature and Arte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Imitated: which you shall (by furder search in waightier studyes) hereafter, vnderstand more of. And so, it is tyme to finish this Annotation, of Tymes distinction, vsed in our common, and priuate affaires: The commoditie wherof, no man would want, that can tell, how to bestow his tyme.</p>
            <p>Zographie, is an Arte Mathematicall, which teacheth and de<g ref="char:EOLhyphen"/>monstrateth, how, the Intersection of all visuall Pyramides, made by any playne assigned, (the Centre, distance, and lightes, beyng de<g ref="char:EOLhyphen"/>termined) may be, by lynes, and due propre colours, represented. A notable Arte, is this and would require a whole Volume, to declare the proper<g ref="char:EOLhyphen"/>ty thereof: and the Commodities ensuyng. Great skill of <hi>Geometrie, Arithme<g ref="char:EOLhyphen"/>tike, Perspectiue,</hi> and <hi>Anthropographie,</hi> with many other particular Art<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s, hath the <hi>Zo<g ref="char:EOLhyphen"/>grapher,</hi> nede of, for his perfection. For, the most excellent Painter, (who is but the propre Mechanicien, &amp; Imitator sensible, of the Zographer) hath atteined to such perfection, that Sense of Man and beast, haue iudged thinges painted, to be things naturall, and not artificiall: aliue, and not dead. This Mechanicall Zographer (com<g ref="char:EOLhyphen"/>monly called the Painter) is meruailous in his skill: and seemeth to haue a certaine diuine power: As, of frendes absent, to make a frendly, present comfort: yea, and of frendes dead, to giue a continuall, silent presence: not onely with vs, but with our posteritie, for many Ages. And so procedyng, Consider, How, in Winter, he can shew you, the liuely vew of Sommers Ioy, and riches: and in Sommer, exhibite the countenance of Winters dolefull State, and nakednes. Cities, Townes, Fortes, Woodes, Armyes, yea whole Kingdomes (be they neuer so farre, or greate) can he, with ease, bring with him, home (to any mans Iudgement) as Paternes liuely, of the thinges rehearsed. In one little house, can he, enclose (with great pleasure of the beholders,) the portrayture liuely, of all visible Creatures, either on earth, or in the earth, liuing: or in the waters lying, Creping, slyding, or swimming: or of any <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>oule, or fly, in the ayre flying. Nay, in respect of the Starres, the Skie, the Cloudes: yea, in the shew of the very light it selfe (that Diuine Creature) can he match our eyes Iudgement, most nerely. What a thing is this? thinges not yet being, he can represent so, as, at their being, the Picture shall seame (in maner) to haue Created them. To what Artificer, is not Picture, a great pleasure and Commoditie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Which of them all, will refuse the Direction and ayde of Picture? The Architect, the Gold<g ref="char:EOLhyphen"/>smith, and the Arras Weauer: of Picture, make great account. Our liuely Herbals, our portraitures of birdes, beastes, and fishes: and our curious Anatomies, which way, are they most perfectly made, or with most pleasure, of vs beholden? Is it not, by Picture onely? And if Picture, by the Industry of the Painter, be thus commo<g ref="char:EOLhyphen"/>dious and meruailous: what shall be thought of <hi>Zographie,</hi> the Scholemaster of Pi<g ref="char:EOLhyphen"/>cture, and chief gouernor? Though I mencion not <hi>Sculpture,</hi> in my Table of Artes Mathematicall: yet may all men perceiue, How, that <hi>Picture</hi> and <hi>Sculpture,</hi> are Si<g ref="char:EOLhyphen"/>sters germaine: and both, right profitable, in a Commo<g ref="char:cmbAbbrStroke">̄</g> wealth. and of <hi>Sculpture,</hi> as<g ref="char:EOLhyphen"/>well as of Picture, excellent Artificers haue written great bokes in commendation. Witnesse I take, of <hi>Georgio Vasari, Pittore Aretino</hi>: of <hi>Pomponius Gauricus</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and other. To these two Artes, (with other,) is a certaine od Arte, called <hi>Althalmasat,</hi> much beholdyng: more, then the common <hi>Sculptor, Entayler, Keruer, Cut<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er, Grauer, Foun<g ref="char:EOLhyphen"/>der,</hi>
               <pb facs="tcp:7062:23"/>
or <hi>Paynter</hi> (<hi>&amp;c</hi>) know their Arte, to be commodious.</p>
            <p>Architecture, to many may seme not worthy, or not mete,<note place="margin">
                  <hi>An obiection.</hi>
               </note> to be reckned among the <hi>Artes Mathematicall</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> To whom, I thinke good, to giue some account of my so doyng. Not worthy, (will they say,) bycause it is but for building, of a house, Pallace, Church, Forte, or such like, grosse workes. And you, also, defined the <hi>Artes Mathematicall,</hi> to be such, as dealed with no Materiall or corruptible thing: and al<g ref="char:EOLhyphen"/>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>o did demonstrat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>uely procede in their faculty, by Number or Magnitude. First, you see, that I count, here, <hi>Architecture,</hi> among those <hi>Artes Mathematicall,</hi>
               <note place="margin">
                  <hi>The Answer.</hi>
               </note> which are Deriued from the Principals: and you know, that such, may deale with Na<g ref="char:EOLhyphen"/>turall thinges and sensib<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>a<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>er. Of which, some draw nerer, to the Simple and absolute Mathematicall Speculation, then other do.
<q>And though, the <hi>Architect</hi>
                  <note place="margin">☜</note> procureth, enformeth, &amp; directeth, the <hi>Mechanicien,</hi> to handworke, &amp; the building actuall, of house, Castell, or Pallace, and is chief Iudge of the same: yet, with him selfe (as chief <hi>Master</hi> and <hi>Architect,</hi>) remaineth the Demonstratiue reason and cause, of the Mechaniciens worke in Lyne, plaine, and Solid: by <hi>Geometricall, A<g ref="char:EOLhyphen"/>rithmeticall, Opticall, Musi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all, Astronomicall, Cosmographicall</hi> (&amp; to be brief) by all the former Deriued <hi>Artes Mathematicall,</hi> and other Naturall Artes, hable to be confir<g ref="char:EOLhyphen"/>med and stablished.</q>
If this be so<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>then, may you thinke, that <hi>Architecture,</hi> hath good and due allowance, in this honest Company of <hi>Artes Mathematicall</hi> Deriuatiue, I will, herein, craue Iudgement of two most perfect <hi>Architect<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s:</hi> the one, being <hi>Vi<g ref="char:EOLhyphen"/>truuius,</hi> the Romaine: who did write ten bookes thereof, to the Emperour <hi>Augu<g ref="char:EOLhyphen"/>stus</hi> (in whose daies our Heauenly Archemaster, was borne): and the other, <hi>Leo Baptista Albertus,</hi> a Florentine: who also published ten bookes therof. <hi>Archi<g ref="char:EOLhyphen"/>tectura</hi> (sayth <hi>Vitruuius</hi>) <hi>est Scientia pluribus disciplinis &amp; varijs eruditionibus ornata<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cuius Iudicio probantur omni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, qua ab cateris Artificibus perficiuntur opera.</hi> That is. Architecture, is a Science garnished with many doctrines &amp; diuerse instructions: by whose Iudgement, all workes, by other workmen finished, are Iudged. It followeth. <hi>Ea nascitur ex Fabric<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. &amp; Ratiocinatione. &amp;c. Ratiocinatio aute<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> est, quae, res fabricatas, Solertia ac ratione proportionis, demonstrare at<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> explicare potest. Architecture, groweth of Framing, and Reasoning. &amp;c. Rea<g ref="char:EOLhyphen"/>soning, is that, which of thinges framed, with forecast, and proportion: can make demonstration, and manifest declaration.</hi> Againe. <hi>Cùm, in omnibus enim re<g ref="char:EOLhyphen"/>bus, tùm maxime etiam in Architectura, haec duo insunt: qoud significatur, &amp; qoud signi<g ref="char:EOLhyphen"/>fica<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Significatur proposita res, de qua dicitur: hanc autem Significat Demonstratio, rati<g ref="char:EOLhyphen"/>onibus doctrinarum explicata. Forasmuch as, in all thinges: therefore chiefly in Architecture, these two thinges are: the thing signified: and that which sig<g ref="char:EOLhyphen"/>nifieth. The thing propounded, whereof we speake, is the thing Signified. But Demonstration, expressed with the reasons of diuerse doctrines, doth signi<g ref="char:EOLhyphen"/>fie the same thing.</hi> After that<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <hi>Vt literatus sit, peritus Graphidos, eruditus Geometriae, &amp; Optices non ignarus: instructus Arithmetica: historias complures nouerit, Philosophos diligenter audiuerit: Musicam sciuerit: Medicinae non sit ignarus, responsa Iurisperitoru<g ref="char:cmbAbbrStroke">̄</g> nouerit: Astrologiam, Caeli<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> rationes cognitas habeat. An Architect</hi> (sayth he) <hi>ought to vnderstand Languages, to be skilfull of Painting, well instructed in Geome<g ref="char:EOLhyphen"/>tri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, not ignorant of Perspectiue, furnished with Arithmetike, haue knowledge of many histories, and diligently haue heard Philosophers, haue skill of Mu<g ref="char:EOLhyphen"/>sike, not ignorant of Physike, know the aunsweres of Lawyers, and haue Astro<g ref="char:EOLhyphen"/>nomie,
<pb facs="tcp:7062:24"/>
and the courses Caelestiall, in good knowledge.</hi> He geueth reason, or<g ref="char:EOLhyphen"/>derly, wherefore all these Artes, Doctrines, and Instructions, are requisite in an ex<g ref="char:EOLhyphen"/>cellent <hi>Architect.</hi> And (for breuitie) omitting the Latin text, thus he hath. <hi>Secondly, it is behofefull for an Architect to haue the knowledge of Painting<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that he may the more easilie fashion out, in patternes painted, the forme of what worke he liketh. And Geometrie, geueth to Architecture many helpes: and first teacheth the Vse of the Rule, and the Cumpasse: wherby (chiefly and easilie) the descriptions of Buildinges, are despatched in Groundplats: and the directions of Squires, Leuells, and Lines. Likewise, by Perspectiue, the Lightes of the hea<g ref="char:EOLhyphen"/>uen, are well led, in the buildinges: from certaine quarters of the world. By Arithmetike, the charges of Buildinges are summed together: the measures are expressed, and the hard questions of Symmetries, are by Geometricall Meanes and Methods discoursed on. &amp;c. Besides this, of the Nature of thinges (which in Greke is called <gap reason="foreign">
                     <desc>〈 in non-Latin alphabet 〉</desc>
                  </gap>) Philosophie doth make declaration. Which, it is necessary, for an Architect, with diligence to haue learned: because it hath ma<g ref="char:EOLhyphen"/>ny and diuers naturall questions: as specially, in Aqueductes. For in their courses, leadinges about, in the leuell ground, and in the mountinges, the natu<g ref="char:EOLhyphen"/>rall Spirites or breathes are ingendred diuers wayes: The hindrances, which they cause, no man can helpe, but he, which out of Philosophie, hath learned the originall causes of thinges. Likewise, who soeuer shall read C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>esibius, or Ar<g ref="char:EOLhyphen"/>chimedes bookes, (and of others, who haue written such Rules) can not thinke, as they do: vnlesse he shall haue receaued of Philosophers, instructions in these thinges. And Musike he must nedes know: that he may haue vnderstanding, both of Regular and Mathematicall Musike: that he may temper well his Ba<g ref="char:EOLhyphen"/>listes, Catapultes, and Scorpions. &amp;c. Moreouer, the Brasen Vessels, which in Theatres, are placed by Mathematicall order, in ambries, vnder the steppes: and the diuersities of the soundes (which y<hi rend="sup">e</hi> Grecians call <gap reason="foreign">
                     <desc>〈 in non-Latin alphabet 〉</desc>
                  </gap>) are ordred according to Musicall Symphonies &amp; Harmonies: being distributed in y<hi rend="sup">e</hi> Circuites, by Di<g ref="char:EOLhyphen"/>atessaron, Diapente, and Diapason. That the conuenient voyce, of the players sound, whe<g ref="char:cmbAbbrStroke">̄</g> it came to these preparations, made in order, there being increased: with y<hi rend="sup">t</hi> increasing, might come more cleare &amp; pleasant, to y<hi rend="sup">e</hi> eares of the lokers on. &amp;c. And of Astronomie, is knowe<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> East, West, South, and North. The fashion of the heauen, the AEquinox, the Solsticie, and the course of the sterres. Which thinges, vnleast one know: he can not perceiue, any thyng at all, the reason of Ho<g ref="char:EOLhyphen"/>rologies. Seyng therfore this ample Science, is garnished, beautified and stored, with so many and sundry skils and knowledges: I thinke, that none can iustly ac<g ref="char:EOLhyphen"/>count them selues Architectes, of the suddeyne. But they onely, who from their childes yeares, ascendyng by these degrees of knowledges, beyng fostered vp with the atteynyng of many Languages and Artes, haue wonne to the high Taber<g ref="char:EOLhyphen"/>nacle of Archicture. &amp;c. And to whom Nature hath giuen such quicke Circum<g ref="char:EOLhyphen"/>spection, sharpnes of witt, and Memorie, that they may be very absolutely skill<g ref="char:EOLhyphen"/>full in Geometrie, Astronomie, Musike, and the rest of the Artes Mathemati<g ref="char:EOLhyphen"/>call:
<pb facs="tcp:7062:24"/>
Such surmount and p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>sse the callyng, and state, of Architectes: and are be<g ref="char:EOLhyphen"/>come Mathematiciens.</hi>
               <note place="margin">
                  <hi>A Mathe<g ref="char:EOLhyphen"/>maticien.</hi>
               </note> 
               <hi>&amp;c. And they are found, seldome. As, in tymes past, was Aristarchus Samius: Philolaus, and Archytas, Tarentynes: Apollonius Pergeus: Eratosthenes Cyreneus: Archimedes, and Scopas, Syracusians. Who also, left to theyr posteritie, many Engines and Gnomonicall workes: by numbers and natu<g ref="char:EOLhyphen"/>rall meanes, inuented and declared.</hi>
            </p>
            <p>Thus much, and the same wordes (in sense) in one onely Chapter of this Inco<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>parable <hi>Architect Vitrunius,</hi>
               <note place="margin">
                  <hi>Vitrunius.</hi>
               </note> shall you finde. And if you should, but take his boke in your hand, and slightly loke thorough it, you would say straight way: This is <hi>Geo<g ref="char:EOLhyphen"/>metrie, Arithmetike, Astronomie, Musike, Anthropographie, Hydragogie, Horometrie &amp;c.</hi> and (to co<g ref="char:cmbAbbrStroke">̄</g>clude) the Storehouse of all workma<g ref="char:cmbAbbrStroke">̄</g>ship. Now, let vs listen to our other Iudge, our Florentine, <hi>Leo Baptista:</hi> and narrowly consider, how he doth determine of <hi>Architecture. Sed ante<expan>
                     <am>
                        <g ref="char:abque"/>
                     </am>
                     <ex>que</ex>
                  </expan> vltra progrediar. &amp;c. But before I procede any further</hi> (sayth he) <hi>I thinke, that I ought to expresse, what man I would haue to bee al<g ref="char:EOLhyphen"/>lowed an Architect. For, I will not bryng in place a Carpenter: as though you might Compare him to the Chief Master<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of other Artes. For the hand of the Carpenter, is the Architectes Instrument.</hi>
               <note place="margin">
                  <hi>VVho is an Architect.</hi>
               </note>
               <q>But I will appoint the Architect to be that man, who hath the skill, (by a certaine and meruailous meanes and way,) both in minde and Imagination to determine: and also in worke to finish:
what workes so euer, by motion of waight, and cuppling and framyng together of bo<g ref="char:EOLhyphen"/>dyes, may most aptly be Commodious for the worthiest. Vses of Man. And that he may be able to performe these thinges, he hath nede of atteynyng and knowledge of the best, and most worthy thynges. &amp;c.</q>
               <hi>The whole Feate of Architecture in buildyng, confisteth in Lineamentes, and in Framyng. And the whole power and skill of Lineamentes, tendeth to this: that the right and absolute way may he had, of Coaptyng and ioyning Lines and angles: by which, the face of the buil<g ref="char:EOLhyphen"/>dyng or frame, may be comprehended and concluded. And it is the property of Lineamentes, to prescribe vnto buildynges, and euery part of them, an apt place, &amp; certaine nu<g ref="char:cmbAbbrStroke">̄</g>ber: a worthy maner, and a semely order: that, so, y<hi rend="sup">e</hi> whole <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orme and figure of the buildyng, may rest in the very Lineamentes. &amp;c. And we may prescribe in mynde and imagination the whole formes,</hi>
               <note n="*" place="margin">
                  <hi>The Im<g ref="char:EOLhyphen"/>materialitie of perfect Ar<g ref="char:EOLhyphen"/>chitecture.</hi>
               </note> 
               <hi>all materiall stuffe be<g ref="char:EOLhyphen"/>yng secluded. Which point we shall atteyne, by Notyng and forepointyng the an<g ref="char:EOLhyphen"/>gles, and lines, by a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ure and certaine direction and connexion. Seyng then, these thinges, are thus: Lineamente,</hi>
               <note place="margin">
                  <hi>What, Linea<g ref="char:EOLhyphen"/>ment is.</hi>
               </note> 
               <hi>shalbe the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ertaine and constant prescribyng, conceiued in mynde: made in lines and angles: and finished with a learned minde and wyt.</hi>
               <q>We thanke you Master <hi>Baptist,</hi> that you haue so aptly brought your Arte, and phrase therof, to haue some Mathematicall perfection:<note place="margin">
                     <hi>Note.</hi>
                  </note> by certaine or<g ref="char:EOLhyphen"/>der, nu<g ref="char:cmbAbbrStroke">̄</g>ber, forme, figure, and <hi>Symmetri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> mentall: all naturall &amp; sensible stuffe set a<g ref="char:EOLhyphen"/>part.</q>
Now, then, it is euident, (Gentle reader) how aptely and worthely, I haue prefer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ed <hi>Architecture,</hi> to be bred and fostered vp in the Dominion of the per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>les <hi>Princesse, Mathematica:</hi> and to be a naturall Subiect of hers. And the name of <hi>Architectur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,</hi> is of the principalitie, which this Science hath, aboue all other Artes. And <hi>Plato</hi> affirmeth, the <hi>Architect</hi> to be <hi>Master</hi> ouer all, that make any worke. Wherupon, he is neither Smith, not Builder: nor, separately, any Artificer: but the
<pb facs="tcp:7062:25"/>
Hed, the Prouost, the Directer, and Iudge of all Artificiall workes, and all Artifi<g ref="char:EOLhyphen"/>cers. For, the true <hi>Architect,</hi> is hable to teach, Demonstrate, distribute, des<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ribe, and Iudge all workes wrought. And he, onely, searcheth out the causes and reasons of all Artificiall thynges. Thus excellent, is <hi>Architecture</hi>: though few (in our dayes) at<g ref="char:EOLhyphen"/>teyne: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ereto: yet may not the Arte, be otherwise thought on, then in very dede it is worthy. Nor we may not, of auncient Artes, make new and imperfect Definiti<g ref="char:EOLhyphen"/>ons in our dayes: for scarsitie of Artificers: No more, than we may pynche in, the Definitions of <hi>Wisedome,</hi> or <hi>Honestie,</hi> or of <hi>Frendeshyp</hi> or of <hi>Iustice.</hi> No more will I consent, to Diminish any whit, of the perfection and dignitie, (by iust cause) al<g ref="char:EOLhyphen"/>lowed to absolute <hi>Architecture.</hi> Vnder the Direction of this Arte, are thre prin<g ref="char:EOLhyphen"/>cipall, necessary <hi>Mechanicall Artes.</hi> Namely, <hi>Howsing, Fortification,</hi> and <hi>Naupegie. Howsing,</hi> I vnderstand, both for Diuine Seruice, and Mans common vsage: publike, and priuate. Of <hi>Fortification</hi> and <hi>Naupegie,</hi> straunge matter might be told you: But perchaunce, some will be tyred, with this Bederoll, all ready rehearsed: and other some, w<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap> nycely nip my grosse and homely discoursing with you: made in post hast: for feare you should wante this true and frendly warnyng, and tast giuyng, of the <hi>Power Mathematicall.</hi> Lyfe is short, and vncertaine: Tymes are perilouse: &amp;c. And still the Printer awayting, for my pen staying: All these thinges, with farder matter of Ingratefulnes, giue me occasion to passe away, to the other Artes remainyng, with all spede possible.</p>
            <p>THe Arte of Nauigation, demonstrateth how, by the shortest good way, by the aptest Directio<g ref="char:cmbAbbrStroke">̄</g>, &amp; in the shortest time, a sufficient Ship, betwene any two places (in passage Nauigable,) assigned: may be co<g ref="char:cmbAbbrStroke">̄</g>ducted: and in all stormes, &amp; naturall disturbances chauncyng, how, to vse the best possible meanes, whereby to recouer the place first assigned. What nede, the <hi>Master Pilote,</hi> hath of other Artes, here before recited, it is easie to know: as, of <hi>Hydrographie, Astronomie, Astrologie,</hi> and <hi>Horome<g ref="char:EOLhyphen"/>trie.</hi> Presupposing continually, the common Base, and foundacion of all: namely <hi>Arithmetike</hi> and <hi>Geometrie.</hi> So that, he be hable to vnderstand, and Iudge his own necessary Instrumentes, and furniture Necessary: Whether they be perfectly made or no: and also can, (if nede be) make them, hym selfe. As Quadrantes, The Astro<g ref="char:EOLhyphen"/>nomers Ryng, The Astronomers staffe, The Astrolabe vniuersall. An Hydrogra<g ref="char:EOLhyphen"/>phicall Globe. Charts Hydrographicall, true, (not with parallell Meridians). The Common Sea Compas: The Compas of variacion: The Proportionall, and Para<g ref="char:EOLhyphen"/>doxall Compasses (of me Inuented,<note place="margin">
                  <hi>Anno.</hi> 1559.</note> for our two Moscouy Master Pilotes, at the re<g ref="char:EOLhyphen"/>quest of the Company) Clockes with spryng: houre, halfe houre, and three houre Sandglasses: &amp; sundry other Instrume<g ref="char:cmbAbbrStroke">̄</g>tes: And also, be hable, on Globe, or Playne to describe the Paradoxall Compasse: and duely to vse the same, to all maner of purposes, whereto it was inuented. And also, be hable to Calculate the Planetes places for all tymes.</p>
            <p>Moreouer, with Sonne Mone or Sterre (or without) be hable to define the Lon<g ref="char:EOLhyphen"/>gitude &amp; Latitude of the place, which he is in: So that, the Longitude &amp; Latitude of the place, from which he sayled, be giuen: or by him, be knowne. whereto, apper<g ref="char:EOLhyphen"/>tayneth expert meanes, to be certified euer, of the Ships way. &amp;c. And by forese<g ref="char:EOLhyphen"/>ing the Rising, Settyng, Nonestedyng, or Midnightyng of certaine tempestuous fixed Sterres: or their Coniunctions, and Anglynges with the Planetes, &amp;c. he ought to haue expert coniecture of Stormes, Tempestes, and Spoutes: and such lyke Meteorologicall effectes, daungerous on Sea. For (as <hi>Plato</hi> sayth,) <hi>Mu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ationes</hi>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>
               <pb facs="tcp:7062:25"/>
               <gap reason="illegible" resp="#PDCC" extent="1 page">
                  <desc>〈1 page〉</desc>
               </gap>
               <pb facs="tcp:7062:26"/>
Some by waight. wherof <hi>Ti<gap reason="illegible" resp="#PDCC" extent="5 letters">
                     <desc>•••••</desc>
                  </gap>
               </hi> speaketh. Some, by Stringes strayned, or Springs, therwith Imitating liuely Motions. Some, by other meanes, as the Images of Mer<g ref="char:EOLhyphen"/>curie: and the brasen hed, made by <hi>Albertus Magn<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>,</hi> which dyd seme to speake. <hi>B<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>thius</hi> was excellent in these feates. To whom, <hi>Cassiod<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>
               </hi> writyng, sayth. <hi>Your pur<g ref="char:EOLhyphen"/>pose is to know profound thynges: and to shew meruayles. By the disposition of your Arte, Metals do low: Diomedes of brasse, doth blow a Trumpet loude: a brasen Serpent hisseth: byrdes made, sing swetely. Small thynges we rehearse of you, who can Imitate the heauen. &amp;c.</hi> Of the straunge Selfmouyng, which, at Saint Denys, by Paris, <note n="*" place="margin">
                  <hi>Anno.</hi> 1551</note> I saw, ones or twise (<hi>Orontius</hi> beyng then with me, in Company) it were to straunge to tell. But some haue written it. And yet, (I hope) it is there, of other to be sene. And by <hi>Perspectiue</hi> also straunge thinges, are done. As partly (before) I gaue you to vnderstand in <hi>Perspectiue.</hi> As, to see in the Ayre, a loft, the lyuely Image of an other man, either walkyng to and fro: or standyng still. Likewise, to come into an house, and there to see the liuely shew of Gold, Siluer or precious stones: and commyng to take them in your hand, to finde nought but Ayre. Hereby, haue some men (in all other matters counted wise) fouly ouershot the<g ref="char:cmbAbbrStroke">̄</g> selues: misdeaming of the meanes.<note place="margin">
                  <hi>De his quae Mundo mi<g ref="char:EOLhyphen"/>rabiliter eue<g ref="char:EOLhyphen"/>niunt. cap.</hi> 8.</note> Therfore sayd <hi>Claudius Caelestinus. Hodie mag<g ref="char:EOLhyphen"/>na liter<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>rae vi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> magna <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>, opera qu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>da<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> quasi miranda, supra Natura <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> de qu<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>s in <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> sa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iliter reddidisse<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.</hi> That is. <hi>Now a dayes, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> see <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>, y<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a of great learnyng and reputation, to Iudge certain workes <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eruaylous, aboue the power of Nature: Of which workes, one that were skillfull in Perspectiue might easely haue giuen the Cause.</hi> Of <hi>Archimedes Sphaere, Cicero</hi> witnesseth.<note place="margin">
                  <hi>Tusc.</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.</note> Which is very straunge to thinke on. <hi>For when Archi<g ref="char:EOLhyphen"/>medes</hi> (sayth he) <hi>did fasten in a Sphere, the mo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ynges of the Sonne, Mone, and of the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e other Planets, he did, as the God, which (in Timaeus of Plato) did make the world. That, on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, should rule motions most vnlike in slownes, and swiftnes.</hi> But a greater cause of meruayling we haue by <hi>Claudianus</hi> report hereof. Who affirme in this <hi>Archimedes w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rke,</hi> to haue <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>en of Glasse. And discourseth of it more at large: which I omit. The Doue of wood, which the <hi>Mathematicien Ar<g ref="char:EOLhyphen"/>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> did make to flye, is by <hi>Agellius</hi> spoken of. Of <hi>Dadalus</hi> straunge Images, <hi>Plat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> reporteth. <hi>H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mere</hi> of <hi>Vul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ans Selfmouers,</hi> (by secret wheles) leaueth in writyng. <hi>Ari<g ref="char:EOLhyphen"/>stotle,</hi> in hys <hi>Politikes,</hi> of both, maketh mention. Mer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>aylous was the workeman<g ref="char:EOLhyphen"/>shyp, of la<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e dayes, performed by good skill of <hi>Tr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>chi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ike. &amp;c.</hi> For in Noremberge, A slye of Iern, heyng l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>t out of the Artifi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ers <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>nd, did (as it were) fly about by the g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>stes, at the table, and at leng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>h, as though <gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> were weary, retourne to his masters ha<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> agayne. Moreouer, an Artificiall E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>le, was ordred, to fly out of the same Towne, a mighty way, and that a lo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>t in the Ayre, toward the Emperour comming thether: and followed hym, beyng come to the gate of the town<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>.<note n="*" place="margin">☞</note> Thus, you see, what<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> Arte Mathematicall an p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>rforme, when Skill, will, Industry, and Habili<g ref="char:EOLhyphen"/>ty, are duely applyed to pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e.</p>
            <p>
               <note place="margin">
                  <hi>A Digression. Apologeticall.</hi>
               </note>ANd for these, and such like marueilous Actes and Feates, Naturally, Mathe<g ref="char:EOLhyphen"/>matically, and Mechanically, wrought and contriued: ought any honest Student, and Modest Christian Philosopher, be counted, &amp; called a Coniurer? Shall the folly of Idio<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>s, and the Mallice of the Scornfull, so much preuaile, that He, who seeketh no worldly gaine or glory at their handes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> But onely, of God, the threasor of heauenly wisedome, &amp; knowledge of p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>re veritie: Shall he (I say) in the meane
<pb facs="tcp:7062:26"/>
space, be robbed and spoiled of his honest name and fame? He that seketh (by S. Paules aduertisement) in the Creatures Properties, and wonderfull vertues, to finde iuste cause, to glorifie the AEternall; and Almightie Creator by: Shall that man, be (in hugger mugger) condemned, as a Companion of the Helhoundes, and a Caller, and Coniurer of wicked and damned Spirites? He that bewaileth his great want of time, sufficient (to his contentation) for learning of Godly wisdome, and Godly Verities in: and onely therin setteth all his delight: Will that ma<g ref="char:cmbAbbrStroke">̄</g> leese and abuse his time, in dealing with the Chiefe enemie of Christ our Redemer: the deadly foe of all mankinde: the subtile and impudent per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>erter of Godly Veritie: the Hypocriticall Crocodile: the Enuious Basiliske, continually desirous, in the twinke of an eye, to destroy all Mankinde, both in Body and Soule, aeternally? Surely (for my part, somewhat to say herein) I haue not learned to make so brutish, and so wicked a Bargaine. Should I, for my xx. or xxv. yeares Studie: for two or three thousand Markes spending: seuen or eight thousand Miles going and trauai<g ref="char:EOLhyphen"/>ling, onely for good learninges sake: And that, in all maner of wethers: in all ma<g ref="char:EOLhyphen"/>ner of waies and passages: both early and late: in daunger of violence by man: in daunger of destruction by wilde beastes: in hunger: in thirst: in perilous heates by day, with toyle on foote: in daungerous dampes of colde, by night, almost be<g ref="char:EOLhyphen"/>reuing life: (as God knoweth): with lodginges, oft times, to small ease: and som<g ref="char:EOLhyphen"/>time to lesse securitie. And for much more (then all this) done &amp; suffred, for Lear<g ref="char:EOLhyphen"/>ning and attaining of Wisedome: Should I (I pray you) for all this, no otherwise, nor more warily: or (by Gods mercifulnes) no more luckily, haue fished, with so large, and costly, a Nette, so long time in drawing (and that with the helpe and ad<g ref="char:EOLhyphen"/>uise of Lady Philosophie, &amp; Queene Theologie): but at length, to haue catched, and drawen vp,<note n="*" place="margin">
                  <hi>A prouerb. Fayre fisht, and caught a Frog.</hi>
               </note> a Frog? Nay, a Deuill? For, so, doth the Common peuish Pratler Imagine and Iangle: And, so, doth the Malicious skorner, secretly wishe, &amp; brauely and boldly face down, behinde my backe. Ah, what a miserable thing, is this kinde of Men? How great is the blindnes &amp; boldnes, of the Multitude, in thinges aboue their Capacitie? What a Land: what a People: what Maners: what Times are these? Are they become Deuils, themselues: and, by false witnesse bearing against their Neighbour, would they also, become Murderers? Doth God, so long geue them respite, to reclaime them selues in, from this horrible slaundering of the gilt<g ref="char:EOLhyphen"/>lesse: contrary to their owne Consciences: and yet will they not cease? Doth the Innocent, forbeare the calling of them, Iuridically to aunswere him, according to the rigour of the Lawes: and will they despise his Charitable pacience? As they, against him, by name, do forge, fable, rage, and raise slaunder, by Worde &amp; Print: Will they prouoke him, by worde and Print, likewise, to Note their Names to the World: with their particular deuises, fables, beastly Imaginations, and vnchristen<g ref="char:EOLhyphen"/>like slaunders? Well: Well. O (you such) my vnkinde Countrey men. O vn<g ref="char:EOLhyphen"/>naturall Countrey men. O vnthankfull Countrey men. O Brainsicke, Rashe, Spitefull, and Disdainfull Countrey men. Why oppresse you me, thus violently, with your slaundering of me: Contrary to Veritie: and contrary to your owne Consciences? And I, to this hower, neither by worde, deede, or thought, haue bene, any way, hurtfull, damageable, or iniurious to you, or yours? Haue I, so long, so dearly, so farre, so carefully, so painfully, so daungerously sought &amp; trauailed for the learning of Wisedome, &amp; atteyning of Vertue: And in the end (in your iudge<g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>t) am I become, worse, then when I bega<g ref="char:cmbAbbrStroke">̄</g>? Worse, the<g ref="char:cmbAbbrStroke">̄</g> a Mad man? A dangerous Member in the Common Wealth: and no Member of the Church of Christ? Call you this, to be Learned? Call you this, to be a Philosopher? and a louer of Wise<g ref="char:EOLhyphen"/>dome? To forsake the straight heauenly way: and to wallow in the broad way of
<pb facs="tcp:7062:27"/>
damnation? To forsake the light of heauenly Wisedome: and to lurke in the dun<g ref="char:EOLhyphen"/>geon of the Prince of darkenesse? To forsake the Veritie of God, &amp; his Creatures: and to fawne vpon the Impudent, Craftie, Obstinate Lier, and continuall disgracer of Gods Veritie, to the vttermost of his power? To forsake the Life &amp; Blisse AEter<g ref="char:EOLhyphen"/>nall: and to cleaue vnto the Author of Death euerlasting? that Murderous Ty<g ref="char:EOLhyphen"/>rant, most gredily awaiting the Pray of Mans Soule? Well: I thanke God and our Lorde Iesus Christ, for the Comfort which I haue by the Examples of other men, before my time: To whom, neither in godlines of life, nor in perfection of learning, I am worthy to be compared: and yet, they sustained the very like Iniu<g ref="char:EOLhyphen"/>ries, that I do: or rather, greater. Pacient <hi>Socrates,</hi> his <hi>Apologie</hi> will testifie: <hi>Apu<g ref="char:EOLhyphen"/>leius</hi> his <hi>Apologies,</hi> will declare the Brutishnesse of the Multitude. <hi>Ioannes Picus,</hi> Earle of Mirandula, his <hi>Apologie</hi> will teach you, of the Raging slaunder of the Ma<g ref="char:EOLhyphen"/>licious Ignorant against him. <hi>Ioannes Trithemius,</hi> his <hi>Apologie</hi> will specifie, how he had occasion to make publike Protestation: as well by reason of the Rude Sim<g ref="char:EOLhyphen"/>ple: as also, in respect of such, as were counted to be of the wisest sort of men.
<q>Ma<g ref="char:EOLhyphen"/>ny could I recite: But I deferre the precise and determined handling of this mat<g ref="char:EOLhyphen"/>ter: being loth to detect the Folly &amp; Mallice of my Natiue Countrey men.<note place="margin">☞</note> Who, so hardly, can disgest or like any extraordinary course of Philosophicall Studies: not falling within the Cumpasse of their Capacitie: or where they are not made priuie of the true and secrete cause, of such wonderfull Philosophicall Feates.</q>
These men, are of fower sortes, chie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ly. The first, I may name, <hi>Vaine pratling bu<g ref="char:EOLhyphen"/>sie bodies:</hi> The second, <hi>Fond Frendes:</hi> The third, <hi>Imperfectly zelous:</hi> and the fourth, <hi>Malicious Ignorant.</hi> To eche of these (briefly, and in charitie) I will say a word <milestone type="tcpmilestone" unit="unspecified" n="1"/> or two, and so returne to my Praeface. <hi>Vaine pratling busie bodies,</hi> vse your idle assemblies, and conferences, otherwise, then in talke of matter, either aboue your Capacities, for hardnesse: or contrary to your Consciences, in Veritie. <hi>Fonde <milestone type="tcpmilestone" unit="unspecified" n="2"/> Frendes,</hi> leaue of, so to commend your vnacquainted frend, vpon blinde affection: As, because he knoweth more, then the common Student: that, therfore, he must needes be skilfull, and a doer, in such matter and maner, as you terme <hi>Coniuring.</hi> Weening, thereby, you aduaunce his fame: and that you make other men, great marueilers of your hap, to haue such a learned frend. Cease to ascribe Impietie, where you pretend Amitie. For, if your tounges were true, then were that your frend, <hi>Vntrue,</hi> both to God, and his Soueraigne. Such <hi>Frendes</hi> and <hi>Fondlinges,</hi> I shake of, and renounce you: Shake you of, your Folly. <hi>Imperfectly zelous,</hi> to you, <milestone type="tcpmilestone" unit="unspecified" n="3"/> do I say: that (perhaps) well, do you Meane: But farre you misse the Marke: If a Lambe you will kill, to feede the flocke with his bloud. Sheepe, with Lambes bloud, haue no naturall sustenaunce: No more, is Christes flocke, with horrible slaunders, duely aedified. Nor your faire pretense, by such rashe ragged Rheto<g ref="char:EOLhyphen"/>rike, any whit, well graced. But such, as so vse me, will finde a fowle Cracke in their Credite. Speake that you know: And know, as you ought: Know not, by Heare say, when life lieth in daunger. Search to the quicke, &amp; let Charitie be your guide. <milestone type="tcpmilestone" unit="unspecified" n="4"/>
               <hi>Malicious Ignorant,</hi> what shall I say to thee? <hi>Prohibe linguam tuam a malo. A de<g ref="char:EOLhyphen"/>tractione parcite linguae. Cause thy toung to refraine fro<g ref="char:cmbAbbrStroke">̄</g> euill. Refraine your toung from slaunder.</hi> Though your tounges be sharpned, Serpent like, &amp; Adders poy<g ref="char:EOLhyphen"/>son lye in your lippes:<note place="margin">
                  <hi>Psal.</hi> 140.</note> yet take heede, and thinke, betimes, with your selfe, <hi>Vir lin<g ref="char:EOLhyphen"/>guosus non stabilietur in terra. Virum violentum venabitur malum, donec praecipitetur.</hi> For, sure I am, <hi>Quia faciet Dominus Iudicium afflicti: &amp; vindictam pauperum.</hi>
            </p>
            <p>Thus, I require you, my assured frendes, and Countrey men (you Mathemati<g ref="char:EOLhyphen"/>ciens, Mechaniciens, and Philosophers, Charitable and discrete) to deale in my
<pb facs="tcp:7062:27"/>
behalf, with the light &amp; vntrue tounged, my enuious Aduersaries, or Fond frends. And farther, I would wishe, that at leysor, you would consider, how <hi>Basilius Mag<g ref="char:EOLhyphen"/>nus,</hi> layeth <hi>Moses</hi> and <hi>Daniel,</hi> before the eyes of those, which count all such Stu<g ref="char:EOLhyphen"/>dies Philosophicall (as mine hath bene) to be vngodly, or vnprofitable. Waye well <hi>S. Stephen</hi> his witnesse of <hi>Moses.</hi>
               <note place="margin">
                  <hi>Act.</hi> 7. <hi>C.</hi>
               </note> 
               <hi>Eruditus est Moses omni Sapientia AEgyptioru<g ref="char:cmbAbbrStroke">̄</g>: &amp; erat potens in verbis &amp; operibus suis. Moses was instructed in all maner of wise<g ref="char:EOLhyphen"/>dome of the AEgyptians: and he was of power both in his wordes, and workes.</hi> You see this Philosophicall Power &amp; Wisedome, which <hi>Moses</hi> had, to be nothing misliked of the Holy Ghost. Yet <hi>Plinius</hi> hath recorded, <hi>Moses</hi> to be a wicked <hi>Magi<g ref="char:EOLhyphen"/>cien.</hi> And that (of force) must be, either for this Philosophicall wisedome, learned, before his calling to the leading of the Children of <hi>Israel:</hi> or for those his won<g ref="char:EOLhyphen"/>ders, wrought before King <hi>Pharao,</hi> after he had the conducting of the <hi>Israelites.</hi> As concerning the first, you perceaue, how <hi>S. Stephen,</hi> at his Martyrdome (being full of the Holy Ghost) in his Recapitulation of the olde Testament, hath made men<g ref="char:EOLhyphen"/>tion of <hi>Moses</hi> Philosophie: with good liking of it: And <hi>Basilius Magnus</hi> also, auou<g ref="char:EOLhyphen"/>cheth it, to haue bene to <hi>Moses</hi> profitable (and therefore, I say, to the Church of God, necessary). But as co<g ref="char:cmbAbbrStroke">̄</g>cerning <hi>Moses</hi> wonders, done before King <hi>Pharao</hi>: God, him selfe, sayd: <hi>Vide vt omnia ostenta, quae posui in manutua, facias coram Pharaone. See that thou do all those wonders before Pharao, which I haue put in thy hand.</hi> Thus, you euidently perceaue, how rashly, <hi>Plinius</hi> hath slaundered <hi>Moses,</hi>
               <note place="margin">
                  <hi>Lib.</hi> 30. <hi>Cap.</hi> 1.</note> of vayne fra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>dulent <hi>Magike,</hi> saying: <hi>Est &amp; alia Magices Factio, a Mose, Iamne, &amp; Iotape, Iu<g ref="char:EOLhyphen"/>daeis pendens: sed multis millibus annorum post Zoroastrem. &amp;c.</hi>
               <q>Let all such, there<g ref="char:EOLhyphen"/>fore,<milestone type="tcpmilestone" unit="unspecified" n="1"/> who, in Iudgement and Skill of Philosophie, are farre Inferior to <hi>Plinie,</hi> take good heede, least they ouershoote them selues rashly,<note place="margin">☜</note> in Iudging of <hi>Philosophers straunge Act<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the Meanes, how they are done.</q>
But, much more, ought they <milestone type="tcpmilestone" unit="unspecified" n="2"/> to beware of forging, deuising, and imagining monstrous feates, and wonderfull workes, when and where, no such were done: no, not any sparke or likelihode, of such, as they, without all shame, do report. And (to conclude) most of all, let <milestone type="tcpmilestone" unit="unspecified" n="3"/> them be ashamed of Man, and afraide of the dreadfull and Iuste Iudge: both Fo<g ref="char:EOLhyphen"/>lishly or Maliciously to deuise: and then, deuilishly to father their new fond Mon<g ref="char:EOLhyphen"/>sters on me: Innocent, in hand and hart: for trespacing either against the lawe of God, or Man, in any my Studies or Exercises, Philosophicall, or Mathematicall: As in due time, I hope, will be more manifest.</p>
            <p>NOw end I, with Archemastrie. Which name, is not so new, as this Arte is rare. For an other Arte, vnder this, a degree (for skill and power) hath bene indued with this English name before. And yet, this, may serue for our purpose, sufficiently, at this present. This Arte, teacheth to bryng to actuall ex<g ref="char:EOLhyphen"/>perience sensible, all worthy conclusions by all the Artes Mathema<g ref="char:EOLhyphen"/>ticall purposed, &amp; by true Naturall Philosophie concluded: &amp; both addeth to them a farder scope, in the termes of the same Artes, &amp; al<g ref="char:EOLhyphen"/>so by hys propre Method, and in peculier termes, procedeth, with helpe of the foresayd Artes, to the performance of complet Expe<g ref="char:EOLhyphen"/>rie<g ref="char:cmbAbbrStroke">̄</g>ces, which of no particular Art, are hable (Formally) to be challen<g ref="char:EOLhyphen"/>ged. If you remember, how we considered <hi>Architecture,</hi> in respect of all com<g ref="char:EOLhyphen"/>mon handworkes: some light may you haue, therby, to vnderstand the Souerain<g ref="char:EOLhyphen"/>ty and propertie of this Science. <hi>Science</hi> I may call it, rather, then an Arte: for the excellency and Mastershyp it hath, ouer so many, and so mighty Artes and
<pb facs="tcp:7062:28"/>
Sciences. And bycause it procedeth by <hi>Experiences,</hi> and searcheth forth the causes of Conclusions, by <hi>Experiences</hi>: and also putteth the Conclusions them selues, in <hi>Experience,</hi> it is named of some, <hi>Scientia Experimentalis.</hi> The <hi>Experimentall Sci<g ref="char:EOLhyphen"/>ence. Nicolaus Cusanus</hi> termeth it so, in hys <hi>Experimentes Statikall,</hi> And an other <hi>Philosopher,</hi>
               <note place="margin">
                  <hi>R. B.</hi>
               </note> of this land Natiue (the floure of whose worthy fame, can neuer dye nor wither) did write therof largely, at the request of <hi>Clement the sixt.</hi> The Arte carrieth with it, a wonderfull Credit: By reason, it certefieth, sensibly, fully, and completely to the vtmost power of Nature, and Arte. This Arte, certifieth by <hi>Ex<g ref="char:EOLhyphen"/>perience</hi> complete and absolute: and other Artes, with their Argumentes, and De<g ref="char:EOLhyphen"/>monstrations, persuade: and in wordes, proue very well their Conclusions. <note place="margin">☞</note> But wordes, and Argumentes, are no sensible certifying: nor the full and finall frute of Sciences practisable. And though some Artes, haue in them, <hi>Experiences,</hi> yet they are not complete, and brought to the vttermost, they may be stretched vnto, and applyed sensibly. As for example: the Naturall Philosopher disputeth and maketh goodly shew of reason: And the Astronomer, and the Optical Mechanicien, put some thynges in <hi>Experience:</hi> but neither, all, that they may: nor yet sufficiently, and to the vtmost, those, which they do, There, then, the <hi>Archemaster</hi> steppeth in, and leadeth forth on, the <hi>Experiences,</hi> by order of his doctrine <hi>Experimentall,</hi> to the chief and finall power of Naturall and Mathematicall Artes. Of two or three men, in whom, this Description of <hi>Archemastry</hi> was <hi>Experimentally,</hi> verified, I haue read and hard: and good record, is of their such perfection. So that, this Art, is no fan<g ref="char:EOLhyphen"/>tasticall Imagination: as some Sophister, might, <hi>Cum suis Insolubilibus,</hi> make a slo<g ref="char:EOLhyphen"/>rish: and dassell your Imagination: and dash your honest desire and Courage, from beleuing these thinges, so vnheard of, so meruaylous, &amp; of such Importance. Well: as you will. I haue forewarned you. I haue done the part of a frende: I haue dischar<g ref="char:EOLhyphen"/>ged my Duety toward God: for my small Talent, at hys most mercyfull handes re<g ref="char:EOLhyphen"/>ceiued. To this Science, doth the <hi>Science Alnirangiat,</hi> great Seruice. Muse nothyng of this name. I chaunge not the name, so vsed, and in Print published by other: beyng a name, propre to the Science. Vnder this, commeth <hi>Ars Sintrillia,</hi> by <hi>Artephius,</hi> briefly written. But the chief Science, of the Archemaster, (in this world) as yet knowen, is an other (as it were) OPTICAL Science: wherof, the name shall be told (God willyng) when I shall haue some, (more iust) occasion, therof, to Discourse.</p>
            <p>Here, I must end, thus abruptly (Gentle frende, and vnfayned louer of honest and necessary verities.) For, they, who haue (for your sake, and vertues cause) re<g ref="char:EOLhyphen"/>quested me, (an old forworne Mathematicien) to take pen in hand: (through the confidence they reposed in my long experience: and tryed sincerity) for the decla<g ref="char:EOLhyphen"/>ryng and reportyng somewhat, of the frute and commodity, by the Artes Ma<g ref="char:EOLhyphen"/>thematicall, to be atteyned vnto: euen they, Sore agaynst their willes, are forced, for sundry causes, to satiffie the workemans request, in endyng forthwith: He, so feareth this, so new an attempt, &amp; so costly: And in matter so slenderly (he<g ref="char:EOLhyphen"/>therto) among the common Sorte of Studentes, considered or estemed.</p>
            <p>And where I was willed, somewhat to alledge, why, in our vulgare Speche, this part of the Principall Science of <hi>Geometrie,</hi> called <hi>Euclides Geometricall Elementes,</hi> is published, to your handlyng: being vnlatined people, and not Vniuersitie Scholers: Verily, I thinke it nedelesse.</p>
            <p>
               <milestone type="tcpmilestone" unit="unspecified" n="1"/> For, the Honour, and Estimation of the Vniuersities, and Graduates, is, hereby, nothing diminished. Seing, from, and by their Nurse Children, you receaue all this Benefite: how great soeuer it be.</p>
            <p>
               <pb facs="tcp:7062:28"/>
               <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                  <desc>〈1 paragraph〉</desc>
               </gap>
            </p>
            <p>And su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ely, the Common and Vulgar Scholer (much more, the Grama<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>an)<milestone type="tcpmilestone" unit="unspecified" n="3"/>
before his comming to the <hi>Vniuersitie,</hi> shall (or may) be, now (according to <hi>Plato</hi>
his Counsell) su<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>iciently ins<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ructed in <hi>Arithmetike,</hi> and <hi>Geometrie,</hi> for the better and
ea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ier learn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ng of all maner of <hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hilosophie, Academicall,</hi> or <hi>Perip<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icall.</hi> And by that
meanes, goe more cherefu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>y, more skil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ully, and spedily fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>warde, in his Studies,
there to be learned. And, so, in lesse time, profite more, then (otherw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>se) he should<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>
or could do.</p>
            <p>
               <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                  <desc>〈1 paragraph〉</desc>
               </gap>
            </p>
            <p>And great Comfort, with good hope, may the Vniuersities haue, by reason of <milestone type="tcpmilestone" unit="unspecified" n="5"/> this <hi>Englishe</hi> Geometrie and Mathematicall Praeface, that they (hereafter) shall be the more regarded, esteemed, and resorted vnto. For, when it shall be knowen and reported, that of the <hi>Mathematicall Sciences</hi> onely, such great Commo<g ref="char:EOLhyphen"/>dities are enfu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ng (as Thaue specified): and that in dede, some of you vnlatined Studentes, can be good witnesse, of such rare fruite by you enioyed (thereby): as either, before this, was not heard of <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> fully credited:
<q>Well, may all men coniecture, that farre greater <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>, to winne to the Perfection of all Philosophie,<note place="margin">
                     <hi>Vniuersities.</hi>
                  </note> may in <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the Storehouses &amp; Threa<g ref="char:EOLhyphen"/>sory of all Sciences, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>,<note place="margin">☜</note> and most noble State of Common Wealthes.</q>
            </p>
            <p>Besides this, how ma<gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> here, in these Realmes of <milestone type="tcpmilestone" unit="unspecified" n="6"/> England and Ireland, tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> &amp; Cumpasse: Who, with their owne Skill and expe<gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap>ble (by these good helpes and informations) to find <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap>s, straunge Engines, and In<g ref="char:EOLhyphen"/>strumentes: for sundry purp<gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> Wealth? or for priuate plea<g ref="char:EOLhyphen"/>sure? and for the better maintayni<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap> 
               <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> owne estate? I will not (therefore)
<pb facs="tcp:7062:29"/>
fight against myne owne shadowe. For, no man (I am s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>re) will open his mouth against this Enterprise. No ma<g ref="char:cmbAbbrStroke">̄</g> (I say) who either hath Charitie toward his brother (and would be glad of his further<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nce in vertuous knowledge): o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> that hath any care &amp; zeale for the bettering of the Co<g ref="char:cmbAbbrStroke">̄</g>mon state of this Realme. Neither any, that make accompt, what the wiser sort of men (Sage and Stayed) do thinke of them. To <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>one (therefore) will I make any <hi>Ap<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>logie,</hi> for a vertuous acte doing: and for co<g ref="char:cmbAbbrStroke">̄</g>mending, or setting forth Profitable Artes to English men, in the English toung.
<q>But, vnto God our Creator, let vs all be thankefull: for that, <hi>As he, of his Good<g ref="char:EOLhyphen"/>nes, by his Powre, and in his wisedome, hath Created all thynges, in Number, Waight,<note place="margin">☞</note> and Measure</hi>: So, to vs, of hys great Mercy, he hath reuealed Meanes, whereby, to atteyne the sufficient and necessary knowledge of the foresayd hy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> three principall Instrumentes: Which Meanes, I haue abundantly proued vnto you, to be the <hi>Sciences</hi> and <hi>Artes Mathematicall.</hi>
               </q>
            </p>
            <p>And though I haue ben pinched with straightnes of tyme: that no way, I could so pen downe the m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>tter (in my Mynde), as I determined<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> hopyng of conuenient laysure: Yet, if vertuous zeale, and honest I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>tent prouoke and bryng you to the readyng and examinyng of this Compendious treatise, I do not doute, but, as the veritie therof (accordyng to our purpose) will be euident vnto you: So the pith and force therof, will persuade you: and the wonderfull frute therof, highly plea<g ref="char:EOLhyphen"/>sure you. And that you may the easier perceiue, and better remember, the prin<g ref="char:EOLhyphen"/>cipall pointes, whereof my Preface treateth, I will giue you the Ground platt of my whole discourse,<note place="margin">
                  <hi>The Ground platt of this Praeface in a Table.</hi>
               </note> in a Table annexed from the first to the last, somewhat Me<g ref="char:EOLhyphen"/>thodically contriued.</p>
            <p>If Hast, hath caused my poore pen, any where, to stumble: You will, (I am sure) in part of recompence<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> (for my earnest and sincere good will to plea<g ref="char:EOLhyphen"/>sure you). Conside<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> the rockish huge mountaines, and the perilous vnbeaten wayes which (both night and day, for the while) it hath <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>oyled and labored through, to bryng you this good Newes and Comfortable profe of Vertues frute.</p>
            <p>So, I Commit you vnto Gods Mercyfull direction, for the rest: hartely besechyng hym, to prosper your Studyes, and honest Intentes: to his Glory &amp; the Commodity of our Countrey. <hi>Amen.</hi>
            </p>
            <closer>
               <dateline>
                  <hi>Written at my poore H<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>se At Mortlake.</hi>
                  <date>
                     <hi>Anno.</hi> 1570<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>February</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</date>
               </dateline>
            </closer>
            <pb facs="tcp:7062:29"/>
         </div>
         <div type="abstract">
            <pb facs="tcp:7062:30"/>
            <head>Here haue you (according to my promisse) the Groundplat of my MATHEMATICALL Praeface: annexed to <hi>Euclide</hi> (now first) published in our Englishe tounge. An. 1570. Febr. 3.</head>
            <p>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ciences, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>nd Artes Mathe<g ref="char:EOLhyphen"/>
               <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>aticall, are, either
<list>
                  <item>Principall, which are two, onely,
<list>
                        <item>
                           <hi>Arithmetike.</hi>
                           <list>
                              <item>
                                 <p>Simple, Which dealeth with Numbers onely: and demonstrateth all their properties and apper<g ref="char:EOLhyphen"/>tenances: where, an Vnit, is Indiuisible.</p>
                                 <p>The vse whereof, is either,
<list>
                                       <item>In thinges Supernatu<g ref="char:EOLhyphen"/>rall, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                             <desc>•</desc>
                                          </gap>ternall, &amp; Diuine: By Application, <hi>Ascen<g ref="char:EOLhyphen"/>ding.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Mathema<g ref="char:EOLhyphen"/>ticall: without farther Application. The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Naturall: both Substa<g ref="char:cmbAbbrStroke">̄</g>tiall, &amp; Ac<g ref="char:EOLhyphen"/>cidentall, Visible, &amp; In<g ref="char:EOLhyphen"/>uisible. &amp;c. By Applica<g ref="char:EOLhyphen"/>tion: <hi>Descending.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                    </list>
                                 </p>
                              </item>
                              <item>
                                 <p>Mixt, Which with aide of Geometrie principall, demonstrateth some Arithmeticall Con<g ref="char:EOLhyphen"/>clusion, or Purpose.</p>
                                 <p>The vse whereof, is either,
<list>
                                       <item>In thinges Supernatu<g ref="char:EOLhyphen"/>rall, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                             <desc>•</desc>
                                          </gap>ternall, &amp; Diuine: By Application, <hi>Ascen<g ref="char:EOLhyphen"/>ding.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Mathema<g ref="char:EOLhyphen"/>ticall: without farther Application. The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Naturall: both Substa<g ref="char:cmbAbbrStroke">̄</g>tiall, &amp; Ac<g ref="char:EOLhyphen"/>cidentall, Visible, &amp; In<g ref="char:EOLhyphen"/>uisible. &amp;c. By Applica<g ref="char:EOLhyphen"/>tion: <hi>Descending.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                    </list>
                                 </p>
                              </item>
                           </list>
                        </item>
                        <item>Geometrie.
<list>
                              <item>
                                 <p>Simple, Which dealeth with Magnitudes, onely: and demonstrat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                       <desc>•</desc>
                                    </gap>th all their properties, passi<g ref="char:EOLhyphen"/>ons, and appertenances: whose Point, is Indiuisible.</p>
                                 <p>The vse whereof, is either,
<list>
                                       <item>In thinges Supernatu<g ref="char:EOLhyphen"/>rall, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                             <desc>•</desc>
                                          </gap>ternall, &amp; Diuine: By Application, <hi>Ascen<g ref="char:EOLhyphen"/>ding.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Mathema<g ref="char:EOLhyphen"/>ticall: without farther Application. The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Naturall: both Substa<g ref="char:cmbAbbrStroke">̄</g>tiall, &amp; Ac<g ref="char:EOLhyphen"/>cidentall, Visible, &amp; In<g ref="char:EOLhyphen"/>uisible. &amp;c. By Applica<g ref="char:EOLhyphen"/>tion: <hi>Descending.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                    </list>
                                 </p>
                              </item>
                              <item>
                                 <p>Mixt, Which with aide of Arithmetike principall, demonstrateth some Geometricall purpose: as <hi>EVCLIDES ELEMENTES.</hi>
                                 </p>
                                 <p>The vse whereof, is either,
<list>
                                       <item>In thinges Supernatu<g ref="char:EOLhyphen"/>rall, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                             <desc>•</desc>
                                          </gap>ternall, &amp; Diuine: By Application, <hi>Ascen<g ref="char:EOLhyphen"/>ding.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Mathema<g ref="char:EOLhyphen"/>ticall: without farther Application. The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                       <item>In thinges Naturall: both Substa<g ref="char:cmbAbbrStroke">̄</g>tiall, &amp; Ac<g ref="char:EOLhyphen"/>cidentall, Visible, &amp; In<g ref="char:EOLhyphen"/>uisible. &amp;c. By Applica<g ref="char:EOLhyphen"/>tion: <hi>Descending.</hi> The like Vses and Appli<g ref="char:EOLhyphen"/>cations are, (though in a degree lower) in the <hi>Artes Mathema<g ref="char:EOLhyphen"/>ticall Deri<g ref="char:EOLhyphen"/>uatiue.</hi>
                                       </item>
                                    </list>
                                 </p>
                              </item>
                           </list>
                        </item>
                     </list>
                  </item>
                  <item>Deriuatiue fro<g ref="char:cmbAbbrStroke">̄</g> the Princi<g ref="char:EOLhyphen"/>palls: o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which, some haue
<list>
                        <item>The names of the Principalls: as,
<list>
                              <item>
                                 <hi>Arithmetike,</hi> vulgar: which considereth
<list>
                                    <item>
                                       <hi>Arithmetike of most vsuall whole Numbers: And of Fractions to them appertaining.</hi>
                                    </item>
                                    <item>
                                       <hi>Arithmetike of Proportions.</hi>
                                    </item>
                                    <item>
                                       <hi>Arithmetike Circular.</hi>
                                    </item>
                                    <item>
                                       <hi>Arithmetike of Radicall Nu<g ref="char:cmbAbbrStroke">̄</g>bers: Simple, Compound, Mixt: And of their Fractions.</hi>
                                    </item>
                                    <item>
                                       <hi>Arithmetike of Cossike Nu<g ref="char:cmbAbbrStroke">̄</g>bers: with their Fractions: And the great Arte of Algiebar.</hi>
                                    </item>
                                 </list>
                              </item>
                              <item>
                                 <hi>Geometrie,</hi> vulgar: which tea<g ref="char:EOLhyphen"/>cheth Measuring
<list>
                                    <item>At hand
<list>
                                          <label>All Lengthes.</label>
                                          <item>
                                             <hi>Mecometrie.</hi>
                                          </item>
                                          <label>All Plaines: As, Land, Borde, Glasse, &amp;c.</label>
                                          <item>
                                             <hi>Embadometrie.</hi>
                                          </item>
                                          <label>All Solids: As, Timber, Stone, Vessels, &amp;c.</label>
                                          <item>
                                             <hi>Stereometrie.</hi>
                                          </item>
                                       </list>
                                    </item>
                                    <item>With dista<g ref="char:cmbAbbrStroke">̄</g>ce from the thing Measured, as,
<list>
                                          <item>
                                             <p>How farre, from the Measurer, any thing is: of him sene, on Land or Water: called <hi>Apomecometrie.</hi>
                                             </p>
                                             <p>Of which are growen the Feates &amp; Artes of
<list>
                                                   <item>
                                                      <hi>Geodesie:</hi> more cunningly to Measure and Suruey Landes, Woods, Waters. &amp;c.</item>
                                                   <item>
                                                      <hi>Geographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Chorographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Hydrographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Stratarithmetrie.</hi>
                                                   </item>
                                                </list>
                                             </p>
                                          </item>
                                          <item>
                                             <p>How high or deepe, from the leuell of the Measurers standing, any thing is: Seene of hym, on Land or Water: called <hi>Hypsometrie</hi>
                                             </p>
                                             <p>Of which are growen the Feates &amp; Artes of
<list>
                                                   <item>
                                                      <hi>Geodesie:</hi> more cunningly to Measure and Suruey Landes, Woods, Waters. &amp;c.</item>
                                                   <item>
                                                      <hi>Geographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Chorographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Hydrographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Stratarithmetrie.</hi>
                                                   </item>
                                                </list>
                                             </p>
                                          </item>
                                          <item>
                                             <p>How broad, a thing is, which is in the Measurers vew: so it be situated on Land or Water: called <hi>Platometrie.</hi>
                                             </p>
                                             <p>Of which are growen the Feates &amp; Artes of
<list>
                                                   <item>
                                                      <hi>Geodesie:</hi> more cunningly to Measure and Suruey Landes, Woods, Waters. &amp;c.</item>
                                                   <item>
                                                      <hi>Geographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Chorographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Hydrographie.</hi>
                                                   </item>
                                                   <item>
                                                      <hi>Stratarithmetrie.</hi>
                                                   </item>
                                                </list>
                                             </p>
                                          </item>
                                       </list>
                                    </item>
                                 </list>
                              </item>
                           </list>
                        </item>
                        <item>Propre names as,
<list>
                              <label>Perspectiue,</label>
                              <item>Which demonstrateth the maners and properties of all Radiations: Directe, Broken, and Reflected.</item>
                              <label>Astronomie,</label>
                              <item>Which demonstrateth the Distances, Magnitudes, and all Naturall motions, Apparences, and Passions, proper to the Planets and fixed Starres: f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>r any time, past, pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>sent, and to come: in respecte of a certaine Horizon, or without respecte of any Horizon.</item>
                              <label>Musike,</label>
                              <item>Which demonstrateth by reason, and teacheth by sense, perfectly to iudge and order the diuersitie of Soundes, hi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> or l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>w.</item>
                              <label>Cosmographie,</label>
                              <item>Which, wholy and perfectly maketh description of the Heauenly, and also Elementall part of the World: and of these partes, maketh h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>gall application, and mutuall collation necessary.</item>
                              <label>Astrologie,</label>
                              <item>Which reasonably demonstrateth the operations and effectes of the naturall bea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>es of light, and <gap reason="illegible" resp="#PDCC" extent="1 word">
                                    <desc>〈◊〉</desc>
                                 </gap> In<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>luence of the Planets, and fixed Starres, <gap reason="illegible" resp="#PDCC" extent="1 word">
                                    <desc>〈◊〉</desc>
                                 </gap> euery Element and Elementall body: at all times, in any Horiz<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>n assigned.</item>
                              <label>Statike,</label>
                              <item>Which demonstrateth the causes of heauines and lightnes of all thinges: and of the motions and properties to heauines and lightnes belonging.</item>
                              <label>Anthropographie,</label>
                              <item>Which describeth the Nu<g ref="char:cmbAbbrStroke">̄</g>ber, Measure, Waight, Figure, Situation, and colour of euery diuers thing contained in the perfect<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> body of <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap> AN: and geueth certaine knowledge of the Figure, Symmetri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>, Waight, Characterization, &amp; due Locall motion of any p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>rcell of the sayd body assigned: and of numbers to the said p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>rcell appertaining.</item>
                              <label>Trochilike,</label>
                              <item>Which demonstrateth the properties of all Circular motions: Simple and Compound.</item>
                              <label>Helicosophie,</label>
                              <item>Which demonstrateth the designing of all Spirall lines: in Plaine, on Cylinder, Co<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>, Sph<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>re, C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>id, and Spharo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>d: and their pro<g ref="char:EOLhyphen"/>perties.</item>
                              <label>Pneumatithmie,</label>
                              <item>Which demonstrateth by close hollow Geometricall figures (Regular and Irregular) the straunge properties (in motion or stay) of the Water, Ayre, Smoke, and Fire, in their Continuiti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>, and as they are ioyned to the Elementes next them.</item>
                              <label>Menadrie,</label>
                              <item>Which demonstrateth, how, about Natures Vertue, and power simple: Vertue and force, may be multiplied: and so to directe, to lif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>, to pull to, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>d to put or cast fro, any multiplied, or simple determined Vertue, Waight, or Force: naturally, not, so, directible, or moueable.</item>
                              <label>Hypogeiodie,</label>
                              <item>Which demonstrateth, how, vnder the Spharicall Superficie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> of the E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>rth, at <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ny depth, to any perpendicular line assigned (whose di<g ref="char:EOLhyphen"/>stance from the perpendicular of the entrance: and the Azi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>uth likewise, <gap reason="illegible" resp="#PDCC" extent="1 word">
                                    <desc>〈◊〉</desc>
                                 </gap> respe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>e of the sayd entrance, is knowen) certaine way, may be prescribed and g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ne, &amp;c.</item>
                              <label>Hydragogie,</label>
                              <item>Which demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>teth the possible leading of water by Natures l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>, and by artificiall helpe, fr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap> any head (being Spring, standing, or running water) to any other place assigned.</item>
                              <label>Horometrie,</label>
                              <item>Which demonstrateth, how, at all times appointed, the precise, vsuall denomination of time, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>y <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>e know<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>n, for any place assigned.</item>
                              <label>Zographie,</label>
                              <item>Which demonstrateth and teacheth, how, the Intersection of all vsuall <gap reason="illegible" resp="#PDCC" extent="1 span">
                                    <desc>〈…〉</desc>
                                 </gap> assigned (the Center, distanc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>, and lightes b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ing determined) may be, by lines, and proper col<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>urs repre<gap reason="illegible" resp="#PDCC" extent="4 letters">
                                    <desc>••••</desc>
                                 </gap>.</item>
                              <label>Architecture,</label>
                              <item>Which is a Sci<gap reason="illegible" resp="#PDCC" extent="4 letters">
                                    <desc>••••</desc>
                                 </gap> gar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ished with many doctrines, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                                    <desc>〈…〉</desc>
                                 </gap>, are iudged.</item>
                              <label>Nauigation,</label>
                              <item>
                                 <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                                    <desc>〈1 paragraph〉</desc>
                                 </gap>
                              </item>
                              <label>Thaumaturgike,</label>
                              <item>
                                 <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                                    <desc>〈1 paragraph〉</desc>
                                 </gap>
                              </item>
                              <label>Archemastrie,</label>
                              <item>
                                 <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                                    <desc>〈1 paragraph〉</desc>
                                 </gap>
                              </item>
                           </list>
                        </item>
                     </list>
                  </item>
               </list>
            </p>
         </div>
      </front>
      <body>
         <div n="1" type="book">
            <pb facs="tcp:7062:31"/>
            <pb facs="tcp:7062:31"/>
            <head>¶The first booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>IN THIS FIRST BOOKE is intreated of the most simple,<note place="margin">
                  <hi>The argum<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> of the first Booke.</hi>
               </note> easie, and first matters and groundes of Geo<g ref="char:EOLhyphen"/>metry, as, namely, of Lynes, Angles, Triangles, Pa<g ref="char:EOLhyphen"/>rallels, Squares, and Parallelogrammes, First of theyr definitions, showyng what they are, After that it tea<g ref="char:EOLhyphen"/>ch<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>th how to draw Parallel lynes, and how to forme diuersly figures of three sides, &amp; foure sides, according to the varietie of their sides, and Angles &amp; co<g ref="char:cmbAbbrStroke">̄</g>pareth them all with Triangles, &amp; also together the one with the other. In it also is taught how a figure of any forme may be chaunged into a Figure of an other forme. And for that it entreateth of these most com<g ref="char:EOLhyphen"/>mon and generall thynges, thys booke is more vniuersall then is the seconde, third, or any other, and therefore iustly occupieth the first place in order: as that without which, the other bookes of <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>clide</hi> which follow, and also the workes of others which haue written in Geometry, cannot be perceaued nor vnderstan<g ref="char:EOLhyphen"/>ded. And forasmuch <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s all the demonstrations and proofes of all the propositi<g ref="char:EOLhyphen"/>ons in this whole booke, depende of these groundes and principles following, which by reason of their playnnes neede no greate declaration, yet to remoue all (be it neuer so litle) obscuritie, there are here set certayne shorte and manifes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> expositions of them.</p>
            <div type="definitions">
               <head>Definitions.</head>
               <div type="section">
                  <head>
                     <hi>1.</hi> A signe or point is that, which hath no part.<note place="margin">
                        <hi>Definition of a poynt.</hi>
                     </note>
                  </head>
                  <p>The better to vnderstand what man<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r of thing a signe or point is, ye must note that the nature and propertie of quantitie (wherof Geometry entreateth) is to be deuided, so that whatsoeuer may be deuided into sund<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y partes, is called quantitie. But a point, although it pertayne to quantitie, and hath his beyng in quantitie, yet is it no quanti<g ref="char:EOLhyphen"/>tie, for that it cannot be deuided. Because (as the definition saith) it hath no partes in<g ref="char:EOLhyphen"/>to which it should be deuided. So that a pointe is the least thing that by minde and vn<g ref="char:EOLhyphen"/>derstanding can be imagined and conceyued: then which, there can be nothing lesse, as the point <hi>A</hi> in the margent.<milestone type="tcpmilestone" unit="unspecified" n="A"/>
                  </p>
                  <p>A signe or point is of <hi>Pithagoras</hi> Scholers after this manner defined.<note place="margin">
                        <hi>Definition of a poynt after</hi> Pithagoras.</note> 
                     <hi>A poynt is an vnitie which hath position.</hi> Nu<g ref="char:cmbAbbrStroke">̄</g>bers are conceaued in mynde without any forme &amp; figure, and therfore without matter wheron to <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> figure, &amp; consequently without place and position. Wherfore vnitie beyng a part of number, hath no position, or determi<g ref="char:EOLhyphen"/>nate place. Wherby it is manifest, that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>umb<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> more simple and pure then is magni<g ref="char:EOLhyphen"/>tude, and also immateriall: and so vnity which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ginning of number, is lesse ma<g ref="char:EOLhyphen"/>teriall then a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igne or poy<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>, which is the beginnyng of magnitude. For a poynt is ma<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eriall, and requireth position and place, and <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>rby differeth from vnitie.</p>
               </div>
               <div type="section">
                  <head>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. A line is length <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ithout breadth.<note place="margin">
                        <hi>Definition of a li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>.</hi>
                     </note>
                  </head>
                  <p>There pertaine to quanti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e three dimensions, length, bredth, &amp; thicknes, or depth: and by these thre are all qua<g ref="char:cmbAbbrStroke">̄</g>titie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> measured &amp; made known. There are also, according
<pb facs="tcp:7062:32"/>
to these three dimensions, three kyndes of continuall quantities: a lyne, a superficies, or plaine, and a body. The first kynde, namely, a line is here defined in these wordes, <hi>A lyne is length without breadth.</hi> A point, for that it is no quantitie nor hath any partes into which it may be deuided, but remaineth indiuisible, hath not, nor can haue any of these three dimensions. It neither hath length, breadth, nor thickenes. But to a line, which is the first kynde of quantitie, is attributed the first dimension, namely, length, and onely that, for it hath neither breadth nor thicknes, but is conceaued to be drawne in, length onely, and by it, it may be deuided into partes as many as ye list, equall, or vnequall. But as touching breadth it remaineth indiuisible. As the lyne AB, which is onely drawen in length, may be deuided in the pointe C equally, or in the
<figure/>
point D vnequally, and so into as many partes as ye list. There are also of diuers other geuen other definitions of a lyne; as these which follow.</p>
                  <list>
                     <item>
                        <note place="margin">
                           <hi>An other defi<g ref="char:EOLhyphen"/>nition of a line.</hi>
                        </note>
                        <hi>A lyne is the mouyng of a poynte,</hi> as the motion or draught of a pinne or a penne to your sence maketh a lyne.</item>
                     <item>
                        <note place="margin">
                           <hi>An other.</hi>
                        </note>Agayne, <hi>A lyne is a magnitude hauing one onely space or dimension,</hi> namely, <hi>length wantyng breadth and thic<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>es.</hi>
                     </item>
                  </list>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The endes of a line.</hi>
                     </note>
                     <hi>3</hi> The endes or limites of a lyne, are pointes.</head>
                  <p>For a line hath his beginning from a point, and likewise endeth in a point: so that by this also it is manifest, that pointes, for their simplicitie and lacke of composition, are neither quantitie, nor partes of quantitie, but only the termes and endes of quantitie. As the pointes <hi>A, B,</hi> are onely the endes of the line <hi>AB,</hi> and no partes thereof. And herein differeth a poynte in quantitie,<note place="margin">
                        <hi>Difference of a point fr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>nity.</hi>
                     </note> from vnitie in number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <figure/>
for that although vnitie be the beginning of nombers, and no number (as a point is the beginning of quantitie, and no quan<g ref="char:EOLhyphen"/>titie) yet is vnitie a part of number.<note place="margin">
                        <hi>Vnitie is a part of number.</hi>
                     </note> For number is nothyng els but a collection of vnities, and therfore may be deuided into them, as into his partes. But a point is no part of quantitie,<note place="margin">
                        <hi>A poynt is no part of quan<g ref="char:EOLhyphen"/>titie.</hi>
                     </note> or of a lyne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> neither is a lyne composed of pointes, as number is of vnities. For things indiuisible being neuer so many added together, can neuer make a thing diuisible, as an instant in time, is neither tyme, nor part of tyme, but only the beginning and end of time, and coupleth &amp; ioyneth partes of tyme together.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a right line.</hi>
                     </note>
                     <hi>4</hi> A right lyne is that which lieth equally betwene his pointes.</head>
                  <p>As the whole line <hi>AB</hi> lyeth straight and equally betwene the poyntes <hi>AB</hi> without any going vp or comming downe on eyther side.</p>
                  <p>
                     <note place="margin">
                        <hi>Definition of a right line after</hi> Campanus.</note>
                     <hi>Campanus</hi> and certain others, define a right line thus:
<figure/>
                  </p>
                  <p>
                     <hi>A right line is the shortest extension or draught, that is or may be from one poynt to an other. Archimedes</hi> defineth it thus.</p>
                  <p>
                     <note place="margin">
                        <hi>Definitio<g ref="char:cmbAbbrStroke">̄</g> therof after</hi> Archimedes.</note>
                     <hi>A right line is the shortest of all lines, which haue one and the self same limites or endes:</hi> which is in maner al one with the definitio<g ref="char:cmbAbbrStroke">̄</g> of <hi>Campanus.</hi> As of all these lines <hi>ABC, ADC, AEC, AFC,</hi> which are all drawen from the point <hi>A,</hi> to the
<figure/>
poynte <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> as <hi>Campanus</hi> speaketh, or which haue the self same limites or endes, as <hi>Archimedes</hi> speaketh, the lyne <hi>ABC,</hi> beyng a right line, is the shortest.</p>
                  <p>
                     <note place="margin">
                        <hi>Definitio<g ref="char:cmbAbbrStroke">̄</g> therof after</hi> Plato.</note>
                     <hi>Plato</hi> defineth a right line after this maner: <hi>A right line is that whose middle part shadoweth the ex<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>reme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> As if you put any thyng in the middle of a right lyne, you shall not see from the one ende to the other, which thyng happeneth not in a crooked lyne. The Ecclipse of the Sunne (say Astronomers) then happeneth, when the Sunne, the Moone, &amp; our eye are in one right line. For the Moone then being in the midst betwene vs and the Sunne, causeth it to be darkened. Diuers other define a right line diuersly,<note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note> as followeth.</p>
                  <p>
                     <hi>A right lyne is that which standeth firme betwene his extremes.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>An other.</hi>
                     </note>Agayne, <hi>A right line is that which with an other line of lyke forme cannot make a figure.</hi>
                  </p>
                  <p>
                     <pb n="2" facs="tcp:7062:32"/>Agayne,<note place="margin">
                        <hi>An other.</hi>
                     </note> 
                     <hi>A right lyne is that which hath not one part in a plaine superficies, and an other erected on high.</hi>
                  </p>
                  <p>Agayne,<note place="margin">
                        <hi>An other.</hi>
                     </note> 
                     <hi>A right lyne is that, all whose partes agree together with all his other partes.</hi>
                  </p>
                  <p>Agayne,<note place="margin">
                        <hi>An other.</hi>
                     </note> 
                     <hi>A right lyne is that, whose extremes abiding, cannot be altered.</hi>
                  </p>
                  <p>
                     <hi>Euclide</hi> doth not here define a crooked lyne,<note place="margin">
                        <hi>VVhy Euclide here defineth not a crooked lyne.</hi>
                     </note> for it neded not. It may easely be vnder<g ref="char:EOLhyphen"/>stand by the definition of a right lyne, for euery contrary is well manifested &amp; set forth by hys contrary. One crooked lyne may be more crooked then an other, and from one poynt to an other may be drawen infinite crooked lynes: but one right lyne cannot be righter then an other, and therfore from one point to an other, there may be drawen but one tight lyne. As by figure aboue set, you may see.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>5</hi> A superficies is that, which hath onely length and breadth.<note place="margin">
                        <hi>Definition of a superficies.</hi>
                     </note>
                  </head>
                  <p>A superficies is the second kinde of quantitie, and to it are attributed two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>imensi<g ref="char:EOLhyphen"/>ons, namely length, and breadth.<note place="margin">
                        <hi>A superficies may be deuided two wayes.</hi>
                     </note> As in the superficies <hi>ABCD.</hi>
                     <figure/>
whose length is taken by the lyne <hi>AB,</hi> or <hi>CD,</hi> and breadth by the lyne <hi>AC.</hi> or <hi>BD</hi>: and by reason of those two dimensions a super<g ref="char:EOLhyphen"/>ficies may be deuided two wayes, namely by his length, and by hys breadth, but not by thicknesse, for it hath none. For, that is attribu<g ref="char:EOLhyphen"/>ted onely to a body, which is the third kynde of quantitie, and hath all three dimensions, length, breadth, and thicknes, and may be de<g ref="char:EOLhyphen"/>uided according to any of them.</p>
                  <p>Others define a superficies thus:<note place="margin">
                        <hi>An other defini<g ref="char:EOLhyphen"/>tion of a super<g ref="char:EOLhyphen"/>ficies.</hi>
                     </note> 
                     <hi>A superficies is the terms or ende of a body.</hi> As a line is the ende and terme of a superficies.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>6</hi> Extremes of a superficies, are lynes.<note place="margin">
                        <hi>The extremes of a superficies.</hi>
                     </note>
                  </head>
                  <p>As the endes, limites, or borders of a lyne, are pointes, inclosing the line: so are lines the limites, borders, and endes inclosing a superficies. As in the figure aforesayde you maye see the superficies inclosed with foure lynes. The extremes or limites of a bodye, are superficiesles. And therfore a superficies is of some thus defined:<note place="margin">
                        <hi>Another defini<g ref="char:EOLhyphen"/>tion of a super<g ref="char:EOLhyphen"/>ficies.</hi>
                     </note> 
                     <hi>A superficies is that, which endeth or inclose<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h a body:</hi> as is to be sene in the sides of a die, or of any other body.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>7</hi> A plaine superficies is that, which lieth equally betwene his lines.<note place="margin">
                        <hi>Definition of a plaine superfi<g ref="char:EOLhyphen"/>cies.</hi>
                     </note>
                  </head>
                  <p>As the superficies <hi>ABCD</hi> lyeth equally and smoothe betwene
<figure/>
the two lines <hi>AB,</hi> and <hi>CD</hi>: or betwene the two lines <hi>AC,</hi> and <hi>BD</hi>: so that no part therof eyther swelleth vpward, or is depres<g ref="char:EOLhyphen"/>sed downward. And this definitio<g ref="char:cmbAbbrStroke">̄</g> much agreeth with the defini<g ref="char:EOLhyphen"/>tion of a right line. A right line lieth equally betwene his points, and a plaine superficies lyeth equally betwene his lynes. Others define a plaine super<g ref="char:EOLhyphen"/>ficies after this maner:</p>
                  <p>
                     <hi>A plaine superficies is the shortest extension or dr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ught from one lyne to an other:</hi>
                     <note place="margin">
                        <hi>Another defini<g ref="char:EOLhyphen"/>tion of a playne superficies.</hi>
                     </note> like as a right lyne is the shortest extension or draught from one point to an other.</p>
                  <p>
                     <hi>Euclide</hi> also leaueth out here to speake of a crooked and hollow superficies, because it may easely be vnderstand by the diffinition of a plaine superficies, being hys contrary. And euen as from one point to an other may be drawen infinite crooked lines, &amp; but one right line, which is the shortest: so from one lyne to an other may be drawen infi<g ref="char:EOLhyphen"/>nite croked superficiesses, &amp; but one plain superficies, which is the shortest. Here must you consider when there is in Geometry mention made of pointes,<note place="margin">
                        <hi>NOTE.</hi>
                     </note> lines, circles, trian<g ref="char:EOLhyphen"/>gles, or of any other figures, ye may not conceyue of them as they be in matter, as in woode, in mettall, in paper, or in any such lyke, for so is there no lyne, but hath some breadth, and may be deuided<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> nor points, but that shal haue some partes, and may also be deuided, and so of others. But you must conceiue them in mynde, plucking them by imagination from all matter so shall ye vnderstande them truely and perfectly, in their owne nature as they are defined. As a lyne to be long, and not broade: and a poynte to
<pb facs="tcp:7062:33"/>
be so little, that it shall haue no part at all.</p>
                  <p>
                     <note place="margin">
                        <hi>Another defini<g ref="char:EOLhyphen"/>tion of a playne superficies.</hi>
                     </note>Others otherwyse define a playne superficies: <hi>A plaine superficies is that, which is firmly set betwene his extremes,</hi> as before was sayd of a right lyne.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>Agayne, <hi>A plaine superficies is that, vnto all whose partes a right line may well be applied.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>An other definition.</hi>
                     </note>Again, <hi>A plaine superficies is that, which is the shortest of al superficies, which haue one &amp; the self extremes:</hi> As a rig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t line was the shortest line that can be drawen betwene two pointes.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>Againe, <hi>A playne superficies is that, whose middle darkeneth the extremes,</hi> as was also sayd of a right lyne.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a playne angle.</hi>
                     </note>
                     <hi>8</hi> A plaine angle is an inclination or bowing of two lines the one to the other and the one touching the other, and not beyng directly ioyned together.</head>
                  <p>As the
<figure/>
two lines <hi>AB,</hi> &amp; <hi>BC,</hi> incline the one to the o<g ref="char:EOLhyphen"/>ther, and touch the one the other in the point <hi>B,</hi> in which point by reason of the inclination of the sayd lines, is made the angle <hi>ABC.</hi> But if the two lines which touch the one the other, be without all inclination of the one to the other, and be drawne directly the one to the other, then make they not any angle at all, as the lines <hi>CD,</hi> and <hi>DE,</hi> touch the one the other in the point <hi>D,</hi> and yet as ye see they make no angle.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ilined an<g ref="char:EOLhyphen"/>gle.</hi>
                     </note>
                     <hi>9</hi> And if the lines which containe the angle be right lynes, then is it called a rightlyned angle.</head>
                  <p>As the angle <hi>ABC,</hi> in the former figures, is a rightlined angle, because it is contai<g ref="char:EOLhyphen"/>ned of right lines: where note, that an angle is for the most part described by thre let<g ref="char:EOLhyphen"/>ters, of which the second or middle letter representeth the very angle, and therfore is set at the angle.</p>
                  <p>By the contrary, a crooked lyned angle, is that which is contained of crooked lines, which may be diuersly figured. Also a mixt angle is that which is caused of them both, namely, of a right line and a crooked, which may also be diuersly figured, as in the fi<g ref="char:EOLhyphen"/>gures before set ye may see.<note place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> of angles.</hi>
                     </note> There are of angles thre kindes, a right angle, an acute an<g ref="char:EOLhyphen"/>gle, and an obtuse angle, the definitions of which now follow.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>VVhat a right angle, &amp; VVhat also a perpendicular lyne i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note>
                     <hi>10</hi> VVhen a right line standing vpon a right line maketh the angles on either side equall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then either of those angles is a right angle. And the right lyne which standeth erected, is called a perpendiculer line to that vpon which it standeth.</head>
                  <p>As vpon the right line <hi>CD,</hi> suppose there do stand an other line
<figure/>
                     <hi>AB,</hi> in such sort, that it maketh the angles on either side therof e<g ref="char:EOLhyphen"/>quall: namely, the angle <hi>ABC</hi> on the one side equall to the angle <hi>ABD</hi> on the other side: then is eche of the two angles <hi>ABC,</hi> and <hi>ABD</hi> a right angle, and the line <hi>AB,</hi> which standeth erected vpon the line <hi>CD,</hi> without inclination to either part is a perpendicular line, commonly called among artificers a plumbe lyne.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>VVhat an ob<g ref="char:EOLhyphen"/>tuse angle <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>.</hi>
                     </note>
                     <hi>11</hi> An obtuse angle is that which is greater then a right angle.</head>
                  <p>
                     <pb n="3" facs="tcp:7062:33" rendition="simple:additions"/>As the angle <hi>CBE</hi> in the example is an obtuse angle, for it is
<figure/>
greater then the angle <hi>ABC,</hi> which is a right angle, because it con<g ref="char:EOLunhyphen"/>tayneth it, and containeth moreouer the angle <hi>ABE.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <hi>12</hi> An acute angle is that, which is lesse then a right angle.<note place="margin">
                        <hi>VVhat an acute angle is.</hi>
                     </note>
                  </head>
                  <p>As the angle <hi>EBD</hi> in the figure before put is an acute angle, for that it is lesse then the angle <hi>ABD,</hi> which is a right angle, for the right angle contai<g ref="char:EOLhyphen"/>neth it, and moreouer the angle <hi>ABE.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <hi>13</hi> A limite or terme, is the ende of euery thing.<note place="margin">
                        <hi>The limite of any thing.</hi>
                     </note>
                  </head>
                  <p>For as much as of thinges infinite (as <hi>Plato</hi> saith) there is no science,<note place="margin">
                        <hi>No science of thinges infinite</hi>
                     </note> therefore must magnitude or quantitie (wherof Geometry entreateth) be finite, and haue borders and limites to inclose it, which are here defined to be the endes therof. As a point is the li<g ref="char:EOLhyphen"/>mite or terme of a line, because it is thend therof: A line likewise is the limite &amp; terme of a superficies: and likewise a superficies is the limite and terme of a body, as is before declared.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>14</hi> A figure is that, which is contayned vnder one limite or terme, or many.<note place="margin">
                        <hi>Definition of a figure.</hi>
                     </note>
                  </head>
                  <p>As the figure <hi>A</hi> is contained vnder one limit,
<figure/>
which is the round line, also the figure <hi>B</hi> is con<g ref="char:EOLunhyphen"/>tayned vnder three right lines. And the figure <hi>C</hi> vnder foure, and so of others, which are their li<g ref="char:EOLhyphen"/>mites or termes.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>15</hi> A circle is a plaine figure, conteyned vnder one line, which is called a cir<g ref="char:EOLhyphen"/>cumference,<note place="margin">
                        <hi>Definition of a circle.</hi>
                     </note> vnto which all lynes drawen from one poynt within the figure and falling vpon the circumference therof are equall the one to the other.</head>
                  <p>As the figure here set is a plaine figure, that is a figure without grossenes or thick<g ref="char:EOLhyphen"/>nes, and is also contayned vnder one line, namely, the crooked lyne
<figure/>
                     <hi>B CD,</hi> which is the circumference therof, it hath moreouer in the middle therof a point, namely, the point <hi>A,</hi> from which, all the lynes drawen to the superficies, are equal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as the lines <hi>AB, AC, AD,</hi> and other how many soeuer.</p>
                  <p>Of all figures a circle is the most perfect,<note place="margin">
                        <hi>A circle the most perfect of all figures.</hi>
                     </note> and therfore is it here first defined.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>16</hi> And that point is called the centre of the circle, as is the point A, which is set in the middes of the former circle.<note place="margin">
                        <hi>The centre of a circle.</hi>
                     </note>
                  </head>
                  <p>For the more easy declaration, that all the lines drawen from the centre of the circle to the circumference, are equall, ye must note, that although a line
<figure/>
be not made of pointes: yet a point, by his motion or draught, de<g ref="char:EOLhyphen"/>scribeth a line. Likewise a line drawen, or moued, describeth a su<g ref="char:EOLhyphen"/>perficies<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> also a superficies being moued maketh a solide or bodie. Now the<g ref="char:cmbAbbrStroke">̄</g> imagine the line <hi>AB,</hi> (the point <hi>A</hi> being fixed) to be mo<g ref="char:EOLhyphen"/>ued about in a plaine superficies, drawing the point <hi>B</hi> continually about the point <hi>A,</hi> till it returne to the place where it began first to moue: so shall the point <hi>B,</hi> by this motion, describe the circum<g ref="char:EOLhyphen"/>ference of the circle, and the point <hi>A</hi> being fixed, is the centre of the circle. Which in
<pb facs="tcp:7062:34"/>
all the time of the motion of the line, had like distance: from the circumference, name<g ref="char:EOLhyphen"/>ly, the length of the line <hi>AB.</hi> And for that al the lines drawn from the centre to the cir<g ref="char:EOLhyphen"/>cumference are described of that line, they are also equal vnto it, &amp; betwene the<g ref="char:cmbAbbrStroke">̄</g>selues.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a diameter.</hi>
                     </note>
                     <hi>17</hi> A diameter of a circle, is a right line, which drawen by the centre thereof, and ending at the circumference on either side, deuideth the circle into two equal partes.</head>
                  <p>As the line <hi>BAC</hi> in this circle present is the diameter, because
<figure/>
it passeth from the point <hi>B,</hi> of the one side of the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the point <hi>C,</hi> on the other side of the circumference, &amp; passeth also by the point <hi>A</hi> being the centre of the circle. And moreo<g ref="char:EOLhyphen"/>uer it deuideth the circle into two equall partes: the one, name<g ref="char:EOLhyphen"/>ly <hi>BDC,</hi> being on the one side of the line, &amp; the other namely, <hi>BEC,</hi> on the other side, which thing did <hi>Thales Miletius</hi> (which brought Geometry out of Egipt into Grece) first obserue and proue. For if a line drawen by the centre, do not deuide a circle into two equall partes; all the lines drawen from the centre to the circumference should not be equall.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a semicircle.</hi>
                     </note>
                     <hi>18</hi> A semicircle, is a figure which is contayned vnder the diameter, and vn<g ref="char:EOLhyphen"/>der that part of the circumference which is cut of by the diametre.</head>
                  <p>As in the circle <hi>ABCD</hi> the figure <hi>BAC</hi> is a semicircle, because
<figure/>
it is contained of the right line <hi>BGC,</hi> which is the diametre, and of the crooked line <hi>BAC,</hi> being that part of the circumference, which is cut of by the diametre <hi>BGC.</hi> So likewise the other part of the circle, namely <hi>BDC</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is a semicircle as the other was.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a section of a cir<g ref="char:EOLhyphen"/>cle.</hi>
                     </note>
                     <hi>19</hi> A section or portion of a circle, is a figure whiche is contayned vnder a right lyne, and a parte of the circumference, greater or lesse then the semicircle.</head>
                  <p>As the figure <hi>ABC,</hi> in the example, is a section of a circle, &amp;
<figure/>
is greater then halfe a circle, and the figure <hi>ADC,</hi> is also a secti<g ref="char:EOLhyphen"/>on of a circle, and is lesse then a semicircle. A section, portion, or part of a circle is all one, and signifieth such a part which is ei<g ref="char:EOLhyphen"/>ther more or lesse, then a semicircle: so that a semicircle is not here called a section or portion of a circle. A right lyne drawen from one side of the circumference of a circle to the other, not passyng by the centre, deuideth the circle into two vnequall partes, which are two sections, of which that which contayneth the centre is called th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> greater section, and the other is called the lesse section. As in the example, the part of the circle <hi>ABC,</hi> which containeth in it the centre <hi>E,</hi> is the greater section, being grea<g ref="char:EOLunhyphen"/>ter then the halfe circle: the other part, namely <hi>ADC,</hi> which hath not the centre in it, is the lesse section of the circle, being lesse then a semicircle.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of r<gap reason="illegible" resp="#PDCC" extent="4 letters">
                              <desc>••••</desc>
                           </gap>lined fi<g ref="char:EOLhyphen"/>gures.</hi>
                     </note>
                     <hi>20</hi> Rightlined figures are such which are contayned vnder right lynes.</head>
                  <p>As are such as followeth, of which some are contayned vnder three right lines, some vnder <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oure, some vnder fiue, and some vnder mo.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of three sided fi<g ref="char:EOLhyphen"/>gures.</hi>
                     </note>
                     <hi>21</hi> Thre sided figures, or figures of thre sydes, are such which are contay<g ref="char:EOLhyphen"/>ned vnder three right lines.</head>
                  <p>
                     <pb n="4" facs="tcp:7062:34"/>As the figure in the example <hi>ABC,</hi> is a figure of three sides
<figure/>
because it is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder thre right lines, namely, vnder the lines <hi>AB, BC, CA.</hi>
                  </p>
                  <p>A figure of three sides, or a triangle, is the first figure in order of all right lined figures, and therfore of all others it is first defined. For vnder lesse then three lines, can no figure be comprehended.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>22.</hi> Foure sided figures or figures of foure sides are such,<note place="margin">
                        <hi>Definition of foure sided figures.</hi>
                     </note> which are contained vnder foure right lines.</head>
                  <p>As the figure here set, is a figure of foure sides, for that it is co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>prehended
<figure/>
vnder foure right lines, namely, AB, BD, DC, CA.</p>
                  <p>Triangles, and foure sided figures serue commonly to many v<g ref="char:EOLhyphen"/>ses in demonstrations of Geometry. Wherfore the nature and properties of them, are much to be obserued, the vse of other fi<g ref="char:EOLhyphen"/>gures is more obscure.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>23.</hi> Many sided figures are such which haue mo sides then foure.<note place="margin">
                        <hi>Definition of many sided figures.</hi>
                     </note>
                  </head>
                  <p>Right lined figures hauing mo sides then fower, by continual adding of sides may be
<figure/>
infinite. Wherfore to define them all seuerally, according to the number of their <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ides, should be very tedious, or rather impossible. Therfore hath <hi>Euclide,</hi> comprehended the<g ref="char:cmbAbbrStroke">̄</g> vnder one name, and vnder one diffinition: calling them many sided figures, as many as haue mo sides then foure: as if they haue fiue sides, sixe, seuen, or mo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Here note ye, that euery rightlined figure hath as many angles, as it hath sides, &amp; taketh his denomi<g ref="char:EOLhyphen"/>nation aswell of the number of his angles, as of the number of his sides. As a figure co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tained vnder three right lines, of the number of his three sides, is called a thre sided fi<g ref="char:EOLhyphen"/>gure <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>euen so of the number of his three angles, it is called a triangle. Likewise a figure contained vnder foure right lines, by reason of the number of his sides, is called a foure sided figure: and by reason of the number of his angles, it is called a quadrangled fi<g ref="char:EOLhyphen"/>gure, and so of others.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>24.</hi> Of three sided figures or triangles,<note place="margin">
                        <hi>Definition of an equilater triangle.</hi>
                     </note> an equilatre triangle is that, which hath three equall sides.</head>
                  <p>Triangles haue their differences partly of their sides, and
<figure/>
partly of their angles. As touching the differences of their sides, there are three kindes. For either all thre sides of the tri<g ref="char:EOLhyphen"/>angle are equall, or two onely are equall, &amp; the third vnequal: <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> three are unequall the one to the other. The first kind of triangles, namely, that which hath three equall sides, is most simple, and easiest to be knowen: and is here first defined, and
<pb facs="tcp:7062:35"/>
is called an equilater triangle, as the triangle <hi>A</hi> in the example, all whose sides are of one length.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of an Isosceles.</hi>
                     </note>
                     <hi>25.</hi> Isosceles, is a triangle, which hath onely two sides equall.</head>
                  <p>The second kinde of triangles
<figure/>
hath two sides of one length, but the third side is either longer or shorter the<g ref="char:cmbAbbrStroke">̄</g> the other two, as are the triangles here figured, <hi>B, C, D</hi>
                  </p>
                  <p>In the triangle <hi>B,</hi> the two sides <hi>AE</hi> and <hi>EF</hi> are equal the one to the other, and the side <hi>AE,</hi> is lo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ger then any of them both: Likewise in the triangle <hi>C</hi> the two sides <hi>GH</hi> and <hi>HK,</hi> are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>quall, and the side <hi>GK</hi> is greater. Also in the triangle <hi>D,</hi> the sides <hi>LM</hi> and <hi>MN,</hi> are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>quall, and the side <hi>LN</hi> is shorter.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of a Scalenum.</hi>
                     </note>
                     <hi>26.</hi> Scalenum is a triangle, whose three sides are all vnequall.</head>
                  <p>As are the triangles <hi>E, F,</hi> in which there is no
<figure/>
one side equall to any of the other. For in the tri<g ref="char:EOLhyphen"/>angle <hi>E,</hi> the side <hi>AC</hi> is greater then the side <hi>BC,</hi> and the side <hi>BC</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the side <hi>AB.</hi> Like<g ref="char:EOLhyphen"/>wise in the triangle <hi>F,</hi> the side <hi>DH,</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the side <hi>DG,</hi> and the side <hi>DG,</hi> is greater then the side <hi>GH.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>An Orthigoni<g ref="char:EOLhyphen"/>um triangle.</hi>
                     </note>
                     <hi>27.</hi> Againe of triangles, an Orthigonium or a rightangled triangle, is a tri<g ref="char:EOLhyphen"/>angle which hath a right angle.</head>
                  <p>As there are three kindes of triangles, by reason of the diuersitie
<figure/>
of the sides, so are there also three kindes of triangles, by reason of the varietie of the angles. For euery triangle either containeth one right angle, &amp; two acute angles: or one obtuse angle, &amp; two acute: or three acute angles: for it is impossible that one triangle should contain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> two obtuse angles, or two right angles, or one obtuse an<g ref="char:EOLhyphen"/>gle, and the other a right angle. All which kindes are here defined. First a rightangled triangle whiche hath in it a right angle. As the triangle <hi>BCD,</hi> of which the angle <hi>BCD,</hi> is a right angle.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>An Ambligoni<g ref="char:EOLhyphen"/>um triangle.</hi>
                     </note>
                     <hi>28.</hi> An ambligonium or an obtuse angled triangle, is a triangle which hath an obtuse angle.</head>
                  <p>As is the triangle <hi>B,</hi> whose angle <hi>AC
<figure/>
D,</hi> is an obtuse angle, and is also a Scale<g ref="char:EOLhyphen"/>non, hauing his three sides vnequall: the triangle <hi>E,</hi> is likewise an Ambligonion, whose angle <hi>FGH,</hi> is an obtuse angle, &amp; is an Isosceles, hauing two of his sides equall, namely <hi>FG</hi> and <hi>GH.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>An Oxigoni<g ref="char:EOLhyphen"/>um triangle.</hi>
                     </note>
                     <hi>29.</hi> An oxigonium or an acute angled triangle, is a triangle which hath all his three angles acute.</head>
                  <p>
                     <pb n="5" facs="tcp:7062:35"/>As the triangles <hi>A, B, C,</hi> in the example, al
<figure/>
whose angles are acute: of which <hi>A</hi> is an e<g ref="char:EOLhyphen"/>quilater triangle, <hi>B,</hi> an Isosceles, and <hi>C</hi> a Scalenon. An equilater triangle is most sim<g ref="char:EOLunhyphen"/>ple, and hath one vniforme construction, and therfore all the angles of it are equall, and neuer hath in it either a right angle, or an obtuse: but the angles of an Isosceles or a Scalenon, may di<g ref="char:EOLhyphen"/>uersly
<figure/>
vary. It is also to be noted that in comparison of any two sides of a triangle, the third is called a base. As of the triangle <hi>ABC</hi> in respect of the two lines <hi>AB</hi> and <hi>AC,</hi> the line <hi>BC,</hi> is the base: and in respect of the two sides <hi>AC</hi> and <hi>CB,</hi> the line <hi>AB,</hi> is the base, and likewyse in respect of the two sides <hi>CB</hi> &amp; <hi>BA,</hi> the line <hi>AC,</hi> is the base.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>30</hi> Of foure syded figures,<note place="margin">
                        <hi>Definition of a square.</hi>
                     </note> a quadrate or square is that, whose sydes are e<g ref="char:EOLhyphen"/>quall, and his angles right.</head>
                  <p>As triangles haue their difference and varietie by reason of their
<figure/>
sides and angles: so likewise do figures of foure sides, take their varie<g ref="char:EOLhyphen"/>tie and difference partly by reason of their sides, &amp; partly by reason of their angles, as appeareth by their definitions. The four sided figure <hi>ABCD</hi> is a square or a quadrate, because it is a right angled figure, al hys angles are right angles, and also all his four sides are equall.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>31</hi> A figure on the one syde longer,<note place="margin">
                        <hi>Definition of a long square.</hi>
                     </note> or squarelike, or as some call it, a long square, is that which hath right angles, but hath not equall sydes.</head>
                  <p>This figure agreeth with a square touching his angles, in
<figure/>
that either of them hath right angles, and differeth from it onely by reason of his sides, in that the sides therof be not e<g ref="char:EOLhyphen"/>quall, as are the sides of a square. As in the example, the an<g ref="char:EOLhyphen"/>gles of the figure <hi>ABCD,</hi> are right angles, but the two sides thereof <hi>AB,</hi> and <hi>CD,</hi> are longer then the other two sides <hi>AC, &amp; BD.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <hi>32</hi> Rhombus (or a diamonde) is a figure hauing foure equall sydes,<note place="margin">
                        <hi>Definition of a Diamond figure</hi>
                     </note> but it is not rightangled.</head>
                  <p>This figure agreeth with a square, as touching the equallitie of
<figure/>
lines, but differeth from it in that it hath not right angles, as hath the square. As of this figure, the foure lines <hi>AB, BC, CD, DA,</hi> be e<g ref="char:EOLhyphen"/>quall, but the angles therof are not right angles. For the two angles <hi>ABC</hi> and <hi>ADC,</hi> are obtuse angles, greater then right angles, &amp; the other two angles <hi>BAD</hi> and <hi>BCD,</hi> are two acute angles lesse then two right angles. And these foure angles are yet equall to foure right angles: for, as much as the acute angle wa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th of a right angle, so much the obtuse angle excedeth a right angle.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:36" rendition="simple:additions"/>
                  <head>
                     <note place="margin">
                        <hi>Definition of a diamondlike fi<g ref="char:EOLhyphen"/>gure.</hi>
                     </note>
                     <hi>33</hi> Rhomb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ides (or a diamond like) is a figure, whose opposite sides are e<g ref="char:EOLhyphen"/>quall, and whose opposite angles are also equall, but it hath neither e<g ref="char:EOLhyphen"/>quall sides, nor right angles.</head>
                  <p>As in the figure <hi>ABCD,</hi> all the foure sides are not
<figure/>
equall, but the two sides <hi>AB</hi> and <hi>CD,</hi> being opposite the one to the other, also the other two sides <hi>AC</hi> and <hi>BD,</hi> being also opposite, are equall the one to the o<g ref="char:EOLhyphen"/>ther. Likewise the angles are not right angles, but the angles <hi>CAB,</hi> and <hi>CDB,</hi> are obtuse angles, and op<g ref="char:EOLhyphen"/>posite and equall the one to the other. Likewyse the angles <hi>ABD,</hi> and <hi>ACD,</hi> are acute angles, and oppo<g ref="char:EOLhyphen"/>site, and also equall the one to the other.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Trapezia or tables.</hi>
                     </note>
                     <hi>34</hi> All other figures of foure sides besides these, are called trapezia, or tables.</head>
                  <p>Such are all figures, in which is obserued no equallitie of sides
<figure/>
nor angles: as the figures <hi>A</hi> and <hi>B,</hi> in the marge<g ref="char:cmbAbbrStroke">̄</g>t, which haue nei<g ref="char:EOLhyphen"/>ther equall sides, nor equal angles, but are described at all aduen<g ref="char:EOLhyphen"/>ture without obseruation of order, and therefore they are called irregular figures.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Definition of Parallel lines.</hi>
                     </note>
                     <hi>35</hi> Parallel or equidistant right lines are such, which
<figure/>
being in one and the selfe same superficies, and pro<g ref="char:EOLhyphen"/>duced infinitely on both sydes, do neuer in any part concurre.</head>
                  <p>As are the lines <hi>AB,</hi> and <hi>CD,</hi> in the example.</p>
                  <p>
                     <figure/>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>Peticions or requestes.</head>
               <div type="section">
                  <head>
                     <hi>1</hi> From any point to any point, to draw a right line.</head>
                  <p>After the definitions, which are the first kind of principles, now follow petitions, which are the second kynd of principles:<note place="margin">
                        <hi>What Petici<g ref="char:EOLhyphen"/>ons are.</hi>
                     </note> which are certain general sentences, so plain, &amp; so perspicuous, that they are perceiued to be true as soone as they are vttered, &amp; no man that hath but common sence, can, nor will deny them. Of which, the first is that, which is here set. As from the point <hi>A,</hi> to the point <hi>B,</hi> who wil de<g ref="char:EOLhyphen"/>ny,
<figure/>
but easily graunt that a right line may be drawn? For two points howsoeuer they be set, are imagined to be in one and the selfe same plaine superficies, wherfore from the one to the other there is some shortest draught, whiche is a right line. Likewise any two right lines howsoeuer they be set, are imagined to be in one superficies, and therefore from any one line to any one line, may be drawen a superficies.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>2</hi> To produce a right line finite, straight forth continually.</head>
                  <p>As to draw in length continually the right line <hi>AB,</hi> who will
<figure/>
not graunt? For there is no magnitude so great, but that there may be a greter, nor any so litle, but that there may be a lesse. And
<pb n="6" facs="tcp:7062:36"/>
a line is a draught from one point to an other, therfore from the point <hi>B,</hi> which is the ende of the line <hi>AB,</hi> may be drawn a line to some other point, as to the point <hi>C,</hi> and from that to an other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and so infinitely.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>3</hi> Vpon any centre and at any distance, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o describe a circle.</head>
                  <p>A playne superficies may in compasse be extended infi<g ref="char:EOLhyphen"/>nitely:
<figure/>
as from any pointe to any pointe may be drawen a right line, by reason wherof it commeth to passe that a cir<g ref="char:EOLhyphen"/>cle may be described vpon any centre and at any space or distance. As vpon the centre <hi>A,</hi> and vpon the space <hi>AB,</hi> ye may describe the circle <hi>BC,</hi> &amp; vpon the same centre, vpon the distance <hi>AD,</hi> ye may describe the circle <hi>DE,</hi> or vppon the same centre <hi>A,</hi> according to the distaunce <hi>AF,</hi> ye may describe the circle <hi>FG,</hi> and so infinitely extendyng your space.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>4</hi> All right angles are equall the one to the other.</head>
                  <p>This peticion is most plaine, and offreth it selfe euen to the
<figure/>
sence. For as much as a right angle is <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ht lyne falling perpendicularly vppon an other, and no one line can fall more perpendicularly vpo<g ref="char:cmbAbbrStroke">̄</g> a line then an other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therfore no one right angle can be greater <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> an other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> neither d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the length or shortenes of the lines alter the greatnes of the angle. For in the example, the right angle <hi>ABC,</hi> though it be made of much lon<g ref="char:EOLhyphen"/>ger lines then the right angle <hi>DEF,</hi> whose lines are much shorter, yet is that angle no greater then the other. For if ye set the point <hi>E</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ust vpon the point <hi>B</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then shal the line <hi>ED,</hi> euenly and iustly fall vpon the line <hi>AB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and the line <hi>EF,</hi> shall also fall equally vpon the line <hi>BC,</hi> and so shal the angle <hi>DEF,</hi> be equall to the angle <hi>ABC,</hi> for that the lines which cause them, are of like inclination.</p>
                  <p>It may euidently also be sene at the centre of a circle. For if
<figure/>
ye draw in a circle two diameters, the one cutting the other in the centre by right angles, ye shall deuide the circle into fowre equall partes, of which eche contayneth one right angle, so are all the foure right angles about the centre of the circle equall.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>5</hi> VVhen a right line falling vpon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wo right lines, doth make <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n one &amp; the selfe same syde, the two inwarde angles lesse then two right angles, then shal these two right lines beyng produced <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> length concurre on that part, in which are the two angles lesse then two right angles.</head>
                  <p>As if the right line <hi>AB,</hi> fall vpon two right lines,
<figure/>
namely, <hi>CD</hi> and <hi>EF,</hi> so that it make the two inward angles on the one side, as the angles <hi>DHI</hi> &amp; <hi>FIH,</hi> lesse then two right angles (as in the example they do) the said two lines <hi>CD,</hi> and <hi>EF,</hi> being drawen forth in le<g ref="char:cmbAbbrStroke">̄</g>gth on that part, wheron the two angles being less<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wo right angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> consist, shal <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> le<g ref="char:cmbAbbrStroke">̄</g>gth concurre and meete together: as in the point <hi>D,</hi> as it is easie to see. For the partes of the lines towardes <hi>DF,</hi> are more enclined the one to
<pb facs="tcp:7062:37"/>
the other, then the partes of the lines towardes <hi>CE</hi> are. Wherfore the more these parts are produced, the more they shall approch neare and neare, till at length they shal mete in one point. Contrariwise the same lines drawn in le<g ref="char:cmbAbbrStroke">̄</g>gth on the other side, for that the angles on that side, namely, the angle <hi>CHB,</hi> and the angle <hi>EIA,</hi> are greater then two right angles, so much as the other two angles are lesse th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n two right angles, shall ne<g ref="char:EOLhyphen"/>uer mete, but the further they are drawen, the further they shalbe distant the one from the other.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>6</hi> That two right lines include not a superficies.</head>
                  <p>If the lines <hi>AB</hi> and <hi>AC,</hi> being right lines, should inclose
<figure/>
a superficies, they muste of necessitie bee ioyned together at both the endes, and the superficies must be betwene the<g ref="char:cmbAbbrStroke">̄</g>. Ioyne them on the one side together in the pointe A, and imagine the point <hi>B</hi> to be drawen to the point <hi>C,</hi> so shall the line <hi>AB,</hi> fall on the line <hi>AC,</hi> and couer it, and so be all one with it, and neuer inclose a space or superficies.</p>
               </div>
            </div>
            <div type="part">
               <head>Common sentences.</head>
               <div type="section">
                  <head>
                     <hi>1</hi> Thinges equall to one and the selfe same thyng: are equall also the one to the other.</head>
                  <p>
                     <note place="margin">
                        <hi>What common sentences are.</hi>
                     </note>After definitions and petitions, now are set common sentences, which are the third and last kynd of principles. Which are certaine general propositio<g ref="char:cmbAbbrStroke">̄</g>s, commonly known of all men, of themselues most manifest &amp; cleare, &amp; therfore are called also dignities not able to be denied of any. Peticions also are very manifest, but not so fully as are the co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>mon sentences,<note place="margin">
                        <hi>Difference be<g ref="char:EOLhyphen"/>twene peticions &amp; common sen<g ref="char:EOLhyphen"/>tences.</hi>
                     </note> and therfore are required or desired to be graunted. Peticions also are more peculiar to the arte whereof they are: as those before put are proper to Geome<g ref="char:EOLhyphen"/>try: but common sentences are generall to all things wherunto they can be applied. Agayne, peticions consist in actions or doing of somewhat most easy to be done: but common sentences consist in consideration of mynde, but yet of such thinges which are most easy to be vnderstanded, as is that before set.
<figure/>
As if the line <hi>A</hi> be equall to the line <hi>B,</hi> And if the line <hi>C</hi> be also equall to the line <hi>B,</hi> then of necessitie the lines <hi>A</hi> and <hi>C,</hi> shalbe equal the one to the other. So is it in all super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iciesses, angles, &amp; numbers, &amp; in all other things (of one kynde) that may be compared together.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>2</hi> And if ye adde equall thinges to equall thinges: the whole shalbe equall.</head>
                  <p>As if the line <hi>AB</hi> be equal to the line <hi>CD,</hi> &amp; to the
<figure/>
line <hi>AB,</hi> be added the line <hi>BE,</hi> &amp; to the line <hi>CD,</hi> be added also an other line <hi>DF,</hi> being equal to the line <hi>BE,</hi> so that two equal lines, namely, <hi>BE,</hi> and <hi>DF,</hi> be added to two equall lynes <hi>AB,</hi> &amp; <hi>CD</hi>: then shal the whole lyne <hi>AE,</hi> be equall to the whole lyne <hi>CF,</hi> and so of all quantities generally.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>3</hi> And if from equall thinges, ye take away equall thinges: the thinges re<g ref="char:EOLhyphen"/>mayning shall be equall.</head>
                  <p>
                     <pb n="7" facs="tcp:7062:37"/>As if from the two lines <hi>AB</hi> and <hi>CD,</hi> being
<figure/>
equal, ye take away two equall lines, namely, <hi>EB</hi> and <hi>FD,</hi> then maye you conclude by this com<g ref="char:EOLhyphen"/>mon sentence, that the partes remayning, name<g ref="char:EOLhyphen"/>ly, <hi>AE,</hi> and <hi>CF</hi> are equall the one to the other and so of all other quantities.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>4</hi> And if from vnequall thinges ye take away equall thinges: the thynges which remayne shall be vnequall.</head>
                  <p>As if the lines <hi>AB,</hi> and <hi>CD,</hi> be vnequall, the line <hi>AB,</hi> beyng
<figure/>
longer then the line <hi>CD,</hi> &amp; if ye take fro<g ref="char:cmbAbbrStroke">̄</g> them two equall lines, as <hi>EB,</hi> and <hi>FD</hi>: the partes remayning, which are the lines <hi>AE</hi> and <hi>CF,</hi> shall be vnequall the one to the other, namely, the lyne <hi>AE,</hi> shall be greater then the line <hi>CF,</hi> which is euer true in all quantities whatsoeuer.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>5</hi> And if to vnequall thinges ye adde equall thinges: the whole shall be vn<g ref="char:EOLhyphen"/>equall.</head>
                  <p>As if ye haue two vnequal lines, namely, <hi>AE</hi> the greater, and
<figure/>
                     <hi>CF</hi> the lesse, &amp; if ye adde vnto the<g ref="char:cmbAbbrStroke">̄</g> two equall lines, <hi>EB</hi> &amp; <hi>FD,</hi> then maye ye conclude that the whole lines composed are vn<g ref="char:EOLhyphen"/>equall: namely, that the whole lyne <hi>AB,</hi> is greater then the whole line <hi>CD,</hi> and so of all other quantities.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>6</hi> Thinges which are double to one and the selfe same thing: are equall the one to the other.</head>
                  <p>As if the line <hi>AB</hi> be double to the line <hi>EF,</hi> and if also the
<figure/>
line <hi>CD,</hi> be double to the same line <hi>EF</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the<g ref="char:cmbAbbrStroke">̄</g> may you by this common sentence conclude, that the two lines <hi>AB,</hi> &amp; <hi>CD,</hi> are equall the one to the other. And this is true in all quanti<g ref="char:EOLhyphen"/>ties, and that not only, when they are double, but also if they be triple or quadruple, or in what proportion soeuer it be of the greater inequallitie.<note place="margin">
                        <hi>What proporti<g ref="char:EOLhyphen"/>on of the grea<g ref="char:EOLhyphen"/>ter inequality i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </note> Which is when the greater quantitie is compared to the lesse.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>7</hi> Thinges which are the halfe of one and the selfe same thing<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are equal the one to the other.</head>
                  <p>As if the line <hi>AB,</hi> be the halfe of the line <hi>EF,</hi> and if the lyne
<figure/>
                     <hi>CD,</hi> be the halfe also of the same line <hi>EF</hi>: then may ye con<g ref="char:EOLhyphen"/>clude by this common sentence, that the two lines <hi>AB</hi> and <hi>CD,</hi> are equall the one to the other. This is also true in all kyndes of quantitie, and that not onely when it is a halfe, but also if it be a third, a quarter, or in what proportion soeuer it be of the lesse in equallitie. Which is when the lesse quantitie is co<g ref="char:cmbAbbrStroke">̄</g>pared to the greater.<note place="margin">
                        <hi>What proporti<g ref="char:EOLhyphen"/>on of the lesse inequalitie i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <hi>8</hi> Thinges which agree together are equall the one to the other.</head>
                  <p>Such thinges are sayd to agree together, whiche when they are applied the one to the other: or set the one vpon the other, the one <gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>deth not the other in any thyng.
<pb facs="tcp:7062:38"/>
As if the two triangles <hi>ABC,</hi> and <hi>DEF,</hi> were applied
<figure/>
the one to the other, and the triangle <hi>ABC,</hi> were set v<g ref="char:EOLhyphen"/>pon the triangle <hi>DEF,</hi> if then the angle <hi>A,</hi> do iustly a<g ref="char:EOLhyphen"/>gree with the angle <hi>D,</hi> and the angle <hi>B,</hi> with the angle <hi>E,</hi> and also the angle <hi>C,</hi> with the angle <hi>F</hi>: and moreouer <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>f the line <hi>AB,</hi> do iustly fall vpon the line <hi>DE,</hi> and the line <hi>AC,</hi> vpon the line <hi>DF,</hi> and also the line <hi>BC,</hi> vppon the line <hi>EF,</hi> so that on euery part of these two triangles, there is iust agreement, then may ye conclude that the two triangles are equall.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>9</hi> Euery whole is greater then his part.</head>
                  <p>As the whole is equal to all his partes taken together, so is it grea<g ref="char:EOLhyphen"/>ter
<figure/>
then any one part therof. As if the line <hi>CB</hi> be a part of the line <hi>AB,</hi> then by this common sentence ye may conclude that the whole line <hi>AB,</hi> is greater then the part, namely, the<g ref="char:cmbAbbrStroke">̄</g> the line <hi>CB.</hi> And this is generall in all thinges.</p>
                  <p>
                     <note place="margin">
                        <hi>What a Propo<g ref="char:EOLhyphen"/>sition is.</hi>
                     </note>THe principles thus placed &amp; ended, now follow the propositions, which are sentences set forth to be proued by reasoning and demonstrations, and therfore they are agayne repeated in the end of the demonstration<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For the proposition is euer the conclusion, and that which ought to be p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oued.</p>
                  <p>
                     <note place="margin">
                        <hi>Propositions of two sortes.</hi>
                     </note>Propositions are of two sortes, the one is called a Probleme, the other a Theoreme.</p>
                  <p>
                     <note place="margin">
                        <hi>What a Pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>bleme is.</hi>
                     </note>A Probleme, is a proposit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>on which requireth some action, or doing: as the makyng of some <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igure, or to deuide a figure or line, to apply figure to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igure, to adde figures to<g ref="char:EOLhyphen"/>gether, or to subtrah one from an other, to describe, to inscribe, to circumscribe one fi<g ref="char:EOLhyphen"/>gure within or without another, and suche like. As of the first proposition of the first booke is a probleme, which is thus<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>Vpon a right line geuen not being infinite, to describe an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>quilater triangle, or a triangle of three equall sides.</hi> For in it, besides the demonstration and contemplation of the mynde, is required somewhat to be done: namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to make an equilater triangle vpon a line geuen. And in the ende of euery probleme, after the de<g ref="char:EOLhyphen"/>monstration, is concluded a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ter this man<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er, <hi>Which is the thing, which was required to be done.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>What a Theo<g ref="char:EOLhyphen"/>reme is.</hi>
                     </note>A Theoreme, is a proposition, which requireth the searching ou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and demonstration of some propertie or passion of some figure: Wherin is onely speculation and contem<g ref="char:EOLhyphen"/>plation of minde, without doing or working of any thing<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> As the fifth proposition of the first booke, which is thus, <hi>An Isosceles or triangle of two equall sides, hath his angles at th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> base, equall the one to the other, &amp;c.</hi> is a Theoreme. For in it is required only to be pro<g ref="char:EOLhyphen"/>ued and made plaine by reason and demonstratio<g ref="char:cmbAbbrStroke">̄</g>, that these two angles be equall, without further working or doing. And in the end<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ry Theoreme, after the demonstration is concluded after this ma<g ref="char:EOLhyphen"/>ner, <hi>Which thyng was required to be demonstrated or proued.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <pb n="8" facs="tcp:7062:38"/>
               <head>The first Probleme. The first Proposition. Vpon a right line geuen not beyng infinite, to describe an e<g ref="char:EOLhyphen"/>quilater triangle, or a triangle of three equall sides.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB. It is required vpon the line AB, to describe an e<g ref="char:EOLhyphen"/>quilater triangle, namely, a triangle of three equall sides.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Now therfore making the centre the point A and the space AB, describe (by the third peticion) a circle BCD: and agayne (by the same) makyng the centre the point B, and the space B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, describe an other circle ACE. And (by the first peticion) from the point C, wherin the circles cut the one the other, draw one right line to the point A, and an
<figure/>
other right line to the point B. And forasmuche as the point A is the centre of the circle CBD,</hi>
                  <note place="margin">
                     <hi>Demonstration</hi>
                  </note> 
                  <hi>therfore (by the</hi> 15. <hi>definition) the line AC is e<g ref="char:EOLhyphen"/>quall to the line AB: Agayne forasmuch as the point B is the centre of the circle CAE, ther<g ref="char:EOLhyphen"/>fore (by the same definition) the line BC is e<g ref="char:EOLhyphen"/>quall to the line BA. And it is proued, that the line AC is equall to the line AB: wherfore ei<g ref="char:EOLhyphen"/>ther of these lines CA and CB, is equall to the line AB: but thinges which are equall to one and the same thing, are also equall the one to the other (by the first common sentence) wherfore the line CA, also is equall to the line CB. VVherfore these three right lines CA, AB, and BC are equal the one to the other. VVherfore the tria<g ref="char:cmbAbbrStroke">̄</g>gle ABC is equilater. VVher<g ref="char:EOLhyphen"/>fore vppon the line AB, is described an equilater triangle ABC. VVherfore vppon a line geuen not being infinite, there is described an equilater triangle, VVhich is the thing, which was required to be done.</hi>
               </p>
               <p>A triangle or any other rectilined figure is then said to be set or described vpon a line, when the line is one of the sides of the figure.</p>
               <p>This first proposition is a <hi>Probleme,</hi> because it requireth acte or doyng, name<g ref="char:EOLhyphen"/>ly, to describe a triangle. And this is to be noted, that euery <hi>Proposition,</hi> whether it be a <hi>Probleme,</hi> or a <hi>Theoreme,</hi> commonly containeth in it <hi>a thing geuen, and a thing requi<g ref="char:EOLhyphen"/>red to be searched out:</hi> although it be not alwayes so. And the <hi>thing geuen,</hi> 
                  <note place="margin">
                     <hi>Thing geuen.</hi>
                  </note> is euer set be<g ref="char:EOLhyphen"/>fore the <hi>thing required.</hi>
                  <note place="margin">
                     <hi>Thing required</hi>
                  </note> In some propositions there are more things geue<g ref="char:cmbAbbrStroke">̄</g> then one, and mo thinges required then one. In some there is nothing geuen at all.</p>
               <p>Moreouer euery <hi>Probleme &amp; Theoreme,</hi> beyng perfect and absolute, ought to haue all these partes, namely. First the <hi>Proposition,</hi>
                  <note place="margin">
                     <hi>Proposition.</hi>
                  </note> to be proued.<note place="margin">
                     <hi>Exposition.</hi>
                  </note> Then the <hi>exposition</hi>
                  <pb facs="tcp:7062:39"/>
which is the explication of the thing geuen. After that followeth the <hi>determinatio<g ref="char:cmbAbbrStroke">̄</g>,</hi>
                  <note place="margin">
                     <hi>Determination.</hi>
                  </note> which is the declaration of the thing required. Then is set the <hi>construction</hi> of suche things which are necessary ether for the doing of the propositio<g ref="char:cmbAbbrStroke">̄</g>,<note place="margin">
                     <hi>Construction.</hi>
                  </note> or for the demo<g ref="char:cmbAbbrStroke">̄</g>stration. Afterward followeth the <hi>demonstration,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> which is the reason and proofe of the proposition. And last of all is put the <hi>conclusion,</hi>
                  <note place="margin">
                     <hi>Conclusion.</hi>
                  </note> which is inferred &amp; proued by the demonstration, and is euer the proposition. But all those partes are not of ne<g ref="char:EOLhyphen"/>cessitie required in euery Probleme and Theoreme. But the <hi>Proposition, demonstra<g ref="char:EOLhyphen"/>tion, and conclusion,</hi> are necessary partes, &amp; can neuer be absent: the other partes may sometymes be away.</p>
               <p>
                  <note place="margin">
                     <hi>Case.</hi>
                  </note>Further in diuers propositions, there happen diuers cases: which are nothing els, but varietie of delineatio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and construction, or chaunge of position, as when pointes, lines super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iciesses, or bodies are chaunged. VVhich thinges happen in diuers propositions.</p>
               <p>
                  <note place="margin">
                     <hi>The thing geuen in this Pro<g ref="char:EOLhyphen"/>bleme.</hi>
                  </note>NOw then in this <hi>Probleme,</hi> the <hi>thing geuen,</hi> is the line geue<g ref="char:cmbAbbrStroke">̄</g>: the <hi>thing required,</hi>
                  <note place="margin">
                     <hi>The thing required.</hi>
                  </note> to be serched out is, how vpo<g ref="char:cmbAbbrStroke">̄</g> that line to describe an equilater triangle. The <hi>Pro<g ref="char:EOLhyphen"/>position</hi> of this <hi>Probleme</hi> is,<note place="margin">
                     <hi>The pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>position.</hi>
                  </note> 
                  <hi>Vpon a right line geuen not beyng infinite, to describe an equilater tri<g ref="char:EOLhyphen"/>angle.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>The exposition.</hi>
                  </note>The <hi>exposition</hi> is, <hi>Suppose that the right line geuen be AB,</hi> and this declareth onely the <hi>thing geuen.</hi> The <hi>determination</hi> is,<note place="margin">
                     <hi>The determi<g ref="char:EOLhyphen"/>nation.</hi>
                  </note> 
                  <hi>It is required vpon the line AB, to describe an equilater triangle:</hi> for therby as you see, is declared onely <hi>the thing required.</hi> The <hi>construction</hi> be<g ref="char:EOLhyphen"/>ginneth at these wo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ds,<note place="margin">
                     <hi>The constructio<g ref="char:cmbAbbrStroke">̄</g>
                     </hi>
                  </note> 
                  <hi>Now therfore making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the point A, &amp; the space AB, describe (by the third pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icion) a circle &amp;c,</hi> and continueth vntil you come to these wordes, <hi>And forasmuch at the point A &amp;c.</hi> For thethe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>to are described circles and lines, necessarye both for the doyng of the proposition, and also for the demonstration therof. VVhich <hi>demonstration</hi> beginneth at these wordes:<note place="margin">
                     <hi>The demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>And forasmuche as the point A is the centre of the circle CBD &amp;c</hi>: And so procedeth till you come to these wordes, <hi>VVher<g ref="char:EOLhyphen"/>fore vpon the line AB is described an equilater triangle ABC.</hi> For vntill you come thether, is, by groundes before set and constructions had, proued, and made euident, that the triangle made, is equilater And then in these wordes,<note place="margin">
                     <hi>The p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rticular conclusion.</hi>
                  </note> 
                  <hi>wherfore vpon the line AB, is described an equilater triangle ABC,</hi> is put the first <hi>conclusion.</hi> For there are common<g ref="char:EOLhyphen"/>ly in euery proposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ion two conclusions: the one perticuler, the other vniuersal: and from the first you go to the last. And this is the first and perticuler conclusi<g ref="char:EOLhyphen"/>on, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or that it conclu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eth, that vpon the lyne AB is described an equilater tria<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>gle, which is according to the exposition. After it,<note place="margin">
                     <hi>The vn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>u<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>sall c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nclus<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>n.</hi>
                  </note> followeth the last and vniuersal <hi>conclusion, wherfore vpon a right line geuen not being infinite is described an equilater triangle.</hi> For whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>her the line geuen be greater or lesse then thys lyne, the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame constructio<g ref="char:cmbAbbrStroke">̄</g>s and demonstrations proue the same conclusion. Last of all is added this clause,<note place="margin">
                     <hi>The note where by it <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nowne <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o be a Pro<g ref="char:EOLhyphen"/>ble<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e.</hi>
                  </note> 
                  <hi>Which is the thing which was required to be done:</hi> wherby as we haue before noted, is de<g ref="char:EOLhyphen"/>clared, that this proposition is a <hi>Probleme</hi> and not a <hi>Theoreme.</hi> As for varietie of ca<g ref="char:EOLhyphen"/>ses in this proposition there is none,<note place="margin">
                     <hi>No cases in thys proposition.</hi>
                  </note> for that the line geuen, can haue no diuersi<g ref="char:EOLhyphen"/>tie of position.</p>
               <p>As you haue in this <hi>Probleme</hi> sene plainelye set foorthe <hi>the thing geuen, and the thing required,</hi> moreouer the <hi>proposition, exposition, determination, construction, demonstration, and conclusion</hi> (which are generall also to many other both <hi>Problemes and Theoremes</hi>) so may you by the example therof distinct them, and searche them out in other <hi>Problemes</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and also <hi>Theoremes.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Thr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> kindes of dem<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>stration.</hi>
                  </note>This also is to be noted, that there are three kyndes of demonstration, The one is called <hi>Demonstratio a priori,</hi> or composition. The other is called <hi>Demonstrati<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb n="9" facs="tcp:7062:39"/>
a posteriori,</hi> or resolution. And the third is a demonstration leadyng to an impos<g ref="char:EOLhyphen"/>sibilitie.</p>
               <p>A demonstration <hi>a priori,</hi>
                  <note place="margin">
                     <hi>Demonstration a priori, or com<g ref="char:EOLhyphen"/>position.</hi>
                  </note> or composition is, when in reaso<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ing, from the prin<g ref="char:EOLhyphen"/>ciples and first groundes, we passe discending continually, till after many reasons made, we come at the leng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h to conclude that, which we first chiefly entend. And this kinde of demonstration vseth Euclide in his booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for the most part.</p>
               <p>A demonstration <hi>a posteriori,</hi>
                  <note place="margin">
                     <hi>Demonstration a posteriori, or resolution.</hi>
                  </note> or resolution is, when contrariwise in reasoning, we passe from the last conclusion made by the premisses, and by the premisses of the premisses, continually ascending, til we come to the first principles and grounds, which are indemonstrable, and for theyr simplicity can suffer no farther resolu<g ref="char:EOLhyphen"/>tion.</p>
               <p>A demonstration leadyng to an impossibilitie is that argument, whose co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>clusion is impossible:<note place="margin">
                     <hi>Demonstration leading to an impossibility.</hi>
                  </note> that is, when it concludeth directly against any principle, or against any proposition before proued by principles, or propositio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore proued.</p>
               <p>Premisses in an argument, are propositions goyng before the conclusion by which the conclusion is proued.<note place="margin">
                     <hi>Premisses what they are.</hi>
                  </note>
               </p>
               <p>Composition passeth from the cause to the effect, or from thinges simple to thinges more compounded. Resolution contrariwise passeth from thinges com<g ref="char:EOLhyphen"/>pounded to thinges more simple, or from the effect to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he cause.</p>
               <p>Composition or the first kynde of demonstration, which passeth from the principles, may easely be sene in this first proposition of Euclide.<note place="margin">
                     <hi>An example of composition in the first proposi<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> The demon<g ref="char:EOLhyphen"/>stration wherof beginneth thus. And forasmuch as the point A is the centre of the circle CBD, therfore the line AC, is equal to the line AB. This reason (you see) taketh his beginnyng of a principle, namely, of the definition of a circle. And this is the first reason.<note place="margin">
                     <hi>First reason.</hi>
                  </note> Agayne forasmuch as B is the centre of the circle CAE, therfore the line BC is equall to the lyne BA: which is the second reason.<note place="margin">
                     <hi>Second reason.</hi>
                  </note> And it was before proued that the lyne AC is equall to the line AB, wherfore either of these lines CA &amp; CB is equal to the lyne AB. And this is the third reaso<g ref="char:cmbAbbrStroke">̄</g>.<note place="margin">
                     <hi>Third reason.</hi>
                  </note> But things which are equall to one &amp; the selfe same thyng, are also equall the one to the other. VVherfore the line CA is equal to the line CB. And this is the fourth argument.<note place="margin">
                     <hi>Fourth reason.</hi>
                  </note> VVherfore these three lines CA, AB, and BC are equall the one to the other which is the conclusion,<note place="margin">
                     <hi>Conclusion.</hi>
                  </note> and the thing to be proued.</p>
               <p>You may also in the same first Propositio<g ref="char:cmbAbbrStroke">̄</g>, easely take an exa<g ref="char:cmbAbbrStroke">̄</g>ple of Resolutio<g ref="char:cmbAbbrStroke">̄</g>:<note place="margin">
                     <hi>Example of re<g ref="char:EOLhyphen"/>solution in the first proposition.</hi>
                  </note> vsing a contrary order passyng backward fro<g ref="char:cmbAbbrStroke">̄</g> the last conclusio<g ref="char:cmbAbbrStroke">̄</g> of the former de<g ref="char:EOLhyphen"/>monstration, til you come to the first principle or ground wheron it began. For the last argument or reason in composition, is the first in Resolution: &amp; the first in composition, is the last in resolution. Thus therfore must ye procede. The tri<g ref="char:EOLunhyphen"/>angle ABC is contained of three equall right lines, namely, AB, AC, and BC, and therfore it is an equilater triangle by the definition of an equilater triangle: and this is the first reason.<note place="margin">
                     <hi>First reason.</hi>
                  </note> That the three lines be equall, is thus proued. The lines AC and CB are equall to the line AB, wherfore they are equall the one to the other: and this is the second reason.<note place="margin">
                     <hi>Second reason.</hi>
                  </note> That the lines AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and BC, are equal is thus proued: The lines AB and AC, are drawen from the centre of the circle ACE, to the circumference of the same: wherfore they are equall by the definitio<g ref="char:cmbAbbrStroke">̄</g> of a circle; and this is the third reason.<note place="margin">
                     <hi>Third reason.</hi>
                  </note> Likewise that the lines AC and AB,<note place="margin">
                     <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ourth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>eason which is th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> end of the whole re<g ref="char:EOLhyphen"/>solution.</hi>
                  </note> are
<pb facs="tcp:7062:40"/>
equall, is proued by the same reason. For the lines AC and AB are drawn from the centre of the circle BCD: wherfore they are equall by the same definition of a circle: this is the fourth reason or sillogisme. And thus is ended the whole resolution: for that you are come to a principle, which is indemo<g ref="char:cmbAbbrStroke">̄</g>strable, &amp; can not be resolued.</p>
               <p>Of a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>emonstration leading to an impossibilitie, or to an absurditie, you may haue an example in the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ourth proposition of this booke.</p>
               <div type="section">
                  <head>An addition of Campanus.</head>
                  <p>BVt nowe if vpon the same line geuen, namely, <hi>AB,</hi> ye wil describe the other two kinds of triangles, namely, an <hi>Isosceles</hi> or a tria<g ref="char:cmbAbbrStroke">̄</g>gle of two equal sides, &amp; a <hi>Scalenon,</hi> or a triangle of three vnequall sides. First for the describing of an <hi>Isosceles</hi> triangle produce the line <hi>AB</hi> on ether side,<note place="margin">
                        <hi>How to describe an Isosceles triangle.</hi>
                     </note> vntill it concur with the circumferences of both the circles in the pointes <hi>D</hi> and <hi>F,</hi> and making the centre the point <hi>A,</hi> describe a circle <hi>HFG</hi> ac<g ref="char:EOLhyphen"/>cording to the qua<g ref="char:cmbAbbrStroke">̄</g>tity of the line <hi>AF.</hi>
                     <figure/>
Likewise making the centre the poynte <hi>B,</hi> describe a circle <hi>HDG,</hi> according to the quantitie of the line <hi>BD.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> these circles shall cut the one the other in two poyntes, which let be <hi>H,</hi> and <hi>G</hi>: And l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s of the line geuen be ioyned with one of the sayd sections by two right lines, which let be <hi>AG</hi> and <hi>BG.</hi> And forasmuche as these two lines <hi>AB</hi> and <hi>AD</hi> are drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre of the circle <hi>CDE</hi> vnto the circumfe<g ref="char:EOLhyphen"/>re<g ref="char:cmbAbbrStroke">̄</g>ce therof, therfore ar they equal. Like<g ref="char:EOLunhyphen"/>wise the lines <hi>BA</hi> and <hi>BF,</hi> for that they are drawen from the centre of the cir<g ref="char:EOLhyphen"/>cle <hi>EACF</hi> to the circumference ther<g ref="char:EOLhyphen"/>of, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e equal. And forasmuch as ether of the lines <hi>AD</hi> and <hi>BF</hi> is equall to the line <hi>AB,</hi> therfore they are equal the one to the other. Wherfore putting the line <hi>AB</hi> co<g ref="char:cmbAbbrStroke">̄</g>mo<g ref="char:cmbAbbrStroke">̄</g> to the<g ref="char:cmbAbbrStroke">̄</g> both, the whole line <hi>BD</hi> shal<g ref="char:EOLhyphen"/>be equall to the whole line <hi>AF.</hi> But <hi>BD</hi> is equal to <hi>BG,</hi> for they are both drawen fro<g ref="char:cmbAbbrStroke">̄</g> the ce<g ref="char:cmbAbbrStroke">̄</g>tre of the circle <hi>HDG</hi> to the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce therof. And likewise by the same rea<g ref="char:EOLhyphen"/>son the line <hi>AF</hi> is equal to the line <hi>AG.</hi> Wherfore by the co<g ref="char:cmbAbbrStroke">̄</g>mo<g ref="char:cmbAbbrStroke">̄</g> sente<g ref="char:cmbAbbrStroke">̄</g>ce the lines <hi>AG</hi> and <hi>BG</hi> are equal the one to the other, and either of them is greater then the line <hi>AB,</hi> for that either of the two lines <hi>BD</hi> and <hi>AF</hi> is greater then the line <hi>AB.</hi> Wherfore v<g ref="char:EOLhyphen"/>pon the line geuen is described an <hi>Isosceles</hi> or triangle of two equall sides.</p>
                  <p>
                     <note place="margin">
                        <hi>How to describe a Scale<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>um.</hi>
                     </note>Ye may also describe vpon the selfe same line a <hi>Scale<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on,</hi> or triangle of three vne<g ref="char:EOLhyphen"/>quall sides, if by two right lines, ye ioyne both the endes of the line g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uen to some one point that is in the circumference of one of the two greater circles, so that that poynt be not in one of the two sections, and that the line <hi>DF</hi> do not concur with it, when it is on either side produced continuallye and directlye. For let the poynte <hi>K</hi> be taken in the circumference of the circle <hi>HDG,</hi> and let it not be in any of the sections, neyther let the line <hi>DF</hi> concur with it, when it is produced continually and directly vnto the circumference therof. And draw these lines <hi>AK</hi> and <hi>BK,</hi> and the line <hi>AK</hi> shal cut the circumference of the circle <hi>HFG.</hi> Let it cut it in the poynte <hi>L</hi>: now then by the common sentence the line <hi>BK</hi> shalbe equal to the line <hi>AL,</hi> for (by the definition of a circle) the line <hi>BK</hi> is equall to the line <hi>BG,</hi> and the line <hi>AL</hi> is equall to the line <hi>AG</hi> which is equal to the line <hi>BG.</hi> Wherfore the line <hi>AK</hi> is greater then the line <hi>BK</hi> and by the same reason maye it be proued that the line <hi>BK</hi> is greater then the line <hi>AB.</hi>
                     <pb n="10" facs="tcp:7062:40"/>
Wherfore the triangle <hi>ABK</hi> consisteth of three vnequal sides. And so haue ye vpon the line geuen, described all the kindes of triangles.</p>
                  <p>This is to be noted, that if a man will mechanically and re<g ref="char:EOLhyphen"/>dely,
<figure/>
not regarding demonstration vpon a line geuen describe a triangle of three equall sides,<note place="margin">
                        <hi>How to describe an equilater tri<g ref="char:EOLunhyphen"/>angle redily &amp; mechanically.</hi>
                     </note> he needeth not to describe the whole foresayd circle, but onely a little part of eche: namely, where they cut the one the other, and so from the point of the section to draw the lines to the endes of the line geuen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> As in this figure here put.</p>
                  <p>And likewise, if vpon the said line he will describe a trian<g ref="char:EOLhyphen"/>gle
<figure/>
of two equall sydes,<note place="margin">
                        <hi>How to describe an Isosceles tri<g ref="char:EOLhyphen"/>angle redily.</hi>
                     </note> let him extende the compasse accor<g ref="char:EOLhyphen"/>ding to the quantitie that he will haue the syde to be, whether longer then the line geuen or shorter: and so draw onely a li<g ref="char:EOLhyphen"/>tle part of eche circle, where they cut the one the other, &amp; fro<g ref="char:cmbAbbrStroke">̄</g> the point of the section draw the lines to the ende of the line geuen. As in the figures here put. Note that in this the two sydes must be such, that beyng ioyned together, they be lon<g ref="char:EOLhyphen"/>ger then the line geuen.</p>
                  <p>And so also if vpon the sayd right line he will describe a
<figure/>
triangle of three vnequal sydes,<note place="margin">
                        <hi>How to describe a Scalenum tri<g ref="char:EOLhyphen"/>angle redily.</hi>
                     </note> let him extend the compasse. First, according to the quantitie that he will haue one of the vnequall sydes to be, and so draw a little part of the circle, &amp; then extend it according to the quantitie that he wil haue the other vnequal syde to be, and draw likewyse a little part of the circle, and that done, from the point of the section dra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the lines to the endes of the line geuen, as in the figure here put. Note that in this the two sides must be such, that the circles described according to their qua<g ref="char:cmbAbbrStroke">̄</g>titie may, cut the one the other.</p>
               </div>
            </div>
            <div type="part">
               <head>The second Probleme. The second Proposition. Fro<g ref="char:cmbAbbrStroke">̄</g> a point geuen, to draw a right line equal to a rightline geuen.</head>
               <p>
                  <hi>SVppose that the point geue<g ref="char:cmbAbbrStroke">̄</g> be A, &amp; let the right line ge<g ref="char:EOLhyphen"/>uen be BC. It
<figure/>
is required fro<g ref="char:cmbAbbrStroke">̄</g> the point A, to drawe a right lyne equall to the line BC. Draw (by the first peticio<g ref="char:cmbAbbrStroke">̄</g>) from the point A to the poynte B a right line AB:</hi>
                  <note place="margin">
                     <hi>Constru<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>.</hi>
                  </note> 
                  <hi>and vpon the line AB describe (by the first propositio<g ref="char:cmbAbbrStroke">̄</g>) an equilater tri<g ref="char:EOLhyphen"/>angle, and let the same be DAB, and exte<g ref="char:cmbAbbrStroke">̄</g>d, by the second peticio<g ref="char:cmbAbbrStroke">̄</g>, the right lines DA &amp; DB, to the poyntes E
<pb facs="tcp:7062:41"/>
and F, &amp; (by the third peticio<g ref="char:cmbAbbrStroke">̄</g>) making the centre B and the space BC describe a circle CGH: &amp; againe (by the same) making the centre D and the space DG describe a circle GKL. And foras<g ref="char:EOLhyphen"/>much
<figure/>
as the pointe B is the centre of the circle CGH,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>therfore (by the</hi> 15. <hi>definitio<g ref="char:cmbAbbrStroke">̄</g>) the line BC is equal to the line BG: and forasmuch as the poynt D is the centre of the circle GKL: therefore (by the same) the line DL is equall to the line DG: of which the line DA is equall to a line DB (by the propositio<g ref="char:cmbAbbrStroke">̄</g> going before) wherfore the residue, namely, the line AL is e<g ref="char:EOLhyphen"/>qual to the residue, namely, to the line BG (by the third common sentence) And it is proued that the line BC is e<g ref="char:EOLhyphen"/>quall to the line BG VVherfore eyther of these lines AL &amp; BC is equal to the line BG. But things which are equall to one and the same thing are also equall the one to the other (by the first commo<g ref="char:cmbAbbrStroke">̄</g> sentence.) VVherfore the line AL is e<g ref="char:EOLhyphen"/>qual to the line BC. VVherfore from the poynt geue<g ref="char:cmbAbbrStroke">̄</g>, namely, A, is drawn a right line AL equall to the right line geuen BC: which was required to be done.</hi>
               </p>
               <p>Of Problemes and Theoremes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> as we haue before noted, some haue no cases at all, which are those which haue onely one position and construction<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and other some haue many and diuers cases: which are such propositions which haue diuers descriptions &amp; constructions, and chaunge their positions. Of which sorte is this second proposition, which is also a Probleme. This proposition hath two thinges geuen:<note place="margin">
                     <p>
                        <hi>Two thinges ge<g ref="char:EOLhyphen"/>uen in this pro<g ref="char:EOLhyphen"/>position.</hi>
                     </p>
                     <p>
                        <hi>Power cases in this proposition.</hi>
                     </p>
                  </note> Namely, a pointe, and a line: the thing re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uired is, that from the pointe geuen wheresoeuer it be put, be drawen a line equall to the line geuen. Now this poynt geuen may haue diuers positio<g ref="char:cmbAbbrStroke">̄</g>s For it may be placed eyther without the right line geuen, or in some point in it. If it be without it, either it is on the side of it, so that the right line drawen from it to the ende of the right line geuen maketh an angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or els it is put directly vnto it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so that the right line geuen being produced shall fall vpon the point geuen which is without. But if it be in the line geuen, then either it is in one of the endes or extreames thereof: or in some place betwene the extremes. So are there foure diuers positions of the poynt in respect of the line. Wherupon follow diuers delineations and constructions, and consequent<g ref="char:EOLhyphen"/>ly varietie of cases.</p>
               <p>
                  <note place="margin">
                     <hi>The first case.</hi>
                  </note>For the first case the figure before put, serueth.
<figure/>
               </p>
               <p>
                  <note place="margin">
                     <hi>The second case.</hi>
                  </note>To the second case the figure here on the side set belongeth. And as touching the or<g ref="char:EOLhyphen"/>der both o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> construction and of demonstra<g ref="char:EOLhyphen"/>tion it is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll one with the first.</p>
               <p>
                  <note place="margin">
                     <hi>The thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>d case.</hi>
                  </note>The third case is easiest of all, namely, whe<g ref="char:cmbAbbrStroke">̄</g> the poynt geuen is in one of the extreames. As for exa<g ref="char:cmbAbbrStroke">̄</g>ple, if it were in the point <hi>C,</hi> which
<pb n="11" facs="tcp:7062:41"/>
is one of the extreames of the line <hi>BC.</hi> Then
<figure/>
making the centre the poynt <hi>C,</hi> and the space <hi>CB</hi> describe a circle <hi>BLG</hi>: and from the cen<g ref="char:EOLhyphen"/>tre <hi>C</hi> drawe a line vnto the circumference, which let the <hi>CL,</hi> which by the definition of a circle, shalbe equall to the line geuen, <hi>BC.</hi>
               </p>
               <p>The fourth case as touching construction herein differeth from the two firste,<note place="margin">
                     <hi>The fourth case.</hi>
                  </note> for th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t whereas in the<g ref="char:cmbAbbrStroke">̄</g> you are willed to draw a right line from the poynt geuen, namely, <hi>A,</hi> to the poynt <hi>B</hi> which is one of the endes of the line geue<g ref="char:cmbAbbrStroke">̄</g>, here you shal not nede to draw that line, for that it is already drawen. As touching the rest, both in construction and demonstration you may proceede as in the two firste<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> As it is manifeste to see in thys figure here on the side put.</p>
               <p>This proposition for the playnes &amp; easi<g ref="char:EOLhyphen"/>nes
<figure/>
thereof,<note place="margin">
                     <hi>This proposition though it be ve<g ref="char:EOLhyphen"/>ry easie to be done mechani<g ref="char:EOLhyphen"/>cally<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> yet is n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> principl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note> seemeth to be as it were a princi<g ref="char:EOLhyphen"/>ple, and may easly mechanically be done. For opening the compasse to the quantitye of the line geuen, and setting on foote of it fixed in the poynt geuen and marking with the other another poynt whersoeuer it fall, &amp; so by the first peticion drawing a right line fro<g ref="char:cmbAbbrStroke">̄</g> the one of those poyntes to the other, the sayd righte line shall be equall to the right line geuen: yet in deede is it no principle, for that it may by demonstration be proued: but principles can not be proued, as we haue before declared.</p>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Probleme. The <hi>3.</hi> Proposition. Two vnequal right lines being geuen, to cut of from the grea<g ref="char:EOLhyphen"/>ter, a right lyne equall to the lesse.</head>
               <p>
                  <hi>SVppose that the two vnequal right lines geuen be AB &amp; C, of which l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the lyne AB be the greater. It is requi<g ref="char:EOLhyphen"/>red from the line AB being the greater, to cut of a right line equal to the right line C, which is the lesse line-draw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> on (by the second proposition) fro<g ref="char:cmbAbbrStroke">̄</g> the point A a right line equall to the line C, and let the same be AD: and making the centre A and the space AD describe (by the third peticion) a circle DEF. And forasmuche as the
<figure/>
point A is the centre of y<hi rend="sup">e</hi> circle DEF, therfore AE is equal to AD, but the line C is equal to the line AD. VVherfore either of these lines AE and C is equall to AD, wherfore the line AE is equall to the line C, wherfore two vnequall right lines being geuen, namely, AB and C, there is cut of from AB being y<hi rend="sup">e</hi> greater, a right line AE equall to the lesse lyne, namely, to C: which was required to be done.</hi>
               </p>
               <p>
                  <pb facs="tcp:7062:42"/>This proposition, which is a Probleme, hath two thinges geuen,<note place="margin">
                     <p>
                        <hi>Two <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> ge<g ref="char:EOLhyphen"/>uen in <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>is pro<g ref="char:EOLhyphen"/>position.</hi>
                     </p>
                     <p>
                        <hi>Diuers cases in <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap>.</hi>
                     </p>
                  </note> namely, two vnequall right lines: the thing required is, from the greater to cut of a line equal to the lesse. It hath also diuers cases. For the lines geuen either are distinct th'one from the other: or are ioyned together at one of their endes: or they cut the one the other, or the one cutteth the other in one of the extreames. VVhich may be two wayes. For ether the greater cutteth the lesse, or the lesse the greater. If they cut the one the other, either <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ch cutteth th'other into equall partes: or into vne<g ref="char:EOLhyphen"/>quall partes: or the one into equall partes, and the other into vnequall partes. VVhich may happ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n in two sorts, for the greater may be cut into equall partes, and the lesse into vnequall partes: or contrariwise.</p>
               <p>
                  <note place="margin">
                     <hi>The first case.</hi>
                  </note>When the vnequall lines geuen are distinct the one from the other, the figure be<g ref="char:EOLhyphen"/>fore put serueth.</p>
               <p>
                  <note place="margin">
                     <hi>The second case.</hi>
                  </note>If they be ioyned together at one of their
<figure/>
ends, it is easie to do. For making the centre that end where they are ioyned together, &amp; the space the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> line, describe a circle: whiche shall of ne<g ref="char:EOLhyphen"/>cessitie (by the definition of a circle) cut of from the greater line a line equall to the lesse line, as it is playne to see in the figure here put.</p>
               <p>But if the one cut the other in one of the ex<g ref="char:EOLhyphen"/>tremes.<note place="margin">
                     <hi>The third case.</hi>
                  </note> As for exa<g ref="char:cmbAbbrStroke">̄</g>ple: Suppose that the vnequall right lines geuen be <hi>AB</hi> and <hi>CD,</hi> of which let the line <hi>CD</hi> be the greater: And let the line <hi>CD</hi> cut the line <hi>AB</hi> in his extreame <hi>C.</hi> Then making the centre <hi>A</hi> and the space <hi>AB,</hi> describe a circle <hi>BE.</hi> And vpon the line <hi>AC</hi> describe an equilater tri<g ref="char:EOLhyphen"/>angle (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst) which let b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>AEC</hi>: &amp; produce the lines <hi>EA</hi> and <hi>EC.</hi> And againe making the ce<g ref="char:cmbAbbrStroke">̄</g>tre <hi>E</hi> and the space <hi>EF</hi> describe a ci<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>e <hi>GF.</hi> Like<g ref="char:EOLhyphen"/>wise making the centre <hi>C</hi> and the space <hi>CG,</hi> des<g ref="char:EOLhyphen"/>cribe a circle <hi>GL.</hi> Now forasmuch as the line <hi>EF</hi> is equall to the line <hi>EG</hi> (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or <hi>E</hi> is the centre) of which the line <hi>EA</hi> is equall to the line <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi>: th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<g ref="char:EOLunhyphen"/>fore the residue <hi>AF</hi> is equal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the residue <hi>CG.</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut the line <hi>AF</hi> is equall to the line <hi>AB,</hi> for <hi>A</hi> is the centre, where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e also the line <hi>CG</hi> is equall to the line <hi>AB.</hi> But the line <hi>CG</hi> is also equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the line <hi>CL,</hi> for the point <hi>C</hi> is the centre. Where<g ref="char:EOLhyphen"/>fore the line <hi>AB</hi> is equall to the line <hi>CL.</hi> Where<g ref="char:EOLhyphen"/>fore from the line <hi>CD</hi> is cut of the line <hi>CL</hi> which is equall to the line <hi>AB.</hi>
               </p>
               <p>But now let <hi>CD</hi> be lesse then <hi>AB,</hi> and let it cut <hi>AB</hi>
                  <figure/>
by his extreame <hi>C.</hi>
                  <note place="margin">
                     <hi>The fourth case</hi>
                  </note> Now then eyther it cutteth it in the middest or not in the middest. First let it cut it in the mid<g ref="char:EOLunhyphen"/>dest: then <hi>CD</hi> is ether the hal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e of <hi>AB,</hi> &amp; so is <hi>AC</hi> equal to <hi>CD.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>The fifth case.</hi>
                  </note>Or it is lesse then the halfe: and then making the centre <hi>C</hi> &amp; the space <hi>CD</hi> describe a circle, which shall cut of from the line <hi>AB</hi> a line equal to the line <hi>CD.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>The s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>xt case.</hi>
                  </note>Or it is greater then the half. And the<g ref="char:cmbAbbrStroke">̄</g> vnto the point <hi>A</hi> put the line <hi>AF</hi> equall to the line <hi>CD,</hi> by the second. And making the centre <hi>A</hi> &amp; the space <hi>AF</hi> describe a cir<g ref="char:EOLhyphen"/>cle, which shall cut of from the line <hi>AB</hi> a line equall to the line <hi>AF,</hi> that is, vnto the line <hi>CD.</hi>
               </p>
               <p>
                  <pb n="12" facs="tcp:7062:42"/>But if the line <hi>CD</hi> do not cut the line <hi>AB</hi> in
<figure/>
the midst:<note place="margin">
                     <hi>The seuenth &amp; eight cases.</hi>
                  </note> 
                  <hi>CD</hi> shal either be the halfe of the line <hi>AB</hi>: or greater then the halfe, or lesse. If <hi>CD</hi> be the halfe of <hi>AB,</hi> or lesse then the half of <hi>AB,</hi> the<g ref="char:cmbAbbrStroke">̄</g> making the centre <hi>C,</hi> and the space <hi>CD</hi> describe a circle whi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he shall cut of from the line <hi>AB</hi> a line equal to the line <hi>CD.</hi>
               </p>
               <p>But if it be greater then the halfe,<note place="margin">
                     <hi>The ninth case.</hi>
                  </note> then againe vnto the point <hi>A</hi> put the line <hi>AF</hi> equal to the line <hi>CD</hi> (by the second propositio<g ref="char:cmbAbbrStroke">̄</g>:) &amp; making the centre <hi>A,</hi> and the space <hi>AF</hi> describe a circle which shall cut of from the line <hi>AB</hi> a line equall to the line <hi>AF,</hi> that is, to the line <hi>CD.</hi>
               </p>
               <p>But if they cut the one the other as the lines <hi>CD</hi> &amp; <hi>AB</hi> do.<note place="margin">
                     <hi>The tenth case.</hi>
                  </note> The<g ref="char:cmbAbbrStroke">̄</g> making the ce<g ref="char:cmbAbbrStroke">̄</g>tre <hi>B</hi> &amp; the space <hi>BA</hi> describe a circle <hi>AF,</hi> &amp; draw a line from the point <hi>B</hi> to the point <hi>C,</hi> &amp; produce it to the point <hi>F.</hi> And forasmuch as the two right lines <hi>BF</hi> and <hi>CD</hi> are vnequall, and the line <hi>CD</hi> cutteth the line <hi>BF</hi> by one of his extreames, therefore it is possible to cut of from <hi>CD</hi> a line equall to the line <hi>BF.</hi> For how to do it we haue before decla<g ref="char:EOLhyphen"/>red, wherefore it is possible from the line <hi>CD</hi> to cut of a line equall to the line <hi>AB</hi>: or <hi>AB</hi> and <hi>BF</hi> are equall the one to the other.</p>
               <p>This is to be noted, that in all these cases,<note place="margin">
                     <hi>In all these ca<g ref="char:EOLhyphen"/>ses the constru<g ref="char:EOLhyphen"/>ction and demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration of the first case will serue.</hi>
                  </note> a man may both as touchyng con<g ref="char:EOLhyphen"/>struction and demonstration, procede as in the first case. For it is possible in any position to put to the ende of the greater lyne a line equall to the lesse lyne, and so makyng the centre the sayd ende, and the space the lesse line, to describe a cir<g ref="char:EOLhyphen"/>cle, which shall cut of fro<g ref="char:cmbAbbrStroke">̄</g> the greater lyne a lyne equall to the line put, namely, to the lesse line geuen, as it is manifest to see in the figures partly here vnder set, and partly in the beginning of the other side put.</p>
               <p>
                  <figure/>
                  <pb facs="tcp:7062:43"/>
                  <figure/>
               </p>
               <p>
                  <note place="margin">
                     <hi>This Proposi<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion, thoughe also it be most easie to be done me<g ref="char:EOLhyphen"/>chanically, yet is no prin<g ref="char:EOLunhyphen"/>ciple.</hi>
                  </note>If a man will me<g ref="char:EOLhyphen"/>chanically
<figure/>
and redi<g ref="char:EOLhyphen"/>ly do this propositi<g ref="char:EOLhyphen"/>on, not regardyng demonstration, hee may extende his co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>passe accordyng to the quantitie of the lesse lyne geuen, &amp; so set on foote ther<g ref="char:EOLhyphen"/>of in one of the ends of the greater lyne geuen, and with the other foote marke a pointe in the said greater line, which shall cutte of from the greater line a line eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all to the lesse. The easines of doing wherof may cause this proposition also to seeme vnto some to be rather a principle, then a proposition. But to that we haue before in the former proposition aunswered.</p>
            </div>
            <div type="part">
               <head>The first Theoreme. The <hi>4.</hi> Proposition. If there be two triangles, of which two sides of th'one be equal to two sides of the other, eche side to his correspondent side, and hauing also on angle of the one equal to one angle of the other, namely, that angle which is contayned vnder the equall right lines: the base also of the one shall be equall to the base of the other, and the one triangle shall be equal to the other triangle, and the other angles remayning shal be equall to the other an<g ref="char:EOLhyphen"/>gles remayning, the one to the other, vnder which are subten<g ref="char:EOLhyphen"/>ded equall sides.</head>
               <p>
                  <pb n="13" facs="tcp:7062:43" rendition="simple:additions"/>
                  <hi>SVppose that there be two triangles ABC, &amp; DEF, ha<g ref="char:EOLhyphen"/>uing two sides of the one, namely AB, and AC, equall to two sides of the other, namely, to DE and DF, the one to the other, that is, AB to DE, and AC to DF: hauyng al<g ref="char:EOLhyphen"/>so the angle BAC, equall to the angle EDF. Then I say that the base also BC is equall to y<hi rend="sup">e</hi> base EF: &amp; y<hi rend="sup">t</hi> the tri<g ref="char:EOLhyphen"/>angle ABC, is equall to the triangle DEF: and y<hi rend="sup">t</hi> the o<g ref="char:EOLhyphen"/>ther angles remainyng are equall to the other angles re<g ref="char:EOLhyphen"/>mayning, the one to the other, vnder which are subtended equall sydes: that is, y<hi rend="sup">t</hi> the angle ABC is equall to the angle DEF, and y<hi rend="sup">t</hi> the angle ACB is equall to to the angle DFE.</hi>
                  <note place="margin">Demonstratio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ing to an absurditie.</note> 
                  <hi>For the triangle ABC ex<g ref="char:EOLhyphen"/>actly
<figure/>
agreyng with the triangle DEF, and the point A being put vpo<g ref="char:cmbAbbrStroke">̄</g> the point D, &amp; the right line AB vpon the right line DE, the pointe B also shall exactly agree with the pointe E: for that (by supposition) the line AB is equal to the line DE. And the line AB exactly agreeyng with the line DE, the right line also AC exact<g ref="char:EOLhyphen"/>ly agreeth with the right line DF, for that (by supposition) the angle BAC is equall to the an<g ref="char:EOLhyphen"/>gle EDF. And forasmuch as the right line AC is also (by supposition) equall to the right line DF, therfore the pointe C exactly agreeth with the pointe F. A<g ref="char:EOLhyphen"/>gaine forasmuch as the pointe C exactly agreeth with the poynte F, and the point B exactly agreeth with the point E: therefore the base BC shall exactly agree with the base EF. For if the point B do exactly agree with the point E, and the point C with the point F, and the base BC do not exactly agre wyth the base EF, then two right lines do include a superficies: which (by the</hi> 10. <hi>co<g ref="char:cmbAbbrStroke">̄</g>mon sentence) is impossible. VVherfore the base BC exactly agreeth w<hi rend="sup">t</hi> the base EF, and therfore is equall vnto it. VVherfore the whole triangle ABC exactly a<g ref="char:EOLhyphen"/>greeth with the whole triangle DEF, &amp; therfore (by the</hi> 8. <hi>common sentence) is equall vnto it. And (by the same) the other angles remayning exactly agree with the other angles remayning, and are equall the one to the other: that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. If therfore there be two triangles, of which two sides of the one, be equall to two sydes of the other, eche to his correspondent side, and hauing also one angle of the one equall to one angle of the other, namely, that angle which is contayned vn<g ref="char:EOLhyphen"/>der the equall right lines: the base also of the one shall be equall to the base of the other, and the one triangle shall be equall to the other triangle, and the other an<g ref="char:EOLhyphen"/>gles remainyng shall be equall to the other angles remayning, the one to the o<g ref="char:EOLhyphen"/>ther, vnder which are subtended eqaull sydes: whiche thing was required to be demonstrated.</hi>
               </p>
               <p>This Proposition which is a Theoreme,<note place="margin">
                     <hi>Two <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                     </hi>
                  </note> hath two things geuen: namely, the
<pb facs="tcp:7062:44"/>
equality of two sides of the one triangle, to two sides of the other triangle, and the equalitie of two angles contayned vnder the equall sydes. In it also are thre thinges required.<note place="margin">
                     <hi>Three thinges required in it.</hi>
                  </note> The equality of base to base: the equality of field to field: and the equality of the other angles of the one triangle to the other angles of the o<g ref="char:EOLhyphen"/>ther triangle, vnder which are subtended equall sides.</p>
               <p>
                  <note place="margin">
                     <hi>How one side i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> equall to an o<g ref="char:EOLhyphen"/>ther, &amp; so gene<g ref="char:EOLunhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ally how one right line is e<g ref="char:EOLhyphen"/>quall to an other.</hi>
                  </note>One side of a playne figure is equall to an other, and so generally one right lyne is equall to an other, when the one being applied to the other, theyr ex<g ref="char:EOLhyphen"/>treames agree together. For otherwise euery righte line applied to any right lyne, agreeth therwith: but equall right lines only, agree in the extremes.</p>
               <p>
                  <note place="margin">
                     <hi>How one rectili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> angle is e<g ref="char:EOLhyphen"/>qual to an other</hi>
                  </note>One rectilined angle is equall to an other rectilined angle, when one of the sides which comprehendeth the one angle, being set vpon one of the sides which comprehendeth the other angle, the other side of the one agreeth with the other syde of the other. And that angle is the greater, whose syde falleth without: and that the lesse, whose syde falleth within.</p>
               <p>
                  <note place="margin">
                     <hi>Why this parti<g ref="char:EOLhyphen"/>cle <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> to his c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>espondent side, is put.</hi>
                  </note>VVhere as in this proposition is put this particle <hi>eche to his correspondent side,</hi> (in<g ref="char:EOLunhyphen"/>stede wherof often times afterward is vsed this phrase <hi>the one to the other)</hi> it is of ne<g ref="char:EOLhyphen"/>cessity so put. For otherwise two sydes of one triangle added together, may be e<g ref="char:EOLhyphen"/>quall to two sydes of an other triangle added together, and the angles also con<g ref="char:EOLhyphen"/>tayned vnder the equall sydes may be equall: and yet the two triangles may not<g ref="char:EOLhyphen"/>withstanding be vnequall.<note place="margin">
                     <hi>How one trian<g ref="char:EOLhyphen"/>gle is equal to an other</hi>
                  </note> VVhere note that a triangle is sayd to be equall to an other triangle, when the field or area of the one is equall to the area of the other. And the area of a triangle,<note place="margin">
                     <hi>What the fielde or area of a tri<g ref="char:EOLhyphen"/>angle is, and so of any rectilined <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>.</hi>
                  </note> is that space, which is contayned within the sydes of a triangle. And the circuite or compasse of a triangle is a line composed of all the sides of a triangle. And so may you think of all o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>her rectilined figures.<note place="margin">
                     <hi>What <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> of a triangle is, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o also of a<g ref="char:EOLhyphen"/>ny <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> fi<g ref="char:EOLhyphen"/>gure.</hi>
                  </note> And now to proue that there may be two triangles, two sydes of one of which being added together, may be equall to two sydes o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the other added together, and the angles contayned vnder the equall sydes may be equall, and yet notwithstanding the two triangles vnequall. Suppose that there be two rectangle triangles: namely, ABC, and DEF, and let their right angles be BAC and EDF. And in the tri<g ref="char:EOLhyphen"/>angle ABC let the syde AB b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 3. and the syde AC 4. which both added toge<g ref="char:EOLhyphen"/>ther make 7.</p>
               <p>And in the
<figure/>
tria<g ref="char:cmbAbbrStroke">̄</g>gle DEF, let the side DE be 2. and the side DF be 5. whiche added toge<g ref="char:EOLhyphen"/>ther make al<g ref="char:EOLunhyphen"/>so 7. &amp; so the sydes of the one triangle added together, are equall to the sides of the other tria<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>gle added together. Yet are both the triangles vnequall, and also their bases. For the area of the triangle ABC is 6 and his base is 5. And the area of the triangle DEF is 5: and his base <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 29. So that to haue the areas of two triangles to be e<g ref="char:EOLhyphen"/>quall, it is requisite that all the sydes of the two triangles be equall, eche to hys correspondent syde. It happeneth also sometymes in triangles, that the areas of them beyng equall, their sydes added together shall be vnequall. And contrari<g ref="char:EOLhyphen"/>wise,
<pb n="14" facs="tcp:7062:44"/>
their sides beyng equall, their areas be vnequall. As in these figures here put it is plaine to see. In the first
<figure/>
example the areas of the two triangles are equal, for they are eche 12. and the sides in ech ad<g ref="char:EOLhyphen"/>ded together are vnequall, for in the one triangle the sides ad<g ref="char:EOLhyphen"/>ded together make 18. and in the other they make 16. But in the second exa<g ref="char:cmbAbbrStroke">̄</g>ple the areas of the two tria<g ref="char:cmbAbbrStroke">̄</g>gles are vnequal, for the one is 12. and th'other is 15. But the sides added to<g ref="char:EOLhyphen"/>gether in eche are equall, for in eche they make 18.</p>
               <p>That angle is said to sub<g ref="char:EOLhyphen"/>tend a side of a tria<g ref="char:cmbAbbrStroke">̄</g>gle,<note place="margin">
                     <hi>How an angle is sayd to subte<g ref="char:cmbAbbrStroke">̄</g>d a side: and a side an angle.</hi>
                  </note> which is placed directly opposite, &amp; against that side. That side also is sayd to subtend an an<g ref="char:EOLhyphen"/>gle, which is opposite to the angle. For euery angle of a triangle is contayned of two sydes of the triangle, and is subtended to the third side.</p>
               <p>This is the first Proposition in which is vsed a demonstration leading to an absurditi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> an impossib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>litie.<note place="margin">
                     <hi>This proposition proued by a de<g ref="char:EOLhyphen"/>monstration lea<g ref="char:EOLunhyphen"/>ding to an absur<g ref="char:EOLunhyphen"/>dity.</hi>
                  </note> VVhich is a demonstration that proueth not di<g ref="char:EOLhyphen"/>rectly the thing entended, by principles, or by thinges before proued by these principles: but proueth the contrary therof to be impossible, &amp; so doth indirect<g ref="char:EOLhyphen"/>ly proue the thing entended.</p>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Theoreme. The <hi>5.</hi> Proposition. An Isosceles, or triangle of two equal sides, hath his angles a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the base equall the one to the other. And those equal sides be<g ref="char:EOLhyphen"/>ing produced, the angles which are vnder the base are also e<g ref="char:EOLhyphen"/>quall the one to the other.</head>
               <p>
                  <hi>SVppose that ABC be a triangle of two
<figure/>
equall sydes called Isosceles, hauing the syde AB equall to the side AC. And (by the second peticio<g ref="char:cmbAbbrStroke">̄</g>) produce the lines AB &amp; AC directly to y<hi rend="sup">e</hi> points D &amp; E. The<g ref="char:cmbAbbrStroke">̄</g> I say, that y<hi rend="sup">e</hi> angle ABC is equal to the angle ACB: and y<hi rend="sup">e</hi> y<hi rend="sup">e</hi> angle CBD is equal to th' angle BCE. Take in the line BD a point at all aduentures,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>and let the
<pb facs="tcp:7062:45"/>
same be F, and (by the third proposition) from the greater line, namely, AE, cut of a line equall to AF being the lesse line, and
<figure/>
let the same be AG: and draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point F to the point C,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>and an other from the point G to the point B. Now then for as muche as AF is equall to AG, and AB is equall to AC, therefore these two lines FA and AC are equall to these two lines GA and AB, the one to the other, and they containe a common angle, namely, that which is co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tained vnder FAG: wherfore (by the fourth pro<g ref="char:EOLhyphen"/>position) the base FC is equall to the base GB: and the triangle AFC is equall to the triangle AGB, and the other angles remai<g ref="char:EOLhyphen"/>ning, are equall to the other angles remaining the one to the other, vnder which are subtended equall sides: that is, the angle ACF is equall to the angle ABG, and the angle AFC is equall to the angle AGB. And forasmuch as the whole line AF is equall to the whole line AG, of which the line AB is equal to y<hi rend="sup">e</hi> lyne AC, therfore the residue of the line AF, namely, the line BF, is equal to the re<g ref="char:EOLhyphen"/>sidue of the line AG, namely, to the line CG (by the third common sentence) And it is proued that CF is equal to BG. Now therfore these two BF &amp; FC are equall to these two CG and GB the one to the other, and the angle BFC is equall to the angle CGB, and they haue one base, namely, BC, common to them both: wherfore (by the</hi> 4. <hi>proposition) the triangle BFC is equall to the tri<g ref="char:EOLhyphen"/>angle CGB, and the other angles remaynyng are equall to the other an<g ref="char:EOLhyphen"/>gles remaining eche to other, vnder which are subtended equall sides. VVher<g ref="char:EOLhyphen"/>fore the angle FBC is equall to the angle GCB, and the angle BCF is equall to the angle CBG. Now forasmuch as the whole angle ABG is equall to the whole angle ACF (as it hath bene proued) of which the angle CBG is equal to the angle BCF. therfore the angle remayning: namely, ABC is equall to the angle remaining, namely, to ACB (by the third common sentence) And they ar the angles at the base of the triangle ABC. And it is proued that the angle FBC is equall to the angle GCB, and they are angles vnder the base. VVherfore a triangle of two equall sides hath his angles at the base equall the one to y<hi rend="sup">e</hi> other. And those equall sides being produced, the angles which are vnder the base are also equall the one to the other: which was required to be proued.</hi>
               </p>
               <p>For that
<figure/>
this pro<g ref="char:EOLunhyphen"/>positio<g ref="char:cmbAbbrStroke">̄</g> is some what hard to perceue by rea<g ref="char:EOLhyphen"/>son the
<pb n="15" facs="tcp:7062:45"/>
sides of the triangles AFC and AGB &amp; also the sydes of the tria<g ref="char:cmbAbbrStroke">̄</g>gle BFC &amp; CGB run so one within an other, therfore I haue here put the distinctly, name<g ref="char:EOLhyphen"/>ly, the triangles FAC and BFC on o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e syde of the figure of the propositio<g ref="char:cmbAbbrStroke">̄</g> &amp; the triangles ACB and CGB on the other syde to that <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>de may with lesse la<g ref="char:EOLhyphen"/>bor see the demonstration playnely.</p>
               <p>That in an Isosceles triangle, the two angles aboue the base are equall, may otherwise be <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> on str<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>red without drawing lines beneath the base somwhat al<g ref="char:EOLhyphen"/>tering the construction. Namely, drawing the lines within the triangle, whiche before were without it after this manner.</p>
               <p>Suppose that <hi>ABC</hi> be an Isosceles triangle and in the line
<figure/>
                  <hi>AB</hi> take a point at all aduentures,<note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration inuen<g ref="char:EOLhyphen"/>ted by Proclus.</hi>
                  </note> and let the same be <hi>D,</hi> And from the line <hi>AC</hi> out of a line equall to the line <hi>AD.</hi> Which let be <hi>AE.</hi> And draw these right lines <hi>BE, DC,</hi> and <hi>DE.</hi> Now forasmuch as in the triangles, <hi>ABE,</hi> and <hi>ACD,</hi> the side <hi>AB</hi> is equall to the side <hi>AC,</hi> by supposition, and the sides <hi>AD</hi> and <hi>AE</hi> are also equall by construction, and the angle at the poynt <hi>A</hi> is common to them both, therfore, by the fourth propositi<g ref="char:EOLhyphen"/>on, the base <hi>BE</hi> is equall to the base <hi>DC.</hi> And, by the same, the angles remayning of the one triangle are equall to the angles remayning of the other triangle. Wherefore the angle <hi>ABE</hi> is equal to the angle <hi>AC</hi>D. Againe forasmuch as in the triangles <hi>BDE,</hi> and <hi>CED</hi> the side <hi>DB</hi> is equall to the side <hi>EC,</hi> and the side <hi>BE</hi> to the side <hi>DC,</hi> and the angle <hi>DBE</hi> is equal to the an<g ref="char:EOLhyphen"/>gle <hi>ECD</hi> and the base <hi>DE</hi> being common to both triangles is equall to it selfe, there<g ref="char:EOLhyphen"/>fore the angles remayning of the one triangle, are equall to the angles remayning of the other triangle. Wherfore the angle <hi>ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> is equall to the angle D<hi>EC</hi>: &amp; the angle <hi>DEB</hi> is equal to the angle <hi>EDC.</hi> And forasmuch as the angle EDB is equal to the an<g ref="char:EOLunhyphen"/>gle DEC, fro<g ref="char:cmbAbbrStroke">̄</g> which are taken away equall angles DEB, &amp; EDC, therfore by the co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon sentence the angles remayning, namely, BDC and <hi>CEB</hi> are equall, And as it was before manifest the sides <hi>B</hi>D and D<hi>C</hi> are equall to the sides <hi>CE</hi> and <hi>EB</hi> the one to the other, that is, ech to his correspondent side: and the base <hi>BC</hi> is common to both the triangles: wherfore the angles remayning are equall to the angles remayning the one to the other, vnder which are subte<g ref="char:cmbAbbrStroke">̄</g>ded equall sides. Wherfore the angle D<hi>BC</hi> is equall to the angle <hi>ECB.</hi> For the line D<hi>C</hi> subtendeth the angle D<hi>BC,</hi> and the line <hi>EB</hi> subte<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>deth the angle <hi>ECB</hi>: which two lines are as we haue before proued equall. Wherfore in an Isosceles triangle, the angles at the base are equall, though the right lines be not produced.</p>
               <p>To proue this also, there is an other demonstration of Pappus much shor<g ref="char:EOLunhyphen"/>ter which needeth no kind of addition of any thing at all: as followeth.</p>
               <p>Suppose that <hi>ABC</hi> be an Isosceles triangle,<note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration inuen<g ref="char:EOLhyphen"/>ted by Pappus.</hi>
                  </note> &amp; let the side
<figure/>
                  <hi>AB</hi> be equall to the side <hi>AC.</hi> Now then vnderstand this one triangle to be as it were two triangles. And thus reason. For<g ref="char:EOLhyphen"/>asmuch as in the two triangles <hi>ABC</hi> and <hi>ACB, AB</hi> is equal to <hi>AC</hi> &amp; <hi>AC</hi> to <hi>AB,</hi> therfore two sides of the one are equall to two sides of the other, ech to his correspondent side, &amp; the angle <hi>BAC</hi> is equall to the angle <hi>CAB,</hi> for it is one and the selfe same angle. Wherfore by the 4. proposition the base <hi>CB</hi> is equall to the base <hi>BC,</hi> and the triangle <hi>ABC</hi> is equall to the triangle <hi>ACB</hi>: and the angle <hi>ABC</hi> is equall to the angle <hi>ACB,</hi> and the angle <hi>ACB</hi> to the angle <hi>ABC</hi> for vnder them are subtended equall sides, namely, the lines <hi>AB</hi> &amp; <hi>AC.</hi> Wher<g ref="char:EOLunhyphen"/>fore in an Isosceles triangle, the angels at the base are equall.</p>
               <p>
                  <pb facs="tcp:7062:46"/>
                  <note place="margin">
                     <hi>Thales Milesius the inuent<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r of this proposition.</hi>
                  </note>The old Philosopher Thales Milesius was the first inuenter of this fifth propositi<g ref="char:EOLhyphen"/>on, as also of many other.</p>
            </div>
            <div type="part">
               <head>The third Theoreme. The sixt Proposition. If a triangle haue two angles equall the one to the other: the sides also of the same, which subtend the equall angles, shalbe equall the one to the other.</head>
               <p>
                  <hi>SVppose that ABC be a triangle, hauing the angle ABC equall to the angle ACB. Then I say that the side AB is equall to the side AC. For if the side AB be not equal to the side AC, then one of them is greater. Let AB be the greater. And by the third proposition,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>from AB be<g ref="char:EOLhyphen"/>ing the greater cut of a line equal to the lesse line, which is AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And let the same be D B. And drawe a line from the poynt D to the poynt C.</hi>
                  <note place="margin">
                     <hi>Demonstration leading to an impossib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ty.</hi>
                  </note> 
                  <hi>Now forasmuch as the
<figure/>
side DB is equall to the syde AC, and the line BC is common to the<g ref="char:cmbAbbrStroke">̄</g> both: therefore these two sydes DB and BC are equall to these two sydes AC &amp; CB the one to the other. And the angle DBC is by supposytion equall to the angle ACB. VVher<g ref="char:EOLhyphen"/>fore (by the 4 proposytion) the base DC is equall to the base AB: &amp; (by the same) the triangle DBC is equall to the triangle ACB: namely, the lesse triangle vnto the greater tria<g ref="char:cmbAbbrStroke">̄</g>gle, which is impossi<g ref="char:EOLunhyphen"/>ble. VVhe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore the syde AB is not vnequal to the side AC. VVherfore it is e<g ref="char:EOLhyphen"/>qual. If therfore a triangle haue two angles equall the one to the other: the sydes also of the same, which subtende the equall angles, shall be equall the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be demonstrated.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>The chiefest and <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> proper kind of conuersion.</hi>
                  </note>In Geometrie is o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tentimes vsed conuersion of propositions. As this propo<g ref="char:EOLhyphen"/>sition is the conuerse of the proposition next before. The chiefest and most pro<g ref="char:EOLhyphen"/>per kind of conuersion is, when that which was the thing supposed in the former proposition, is the conclusion of the conuerse and second proposition: and con<g ref="char:EOLhyphen"/>trary wise that which was concluded in the first, is the thing supposed in the se<g ref="char:EOLhyphen"/>cond As in the fifth proposition it was supposed the two sides of a triangle to be equal, the thing concluded is, that the two angles at the base are equall &amp; in this proposition, which is the conuerse therof is supposed that the angles at the base be equall. VVhich in the former proposition was the conclusion. And the con<g ref="char:EOLhyphen"/>clusion is, that the two sydes subtending the two angles are equall, which in the former proposition was the supposition. This is the chiefest kind of conuersion vniforme and certayne.</p>
               <p>
                  <pb n="16" facs="tcp:7062:46"/>There is an other kind of conuersion,<note place="margin">
                     <hi>An other kind of conuersion not so perfect as the first.</hi>
                  </note> but not so full a conuersion nor so per<g ref="char:EOLhyphen"/>fect as the first is. VVhich happeneth in composed propositions, that is, in such, which haue mo suppositions then one, and passe from these suppositions to one conclusion. In the co<g ref="char:cmbAbbrStroke">̄</g>nuerses of such propositio<g ref="char:cmbAbbrStroke">̄</g>s, you passe from the conclusion of the first proposition, with one or mo of the suppositions of the same: &amp; con<g ref="char:EOLhyphen"/>clude some other supposition of the selfe first proposition: of this kinde there are many in Euclide. Therof you may haue an example in the 8. proposition be<g ref="char:EOLhyphen"/>ing the conuerse of the four<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h. This conuersion is not so vniforme as the other; but more diuers and vncertaine according to the multitude of the things geuen, or suppositions in the proposition.</p>
               <p>But because in the fifth proposition there are two conclusions,<note place="margin">
                     <hi>Two conclusions in the fifth pro<g ref="char:EOLhyphen"/>position.</hi>
                  </note> the first, that the two angles at the ba<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e be equall: the second, that the angles vnder the ba<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e are equall: this is to be noted, that this sixt proposition is the conuerse of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame fifth as touching the first conclusion onely.<note place="margin">
                     <hi>The sixt propo<g ref="char:EOLhyphen"/>sition i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the con<g ref="char:EOLhyphen"/>uerse as tou<g ref="char:EOLhyphen"/>ching the first conclusion onely. The conuerse as touching the se<g ref="char:EOLhyphen"/>cond conclusion.</hi>
                  </note> You may in like maner make a con<g ref="char:EOLhyphen"/>uerse of the same proposition touching the second conclusion therof. And that after this maner.</p>
               <p>
                  <hi>THe two sides of a triangle beyng produced, if the angles vnder the base be equall, the said triangle shall be an Isosceles triangle.</hi> In which propositio<g ref="char:cmbAbbrStroke">̄</g> the supposition is, that the angles vnder the base are equall: which in the fifth proposition was the conclusion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp; the conclusion in this proposition is, that the two sides of the triangle are equal, which in the fift proposition was the supposition. But now for proofe of the said proposition:</p>
               <p>Suppose that there be a triangle <hi>ABC,</hi> &amp; let the
<figure/>
sides <hi>AB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and <hi>AC</hi> be produced to the poyntes <hi>D</hi> and <hi>G,</hi> and let the angles vnder the base be equall, namely, the angles <hi>DBC,</hi> and <hi>GCB.</hi> Then I say that the triangle <hi>ABC</hi> is an <hi>Isosceles</hi> triangle.<note place="margin">
                     <hi>Construction.</hi>
                  </note> For take in the line <hi>AD</hi> a point which let be <hi>E.</hi> And vnto the line <hi>BE</hi> put the line <hi>CF</hi> equall (by the 3. propositio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>).<note place="margin">
                     <hi>Demonstration<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                  </note> And draw these lines <hi>EC, BF,</hi> and <hi>EF.</hi> Now forasmuch as <hi>BE</hi> is equall to <hi>CF,</hi> and <hi>BC</hi> is common to the<g ref="char:cmbAbbrStroke">̄</g> both<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore these two lines <hi>BE</hi> &amp; <hi>BC,</hi> are equall to these two lines <hi>CF</hi> and <hi>CB</hi> the one to the other, &amp; the an<g ref="char:EOLhyphen"/>gle <hi>EBC</hi> is equall to the angle <hi>FCB</hi> by supposition. Wherfore (by the 4. proposition) the base of the one is equall to the base of the other, and the one triangle is equall to the other triangle, &amp; the other angles re<g ref="char:EOLhyphen"/>mayning are equal vnto the other angles remayning, the one to the other, vnder which are subtended equall sides<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore the base <hi>EC</hi> is equall to the base <hi>FB,</hi> and the angle <hi>BEC</hi> to the angle <hi>CFB,</hi> and the angle <hi>CBF</hi> to the angle <hi>BCE.</hi> But the whole angle <hi>EBC</hi> is equall to the whole angle <hi>FCB,</hi> of which the angle <hi>FBC</hi> is equa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>l to the angle <hi>ECB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore the angle remayning <hi>EBF</hi> is e<g ref="char:EOLhyphen"/>quall to the angle remayning <hi>FCE.</hi> But the line <hi>BE</hi> is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall to the line <hi>CF,</hi> &amp; the line <hi>BF</hi> to the line <hi>CE,</hi> and they contayne equall angles: wherfore by the same fourth pro<g ref="char:EOLhyphen"/>position the angle <hi>BEF</hi> is equall to the angle <hi>CFE.</hi> Wherfore by this sixt proposition the side <hi>AE</hi> is equall to the side <hi>AF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> whiche <hi>B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> is equall to <hi>CF,</hi> by construction: wherfore (by the third common sentence) the residue <hi>AB</hi> is equall to the residue <hi>AC</hi> Wherfore the triangle <hi>ABC</hi> is an <hi>Isosceles</hi> triangle. If therfore the two sides of a trian<g ref="char:EOLhyphen"/>gle being produced, the angles vnder the base be equall, the sayd triangle shall be an <hi>I<g ref="char:EOLhyphen"/>sosceles</hi> triangle: which was required to be proued.</p>
               <p>This moreouer is to be noted, that in this proposition there may be an other case<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for in taking an equall line to <hi>AC</hi> from <hi>AB,</hi>
                  <note place="margin">
                     <hi>An other case in this sixt proposi<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> you may take it from the poynte <hi>A</hi> and
<pb facs="tcp:7062:47"/>
not from the poynt <hi>B.</hi> And yet though this supposition
<figure/>
also be put the selfe same absurdity will follow.</p>
               <p>For suppose that <hi>AC</hi> be equall to <hi>AD:</hi> and produce the line <hi>CA</hi> to the poynt <hi>E</hi>: and put the line <hi>AE</hi> equall to the line <hi>DB</hi> (by the third proposition) wherefore the whole line <hi>CE</hi> is equall to the whole line <hi>AB</hi> (by the se<g ref="char:EOLhyphen"/>cond common sente<g ref="char:cmbAbbrStroke">̄</g>ce) Draw a line from the poynt <hi>E</hi> to the point <hi>B.</hi> And forasmuch as the line <hi>AB</hi> is equall to the line <hi>EC,</hi> and the line <hi>BC</hi> is common to them both, and the angle <hi>ACB</hi> is supposed to be equall to the an<g ref="char:EOLhyphen"/>gle <hi>ABC</hi>: Wherfore (by the fourth proposition) the tri<g ref="char:EOLhyphen"/>angle <hi>EBC</hi> is equall to the triangle <hi>ABC,</hi> namelye, the whole to the part: which is impossible.</p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Theoreme. The <hi>7.</hi> Proposition. If from the endes of one line, be drawn two right lynes to any pointe: there can not fro<g ref="char:cmbAbbrStroke">̄</g> the self same endes on the same side, be drawn two other lines equal to the two first lines, the one to the other, vnto any other point.</head>
               <p>FOr if it be possible: then from the endes of one &amp; the self same right line, namely, AB, from the pointes (I say) A and B, let there be drawn two right lines AC and CB to the point C: and from the same endes of the line AB, let there be drawen two other right right lines AD and DB equall to the lines AC and CB the one to the other, is, eche to his correspondent line, and on one and the same side, and to an other pointe, namely, to D: so that
<figure/>
let CA be equall to DA beyng both drawen from one end, that is, A: &amp; let CB be equall to DB, be<g ref="char:EOLhyphen"/>yng both also drawn from one ende, that is, B. And (by the first peticion) draw a right line from the point C to the point D.<note place="margin">Demonstration le<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to an absurditie.</note> Now forasmuch as AC is e<g ref="char:EOLhyphen"/>qual to AD, the angle ACD also is (by the 5. pro<g ref="char:EOLhyphen"/>position) equall to the angle ADC: wherfore the angle ACD is lesse the<g ref="char:cmbAbbrStroke">̄</g> the angle BDC. VVher<g ref="char:EOLhyphen"/>fore the angle BCD is much lesse then the angle BDC. Againe forasmuch as BC is equall to BD, and therfore also the angle BCD is equall to the angle BDC. And it is proued that it is much lesse then it: which is impossible. If therfore from the endes of one line, be drawen <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wo right lines to any pointe: there can not from the selfe same endes on the same side, be drawen two other lines equall to the two first lines, the one to the other, vnto a<g ref="char:EOLhyphen"/>ny other point: VVhich was required to be demonstrated.</p>
               <p>
                  <pb n="17" facs="tcp:7062:47"/>
                  <hi>In this proposition the conclusion is a negation, which very rarely happe<g ref="char:EOLhyphen"/>neth in the mathematicall artes.<note place="margin">Negatiue conclu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>sions rarely vsed in the mathema<g ref="char:EOLunhyphen"/>ticall artes.</note> For they euer for the most part vse to conclude affirmatiuely, &amp; not negatiuely. For a propositio<g ref="char:cmbAbbrStroke">̄</g> vniuersall affirmatiue is most agreable to sciences, as saith</hi> Aristotle, <hi>and is of it selfe strong, and nedeth no nega<g ref="char:EOLhyphen"/>tiue to his proofe. But an vniuersall proposition negatiue must of necessitie haue to his proofe an affirmatiue, For of onely negatiue propositions there can be no demonstrations. And therfore sciences vsing demonstration, conclude affirma<g ref="char:EOLhyphen"/>tiuely, and very seldome vse negatiue conclusions.</hi>
               </p>
               <div type="section">
                  <head>An other demonstration after Campanus.</head>
                  <p>Suppose that there be a line <hi>AB,</hi> from whose ends <hi>A</hi> and <hi>B,</hi>
                     <figure/>
let there be drawen two lines <hi>AC</hi> and <hi>BC</hi> on one side, which let concur in the poynt <hi>C.</hi> Then I say that on the same side there can<g ref="char:EOLhyphen"/>not be drawen two other lines, from the endes of the line <hi>AB,</hi> which shall concur at any other poynt, so that that which is drawe<g ref="char:cmbAbbrStroke">̄</g> from the point <hi>A</hi> shall be equall to the line <hi>AC,</hi> and that which is drawen from the point <hi>B</hi> shalbe equall to the line <hi>BC.</hi> For if it be possible, let there be drawn two other lines on the selfe same side, which let concurre in the point <hi>D,</hi> and let the line <hi>AD</hi> be equall to the line <hi>AC,</hi> &amp; the line <hi>BD</hi> equall to the line <hi>BC.</hi>
                     <note place="margin">
                        <hi>Diuers cases in this demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> Wherfore the poynt <hi>D</hi> shall fall either within the trian<g ref="char:EOLhyphen"/>gle <hi>ABC,</hi> or without. For it cannot fall in one of the sides, for then a parte should be e<g ref="char:EOLhyphen"/>quall to his whole. If therfore it fall without<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then either one of the lines <hi>AD</hi> and <hi>DB</hi> shall cut one of the lines <hi>AC</hi> and <hi>CB,</hi> or els neither shall cut neyther.<note place="margin">
                        <hi>First case.</hi>
                     </note> Firste let one cut the other and draw a right line from <hi>C</hi> to <hi>D.</hi> Now forasmuch as in the triangle <hi>ACD,</hi> the two sides <hi>AC</hi> and <hi>AD</hi> are equall, therfore the angle <hi>ACD</hi> is equall to the angle <hi>ADC,</hi> by the fifth propositio<g ref="char:cmbAbbrStroke">̄</g>: likewise forasmuch as in the tria<g ref="char:cmbAbbrStroke">̄</g>gle <hi>BCD,</hi> the two sides <hi>BC</hi> and <hi>BD</hi> are equall, therfore by the same, the angles <hi>BCD</hi> &amp; <hi>BDC</hi> are also equall. And forasmuch as the angle <hi>BDC</hi> is greter the<g ref="char:cmbAbbrStroke">̄</g> the angle <hi>ADC,</hi>
                     <figure/>
it followeth that the angle <hi>BCD</hi> is greater then the angle <hi>ACD,</hi> namely, the part greater then the whole: which is impossible.</p>
                  <p>But if the point <hi>D</hi> fal without the triangle <hi>ABC,</hi>
                     <note place="margin">
                        <hi>Second case.</hi>
                     </note> so that the lines cut not the one the other, draw a line from <hi>D</hi> to <hi>C.</hi> And produce the lines <hi>BD</hi> &amp; <hi>BC</hi> beyond the base <hi>CD,</hi> vnto the points <hi>E</hi> &amp; <hi>F.</hi> And forasmuch as the lines <hi>AC</hi> and <hi>AD</hi> are equall, the angles <hi>ACD</hi> and <hi>ADC</hi> shall also be equall, by the fifth proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> like<g ref="char:EOLhyphen"/>wise for asmuch as the lines <hi>BC</hi> and <hi>BD</hi> are equal, the angles vn<g ref="char:EOLhyphen"/>der the base, namely, the angles <hi>FDC</hi> and <hi>ECD</hi> are equall, by the seconde part of the same proposition. And for as much as the angle <hi>ECD</hi> is lesse then the angle <hi>ACD:</hi> It followeth that the angle <hi>FDC</hi> is lesse the<g ref="char:cmbAbbrStroke">̄</g> the angle <hi>ADC</hi>: which is impossible: for that the angle <hi>A</hi>D<hi>C</hi> is a part of the angle <hi>F</hi>D<hi>C.</hi> And the same inconuenience will follow if the poynt D fall within the triangle <hi>ABC,</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The fift Theoreme. The <hi>8.</hi> Proposition. If two triangles haue two sides of th'one equall to two sides of the other, eche to his correspondent side, &amp; haue also the base of the one equall to the base of the other: they shall haue also the angle contained vnder the equall right lines of the one, e<g ref="char:EOLhyphen"/>quall to the angle contayned vnder the equall right lynes of the other.</head>
               <p>
                  <pb facs="tcp:7062:48"/>
                  <hi>SVppose that there be two triangles ABC and DEF: &amp; let these two sides of the one AB and AC, be equall to these two sides of the other DE, and DF, ech to his cor<g ref="char:EOLhyphen"/>respondent side, that is, AB to D E, and AC, to DF, &amp; let the base of the one, namely, BC be equal to the base of the other, namely, to EF. Then I say, that the angle BAC is equall to the angle EDF. For the triangle ABC ex<g ref="char:EOLhyphen"/>actly agreing with the triangle DE
<figure/>
F,</hi>
                  <note place="margin">
                     <hi>Demonstration leading to an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>mpossibility.</hi>
                  </note> 
                  <hi>and the point B being put vpon the point E, and the right line BC vpon the right line EF: the point C shall exactly agree with the point F (for the line BC is equall to the line EF) And BC exactly agreeing with EF the lines also BA and AC shall ex<g ref="char:EOLhyphen"/>actly agree with the lines ED &amp; DF. For if the base BC do exactly a<g ref="char:EOLhyphen"/>gree with the base FE, but the sides BA &amp; AC doo not exactly agree w<hi rend="sup">t</hi> the sides ED &amp; DF, but differ as FG &amp; GF do: the<g ref="char:cmbAbbrStroke">̄</g> from y<hi rend="sup">e</hi> endes of one lyne shalbe drawn two right lines to a poynt, &amp; from the self same endes on the same side shalbe drawn two other lines, equal to the two first lines, y<hi rend="sup">e</hi> one to the o<g ref="char:EOLhyphen"/>ther, and vnto an other poynt: but that is impossible, (by the seuenth propositio<g ref="char:cmbAbbrStroke">̄</g>) VVherfore the base BC exactly agreeing with the base EF, the sides also BA and AC do exactly agre with the sides ED and DF. VVherfore also the angle BAC shall exactly agre w<hi rend="sup">t</hi> the angle EDF, and therfore shall also be equal to it. If the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore two triangles haue two sides of the one equall to two sides of the other, ech to his correspondent side, and haue also the base of y<hi rend="sup">e</hi> one equall to the base of the other: they shall haue also the angle contayned vnder the equall right lines of the one, equall to the angle contayned vnder the equall right lines of the other: which was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>This proposition is the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ourth, b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t not the che<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t kind of con<g ref="char:EOLhyphen"/>uersion.</hi>
                  </note>This Theoreme is the conuerse of the fourth, but it is not the chiefest and principall kind o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> conuersion. For it turneth not the whole supposition into the conclusion, and the whole conclusion into the supposition. For the fourth pro<g ref="char:EOLhyphen"/>positio<g ref="char:cmbAbbrStroke">̄</g> whose conuerse this is, is a co<g ref="char:cmbAbbrStroke">̄</g>pound <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>heoreme, hauing two things geue<g ref="char:cmbAbbrStroke">̄</g> or s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>pposed, which are these: the one, that two sides of the one tria<g ref="char:cmbAbbrStroke">̄</g>gle be equal to two sides of the other tria<g ref="char:cmbAbbrStroke">̄</g>gle: th'other, that the angle co<g ref="char:cmbAbbrStroke">̄</g>tained of the two sides of th'one is equal to the angle contained of the two sides of th'one: but hath amon<g ref="char:EOLhyphen"/>gest other one thing required, whiche is, that the base of the one, is equal to the base of the other. Now in this 8. propositio<g ref="char:cmbAbbrStroke">̄</g>, being the conuerse therof<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that the base of the one is equal to the base of th'other, is the supposition, or the thing ge<g ref="char:EOLhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g>: which in the former propositio<g ref="char:cmbAbbrStroke">̄</g> was the conclusio<g ref="char:cmbAbbrStroke">̄</g>. And this, that two sides of
<pb n="18" facs="tcp:7062:48"/>
the one are equall to two sides of the other, is in this proposition also a sup<g ref="char:EOLhyphen"/>position, like as it was in the former proposition: so that it is a thing geuen in ei<g ref="char:EOLhyphen"/>ther proposition. The conclusion of this proposition is that the angle enclosed of the two equall sides of the one triangle is equall to the angle enclosed of the two equall sides of the other triangle: which in the former proposition was one of the things geuen.</p>
               <p>Philo and his scholas demonstrate this proposition without the helpe of the former proposition, in this maner.</p>
               <p>Suppose that there be two triangles <hi>ABC</hi> and <hi>DEF,</hi> hauing two sydes of the one equall to two sydes of the other, namely, <hi>AB</hi> and <hi>AC</hi> equall to DE and D<hi>F,</hi>
                  <note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration inuen<g ref="char:EOLhyphen"/>ted by Philo.</hi>
                  </note> the one to the other, &amp; the base <hi>BC</hi> equal to the base <hi>EF.</hi> And for that the base <hi>BC</hi> is equall to the base E<hi>F,</hi> therfore the one being applied to the other they agree. Place the two tria<g ref="char:cmbAbbrStroke">̄</g>gles <hi>ABC</hi> &amp; D<hi>EF</hi> in one &amp; the self same plaine superficies, &amp; apply the base of the one to the base of the other: But yet so that the tria<g ref="char:cmbAbbrStroke">̄</g>gle <hi>ABC</hi> be set one the other side of the right line <hi>EF,</hi> that the top of the one may be opposite to the top of the other. And in stead of the triangle <hi>ABC</hi> put the
<figure/>
triangle <hi>EFG</hi> as in the figure. And let DE be equall to <hi>EG,</hi> and <hi>DF</hi> to <hi>FG.</hi>
                  <note place="margin">
                     <hi>After this de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> thre cases in this pro<g ref="char:EOLhyphen"/>position.</hi>
                  </note> Nowe then by this meanes shall hap<g ref="char:EOLhyphen"/>pen diuers cases. For the line F<hi>G</hi> may fall directly vpo<g ref="char:cmbAbbrStroke">̄</g> the line DF, or it may so fall that it may make with the line DF an angle within the figure, or with out.</p>
               <p>First let it fall directlye.<note place="margin">
                     <hi>The first case.</hi>
                  </note> And foras<g ref="char:EOLhyphen"/>muche as the line DE is equall to the line E<hi>G,</hi> and DFG, is one righte line: therfore DEG is an Isosceles tria<g ref="char:cmbAbbrStroke">̄</g>gle: and so, by the fifth proposition, the an<g ref="char:EOLhyphen"/>gle at the point D is equal to the angle at the poynt <hi>G</hi>: which was required to be proued.</p>
               <p>But if it fall not directly, but make with the line D
<figure/>
F an angle within the figure, drawe a line from D to G.<note place="margin">
                     <hi>The second case.</hi>
                  </note> Now forasmuch as ED and EG <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re equall, and the line DG is the base: therfore by the fifth propositio<g ref="char:cmbAbbrStroke">̄</g>, the an<g ref="char:EOLhyphen"/>gle EDG is equall to the angle EGD. Agayne foras<g ref="char:EOLhyphen"/>much as DF is equall to FG, and DG is the base: ther<g ref="char:EOLhyphen"/>fore by the same, the angle FDG is equal to the FGD: and it was proued that the angle EDG is equall to the angle EGD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the whole angle EDF is equal to the whole angle FGE: whiche was required to be pro<g ref="char:EOLhyphen"/>ued.</p>
               <p>But if the line FG make with the line DF an angle without the figure:<note place="margin">
                     <hi>The third case.</hi>
                  </note> draw a right line without the figure
<pb facs="tcp:7062:49"/>
from the poynt D to the poynt G. And forasmuch as DE and EG are equall, and DG is the base, therfore by the fifth proposition, the angles EDG and DGE are equall. A<g ref="char:EOLhyphen"/>gaine
<figure/>
forasmuch as DF is equall to FG, and DG is the base, therfore, by the same, the angle FDG is equall to the angle FGD. And it was proued that the whole angles EDG &amp; DGE are equall the one to the other: wherfore the angles remayning EDF &amp; EGF are equall the one to the other, which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Probleme. The <hi>9.</hi> Proposition. To deuide a rectiline angle geuen, into two equall partes.</head>
               <p>
                  <hi>SVppose that the rectiline angle geuen be BAC. It is re<g ref="char:EOLhyphen"/>quired to deuide the angle BAC into two equal partes.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>In the line AB take a point at all aduentures, &amp; let the same be D. And (by the third proposition) from the lyne AC cutte of the line AE equall to AD. And (by the first peticion) draw a right line from the point D to the point E. And (by the first proposition) vpon the line DE describe an equilater triangle and let the same be DFE, and (by the first peti<g ref="char:EOLhyphen"/>cion) drawe a right line from the poynte A to
<figure/>
the point F. Then I say that the angle BAC is by y<hi rend="sup">e</hi> line AF deuided into two equal partes.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>For, for<g ref="char:EOLhyphen"/>asmuch as AD is equall to AE, and AF is co<g ref="char:cmbAbbrStroke">̄</g>mon to them both: therfore these two DA and AF, are equall to these two EA and AF, the one to the o<g ref="char:EOLhyphen"/>ther. But (by the first proposition) the base DF is equall to the base EF: wherfore (by the</hi> 8. <hi>propo<g ref="char:EOLhyphen"/>sition) the angle DAF is equal to the angle FAE. VVherfore the rectiline angle geuen, namely, B AC is deuided into two equal partes by the right line AF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> VVhich was requi<g ref="char:EOLhyphen"/>red to be done.</hi>
               </p>
               <p>In this proposition is not taught to deuide a right lined angle into mo partes then two: albeit to deuide an angle, so it be a right angle, into three partes, it is
<pb n="19" facs="tcp:7062:49"/>
not hard. And it is taught of <hi>Vi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ellio</hi> in his first boke of Perspectiue,<note place="margin">
                     <hi>It is impos<g ref="char:EOLhyphen"/>sible to deuide an acute re<g ref="char:EOLhyphen"/>ctiline angle in<g ref="char:EOLhyphen"/>to three equall partes without the helpe of lines which are of a mixt na<g ref="char:EOLhyphen"/>ture.</hi>
                  </note> the 28. Propo<g ref="char:EOLhyphen"/>sition. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or to deuide an acute angle into three equal partes, is (as saith <hi>Proclus</hi>) im<g ref="char:EOLhyphen"/>possible: vnles it be by the helpe of other lines which are of a mixt nature. Which thing <hi>Nicomedes</hi> did by such lines which are called <hi>Concoide<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> linea,</hi> who first serched out the inuention, nature, &amp; properties of such lines. And others did it by other meanes as by the helpe of quadrant lines inuented by <hi>Hippias &amp; Nicomedes.</hi> Others by <hi>Helices</hi> or Spiral lines inuented of <hi>Archimedes.</hi> But these are things of much dif<g ref="char:EOLhyphen"/>ficulty and hardnes, and not here to be intreated of.</p>
               <p>Here against this proposition may of the aduersary be brought an <note n="*" place="margin">
                     <hi>An instance is an obiection or a doubt, wherby is letted or trou<g ref="char:EOLhyphen"/>bled the con<g ref="char:EOLhyphen"/>struction, or de<g ref="char:EOLhyphen"/>monstration, &amp; contayneth an vnruth, and an impossibility: and therfore it must of necessity be answered vn<g ref="char:EOLhyphen"/>to, and the false<g ref="char:EOLhyphen"/>hode thereof made manifest.</hi>
                  </note> instance. For he may cauill that the hed of the equilater triangle shall not fall betwene the two right lines, but in one of them, or without them both. As for example.</p>
               <p>Suppose that the angle to be deuided into two
<figure/>
equall partes be <hi>BAC,</hi> and in the line <hi>BA</hi> take the poynt <hi>D,</hi> and vnto the line <hi>DA</hi> put the line <hi>AE</hi> equal (by the third proposition.) And draw a line fro<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>D</hi> to <hi>E.</hi> And vpon the line <hi>DE</hi> describe (by the first) an equilater triangle, which let be <hi>DFE.</hi> Now then if it be possible that the point F do not fal betwene the lines <hi>AB</hi> &amp; <hi>AC,</hi> then it shal fal e<g ref="char:EOLhyphen"/>ther in the line <hi>AB</hi> or <hi>AC,</hi> or without them both. Suppose that the point F be fall vpon line <hi>AB,</hi> so that let DFE be an equilater triangle. Wherfore the line DF is equal to the line FE: &amp; the angles at the base are equall, namely, the angles EDF and DEF. Wherefore the whole angle <hi>DEC</hi> is greater then the angle EDF. Againe forasmuch as <hi>AD</hi> is equall to <hi>AE,</hi> therefore <hi>ADE</hi> is an Isosceles triangle. Wherefore (by the fifth proposition) the angles vn<g ref="char:EOLhyphen"/>der the base are equall. Wherfore the angle <hi>DEC</hi> is equall to the angle <hi>EDB.</hi> But it was also greater: which is impossible. Wherfore the top of the equilater triangle ca<g ref="char:cmbAbbrStroke">̄</g>not be in the right line <hi>AB.</hi> And in like sort also may we proue that it ca<g ref="char:cmbAbbrStroke">̄</g>not be in the right line <hi>AC.</hi> Wherfore suppose that it be without them both, if it be possible. And foras<g ref="char:EOLhyphen"/>much as DF is equal to FE, the angles at the base are equal, namely, the angles DEF &amp; EDF. Wherfore the angle DEF is greater then the angle <hi>ED</hi>F. Wherfore the angle DEC is much greater then the angle EDF. But it is also equal vnto it. For they are angles vnder the base <hi>DE</hi> of the Isosceles triangle <hi>A</hi>D<hi>E.</hi> Which is impossib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e. Wherfore the poynt F shall not fall without the two right lines on that side. And in like sorte may we proue that it shall not fall without them on the other side. Wherfore it shall of necessi<g ref="char:EOLhyphen"/>ty fall betwene them: which was required to be proued.</p>
               <p>There may also in this proposition be diuers cases<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">
                     <hi>Diuers cases in this proposition.</hi>
                  </note> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>f it so happen that there be no space vnder the base DE to describe an equilater triangle, but that of neces<g ref="char:EOLhyphen"/>sitie you must describe it on the same side that the lines AB and AC are. For then the sides of the equilater triangle either exactly agree with the lines AD and AE, if the said lines AD and AE be equall with the base DE. Or they fall without them, if the lines AD and AE be lesse then the base DE. Or they fall within them, if the said lines be greater the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the base DE.</p>
               <p>First let them exactly agree<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And let <hi>DAE</hi> be an equilater triangle.<note place="margin">
                     <hi>The first case.</hi>
                  </note> And in the side <hi>AD</hi> take the poynt <hi>G.</hi> And from the side <hi>AE</hi> cut of a line equal to the line <hi>AG</hi> (by the third proposition) which let be <hi>AH</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And draw these right lines <hi>GE, HD</hi> and <hi>GH,</hi> and AF. Now forasmuch as <hi>AD</hi> is equal to <hi>AE</hi> and <hi>AG</hi> vnto <hi>AH</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore these two lines
<pb facs="tcp:7062:50"/>
                  <hi>DA</hi> and AH are equall to these two lines <hi>EA</hi>
                  <figure/>
and AG: and they contayne one and the selfe same angle, Wherfore by the fourth proposi<g ref="char:EOLhyphen"/>tion, the angle GDH is equal to the angle HEG. And the base <hi>D</hi>H is equall to the base <hi>EG.</hi> But the line <hi>D</hi>G is equal to the line EH: wher<g ref="char:EOLunhyphen"/>fore againe by the fourth proposition, the an<g ref="char:EOLhyphen"/>gle EGH is equall to the angle <hi>D</hi>HG. Wher<g ref="char:EOLhyphen"/>fore by the sixt proposition, the base GF is e<g ref="char:EOLhyphen"/>qual to the base HF. And forasmuch as AH is equall to AG, and AF is common to the<g ref="char:cmbAbbrStroke">̄</g> both, and the base GF is equall to the base HF, ther<g ref="char:EOLunhyphen"/>fore the angle GAF is equall to the angle HAF. Wherefore the angle GAH is deuided into two equall partes: which was required to be done.</p>
               <p>
                  <note place="margin">
                     <hi>The second case.</hi>
                  </note>But if the sides of the equilater triangle fall without the right lines <hi>BA</hi> &amp; <hi>AC,</hi> as do the lines <hi>D</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp; <hi>EF,</hi> the<g ref="char:cmbAbbrStroke">̄</g> draw a line from F to A &amp; produce the line FA to the point G. Now forasmuch as the lines DF and FE are equal,
<figure/>
&amp; the line FA is common to them both, &amp; the ba<g ref="char:EOLhyphen"/>ses DA and AE are equall: therfore (by the eight) the angle DFA is equal to the angle EFA. Againe forasmuch as DF and FE are equall, and FG is common to them both, and they containe equall angles (as it hath bene proued) therefore (by the fourth) the base DG is equall to the base GE. And forasmuch as AD is equall to AE, and AG is com<g ref="char:EOLhyphen"/>mon to them both. Therfore (by the eight) the an<g ref="char:EOLhyphen"/>gle DAG is equall to the angle EAG. wherefore the angle DAE is deuided into two equall partes: Which was required to be done.</p>
               <p>But if the sides of the equilater triangle fal with<g ref="char:EOLhyphen"/>in
<figure/>
the right lines BA and AC,<note place="margin">
                     <hi>The third case.</hi>
                  </note> as do the lines DF and FE, then againe draw a line from A to F. And forasmuch as DA is equall to AE, and AF is com<g ref="char:EOLhyphen"/>mon to them both, and the base DF is equal to the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se FE: therfore the angle DAF is (by the eight) equall to the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>AF. Wherefore the angle at the point A is deuided into two equall partes, how soeuer the equilater triangle be placed: which was required to be done.</p>
               <p>This is to be noted, that if a man will me<g ref="char:EOLhyphen"/>chanically
<figure/>
or readily,<note place="margin">
                     <hi>To devide a <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> angle <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> equall <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>.</hi>
                  </note> not regardyng demon<g ref="char:EOLhyphen"/>stration, deuide the foresaid rectiline angle BAC, and so any other rectiline angle geuen whatsoeuer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> into two equall partes, he shall neede onely with one o<g ref="char:EOLhyphen"/>pening of the compasse taken at all aduentures to marke the two pointes D and E, which cut of equal partes of the lines AB and AC, howsoeuer they happen, and so ma<g ref="char:EOLhyphen"/>king the centres the two points D and E, to describe two circles according to the openyng of the compa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e: and from the point A to their intersection, which let be the point F to draw a right line: which shall deuide the an<g ref="char:EOLhyphen"/>gle BAC into two equall partes. And here note, that you shall not nede to draw the circles all whole, but one,
<pb n="20" facs="tcp:7062:50"/>
ly a portion where they cut the one the other: As in the figure here in the end of the other side put.</p>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Probleme. The <hi>10.</hi> Proposition. To deuide a right line geuen being finite, into two equall partes.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB. It is required to deuide the line AB into two equal partes.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Describe (by the first proposition) vp<g ref="char:EOLhyphen"/>on the line AB an equilater triangle and let the same be ABC. And (by the former proposition) deuide the angle ACB into two equall partes by the right line CD. Then I say that the
<figure/>
right line AB is deuided into two equall partes in the poynt D.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>For forasmuch as (by the first propo<g ref="char:EOLhyphen"/>sition) AC is equall to CB, and CD is common to the<g ref="char:cmbAbbrStroke">̄</g> both: therfore these two lines AC &amp; CD are e<g ref="char:EOLhyphen"/>qual to these two lines BC &amp; CD, y<hi rend="sup">e</hi> one to y<hi rend="sup">e</hi> other, and the angle ACD is equall to the angle BCD. VVherfore (by the</hi> 4. <hi>proposition) the base AD is equall to the base BD. VVherefore the righte line geuen AB, is deuided into two equall partes in the poynt D: which was required to be done.</hi>
               </p>
               <p>Apollonius <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eacheth to deuide a right line being finite into two equall partes after this manner.<note place="margin">
                     <hi>An other way to deuide a right line being finis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, inue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ted by A<g ref="char:EOLhyphen"/>pollonius.</hi>
                  </note>
               </p>
               <p>Suppose that the right line being finite
<figure/>
be <hi>AB</hi>: which it is required to deuide in to two equal parts. Now the<g ref="char:cmbAbbrStroke">̄</g> making the centre the point <hi>A</hi> &amp; the space <hi>AB</hi> de<g ref="char:EOLhyphen"/>scribe a circle. Again making the centre the poynt <hi>B</hi> &amp; the space BA describe an other circle: and from the co<g ref="char:cmbAbbrStroke">̄</g>mon secti<g ref="char:EOLhyphen"/>ons draw the right line <hi>CD,</hi> which let cut the line AB in the poynt E. Then I say, that the right line <hi>CD</hi> deuideth the line <hi>AB</hi> into two equall partes in the point E. For draw these right lines D<hi>A, DBCA</hi> and <hi>CB.</hi> Which shal be equal the one to the other, for that they are drawne fro<g ref="char:cmbAbbrStroke">̄</g> the centres to the circumferences of equall circles. And forasmuch as the lines <hi>CA</hi> &amp; <hi>AD</hi> are equall to the lines <hi>CB</hi> and <hi>B</hi>D, and the line <hi>CD</hi> is common to either of them: therefore, by the eight, the angle <hi>ACD</hi> is equall to the angle <hi>BC</hi>D. Again forasmuch as the line <hi>CE</hi> is common to the lines <hi>AC</hi> and <hi>CB,</hi> which are equall the one to the o<g ref="char:EOLhyphen"/>ther, therfore these two lines <hi>AC</hi> and <hi>CE</hi> are equal to these two lines <hi>BC</hi> and <hi>CE,</hi> the one to the other, and they contayne equall angles, as it hath bene proued. Wherefore by the 4. proposition the base <hi>AE</hi> is equall to the base <hi>BE.</hi> Wherefore the line <hi>AB</hi> is deuided into two equall partes in the poynt <hi>E</hi>: which was required to be done.</p>
               <p>
                  <pb facs="tcp:7062:51"/>By this way of deuiding a right line, into two
<figure/>
equall parts inuented by Apollonius, it is manifest, that if a man wil mechanically, or redely, not consi<g ref="char:EOLhyphen"/>dering the demonstratio<g ref="char:cmbAbbrStroke">̄</g>, deuide the said rightline, and so any right line geuen whatsoeuer, into two e<g ref="char:EOLhyphen"/>quall: partes he nede onely to marke the poynts of the intersections of the circles, &amp; to draw a line fro<g ref="char:cmbAbbrStroke">̄</g> the sayd intersections, which shall deuide the right line geuen into two equall partes: as in the figure here put.</p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Probleme. The <hi>11.</hi> Proposition. Vpon a right line geuen, to rayse vp from a poynt geuen in the same line a perpendicular line.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB, &amp; let the point in it geuen be C. It is required from the poynte C to rayse vp vnto the right line AB a perpendicular line. Take in the line AC a poynt at all aduentures,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>&amp; let the same be D, and (by the</hi> 3. <hi>proposition) put vnto DC an equall line CE. And by the first proposition) vpon the line DE describe an equilater triangle FDE, &amp; draw a line fro<g ref="char:cmbAbbrStroke">̄</g> F to C. Then I say that vnto the right line geuen AB, and from the poynte in it geuen, namely, C is raysed vp a perpendicular line FC. For forasmuch
<figure/>
as DC is equal to CE,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>&amp; the line CF is co<g ref="char:cmbAbbrStroke">̄</g>mon to them both: therfore these two DC and CF, are e<g ref="char:EOLhyphen"/>qual to these two EC &amp; CF, the one to the other: and (by the first proposition) the base DF is equal to the base EF: wherefore (by the</hi> 8. <hi>proposition) the angle DCF is equall to the angle ECF: and they be side angles. But whe<g ref="char:cmbAbbrStroke">̄</g> a right line standing vpon a right line doth make the two side angles equall the one to the other, ether of those equall angles is (by the.</hi> 10. <hi>definition) a right angle: &amp; the line standing vpon the right line is called a perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line. VVherfore the angle DCF, &amp; thangle FCE are right angles. VVherfore vnto the right line geue<g ref="char:cmbAbbrStroke">̄</g> AB, &amp; fro<g ref="char:cmbAbbrStroke">̄</g> the poynt in it C, is raysed vp a perpendicular line CF: which was required to be done.</hi>
               </p>
               <p>Although the poynte geuen should be set in one of the endes of the righte line geuen, it is easy so do it as it was before. For producing the line in length from the poynt by the second peticion, you may worke as you did before. But if one require to erect a right line perpendicularly from the poynt at the end of the
<pb n="21" facs="tcp:7062:51"/>
lyne, without producing the rightlyne, that also may well bee done after thys maner.</p>
               <p>Suppose that the right line geuen be <hi>AB,</hi> &amp; let
<figure/>
the point in it geuen be in one of the endes therof,<note place="margin">
                     <hi>An other case in this proposition.</hi>
                  </note> namely, in <hi>A.</hi> And take in the line <hi>AB</hi> a point at all aduentures,<note place="margin">
                     <hi>Construction.</hi>
                  </note> and let the same be <hi>C.</hi> And from the said point raise vp (by the foresaid proposition) vn<g ref="char:EOLhyphen"/>to <hi>AB</hi> a perpendiculer line, which let <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e <hi>OE.</hi> And (by the 3. proposition) from the line <hi>CE</hi> cut of the line <hi>CD</hi> equall to the line <hi>CA,</hi> And (by the <hi>9.</hi> Pro<g ref="char:EOLhyphen"/>position) deuide the angle <hi>ACD</hi> into two equall partes by the line <hi>CF.</hi> And from the point <hi>D</hi> raise vp vnto the line <hi>CE</hi> a perpe<g ref="char:cmbAbbrStroke">̄</g>diculer line, <hi>DF,</hi> which let concurre with the line <hi>CF</hi> in the point <hi>F.</hi> And drawe a right line from <hi>F</hi> to <hi>A.</hi> Then I say that the angle at the point <hi>A</hi> is a right angle. For, foras<g ref="char:EOLhyphen"/>much as <hi>DC</hi> is equall to <hi>CA,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> and <hi>CF</hi> is common to them both, and they containe <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>quall angles (for the angle at the point <hi>C</hi> is deuided into two equall partes) therefore (by the 4. Proposition) the line <hi>DF</hi> is equall to the line <hi>FA,</hi> and so the angle at the point <hi>A</hi> is equal to the angle at the point <hi>D.</hi> But the angle at the point <hi>D</hi> is a right an<g ref="char:EOLhyphen"/>gle. Wherfore also the angle at the point <hi>A</hi> is a right angle. Wherefore from the point <hi>A</hi> vnto the line <hi>AB,</hi> is raised vp a perpendiculer line <hi>AF,</hi> without producing the line <hi>AB.</hi> Which was required to be done.</p>
               <p>Appollonius teacheth to rayse vp vnto a line geuen, from a point in it geuen, a perpendiculer line, after this maner.<note place="margin">
                     <p>
                        <hi>An other way to erect a perpe<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>dicular line in<g ref="char:EOLhyphen"/>uented by Ap<g ref="char:EOLhyphen"/>polonius.</hi>
                     </p>
                     <p>
                        <hi>Construction<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </p>
                  </note>
               </p>
               <p>Suppose that the right
<figure/>
line geuen be <hi>AB.</hi> And let the point in it geue<g ref="char:cmbAbbrStroke">̄</g>, be <hi>C.</hi> And in the line <hi>AC,</hi> take a point at all ad<g ref="char:EOLhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g>tures, &amp; let the same be <hi>D.</hi> And fro<g ref="char:cmbAbbrStroke">̄</g> the lyne <hi>CB,</hi> cut of a line equall to the line <hi>CD,</hi> whiche let be <hi>CE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> and makyng the centre <hi>D,</hi> and the space <hi>DE,</hi> describe a circle. And againe ma<g ref="char:EOLhyphen"/>king the centre <hi>C,</hi> &amp; the space <hi>ED,</hi> describe an oth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r circle, and let the point of their intersection be <hi>F.</hi> And draw a right line from <hi>F</hi> to C. Then I say that the line <hi>FC</hi> is erected perpendiculerly vnto the line <hi>AB.</hi> For drawe these lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s <hi>ED</hi> and <hi>FE</hi>: which shal by the definition of a circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be either of them equal to the line <hi>DE.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> and therfore (by the first common sentence) are equall the one to the other. But the lines <hi>DC</hi> and <hi>CE</hi> are by construction equall, and the line <hi>FC</hi> is common to them both. Wherfore the angles also at the point <hi>C</hi> are equal (by the 8. propositio<g ref="char:cmbAbbrStroke">̄</g>:) wherfore they are right angles. Wherfore the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <hi>CE</hi> is erected perpendiculerly vnto the line <hi>AB</hi> from the point <hi>C</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be done.</p>
               <p>By this way of erecting a perpendiculer line inuented by Appollonius,<note place="margin">
                     <hi>How to erect a perpendicular line mechani<g ref="char:EOLhyphen"/>cally.</hi>
                  </note> it is also manifest, that if a man will mechanically, without d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>monstration, erect vnto
<pb facs="tcp:7062:52" rendition="simple:additions"/>
a line geuen from a point geue<g ref="char:cmbAbbrStroke">̄</g> in it a perpendi<g ref="char:EOLhyphen"/>culer
<figure/>
line: he neede onely on either side of the pointe geuen, to cut of equall lines: and so ma<g ref="char:EOLhyphen"/>king either of the endes of the said lines (either of th'endes I say, which haue not one point co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon to them both) the centres, and the space both the lynes added together, or wider then both, or at the lest wider the<g ref="char:cmbAbbrStroke">̄</g> one of them, to de<g ref="char:EOLhyphen"/>scribe those portions of the circles where they cut the one the other, and from the point of the intersection to the point geuen, to draw a lyne, which shall be perpendicular vnto the lyne ge<g ref="char:EOLhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g>: as in the figure here put it is manifest to see.</p>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Probleme. The <hi>12.</hi> Proposition. Vnto a right line geuen being infinite, and from a point geuen not being in the same line, to draw a perpendicular line.</head>
               <p>
                  <hi>LEt the right line geuen be<g ref="char:EOLhyphen"/>ing
<figure/>
infinite be AB, &amp; let y<hi rend="sup">e</hi> point geuen not being in the said line AB, be C. It is re<g ref="char:EOLhyphen"/>quired from the point geue<g ref="char:cmbAbbrStroke">̄</g>, namely, C, to draw vnto the right line geuen AB, a per<g ref="char:EOLhyphen"/>pendiculer line.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Take on the other syde of the line AB (namely, on that syde wherein is not the pointe C) a pointe at all aduen<g ref="char:EOLhyphen"/>tures, and let the same be D. And making the centre C, and the space CD, de<g ref="char:EOLhyphen"/>scribe (by the third peticion) a circle, and let the same be EFG, which let cutte the line AB in the pointes E and G. And (by the x. proposition) deuide the lyne EG into two equal partes in the point H. And (by the first peticion) draw these right lines, CG, CH, and CE. Then I say, that vnto the right line geuen AB, &amp; from the point geuen not being in it, namely, C, is drawen a perpendiculer lyne CH.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>For forasmuch as GH is equall to HE, and HC is common to them both; therfore these two sydes GH and HC, are equall to these two sydes EH &amp; HC, the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and (by the</hi> 15 <hi>definitio<g ref="char:cmbAbbrStroke">̄</g>) the base CG is equal to the base CE: wherfore (by the</hi> 8. <hi>proposition) the angle CHG is equall to the angle CHE: and they are syde angles: but when a right line standing vpon a right line maketh the two syde angles equall the one to the other, either of those equall an<g ref="char:EOLhyphen"/>gles is (by the</hi> 10. <hi>definition) a right angle, and the line standing vpon the sayde right line is called a perpendiculer line. VVherfore vnto the right line geue<g ref="char:cmbAbbrStroke">̄</g> AB, and from the point geuen C, which is not in the line AB, is drawn a perpendicu<g ref="char:EOLhyphen"/>ler line CH: which was required to be done.</hi>
               </p>
               <p>
                  <pb n="22" facs="tcp:7062:52" rendition="simple:additions"/>This Probleme did <hi>Oen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>pides</hi> first finde out,<note place="margin">
                     <hi>Oenopides the first inuenter of this probleme.</hi>
                  </note> considering the necessary vse ther<g ref="char:EOLhyphen"/>of to the study of Astronomy.</p>
               <p>There ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two kindes of perpendiculer lines:<note place="margin">
                     <hi>Two kindes of perpendiculer lines, namely, a plaine perpendi<g ref="char:EOLhyphen"/>culer line and a solide.</hi>
                  </note> wherof one is a plaine perpen<g ref="char:EOLhyphen"/>diculer lyne, the other is a solide. A plaine perpendiculer line is, when the point from whence the perpendi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uler line i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> dra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>en, is in the same plaine superficies with the line wherunto it is a perpendicular. A solide perpendiculer line is, whe<g ref="char:cmbAbbrStroke">̄</g> the point, from whence the perpendiculer is drawne, is on high, and wi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hout the plaine superficies. So that a plaine perpendiculer line is drawen to a right line: &amp; a solide perpendiculer line is drawn to a superficies. A plain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> perpendiculer line causeth right angles with one onely line, namely, with that vpon whome it fal<g ref="char:EOLhyphen"/>leth. But a solide perpendiculer line causeth right an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, not only with one line, but with as many lynes as may be drawn in that superficies, by the touch therof.<note place="margin">
                     <hi>This proposition teacheth to draw a playne perpendiculer line.</hi>
                  </note> This proposition teacheth to draw a plaine perpendiculer line, for it is drawn to one line, and supposed to be in the selfe same plaine superficies.</p>
               <p>There may be in this proposition an
<figure/>
other case.<note place="margin">
                     <hi>An other case in this proposition.</hi>
                  </note> For if it be so, that on the o<g ref="char:EOLhyphen"/>ther side of the line <hi>AB,</hi> there be no space to take a pointe in but onely on that side wherein is the point C.<note place="margin">
                     <hi>Construction<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                  </note> Then take some certaine point in the line <hi>AB,</hi> which let be D. And making the cen<g ref="char:EOLhyphen"/>tre the point <hi>C,</hi> and the space CD, de<g ref="char:EOLhyphen"/>scribe a part of the circumference of a circle, which let be DEF: which let cut the line AB in the two pointes D and F. And deuide the line DF into two e<g ref="char:EOLhyphen"/>quall partes in the poynt H. And draw these lines CD, CH and CF. And for<g ref="char:EOLhyphen"/>asmuch as DH is equal to HF,<note place="margin">
                     <hi>Demonstration.</hi>
                  </note> and CH is common to them both, and CD is equall to CF (by the 15. de<g ref="char:EOLhyphen"/>finition:) therfore the angles at the point H are equal the one to the o<g ref="char:EOLhyphen"/>ther (by the 8. proposition:) &amp; they are side angles, wherefore they are right angles. Wherfore the line CH is a perpendiculer to the line DF. But if it happen so that the circle which is described do not cutte the lyne, but touche it, then takyng a point without the point E, name<g ref="char:EOLhyphen"/>ly, the point G, and making the centre the point C, and the space CG, describe a part of the circumference of a circle: which shall of necessitie cut the line AB: and so may you proceede as you did before. As you see in the second figure.</p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Theoreme. The <hi>13.</hi> Proposition. When a right line standing vpon a right line maketh any an<g ref="char:EOLhyphen"/>gles: those angles shall be either two right angles, or equall to two right angles.</head>
               <p>
                  <pb facs="tcp:7062:53"/>SVppose that the right line AB standing vppon the right line CD do make these angles CBA and ABD. Then I say, that the angles CBA and ABD are eyther two right angles, or its <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall to two right angles. If the angle CBA be equall to the angle ABD: then are they two right angles (by the tenth difinition.<note place="margin">Construction,</note>) But if not, raise vp (by the <hi>11.</hi> proposition) vnto the right line CD, and from the pointe geuen in it, namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">Demonstration</note> B, a perpendiculer line BE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> VVherfore (by the x. definition) the angle CBE and EBD ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> right angles. Now forasmuch as the angle CBE, is equall to these two angles CBA and ABE, put the angle EBD common to them both<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the angles CBE and EBD, are equal to thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> three angles CBA, ABE, and EBD. Agayne forasmuch as the angle DBA is equall vnto these two angles DBE and EBA, put the angle ABC common to them both: wherfore the angles DBA
<figure/>
and ABC, are equal to these three angles, DBE, EBA, and ABC. And it is proued that the an<g ref="char:EOLhyphen"/>gles CBE and EBD are equal to the selfe same three angles: but thinges equall to one &amp; the self same thing<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are also (by the first commo<g ref="char:cmbAbbrStroke">̄</g> sentence) equall the o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e to the othe. VVherfore the angles CBE and EBD are equall to the angles DBA &amp; ABC. But the angles CBE and EBD are two right angles: wherfore also the angles DBA and ABC are equall to two right angles. VVherfore when a right line standing v<g ref="char:EOLhyphen"/>pon a right line maketh any angles: those angles shalbe either two right angles, or equall to two right angles: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An othe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> demonstration after <hi>Pelitarius.</hi>
                  </head>
                  <p>Suppose that the right line <hi>AB</hi> do stand vpon the right line <hi>CD.</hi> Then I say, that the two angles <hi>ABC</hi> and <hi>ABD,</hi>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Peli<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>riu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> are either two right angles, or equal to two right angles. For if <hi>AB</hi> be perpe<g ref="char:cmbAbbrStroke">̄</g>diculer vnto <hi>CD</hi>: the<g ref="char:cmbAbbrStroke">̄</g> is it manifest, that they are right angles (by the conuersion of the definition) But if it incline towardes the end <hi>C,</hi> then (by the 11. pro<g ref="char:EOLhyphen"/>position) from the point <hi>B,</hi> erect vnto the line <hi>CD</hi> a perpendicul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r line <hi>BE.</hi> By which<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> construction the propositio<g ref="char:cmbAbbrStroke">̄</g> is very manifest. For forasmuch as the angle <hi>ABD</hi> is gre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>ter then the right angle <hi>DBE</hi> by the angle <hi>ABE</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and
<figure/>
the other angle <hi>ABC</hi> is lesse then the right angle <hi>CBE</hi> by the selfe same angle <hi>ABE</hi>: if from the great<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e taken away the excesse, and the same bee added to the lesse, they shall be made two right angles. That is, if from the obtuse angle <hi>ABD</hi> be taken away the angl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>ABE,</hi> there shal remayne the right angle <hi>DBE.</hi> And then if the same angle <hi>ABE</hi> be added to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cute an<g ref="char:EOLhyphen"/>gle <hi>CBA,</hi> there shall bee made the right angle <hi>CBE.</hi> Wherefore it is manifest, that the two angles, namely, the obtuse angle <hi>ABD,</hi> &amp; the acute angle <hi>ABC,</hi> are equall to the two right angles <hi>CBE</hi> and <hi>DBE</hi>: which was requ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>red to be proued.</p>
               </div>
            </div>
            <div type="part">
               <pb n="23" facs="tcp:7062:53" rendition="simple:additions"/>
               <head>The <hi>7.</hi> Theoreme. The <hi>14.</hi> Proposition. If vnto a right line, and to a point in the same line, be drawn two right lines, no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> both on one and the same side, making the side angles equall to two right angles: those two right lynes shall make directly one right line.</head>
               <p>VNto the right line AB, &amp;
<figure/>
to y<hi rend="sup">•</hi> point in it B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> let ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be drawn two right lines BC, and BD, vnto contrary sides, making the syde an<g ref="char:EOLhyphen"/>gles, namely, ABC &amp; ABD, equall to two right an<g ref="char:EOLhyphen"/>gles. Then I say, that y<hi rend="sup">e</hi> right lines BD and BC make both one right line.<note place="margin">Demonstration leading to an absurdit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</note> For if CB and BD do not make both one right line, let the right line BE be so drawn to BC, that they both make one right line. Now forasmuch as the right line AB standeth vpon the right line CBE, therfore the angles ABC and ABE are equall to two right angles (by the <hi>13.</hi> proposition) But (by supposition) the angles ABC and ABD are equall to two right an<g ref="char:EOLhyphen"/>gles: wherfore the angles CBA, and ABE, are equall to the angles CBA, and ABD: take away the angle ABC, which is common to them both. VVherfore the angle remayning ABE, is equall to the angle remaining ABD, namely, the lesse to the greater, which is impossible. VVherefore the line BE is not so di<g ref="char:EOLhyphen"/>rectly drawen to BC, that they both make one right line. In like sorte may we proue, that no other line, besides BD, can so be drawne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> VVherfore the lines CB and BD, make both one right line. If therfore vnto a right line, &amp; to a point in the same line, be drawn two right lines, not both on one and the same side, ma<g ref="char:EOLhyphen"/>king the side angles equall to two right angles: those two lines shal make direct<g ref="char:EOLhyphen"/>ly one right line: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration after <hi>Pelitarius.</hi>
                  </head>
                  <p>Suppose that there be a right line <hi>AB,</hi> vnto whose poi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>te <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,</hi>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Pelitarius.</hi>
                     </note> let there be drawen two right lines <hi>CB</hi> and <hi>BD,</hi> vnto contrary sides: and let the two angles <hi>CBA,</hi> and <hi>DBA,</hi> be either two right angles, or equall to two right angles. Then I say, that the two lines <hi>CB</hi> and <hi>BD,</hi> do make directly one right line,
<figure/>
namely, <hi>CD.</hi> For if they do not, the<g ref="char:cmbAbbrStroke">̄</g> l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>E</hi> b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>awn vnto <hi>CB,</hi> that they both make directly one right line <hi>C<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>
                     </hi>: which shall passe <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ither aboue the line <hi>BD,</hi> or vnder it. First l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t it passe aboue it. And for a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> much as the two angles <hi>CBA</hi> and <hi>ABE,</hi> are (by the former pro<g ref="char:EOLhyphen"/>position) equall to two right angles, and are a part of the two angles, <hi>CBA</hi> and <hi>ABD</hi>: but the angles <hi>CBA</hi> and <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>D</hi> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e by (supposition) equall also to two right angles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore the par<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> is equall to the whole which is im<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ossible. And the like absurditie will follow if <hi>CB
<pb facs="tcp:7062:54"/>
E</hi> passe vnder the line <hi>BD:</hi> namely, that the whole shalbe equall to the part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which is also impossible. Wherefore <hi>CD</hi> is one right line: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Theoreme. The <hi>15.</hi> Proposition. If two right lines cut the one the other: the hed angles shal be equal the one to the other.</head>
               <p>
                  <hi>SVppose that these two right lines AB and CD, do cut the one the o<g ref="char:EOLhyphen"/>ther in the point E. Then I say, that the angle AEC, is equall to the angle DEB.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>For forasmuch as the right line AE, standeth vpon the right line DC, making these angles CEA, and AED: therefore (by the</hi> 13. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>) the angles CEA, and AED, are equall to two right angles. Agayne forasmuch as the right line DE, standeth
<figure/>
vpon the right line AB, making these angles AED, and DEB: therfore (by the same propositi<g ref="char:EOLhyphen"/>on) the angles AED, and DEB, are equall to two right angles: and it is proued, that the angles CEA, and AED, are also equall to two right an<g ref="char:EOLhyphen"/>gles. VVherfore the angles CEA, and AED, are equall to the angles AED, and DEB. Take a<g ref="char:EOLhyphen"/>way the angle AED, which is common to them both. VVherefore the angle remayning CEA, is equall to the angle remayning DEB. And in like sort may it be proued, that the angles CEB, and DEA, are equall the one to the other. If therefore two right lines cut the one the other, the hed angles shalbe equall the one to the o<g ref="char:EOLhyphen"/>ther: which was required to be demonstrated.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Thales Milesius the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>irst inuen<g ref="char:EOLhyphen"/>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r of this pro<g ref="char:EOLhyphen"/>position.</hi>
                  </note>
                  <hi>Thales Milesius</hi> the Philosopher was the first inuenter of this Proposition, as witnesseth <hi>Eudemius,</hi> but yet it was first demonstrated by <hi>Euclide.</hi> And in it there is no construction at all.<note place="margin">
                     <hi>No construction in this proposi<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For the exposition of the thing geue<g ref="char:cmbAbbrStroke">̄</g>, is sufficient inough for the demonstration.</p>
               <p>
                  <hi>Hed Angles,</hi>
                  <note place="margin">
                     <hi>What hed an<g ref="char:EOLhyphen"/>gles are.</hi>
                  </note> are apposite angles, caused of the intersection of two right lines: and are so called, because the heddes of the two angles are ioyned together in one pointe.</p>
               <div type="section">
                  <head>The conuerse of this proposition after Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>The conuerse of this propositio af<g ref="char:EOLunhyphen"/>ter Pelitarius.</hi>
                     </note>
                     <hi>If fower right lines being drawen from one point, do make fower angles, of which the two oppo<g ref="char:EOLhyphen"/>site angles are equall: the two opposite lines shalbe drawen directly, and make one right line.</hi>
                  </p>
                  <p>Suppose that there be fower right lines AB, AC, AD, and AE, drawen from the poynt A, making fower angles at the point A: of which let the angle BAC be equall to the angle DAE, and the angle BAD to the angle CAE. Then I say, that BE and CD are onely two right lines: that is, the two right lines BA and AE are drawen directly,
<pb n="24" facs="tcp:7062:54"/>
and doo make one right line, and likewise the two right
<figure/>
lines <hi>CA</hi> and <hi>AD</hi> are drawen directly, and do make one right line. For otherwise if it be possible,<note place="margin">
                        <hi>Demonstration leading to an absurditie.</hi>
                     </note> let <hi>EF</hi> be one right line, and likewise let <hi>CG</hi> be one right line. And for<g ref="char:EOLhyphen"/>asmuch as the right line <hi>EA</hi> standeth vpon the right line <hi>CG,</hi> therefore the two angles <hi>EAC</hi> and <hi>EAG,</hi> are (by the 13 proposition) equall to two right angles. And for<g ref="char:EOLhyphen"/>asmuch as the right line <hi>B A</hi> standeth vpon the right line <hi>EF</hi>: therefore (by the selfe same) the two angles <hi>EAG</hi> and <hi>FAG</hi> are also equal to two right angles. Wherefore taking away the angle <hi>EAG,</hi> which is common to them both, the angle <hi>EAC,</hi> shall (by the thirde common sen<g ref="char:EOLhyphen"/>tence) be equall to the angle <hi>FAG</hi>: but the angle <hi>EAC</hi> is supposed to be equall to the angle <hi>BAD.</hi> Wherefore the angle <hi>BAD</hi> is equall to the angle <hi>FAG,</hi> namely a part to the whole: which is impossible. And the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e same absurditie will follow, on what side soeuer the lines be drawen. Wherefore <hi>BE</hi> is one line, and <hi>CD</hi> also is oue line: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The same conuerse after Proclus.</head>
                  <p>
                     <hi>If vnto a right line, and to a point thereof be drawen two right lines, not on one and the same side, in such sort that they make the angles at the toppe equall:</hi>
                     <note place="margin">
                        <hi>The same con<g ref="char:EOLhyphen"/>uerse after Peli<g ref="char:EOLhyphen"/>tarius, which is demonstrated directly.</hi>
                     </note> 
                     <hi>those right lines shalbe drawen directly one to the other, and shal make one right line.</hi>
                  </p>
                  <p>Suppose that there be a right line AB, and take a point in in <hi>C.</hi> And vnto the point in it <hi>C,</hi> draw these two right lines <hi>CD</hi> and <hi>CE</hi> vnto contrary sides, making the angles at the hed equal, namely, the angles <hi>ACD</hi> and <hi>BCE.</hi> Then I say, that the lines <hi>CD</hi> and <hi>CE</hi> are drawen directly, and do make one right line. For forasmuch as the right line <hi>CD</hi> standing vpo<g ref="char:cmbAbbrStroke">̄</g> the right line <hi>AB,</hi> doth make the angles <hi>DCA</hi> and <hi>DCB</hi> equall to two right angles (by the 13 proposition:) and the angle <hi>DCA</hi>
                     <figure/>
is equall to the angle <hi>BCE</hi>: therefore the angles <hi>DCB</hi> and <hi>BCE</hi> are equal to two right angles. And forasmuch as vnto a certayne right line <hi>BC,</hi> and to a point thereof <hi>C,</hi> are drawen two right lines not both on one and the same side, making the side angles equall to two right angles, therefore (by the 14 proposition) the lines <hi>CD</hi> and <hi>CE</hi> are drawen directly, &amp; do make one right line, which was required to be proued.</p>
                  <p>The same may also be demonstrated by an argument lea<g ref="char:EOLhyphen"/>ding
<figure/>
to an absurditie.<note place="margin">
                        <hi>The same con<g ref="char:EOLhyphen"/>uerse after Pro<g ref="char:EOLhyphen"/>clus demonstra<g ref="char:EOLhyphen"/>ted indirectly.</hi>
                     </note> For if <hi>CE</hi> be not drawen directly to <hi>CD,</hi> so that they both make one right line, then (if it bee possible) let <hi>CF</hi> bee drawne directly vnto it. So that let <hi>DCF</hi> be one right line. And forasmuch as the tw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> right lines <hi>AB</hi> and <hi>DF</hi> do cutte the one the other, they make the hed angles equall (by the 15. proposition) Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore the an<g ref="char:EOLhyphen"/>gles <hi>ACD</hi> and <hi>BCF</hi> are equall: but (by supposition) the angles <hi>ACD</hi> and <hi>BCE</hi> are also equall. Wherefore (by the first common sentence) the angle <hi>BCE</hi> is equall to the angle <hi>BCF</hi>: namely, the greater to the lesse: which is im<g ref="char:EOLhyphen"/>possible. Wherefore no other right line besides <hi>CE</hi> is dra<g ref="char:EOLhyphen"/>wen directly to <hi>CD.</hi> Wherefore the lines, <hi>CD</hi> and <hi>CE</hi> are <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>, and make one right line: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
                  <p>
                     <pb facs="tcp:7062:55"/>
                     <note place="margin">
                        <hi>What a Corrol<g ref="char:EOLhyphen"/>lary is.</hi>
                     </note>Of this fiuetenth Proposition followeth a <hi>Corrollary.</hi> VVhere note that a <hi>Corollary</hi> is a Proposition, whose demonstration dependeth of the demonstration of an other Proposition, and it appeareth sodenly, as it were by chance offering it selfe vnto vs: and therefore is reckoned as lucre or gayne. The <hi>Corollary</hi> which followeth of this proposition, is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>A Corollary following of this proposition.</hi>
                     </note>
                     <hi>If fower right lines cut the one the other: they make fower angles equall to fower right angles.</hi>
                  </p>
                  <p>This <hi>Corollary</hi> gaue great occasion to finde out that wonderful proposition in<g ref="char:EOLhyphen"/>uented of Pithagoras, which is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>A wonderfull proposition in<g ref="char:EOLhyphen"/>uented by P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hagoras.</hi>
                     </note>
                     <hi>Only three kindes of figures of many angles, namely, an equilater triangle, a right angled figure of fower sides, and a figure of sixe sides, hauing equall sides and equal angles, can fill the whole space about a point, their angles touching the same point.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Euery angle o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> an equilater tri<g ref="char:EOLunhyphen"/>angle i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> equal <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>o <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>wo <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hird par<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap> of a right angle</hi>
                     </note>Euery angle of an equilater triangle contay<g ref="char:EOLhyphen"/>neth
<figure/>
two third partes of a right angle: sixe tymes two thirdes of a right angle make fower right an<g ref="char:EOLhyphen"/>gles. VVherefore sixe equilater triangles fill the whole space about a point which is equal to fower right angles, as in the 1. figure. Also euery angle of a rectangle quadrilater figure is a right angle: wher<g ref="char:EOLunhyphen"/>fore fower of them fill the whole space as in the 2. figure. Euery angle of a sixe angled figure is equal to a right angle,<note place="margin">
                        <hi>Euery angle of a si<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e angled fi<g ref="char:EOLhyphen"/>gu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> equall to a right angle, and to a third part of a right angle.</hi>
                     </note> and moreouer to a third part of a right angle. But a right angle, and a third part of a right angle, take<g ref="char:cmbAbbrStroke">̄</g> thre times, make 4. right angles: wherefore three equilater sixe angled figures fill the whole space about a point: which space (by this <hi>Corrollary</hi>) is equall to fower right angles: as in the third figure. Any other figure of many sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, howsoeuer you ioyne the<g ref="char:cmbAbbrStroke">̄</g> together at the angles, shal either want of fower angles, or exceede them. By this <hi>Corrollary</hi> also it is mani<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>est that if mo then two lines, that is, three, or fower, or how many soeuer do cut the one the other in one point, all the an<g ref="char:EOLhyphen"/>gles by them made at the point shalbe equall to fower right angles. For they fill the place of fower right angles. And it is also many sest, that the angles by those right lines made are double in number to the right lines which cutte the one the other. So that if there b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> two lines which cut the one the other, the<g ref="char:cmbAbbrStroke">̄</g> are there made fower angles equall to fower right angles: but if thre, then are there made sixe angles: if fower, eight angles, and so infinitly. Foreuer the mul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>itude, or number of of the angles is dubled to the multitude of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lines which cut the one the other. And as the angles increase in multitude, so
<pb n="25" facs="tcp:7062:55" rendition="simple:additions"/>
diminish they in magnitude. For that that which is deuided is alwayes one and the selfe same thing, namely, fower right angles.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Theoreme. The <hi>16.</hi> Proposition. Whensoeuer in any triangle, the lyne of one syde is drawen forth in length: the outwarde angle shall be greater then any one of the two inwarde and opposite angles.</head>
               <p>SVppose that ABC be a triangle:
<figure/>
&amp; let one of y<hi rend="sup">e</hi> sides therof, name<g ref="char:EOLhyphen"/>ly, BC be produced vnto the point D. Then I say, that the outwarde angle ACD, is greater then any one of y<hi rend="sup">e</hi> two inward and opposite angles, that is, then the angle CBA,<note place="margin">Cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ru<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ion.</note> or then the angle BAC. Deuide the line AC (by the <hi>10.</hi> proposition) into two equall partes, in the point E. And draw a line from the point B to the point E. And (by the <hi>2.</hi> peticion) extend BE to the point F. And (by the <hi>2.</hi> proposition) vnto the line BE put an equall line EF. And (by the first peticion) draw a line from F to C: and (by the <hi>2.</hi> petici<g ref="char:EOLhyphen"/>on) extend the line AC to the point G.<note place="margin">Demonstration.</note> Now forasmuch as the line AE, is equall to the line EC, and BE is equall to EF. therfore these two sides AE and EB, are equall to these two sides CE and EF, the one to the other: and the angle AEB, is (by the <hi>15.</hi> proposition) equall to the angle FEC, for they are hed an<g ref="char:EOLhyphen"/>gles: wherefore (by the <hi>4.</hi> proposition) the base AB is equall to the base FC: And the triangle ABE is equall to the triangle FEC: and the other angles re<g ref="char:EOLhyphen"/>mayning are equall to the other angles remayning, the one to the other, vnder which are subtended equall sides. VVherefore the angle BAE is equall to the angle ECF. But the angle ECD is greater then the angle ECF. VVherefore the angle ACD, is greater then the angle BAC. In like sort also if the line BC be deuided into two equall partes, may it be pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ed, that the angle BCG, that is, the angle ACD, is greater then the angle ABC. VVhensoeuer therfore in any triangle, the line of one side is drawen forth in length: the ou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ward angle shalbe greater then any one of the two inward and opposite angles: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration after Pelitarius.</head>
                  <p>Suppose that the triangle geuen be <hi>ABC.</hi>
                     <note place="margin">
                        <hi>An other De<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Pelitarius.</hi>
                     </note> Whose side <hi>AB</hi> let be produced vnto
<pb facs="tcp:7062:56"/>
the point <hi>D.</hi> Then I say, that the angle <hi>DBC</hi> is greater then either of the angles <hi>BAC</hi> and <hi>ACB.</hi> For forasmuch as the two lines <hi>AC</hi> and <hi>BC</hi> do concurre in the point <hi>C,</hi> and vpon them falleth the line <hi>AB</hi>: therefore (by the conuerse of the first peticion) the two inward angles on one and the selfe same side, are lesse
<figure/>
then two right angles. Wherefore the angles <hi>ABC</hi> and <hi>CAB</hi> are lesse then two right angles: but the angles <hi>ABC</hi> and <hi>DBC</hi> are (by the 13 proposition) equal to two right angles. Wherefore the two angles <hi>ABC</hi> and <hi>DBC</hi> are greater then the two angles <hi>ABC</hi> and <hi>BAC.</hi> Wherefore taking away the angle <hi>ABC,</hi> which is com<g ref="char:EOLhyphen"/>mon to them both, there shall be left the angle <hi>DBC</hi> greater then the angle <hi>BAC.</hi> And by the same reason, forasmuch as the two lines <hi>BA</hi> and <hi>CA</hi> concurre in the point <hi>A,</hi> and vppon them falleth the right line <hi>CB,</hi> the two inward angles <hi>ABC</hi> and <hi>ACB</hi> are lesse then two right angles. But the angles <hi>ABC</hi> and <hi>DBC</hi> are equall to two right angles. Wherfore the two angles <hi>ABC</hi> and <hi>DBC,</hi> are greater then the two angles <hi>ABC</hi> &amp; <hi>ACB.</hi> Wher<g ref="char:EOLhyphen"/>fore taking away the angle <hi>ABC,</hi> which is common to them both, there shal remaine the angle <hi>DBC</hi> greater then the angle <hi>ACB</hi>: which was required to be proued.</p>
                  <p>Here is to be noted, that when the side of a triangle is drawen forth, the angle of the triangle which is next the outwa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d angle, is called an angle in order vnto it: and the other two angles of the triangle are called opposite angles vnto it.</p>
                  <p>
                     <note place="margin">
                        <hi>A Corrollary following of this proposition.</hi>
                     </note>Of this Proposition followeth this <hi>Corrollary,</hi> that it is not possible that from one &amp; the selfe same point should be drawen to one and the selfe same right line, three equall right lines. For from one point, namely, <hi>A,</hi> if it be
<figure/>
possible, let there be drawen vnto the right line <hi>BD,</hi> these three equall right lines <hi>AB, AC,</hi> &amp; <hi>AD.</hi> And forasmuch as <hi>AB</hi> is equall to <hi>AC,</hi> the angles at the base are (by the fifth proposition) equall. Wherfore the angle <hi>ABC</hi> is equal to the angle <hi>ACB.</hi> Agayne forasmuch as <hi>AB</hi> is equall to <hi>AD,</hi> the angle <hi>ABD</hi> is (by the same) equall to the angle <hi>ADB</hi>: but the angle <hi>ABC</hi> was equall to the angle <hi>ACB.</hi> Where<g ref="char:EOLhyphen"/>fore the angle <hi>ACB</hi> is equall to the angle <hi>ADB</hi>: namely, the outward angle to the inwarde &amp; oppo<g ref="char:EOLhyphen"/>site angle: which is impossible. Wherfore from one and the selfe same point, can not be drawn to one &amp; the selfe same right line three equall right lynes: which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>An other Cor<g ref="char:EOLhyphen"/>rollary <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ollo<g ref="char:EOLhyphen"/>wing also of the same.</hi>
                     </note>By this Proposition also may this be demonstrated, that if a right line falling vpon two right lines, do make the outward angle equall to the inward and oppo<g ref="char:EOLhyphen"/>site angle, those right lines shall not make a triangle, neither shal they concurre. For otherwis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> one &amp; the selfe same angle should be both greater, and also equal: which is impossible. As for example.</p>
                  <p>Suppose that there be two right lines <hi>AB</hi> and <hi>CD,</hi> and vpon them let the right line <hi>BE</hi> fall, making the angles <hi>ABD</hi> and <hi>CDE</hi> equall. Then I say, that the right lines <hi>AB</hi> and <hi>CD</hi> shall not concurre. For if they concurre, the foresaide angles abidyng equall, namely, the angles <hi>CDE</hi> and <hi>ABD</hi>: Then forasmuch as the angle <hi>CDE</hi> is the out<g ref="char:EOLhyphen"/>ward angle it is of necessitie greater then the inward and opposite angle, &amp; it is also e<g ref="char:EOLhyphen"/>qual vnto it: which is impossible. Wherfore if the said lines co<g ref="char:cmbAbbrStroke">̄</g>curre, the<g ref="char:cmbAbbrStroke">̄</g> shal not the an<g ref="char:EOLhyphen"/>gles remayne equall, but the angle at the point <hi>D</hi> shall be encreased. For whether <hi>AB</hi>
                     <pb n="26" facs="tcp:7062:56"/>
abiding fixed you suppose the line <hi>CD</hi> to be moued
<figure/>
vnto it, so that they concurre, the space and distance in the angle will be greater: for how much more <hi>CD</hi> approcheth to <hi>AB,</hi> so much farther of goeth it from <hi>DE.</hi> Or whether <hi>CD</hi> abiding fixed, you ima<g ref="char:EOLhyphen"/>gine the line <hi>AB</hi> to be moued vnto i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, so that they concurre, the angle <hi>ABD</hi> will be lesse, for there<g ref="char:EOLhyphen"/>with all it comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th nere vnto the lines <hi>CD</hi> &amp; <hi>BD.</hi> Or whether you imagine either of them to be mo<g ref="char:EOLhyphen"/>ued the one to the other, you shall finde that the line <hi>AB</hi> comming neere to <hi>CD,</hi> maketh the angle <hi>ABE</hi> lesse, and <hi>CD</hi> going farther from <hi>DE</hi> by reason of his motion to the line <hi>BD</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> maketh the angle <hi>CDE</hi> to increase. Wherefore it followeth of necessitie, that if it be a triangle, and that the right lines <hi>AB</hi> and <hi>CD</hi> do conc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rre, the outward angle also shall be greater<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then the inward and opposite angle. For either the inward and opposite angle abiding fixed, the outward i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> increased: or the outwarde a<g ref="char:EOLhyphen"/>biding fixed, the inward and opposi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> is diminished<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or els both of them being moued till they concurre, the inwarde is diminished, and the outwarde is more increased. And the caus<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> hereof is the motion of the right lines the one tending to that parte where it diminisheth the inwarde angle, the other tending to that part where it increaseth th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> outward angle.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Theoreme. The <hi>17.</hi> Proposition. In euery triangle, two angles, which two soeuer be taken, are lesse then two right angles.</head>
               <p>
                  <hi>SVppose that ABC be a
<figure/>
triangle. Then I saye that two angles of the sayd tri<g ref="char:EOLhyphen"/>angle ABC, which two angles soeuer be taken, are lesse the<g ref="char:cmbAbbrStroke">̄</g> two right angles Extend (by the</hi> 2. <hi>peticio<g ref="char:cmbAbbrStroke">̄</g>) the line BC,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>to the point D.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>And forasmuch as (by the proposition going before) the outward angle of the triangle ABC, namely, the angle ACD is greater then the inward and opposite angle ABC: put the angle ACB commo<g ref="char:cmbAbbrStroke">̄</g> to them both: wherefore the angles ACD and ACB are greater then the an<g ref="char:EOLhyphen"/>gles ABC and BCA. But (by the</hi> 13 <hi>proposition) the angles ACD and ACB are equall to two right angles. VVherefore the angles ABC and BCA are lesse then two right angles. In like sort also may we proue, that the angles BAC and ACB are lesse then two right angles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and also that the angles CAB &amp; ABC are lesse then two right angles. VVherefore in euery triangle, two an<g ref="char:EOLhyphen"/>gles, which two soeuer be taken, are lesse then two right angles: which was re<g ref="char:EOLhyphen"/>quired to be proued.</hi>
               </p>
               <p>
                  <pb facs="tcp:7062:57"/>This may also be demonstrated without the helpe of the former propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ition, by the conuerse of the fifth petition, and by the 13. proposition as you saw was done in the former after <hi>Politarius.</hi>
               </p>
               <p>It may also be demonstrated without producing any of the sides of the tri<g ref="char:EOLhyphen"/>angle, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter this maner.</p>
               <p>
                  <note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration inuen<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ed by Proclus.</hi>
                  </note>Suppose that there a be triangle <hi>ABC.</hi> And in the side <hi>BC</hi> take a point at all aduen<g ref="char:EOLhyphen"/>tures, and let the same be <hi>D,</hi> and draw the line <hi>AD.</hi> And forasmuch as in the triangle <hi>ABD,</hi> the side <hi>BD</hi> is produced, therefore (by the former proposition) the outward angle <hi>ADC,</hi> is greater then the inward and opposite angle <hi>ABD.</hi> Agayne forasmuch as in the triangle <hi>ADC,</hi> the syde <hi>CD</hi> is produced, therefore (by the same) the outward angle <hi>ADB,</hi> is greater then the inwarde and opposite
<figure/>
angle <hi>ACD</hi>: but the angles at the point <hi>D</hi> are equall to two right angles (by the 13. proposition:) wherfore the angles <hi>ABC</hi> and <hi>ACB</hi> are lesse then two right an<g ref="char:EOLhyphen"/>gles. And by the same reason may we proue that the an<g ref="char:EOLunhyphen"/>gles <hi>BAC</hi> and <hi>BCA</hi> are lesse then two right angles, if we take a poynt in the line <hi>AC,</hi> and draw a right line from it to the point <hi>B</hi>: and so also may it be proued that the angles <hi>CAB</hi> and <hi>ABC</hi> are lesse the two ryght angles, if there be taken in the lyne <hi>AB</hi> a point, and from it be a line drawen to the point <hi>C.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>A Corrollary fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>wi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>g this Proposition.</hi>
                  </note>By this proposition also may be proued this Corrollary, that from one and the selfe same point to one and the selfe same right line, can not be drawen two perpendicular lines.</p>
               <p>For if it be possible, from the point <hi>A,</hi> let there be drawen
<figure/>
vnto the right line <hi>BC,</hi> two perpendicular lines <hi>AB,</hi> and <hi>AC</hi>: wherefore the angles <hi>ABC</hi> and <hi>ACB</hi> are right angles. But forasmuch as <hi>ABC</hi> is a triangle. therefore any two angles ther<g ref="char:EOLhyphen"/>of are (by this proposition) lesse then two right angles. Where<g ref="char:EOLhyphen"/>fore the angles <hi>ABC</hi> and <hi>ACB</hi> are lesse then two right angles: but they are also equall to two right angles, by reason <hi>AB</hi> and <hi>AC</hi> are perpendicular lines vpon <hi>BC</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which is impossible. Wherefore from one and the selfe same point cannot be drawe<g ref="char:cmbAbbrStroke">̄</g> to one and the selfe same line two perpendicular lines: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>11.</hi> Theoreme. The <hi>18.</hi> Proposition. In euery triangle, to the greater side is subtended the grea<g ref="char:EOLhyphen"/>ter angle.</head>
               <p>
                  <hi>SVppose that ABC be a
<figure/>
triangle, hauing the side AC greater then the side AB. Then I say that the an<g ref="char:EOLhyphen"/>gle ABC is greater then the angle BCA.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>For for<g ref="char:EOLhyphen"/>asmuch as AC is greater the<g ref="char:cmbAbbrStroke">̄</g> AB, put (by the</hi> 3. <hi>pro<g ref="char:EOLhyphen"/>position) vnto AB an equall line AD. And (by
<pb n="27" facs="tcp:7062:57"/>
the first peticion) draw a line from the point B to the point D.</hi>
                  <note place="margin">
                     <hi>Demonstration<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                  </note> 
                  <hi>And forasmuch as the outward angle of the triangle DBC, namely, the angle ADB is grea<g ref="char:EOLhyphen"/>ter then the inward and opposite angle DCB (by the</hi> 16. <hi>proposition,) but (by the</hi> 5. <hi>proposition) the angle ADB is equall to the angle ABD, for the syde AB is equall to the syde AD: therefore the angle ABD is greater then the angle AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. VVherefore the angle ABC is much greater then the angle ACB. VVherefore in euery triangle, to the greater syde is subtended the greater angle: which was required to be proued.</hi>
               </p>
               <p>You may also proue the angle at the point <hi>B</hi> greater then the angle at the point <hi>C</hi> (the side <hi>AC</hi> being greater then the side <hi>AB</hi>) if from the line <hi>AC</hi> you cut of a line e<g ref="char:EOLhyphen"/>quall to the line <hi>AB,</hi>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Prophyrius.</hi>
                  </note> beginning at the point <hi>C,</hi> as before you begaune at the point <hi>A</hi>: and that after this manner. Let the line DC be equall to the line
<figure/>
                  <hi>AB</hi> and draw the line <hi>BD:</hi> and produce <hi>AB</hi> to the point <hi>E</hi>: and put the line <hi>BE</hi> equal to the line <hi>AD.</hi> Wherefore the whole line <hi>AE</hi> is equall to the whole line <hi>AC</hi>: draw a line from <hi>E</hi> to <hi>C.</hi> And forasmuch as <hi>AE</hi> is equal to <hi>AC,</hi> therfore the angle <hi>AEC</hi> is also equall to the angle <hi>ACE</hi> (by the 5. proposition:) but the angle <hi>ABC</hi> is greater then the angle <hi>AEC.</hi> For one of the sides of the triangle <hi>CBE,</hi> namely, the side <hi>BE</hi> is produced, and so the outward angle <hi>ABC</hi> is greater then the inward and oppo<g ref="char:EOLhyphen"/>site <hi>BEC</hi> (by the 16 proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) wherefore the angle <hi>ABC</hi> is much greater then the angle <hi>ACB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
               <p>Note that that which is here spoken in this propositi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>on,<note place="margin">
                     <hi>That which is spoken in this Proposition is to be vnderstanded in one and the selfe same tri<g ref="char:EOLhyphen"/>angle.</hi>
                  </note> is to be vnderstanded in one and the self same triangle. For it is possible that one and the selfe same angle may be subtended of a greater line, and of a lesse line: and one and the selfe same right line may subtend a greater angle, and a lesse an<g ref="char:EOLhyphen"/>gle. As for example.</p>
               <p>Suppose that there be an Isosceles triangle <hi>ABC</hi> &amp;,
<figure/>
in the side <hi>AB</hi> take the point <hi>D</hi> at all aduentures: &amp; fro<g ref="char:cmbAbbrStroke">̄</g> the line <hi>AC</hi> cut of (by the 3. proposition) the lyne <hi>AE</hi> equall to the line <hi>AD.</hi> And draw a right line from <hi>D</hi> to <hi>E.</hi> Wherfore the right lines <hi>DE</hi> and <hi>BC</hi> do subtend the angle at the point <hi>A,</hi> &amp; of them the one is greater, and the other lesse. And after the selfe same manner a man may put infinite right lines greater &amp; lesse, subtending the angle at the point <hi>A.</hi>
               </p>
               <p>Agayne suppose that <hi>ABC</hi> be an Isosceles triangle. And let <hi>BC</hi> be lesse then either of the lines <hi>BA</hi> and <hi>AC.</hi> And vpon <hi>BC</hi> describe (by the first) an equilater trian<g ref="char:EOLhyphen"/>gle <hi>BCD.</hi> And draw a line from <hi>A</hi> to <hi>D,</hi> and produce it to the point <hi>E.</hi> And forasmuch as in the triangle <hi>ABD,</hi> the outward angle <hi>BDE,</hi> is greater <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hen the inward &amp; opposite angle <hi>BAD</hi> (by the 16. proposition) And by the same in the tria<g ref="char:cmbAbbrStroke">̄</g>gle <hi>ACD,</hi> the outward angle <hi>CDE,</hi> is greater then the inward &amp; opposite angle <hi>CAD</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> ther<g ref="char:EOLhyphen"/>fore the whole angle <hi>BDC</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the whole an<g ref="char:EOLhyphen"/>gle <hi>BAC.</hi> And one and the selfe same right line subten<g ref="char:EOLhyphen"/>deth both these angles, namely, the greater angle &amp; the lesse. And it is also proued, that greater right lines &amp; lesse subtende one and the selfe same angle. But in
<pb facs="tcp:7062:58"/>
one and the selfe same triangle one right line subtendeth one angle, and the great right line euer subtendeth the great angle, and the lesse the lesse, as it was proued in the proposition.</p>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Theoreme. The <hi>19.</hi> Proposition. In euery triangle, vnder the greater angle is subtended the greater side.</head>
               <p>
                  <hi>SVppose that ABC be a triangle, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>auyng
<figure/>
the angle ABC greater then the angle BCA. Then I say that the side AC is greater then y<hi rend="sup">e</hi> side AB. For if no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>,</hi>
                  <note place="margin">
                     <hi>Demonstration lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to an i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>possibilitie.</hi>
                  </note> 
                  <hi>the the side AC is ether equal to y<hi rend="sup">e</hi> side AB, or els it is lesse the<g ref="char:cmbAbbrStroke">̄</g> it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> The side AC is not equal to y<hi rend="sup">e</hi> side AB, for then (by the</hi> 5<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>pro<g ref="char:EOLhyphen"/>position) y<hi rend="sup">e</hi> angle ABC should be equall to the an<g ref="char:EOLhyphen"/>gle ACB: but (by suppositio<g ref="char:cmbAbbrStroke">̄</g>) it is not. VVhere<g ref="char:EOLhyphen"/>fore the side AC is not equall to the side AB. And the side AC can not be lesse then the side AB, for then the angle ABC shoulde be lesse then the angle ACB (by the proposition next goyng before). But (by supposition it is not) VVherefore the side AC is not lesse then the side AB. VVherefore the side AC is greater then the side AB. VVherefore in euery tri<g ref="char:EOLhyphen"/>angle, vnder the greater angle is subtended the greater side: which was requi<g ref="char:EOLhyphen"/>red to be demonstrated.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>This proposition <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> the conuerse the former.</hi>
                  </note>This proposition is the conuerse of the proposition next going before. VVher<g ref="char:EOLhyphen"/>fore as you see, that which was the conclusion in the former, is in this the suppo<g ref="char:EOLhyphen"/>sition, or thing geuen: and that which was there the thing geuen, is here the thing required or conclusion. And it is proued by an argument leading to an impossi<g ref="char:EOLhyphen"/>bilitie, as commonly all conuerses are.</p>
               <p>
                  <hi>P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oclus</hi> demonstrateth this proposition after an other way: but first he putteth this <note n="*" place="margin">
                     <hi>An Assumpt is <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> of necessitie in the helpe of a demonstration, the certainty whereof is not so pl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e, and the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>fore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h is selfe first to be demonstrated. An assumpt put by Pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>er the demonstra<g ref="char:EOLhyphen"/>tion of this Pro<g ref="char:EOLhyphen"/>po<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion.</hi>
                  </note> Assumpt following.</p>
               <p>
                  <hi>If an angle of a triangle be deuided into two equall partes, and if the line which deuideth it being drawen to the base, do deuide the same into two vnequall partes: the sides which contayne that angle shalbe vnequall, and that shalbe the greater side, which falleth on the grater side of the base, and that the lesse which falleth on the lesse side of the base.</hi>
               </p>
               <p>Suppose <hi>ABC</hi> to be a triangle, and (by the 9. proposition) deuide the angle at the point <hi>A,</hi> into two equall partes, by the right line <hi>AD.</hi> And let the line <hi>AD</hi> deuide the base BC, into two vnequall partes, and let the part <hi>CD</hi> be greater then the parte <hi>BD.</hi> Then I say, that the side <hi>AC</hi> is greater then the side <hi>AB.</hi> Produce the line <hi>AD</hi> to the point <hi>E,</hi> and (by the third) put the line <hi>DE</hi> equall to the line <hi>DA.</hi> And forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>DC</hi> is by supposition greater then <hi>DB,</hi> put (by the 3. proposition) <hi>DF</hi> equal to <hi>BD,</hi> and draw a line fro<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>E</hi> to <hi>F,</hi> and produce it to the point <hi>G.</hi> Now forasmuch as <hi>AD</hi> is e<g ref="char:EOLhyphen"/>quall to <hi>ED</hi> and <hi>DB</hi> is equall to <hi>DF,</hi> therfore in the two triangles <hi>ABD,</hi> and <hi>EFD,</hi> two sides of the one are equall to two sides of the other, eche to his correspondent side. and (by the 15. proposition) they contayne equall angles, namely, the hed an<g ref="char:EOLhyphen"/>gles:
<pb n="28" facs="tcp:7062:58"/>
wherfore (by the fourth proposi<g ref="char:EOLhyphen"/>tion)
<figure/>
the base <hi>BA</hi> is equall to the base <hi>BF:</hi> and the angle <hi>DEF</hi> is equall to the angle <hi>DAB.</hi> But the angle <hi>DAG</hi> is by construction equall to the same angle <hi>DAB</hi>: wherefore (by the first common sentence) the angles <hi>EAG</hi> and <hi>AEG</hi> are equall. Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> propo<g ref="char:EOLhyphen"/>sion) the side <hi>AG</hi> is equall to the side <hi>EG.</hi> Wherfore the side <hi>AC</hi> is greater then the side <hi>EG.</hi> Wherefore it is much grea<g ref="char:EOLhyphen"/>ter then the side <hi>EF.</hi> But the side <hi>EF</hi> is equall to the side <hi>AB,</hi> as it hath bene proued. Wherefore the side <hi>AC</hi> is grea<g ref="char:EOLhyphen"/>ter then the side <hi>AB</hi>: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               <p>This assumpt being put, this Proposition is of Proclus thus demonstrated.</p>
               <p>Suppose <hi>ABC</hi> to be a triangle, hauing his angle at the point <hi>B</hi> greater then the an<g ref="char:EOLunhyphen"/>gle at the point <hi>C.</hi>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Proclu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note> Then I say that the side <hi>AC</hi> is greater then the side <hi>AB.</hi> Deuide the line <hi>BC</hi> into two equall partes in the point <hi>D,</hi> and draw a line from <hi>A</hi> to <hi>D.</hi> And pro<g ref="char:EOLhyphen"/>duce the line <hi>AD</hi> to the point <hi>E</hi>: and put the line <hi>DE</hi> equall to the line <hi>AD,</hi> and draw a line from <hi>B</hi> to <hi>E.</hi> Now forasmuch as <hi>BD</hi> is equall to <hi>DC,</hi> and <hi>AD</hi> is equall to <hi>DE</hi> therefore in the two triangles <hi>ADC</hi> and <hi>BDE,</hi> two
<figure/>
sides of the one are equall to two sides of the other, ech to his correspondent side, and they containe equall an<g ref="char:EOLhyphen"/>gles (by the 15. proposition): wherefore (by the fourth proposition) the base <hi>BE</hi> is equall to the base <hi>AC,</hi> and the angle <hi>DBE</hi> is equal to the angle at the point <hi>C.</hi> De<g ref="char:EOLhyphen"/>uide also th'angle <hi>ABE</hi> into two equal parts by the line <hi>BF</hi>: wherfore the line <hi>EF</hi> is greater then the line <hi>FA.</hi> And forasmuch as in the triangle <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>E,</hi> the angle at the point <hi>B</hi> is deuided into two equall partes by the right line <hi>BF,</hi> and the line <hi>EF</hi> is greater then the line <hi>AF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore by the former <hi>Assumps</hi> the side <hi>RE</hi> is greater then the side <hi>BA</hi>: but the line <hi>BE</hi> is equall to the line <hi>AC.</hi> Wherfore the syde <hi>AC</hi> is greater then the side <hi>AB:</hi> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Theoreme. The <hi>20.</hi> Proposition. In euery triangle two sides, which two sides soeuer be taken, are greater then the side remayning.</head>
               <p>
                  <hi>SVuppose that ABC be a
<figure/>
triangle. Then I say that two sides of the triangle ABC, which two sides soe<g ref="char:EOLhyphen"/>uer be taken, are greater<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then the side remayning that is, the sides BA and AC are greater then the
<pb facs="tcp:7062:59"/>
side BC: and the sides AB and BC then the side AC: and the sides AC and BC then the side BA.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Produce (by the</hi> 2. <hi>peticion) the line BA to the point D. And (by the third proposition) vnto the line AC put an equall line AD. and draw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a line from the point D to the pointe C.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>And foras<g ref="char:EOLhyphen"/>much
<figure/>
as the line DA is equall to the line AC, there<g ref="char:EOLhyphen"/>fore (by the</hi> 5. <hi>proposition) the angle ADC, is equall to the angle ACD. But the angle BCD is greater then the angle ACD, therefore the angle BCD is greater then the angle ADC. And forasmuch as DCB is a triangle, hauing the angle BCD greater then the angle ADC, but (by the</hi> 18. <hi>proposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion) vn<g ref="char:EOLhyphen"/>der the greater angle is subtended the greater side: wherfore DB is greater then BC. But the line DB is equall to the lines AB and AC (for the line AD is equall to the line AC) wherfore the sides BA and AC, are greater then the side BC. And in like <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>orte may we proue, that the sides AB and BC are greater then the side AC: &amp; that the sides BC and CA are greater then the side AB. VVherfore in euery trian<g ref="char:EOLhyphen"/>gle two sides, which two sides soeuer be taken, are greater then the side remay<g ref="char:EOLhyphen"/>ning<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be demonstrated.</hi>
               </p>
               <p>This Proposition may also be demonstrated without producing any of the sides, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter this maner.</p>
               <p>
                  <note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion with<g ref="char:EOLhyphen"/>out producing one of the sides.</hi>
                  </note>Suppose <hi>ABC</hi> to be triangle. Then I say, that the two sides <hi>AB</hi> and <hi>AC</hi> are grea<g ref="char:EOLhyphen"/>ter then the side <hi>BC</hi>: deuide the angle at the point <hi>A</hi> (by the 9. proposition) into two equall partes by the right line <hi>AE.</hi> And forasmuch as in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> triangl <hi>ABE,</hi> the out<g ref="char:EOLhyphen"/>ward angle <hi>AEC</hi> is greater then the angle <hi>BAE</hi> (by
<figure/>
the 16 proposition), and the angle <hi>BAE</hi> is put to be equall to the angle <hi>EAC,</hi> therefore the side <hi>AC</hi> is grea<g ref="char:EOLhyphen"/>ter then the side <hi>CE.</hi> And by the same reason the side <hi>AB</hi> is greater thé the side <hi>BE.</hi> For in the triangle <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>C</hi> the outward angle <hi>AEB,</hi> is greater then the angle <hi>CAE,</hi> that is then the angle <hi>EAB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore also the side <hi>AB</hi> is greater then the side <hi>BE.</hi> Wherfore the sides <hi>AB</hi> and <hi>AC</hi> are greater then the whole side <hi>BC.</hi> And after the same maner may you proue touching the other sides also.</p>
               <p>The same may yet also be demonstrated an other way.</p>
               <p>
                  <note place="margin">
                     <hi>An other De<g ref="char:EOLhyphen"/>monstration.</hi>
                  </note>Suppose <hi>ABC</hi> to be a triangle. Now if <hi>ABC</hi> be an
<figure/>
equilater triangle, then without doubt any, two sides thereof are greater then the third. For the three sides being equall any two sides of them are double to the third. But if it be an Isosceles triangle, either the base is lesse then either of the equall sides or it is greater. If the base be lesse, then againe two of them are greater then the thirde, but if the base be greater<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> let <hi>BC</hi> being the base of the Isosceles triangle <hi>ABC</hi> be greater the<g ref="char:cmbAbbrStroke">̄</g> either of the sides <hi>AB</hi> &amp; <hi>AC</hi> and from it <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut of (by the 3. pro<g ref="char:EOLhyphen"/>position)
<pb n="29" facs="tcp:7062:59"/>
a line equall to any one of the other sides, whiche let bee <hi>BE,</hi> and drawe a line from <hi>A</hi> to <hi>E.</hi> And forasmuch as in the triangle <hi>AEB,</hi> the angle <hi>AEC</hi> is an out<g ref="char:EOLhyphen"/>ward angle: therefore it is greater then the angle <hi>BAE</hi> (by the 16. proposition). And by the same reason, the angle <hi>AEB</hi> is greater then the angle <hi>CAE.</hi> Wherefore the an<g ref="char:EOLhyphen"/>gles at the point <hi>E</hi> are greater then the whole angle at the pointe <hi>A.</hi> But the angle <hi>BEA</hi> is equall to the angle <hi>BAE</hi> (by the 5. proposition) for <hi>AB</hi> is put to be equall to <hi>BE.</hi> Wherefore the angle remayning <hi>AEC</hi> is greater then the angle <hi>CAE.</hi> Wherefore also the side <hi>AC</hi> is greater then the side <hi>EC.</hi> But the side <hi>AB</hi> is equall to the side <hi>BE.</hi> Wherefore the sides <hi>AB</hi> and <hi>AC</hi> are greater then the side <hi>BC.</hi>
               </p>
               <p>But if the triangle <hi>ABC</hi> be a S<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>lenum, let the side
<figure/>
                  <hi>AB</hi> be the greatest, and let <hi>AC</hi> be the meane, and <hi>BC</hi> the least. Wherefore the greatest side being added to a<g ref="char:EOLhyphen"/>ny one of the two sides must <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>edes be greater then the third. For of it selfe it is greater then any of them. But if <hi>AB</hi> being the greatest<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> you would proue the sides <hi>AC</hi> and <hi>CB</hi> to be greater then it. Then as you did in the I<g ref="char:EOLhyphen"/>soceles triangle, cut of from the greatest a line equall to one of them, and from the point <hi>C</hi> to the point of the intersection draw a right line, and reason as you did be<g ref="char:EOLunhyphen"/>fore by the outward angles of the triangle, and you shal haue your purpose.</p>
               <p>This proposition may yet moreouer be demonstrated by an argument lea<g ref="char:EOLhyphen"/>ding to an absurditie, and that after this manner.</p>
               <p>Suppose <hi>ABC</hi> to be a triangle.<note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration leading to an absurditi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                  </note> Then I say that the
<figure/>
sides <hi>AB</hi> and <hi>AC,</hi> are greater then the side <hi>BC.</hi> For if they be not greater, they are either equall or lesse. First let them be equall, and from the line <hi>BC</hi> cut of the line <hi>BE</hi> equall to the line <hi>AB</hi> (by the 3. proposition) wher<g ref="char:EOLhyphen"/>fore the residue <hi>EC</hi> is equall to <hi>AC.</hi> Now forasmuch as <hi>AB</hi> is equall to <hi>BE</hi> they subtend equall angles. Like<g ref="char:EOLhyphen"/>wise forasmuch as <hi>AC</hi> is equall to <hi>CE</hi> they subtend e<g ref="char:EOLhyphen"/>qual angles. Wherfore the angles which are at the point <hi>E</hi> are equall to the angles whiche are at the pointe <hi>A,</hi> which is impossible (by the 16. proposition).</p>
               <p>But now let the sides <hi>AB</hi> and <hi>AC</hi> be lesse then the
<figure/>
side <hi>BC,</hi> and from the line <hi>BC</hi> cut of (by the 3. proposi<g ref="char:EOLhyphen"/>tion) the line <hi>BD</hi> equall to the line <hi>AB,</hi> and likewise fro<g ref="char:cmbAbbrStroke">̄</g> the same line <hi>BC</hi> cut of the line <hi>CE</hi> equall to the line <hi>AC.</hi> And forasmuch as <hi>AB</hi> is equall to <hi>BD,</hi> the angle <hi>BDA</hi> also is equall to the angle <hi>BAD</hi> (by the fifth pro<g ref="char:EOLhyphen"/>position). Againe forasmuch as <hi>AC</hi> is equall to <hi>CE,</hi> therefore (by the same) the angle <hi>CEA</hi> is equall to the angle <hi>EAC.</hi> Wherefore these two angles <hi>BDA</hi> and <hi>CEA</hi> are equall to these two angles <hi>BAD</hi> and <hi>EAC.</hi> Agayne forasmuch as the angle <hi>BDA</hi> is the outward angle of the triangle <hi>ADC,</hi> therefore it is greater then the angle <hi>EAC.</hi> For it is greater then the angle <hi>DAC</hi> (by the 16. proposition). And by the same reason, forasmuch as <hi>CEA</hi> is the the outward angle of the triangl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>ABE,</hi> therefore it is greater then the angle <hi>BAD</hi> (for it is greater then the angle <hi>BAE</hi>). Wherfore the angles <hi>BDA</hi> and <hi>CEA</hi> are greater then the two angles <hi>BAD</hi> and <hi>EAC.</hi> But they were also proued equall vnto them: which is impossible. Wherefore th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> sides <hi>AB</hi> and <hi>AC</hi> are neither equall to the side <hi>BC,</hi> nor lesse then it, but greater. And so also may it be proued of the rest.</p>
               <p>
                  <pb facs="tcp:7062:60"/>
                  <note place="margin">
                     <hi>A briefe demon<g ref="char:EOLhyphen"/>stration by the definition of a right line.</hi>
                  </note>A man may also more briefely demonstrate this proposition by Campanus definition of a right line, which as we haue before declared is thus: <hi>A right line is the shortest extension or drawght that is or may be from one point to another.</hi> Wherfore any one side of a triangle, for that it is a right line drawen from some one point to some other one point, is of necessitie shorter then the other two sides drawen from and to the same pointes.</p>
               <p>
                  <note place="margin">
                     <hi>Not all thinges manifest to the sese, are straight way manifest to reason and vn<g ref="char:EOLhyphen"/>derstanding.</hi>
                  </note>
                  <hi>Epicurus</hi> and such as followed him derided this proposition, not counting it worthy to be added in the number of propositions of Geometry for the easines thereof, for that it is manifest euen to the sense. But not all thinges manifest to sense, are straight wayes manifest to reason and vnderstanding. It pertayneth to one that is a teacher of sciences, by profe and demonstration to render a cer<g ref="char:EOLhyphen"/>tayne and vndoubted reason, why it so appeareth to the sense<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and in that onely consisteth science.</p>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Theoreme. The <hi>21.</hi> Proposition. If from the endes of one of the sides of a triangle, be drawen to any point within the sayde triangle two right lines. those right lines so drawen, shalbe lesse then the two other sides of the triangle, but shall containe the greater angle.</head>
               <p>
                  <hi>SVppose that ABC be a
<figure/>
triangle: and fro<g ref="char:cmbAbbrStroke">̄</g> the endes of the side BC, namely, fro<g ref="char:cmbAbbrStroke">̄</g> the pointes B and C, let there be drawen within y<hi rend="sup">e</hi> triangle two right lines BD and CD to y<hi rend="sup">e</hi> point D. Then I say, that the lines BD and CD are lesse then the other sides of the triangle, namely, then the sides BA and AC: and that the angle which they contayne, namely, BDC, is greater then the angle BAC. Extend (by the second peticion) the line BD to the point E.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>And forasmuch as (by the</hi> 20. <hi>pro<g ref="char:EOLhyphen"/>position) in euery triangle the two sides are greater then the side remayning, therefore the two sides of the triangle ABE, namely, the sides AB and AE, are greater then the side EB. Put the line EC common to them both. VVhere<g ref="char:EOLhyphen"/>fore the lines BA and AC, are greater then the lines BE and EC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Againe for<g ref="char:EOLhyphen"/>asmuch as (by the same) in the triangle CED, the two sides CE and ED, are greater then y<hi rend="sup">e</hi> side DC, put y<hi rend="sup">e</hi> line DB common to them both<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wherfore y<hi rend="sup">e</hi> lines CE and ED, are greater then the lines CD and DB. But it is proued that the lines BA and AC, are greater then the lines BE and EC. VVherefore the lines BA and AC, are much greater then the lines BD and DC. Agayne forasmuch
<pb n="30" facs="tcp:7062:60"/>
as (by the</hi> 16. <hi>proposition) in euery triangle, the outward angle is greater then the inward and opposite angle, therefore the outward angle of the triangle CDE, namely, BDC, is greater then the angle CED. VVherefore also (by the same) the outward angle of the triangle ABE, namely, the angle CEB is greater then the angle BAC. But it is proued, that the angle BDC is greater then the angle CEB. VVherfore the angle BDC is much greater then the an<g ref="char:EOLhyphen"/>gle BAC. VVherefore if from the endes of one of the sides of a triangle, be dra<g ref="char:EOLhyphen"/>wen to any point within the sayde triangle two right lines: those right lines so drawn shalbe lesse then the two other sides of the triangle, but shall contayne the greater angle: which was required to be demonstrated.</hi>
               </p>
               <p>In this proposition is expressed, that the two right lines drawen within the triangle, haue their beginning at the extremes of the side of the triangle. For fro<g ref="char:cmbAbbrStroke">̄</g> the one extreme of the side of the triangle, and from some one point of the same side, may be drawen two right lines within the triangle, which shall be longer the<g ref="char:cmbAbbrStroke">̄</g> the two outward lines: which is wonderfull and seemeth straunge, that two right lines drawen vpon a parte of a line, should be greater then two right lines drawen vpon the whole line. And agayne it is possible from the one extreme of the side of a triangle, and from some one point of the same side to drawe two right lynes within the triangle which shall containe an angle lesse then the angle contayned vnder the two outward lines.</p>
               <p>As touching the first part.</p>
               <p>Suppose <hi>ABC</hi> to be a rectan<g ref="char:EOLhyphen"/>gle
<figure/>
triangle, whose right angle let be at the point <hi>B.</hi> And in the side <hi>BC</hi> take a point at al aduentures, which let be <hi>D</hi>: and draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>A</hi> to <hi>D</hi> Wherfore the line <hi>AD</hi> is greater then the line <hi>AE</hi> (by the 19. propositio<g ref="char:cmbAbbrStroke">̄</g>) From the line <hi>AD</hi> cut of (by the thirde) a lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equall to the line <hi>AB,</hi> which let be <hi>DE.</hi> And deuide the line <hi>EA</hi> into two equall partes in the point <hi>F</hi> (by the 10. proposition) And draw a line fro<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>F</hi> to <hi>C.</hi> Now forasmuch as <hi>AFC</hi> is a triangle, therfore the lines <hi>AF</hi> and <hi>FC</hi> are greater the<g ref="char:cmbAbbrStroke">̄</g> the line <hi>AC</hi> (by the former proposition): but <hi>AF</hi> is equal to <hi>FE</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the right lines <hi>FE</hi> and <hi>FC</hi> are greater then the line <hi>AC.</hi> And the line <hi>DE</hi> is equall to the line <hi>AB.</hi> Wher<g ref="char:EOLhyphen"/>fore the right lines <hi>FC</hi> and <hi>FD</hi> are greater then the right lines <hi>AB</hi> and <hi>AC</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and they are drawen within the triangle <hi>ABC,</hi> the one from one extreme of the side <hi>BC,</hi> and the oth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r from a point in the same side <hi>BC</hi>: whiche was required to be proued.</p>
               <p>As touchyng the seconde parte. Suppose <hi>ABC</hi> to bee an Isosce<g ref="char:EOLhyphen"/>les triangle, and let the base thereof namelye, <hi>BC</hi> bee greater then either of the equall sydes, and from the lyne <hi>AC</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>utte of a line equall to the lyne <hi>AB</hi> (by the thirde proposition) whiche let bee <hi>BD:</hi> and drawe a line from <hi>A</hi> to <hi>D</hi>: and in the
<pb facs="tcp:7062:61"/>
line <hi>AD</hi> take a point at al aduentures, which
<figure/>
let be <hi>E,</hi> &amp; draw a line from <hi>C</hi> to <hi>E.</hi> Now for<g ref="char:EOLhyphen"/>asmuch as the line <hi>AB,</hi> is equal to the lyne <hi>BD,</hi> therfore (by the fift proposition) the angle <hi>BAD</hi> is equall to the angle <hi>BDA.</hi> And for<g ref="char:EOLhyphen"/>asmuch as in the triangle <hi>EDC</hi> the angle <hi>EDB</hi> is an outwarde angle, therefore (by the 16. proposition) it is greater then the inward and opposite angle <hi>DEC.</hi> Wherefore the angle <hi>BAD</hi> is greater then the angle <hi>DEC.</hi> Wherfore the angle <hi>BAC</hi> is much greater then the an<g ref="char:EOLhyphen"/>gle <hi>DEC</hi>: and the angle BAC is contained of the outward right lines <hi>BA</hi> and AC, and the angle <hi>DEC</hi> is contayned of the inward right lines <hi>DE</hi> and <hi>EC</hi>: which was required to be proued.</p>
               <p>By meanes of this proposi<g ref="char:EOLhyphen"/>tion
<figure/>
also is described that kynd of triangles, which contayneth fouresides. As for example, this figure ABC. For it is co<g ref="char:cmbAbbrStroke">̄</g>tained of fower sides BA, AC, CE, &amp; EB. But it hath onely three an<g ref="char:EOLhyphen"/>gles, one at the point B, an other at the point A, and the third at the point C. VVherefore this present figure ABC is a qua<g ref="char:EOLhyphen"/>drilater triangle: which of olde philosophers hath euer bene counted wonderfull. And here is to be noted, that there is difference be<g ref="char:EOLhyphen"/>twene a three sided figure, and a three angled figure. For not euery figure hauing three angles hath also onely three sides as it is plaine to see in this figure. Likewise it is not all on, a figure to haue lower sides, and fower angles. For a foure sided figure may haue onely thre angles, as in the former figure: and a foure angled figure may haue fiue sides, as in this figure folowing. And so of all other figures.</p>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Probleme. The <hi>22.</hi> Proposition. Of thre right lines, which are equall to thre right lines geue<g ref="char:cmbAbbrStroke">̄</g>, to make a triangle. But it behoueth two of those lines, which two soeuer be taken, to be greater then the third. For that in euery triangle two sides, which two sides soeuer be taken, are
<pb n="31" facs="tcp:7062:61"/>
greater then the side remayning.</head>
               <p>SVppose that the three right
<figure/>
lines geue<g ref="char:cmbAbbrStroke">̄</g> be A, B, C: of which let two of them, which two soeuer be taken, be greater then the third, that is, let the lines A, B, be greater then the line C, and the lines A, C, then the line B, and the lines B, C, then the line A. It is required of three right lines equall to the right lines A, B, C, to make a triangle.<note place="margin">Construction.</note> Take a right line hauing an appointed ende on the side D, and being infinite on the side E. And (by the <hi>3.</hi> propositi<g ref="char:EOLhyphen"/>on) put vnto the line A, an equall line DF, and put vnto the line B, an equall line FG, and vnto y<hi rend="sup">e</hi> line C, an equall line GH. And making the centre F, and the space DF, describe (by the <hi>3.</hi> peticio<g ref="char:cmbAbbrStroke">̄</g>) a circle DKL. Agayne making the centre G, and the space G H, describe (by the same) a circle HKL; and let the point of the intersection of the sayd circles be K, and (by the first peticion) draw a right line from the point K to y<hi rend="sup">e</hi> point F, &amp; an other from the point K to the point G. Then I say, that of thre right lines equall to the lines A, B, C, is made a triangle KFG.<note place="margin">Demonstration.</note> For forasmuch as the point F is the centre of the circle DKL. therefore (by the <hi>15.</hi> definition) the line FD is equall to the line FK. But the line A is equall to the line FD VVherfore (by the first common sentence) the line FK is equall to the line A. Agayne for<g ref="char:EOLhyphen"/>asmuch as the point G, is the centre of the circle LKH, therefore (by the same definition) the line GK is equall to the line GH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But the line C is equall to the line GH: wherefore (by the first common sentence) the line KG is equall to the line C. But the line FG is by supposition equal to the line B wherefore these three right lines GF, FK, and KG, are equall to these three right lines A, B, C. VVherefore of three right lines, that is, KF, FG, and GK, which are equall to the thre right lines geuen, that is to A, B, C, is made a triangle KFG: which was required to be done.</p>
               <div type="section">
                  <head>An other construction, and demonstration after Flussates.</head>
                  <p>Suppose that the three right lines be <hi>A, B, C.</hi> And vnto some one of them, namely,<note place="margin">
                        <hi>An other con<g ref="char:EOLhyphen"/>struction and demonstration after Flussates.</hi>
                     </note> to <hi>C,</hi> put an equall line <hi>DE,</hi> and (by the second proposition) from the point <hi>E,</hi> draw the line <hi>EG,</hi> equall to the line <hi>B</hi>: and (by the same) vnto the point <hi>D</hi> put the line <hi>DH</hi> e<g ref="char:EOLhyphen"/>qual to the line <hi>A.</hi> And making the centre the point <hi>E,</hi> &amp; the space <hi>EG,</hi> describe a cir<g ref="char:EOLhyphen"/>cle <hi>FG</hi>: likewise making the centre the point <hi>D,</hi> and the space <hi>DH,</hi> describe an o<g ref="char:EOLhyphen"/>ther circle <hi>HF</hi>: which circles let cutte the one the other in the point <hi>F.</hi> And draw
<pb facs="tcp:7062:62"/>
these lines <hi>DF</hi> and <hi>EF.</hi> Then I saye
<figure/>
that <hi>DFE</hi> is a triangle described of 3. right lines equal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the right lines <hi>A, B, C.</hi> For forasmuch as the line <hi>DH</hi> is equall to the right line <hi>A,</hi> the line <hi>DF,</hi> shall also be equall to the same right line <hi>A.</hi> (For that the lines <hi>DH</hi> and <hi>DF,</hi> are drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre to the circumference). Likewise foras<g ref="char:EOLhyphen"/>much as <hi>EG</hi> is equall to <hi>EF</hi> (by the 15. definition) and the right line <hi>B</hi> is equall to the same right line <hi>EG</hi>: ther<g ref="char:EOLunhyphen"/>fore the right line <hi>EF</hi> is equall to the right line <hi>B</hi>: but the right line <hi>DE,</hi> was put to be equall to the right line <hi>C.</hi> Wherfore of three right lines <hi>ED, DF,</hi> and <hi>FE,</hi> which are equal to three right lines geuen, <hi>A, B, C,</hi> is described a triangle: which was <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>equired to be done.</p>
                  <p>
                     <note place="margin">
                        <hi>Instances in this Probleme.</hi>
                     </note>In this proposition the aduersary paraduenture will cauill, that the circles shall not cut the one the other (which thing <hi>Euclide</hi> putteth them to do) But now if they cutte not the one the other, either they touch the one the other, or they are dista<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nte the one from the other. First if it be pos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ible let them tooche the one the other: as in the figure here put (the construction whereof answereth to the construc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion of <hi>Euclide</hi>).</p>
                  <p>
                     <note place="margin">
                        <hi>First instance.</hi>
                     </note>And forasmuch as <hi>F</hi> is the cen<g ref="char:EOLhyphen"/>tre
<figure/>
of the circle <hi>DK,</hi> therfore the line <hi>DF</hi> is equal to the line <hi>FN.</hi> And forasmuch as the point <hi>G,</hi> is the centre of the circle <hi>HL,</hi> ther<g ref="char:EOLhyphen"/>fore the line <hi>HG,</hi> is equall to the line <hi>GM.</hi> Wherefore these two lines <hi>DF,</hi> and <hi>GH,</hi> are equall to one line, namely, to <hi>FG.</hi> But they were put to be greater then it: for the lines <hi>DF, FG,</hi> and <hi>GH,</hi> were put to be equall to the lines <hi>A, B, C,</hi> euery two of which are supposed to be greater then the thirde: wherefore they are both greater, and also equall, which is impossible.<note place="margin">
                        <hi>Second instance</hi>
                     </note> Agayne if it be possible, let the circles be di<g ref="char:EOLhyphen"/>stant the one from the other, as are the circles <hi>DK</hi> and <hi>HL.</hi> And forasmuch as <hi>F</hi> is the centre of the circle <hi>DK,</hi> therfore the line <hi>DF</hi> is equal to the line <hi>FN.</hi> And forasmuch as <hi>G</hi> is the centre of the circle <hi>LH,</hi> therefore the line <hi>HG</hi> is equall to the line <hi>GM</hi>: wherefore the whole line <hi>FG</hi> is greater then the two lines <hi>DF,</hi> and <hi>GH,</hi> (for the line <hi>FG,</hi> exce<g ref="char:EOLhyphen"/>deth the lines <hi>DF,</hi> and <hi>GH,</hi> by the line <hi>NM</hi>) but it was suppo<g ref="char:EOLhyphen"/>sed
<pb n="32" facs="tcp:7062:62"/>
that the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>DF</hi> and <hi>HG</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <hi>FG</hi>: a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> also i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> suppos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d that the lines <hi>A</hi> and <hi>C,</hi> were <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> line <hi>B</hi> (for th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <hi>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is put to be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>qu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll to the line <hi>A,</hi> and the line <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> to the line <hi>B</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and the line <hi>HG</hi> to the line <hi>G</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>) Wh<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>fo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> they are both greater and also equall: which is impossible. Wherefore the circles ney<g ref="char:EOLhyphen"/>ther tooch the one the other, nor are distant the one from the other. Wherefore of ne<g ref="char:EOLhyphen"/>cessitie they cut the one the other: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Probleme. The <hi>23.</hi> Proposition. Vpon a right line geuen, and to a point in it geuen: to make a rectiline angle equall to a rectiline angle geuen.</head>
               <p>SVppose y<hi rend="sup">t</hi> the right line ge<g ref="char:EOLhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g>
                  <figure/>
be AB, &amp; let y<hi rend="sup">e</hi> point in it geuen be A. And let also the rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g> be DCH. It is required vpon the right line geuen AB, and to the point in it geuen A, to make a rectil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne angle e<g ref="char:EOLhyphen"/>quall to the rectiline angle geuen DCH.<note place="margin">Construction.</note> Take in either of the lines CD and CH a point at all aduentures, &amp; let the same be D and E. And (by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> first peticion) draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> D to E. And of thre right lines, AF, FG and GA, which let be equall to the three right lines geuen, that is, to CD, DE, and EC, make (by y<hi rend="sup">e</hi> proposition goyng before) a triangle, and let the same be AFG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so that let the line CD be e<g ref="char:EOLhyphen"/>quall to the line AF, and the line CE to the line AG, and moreouer the lyne DE to the line FG.<note place="margin">Demonstration.</note> And forasmuch as these two lines DC and CE are equall to these two lines FA and AG, the one to the other, and the base DE is equall to the base FG: therfore (by the <hi>8.</hi> proposition) the angle DCE is equall to the angle FAG. VVherfore vpon the right line geuen AB, and to the poi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> ge<g ref="char:EOLhyphen"/>uen namely A, is made a rectiline angle FAG, equal to the rectiline angle geuen DCH: which was required to be done.</p>
               <div type="section">
                  <head>An other construction and demons<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ation after Proclus.</head>
                  <p>Suppose that the right line geuen be <hi>AB</hi>:<note place="margin">
                        <hi>An other con<g ref="char:EOLhyphen"/>struction and de<g ref="char:EOLunhyphen"/>monstration af<g ref="char:EOLunhyphen"/>ter Proclus.</hi>
                     </note> &amp; let the point in it geuen be <hi>A,</hi> &amp; let the rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g> be <hi>CDE.</hi> It is required vpo<g ref="char:cmbAbbrStroke">̄</g> the right line geue<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>AB,</hi> &amp; to the point in it geue<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>A,</hi> to make a rectiline angle equal to the rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>CDE.</hi> Drawe a line fro<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>C</hi> to <hi>E.</hi> And produce the line <hi>AB</hi> on either side to the points <hi>F</hi> and <hi>G.</hi> And vn<g ref="char:EOLhyphen"/>to
<pb facs="tcp:7062:63"/>
the line <hi>CD,</hi> put the line <hi>FA</hi> equal, &amp; vnto the line <hi>DE</hi> let the line <hi>AB</hi> be equal, &amp; vn<g ref="char:EOLhyphen"/>to the line <hi>EC</hi> put the line <hi>BG</hi> equal. And making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the point <hi>A,</hi> &amp; the space <hi>AF,</hi> describe a circle KF. And agayne making the centre the point <hi>B</hi> and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> space <hi>BG</hi> de<g ref="char:EOLhyphen"/>scribe
<figure/>
an other circle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>L: which shal of necessitie cut the one the other, as we haue be<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore proued. Let them cut the one the other in the pointes M &amp; N. And draw these right lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s AN, AM, BN, and BM. And forasmuch as FA is equall to AM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and also to AN (by the definition of a circle) but CD is equall to FA, wherfore the lines AM and AN are ech<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall to the line <hi>DC.</hi> Ag<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>yne forasmuch as <hi>BG,</hi> is equall to BM, and to BN, and <hi>BG</hi> is equall to <hi>CE</hi>: therfore either of these lines BM and BN is equall to the line CE. But the line BA is equall to the line DE. Wherfore these two lines BA &amp; AM, are equall to these two lines <hi>D</hi> E and DC, the one to the other, and the base BM is equal to the base <hi>CE.</hi> Wherfore (by the 8. proposition) the angle MAB, is equall to the angle at the point <hi>D.</hi> And by the same reason the angle NAB, is equall to the same angle at the point <hi>D.</hi> Wherfore vpon the right line geuen <hi>AB,</hi> and to the point in it geuen <hi>A,</hi> is de<g ref="char:EOLhyphen"/>scribed a rectiline angle on either side of the line <hi>AB</hi>: namely, on one side the rectiline angle NAB, and on the other side the rectiline angle MAB, either of which is equall to the rectiline angle geuen <hi>CDE</hi>: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>An other construction also, and demonstration after Pelitar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us.</head>
                  <p>
                     <note place="margin">
                        <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> other D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>mons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion af<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>er <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>eli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>arius.</hi>
                     </note>Suppose that the right line geuen be <hi>AB</hi>: and let the point in it geuen be <hi>C,</hi> and let the rectiline angle geuen be <hi>DEF.</hi> It is required vpon the line geuen <hi>AB,</hi> and to the point in it geuen <hi>C,</hi> to describe a rectiline angle equall to the rectiline angle geuen <hi>DEF.</hi> Produce the line <hi>FE</hi> to the point <hi>G</hi>: and from the point <hi>E</hi> erect (by the 11. proposi<g ref="char:EOLhyphen"/>tion) vnto the line <hi>GF</hi> a perpendiculer line <hi>EH,</hi> which if it exactly agree with the lyne <hi>ED,</hi> then was the angle geue<g ref="char:cmbAbbrStroke">̄</g> a right
<figure/>
angle. Wherfore if from the pointe <hi>C</hi> you erecte a perpendiculer line vnto the line <hi>AB,</hi> that shall be done which was req<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ired to be done But if it do not, then from the point <hi>H,</hi> ere<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t vn<g ref="char:EOLhyphen"/>to the line <hi>HE,</hi> a perpendiculer lyne <hi>HD,</hi> whiche being produced shall (by the fift<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> peticion) concurre with the line <hi>FD</hi> being also produced: for the angle <hi>DEH</hi> is lesse then a right angle (when as <hi>GEH</hi> is a right angle). Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore let them concurre in the point <hi>D,</hi> and so is made the triangle <hi>DEH.</hi> After the same maner fro<g ref="char:cmbAbbrStroke">̄</g> the point geuen <hi>C,</hi> erect vnto the line <hi>AB</hi> a perpendiculer line <hi>CK</hi>: which let be equall to
<pb n="33" facs="tcp:7062:63"/>
the perpendicular line <hi>EH</hi> (by the 3. proposition): and from the point <hi>K</hi> erect vnto the line K<hi>C,</hi> a perpendiculer line <hi>KL,</hi> whiche let be equall to the perpendiculer lyne <hi>HD.</hi> And draw a line from <hi>C</hi> to <hi>L.</hi> Then I s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y that the angle <hi>LCB,</hi> is equall to the angle ge<g ref="char:EOLhyphen"/>uen <hi>DEF.</hi> For the two triangles <hi>HED</hi> and <hi>KCL,</hi> are (by the fourth proposition) e<g ref="char:EOLhyphen"/>qual, and equilater the one to the other: and the two angles <hi>LCK</hi> and <hi>DEH</hi> are equal. And the two angles <hi>BCK</hi> and <hi>FEH</hi> are equal, for either of them is a right angle. Wher<g ref="char:EOLhyphen"/>fore (by the 2. common sentence) the whole angle <hi>LCB,</hi> is equall to the whole angle <hi>DEF.</hi> Which was required to be done.</p>
                  <p>And if the perpendiculer line chaunce to fall without the angle geuen, namely, if the angle geuen be an acute angle, the selfe same manner of demonstration will serue: but onely that in stede of the second common sentence, must be vsed the 3. common sentence.</p>
                  <p>Appollonius putteth another construction &amp; demonstration of this propositio<g ref="char:cmbAbbrStroke">̄</g>: which (though the demonstration thereof depende of propositions put in the third booke, yet for that the construction is very good for him that wil redely, and mechanically, without demonstration, describe vpon a line geuen, and to a point in it geuen, a rectiline angle equall to a rectiline angle geuen) I thought not amisse here to place it. And it is thus.</p>
                  <p>Suppose that the rectiline angle geuen be <hi>CDE,</hi>
                     <note place="margin">
                        <hi>An other con<g ref="char:EOLhyphen"/>struction &amp; d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>monstration af<g ref="char:EOLhyphen"/>ter Appollonius.</hi>
                     </note> and let the right line geuen be <hi>AB,</hi> and let the point in it
<figure/>
geuen be <hi>A.</hi> Take in the line <hi>CD,</hi> a po<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> at all ad<g ref="char:EOLhyphen"/>uentures, which let be <hi>F.</hi> And making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the point <hi>D,</hi> and the space <hi>DF,</hi> describe a circle <hi>FG,</hi> cut<g ref="char:EOLunhyphen"/>ting the line <hi>DE,</hi> in the point <hi>G,</hi> and draw a ryght line from <hi>F</hi> to <hi>G.</hi> Likewise from the line <hi>AB,</hi> cut of a line equall to the line <hi>DF,</hi> which let be <hi>AH.</hi> And making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the point <hi>A,</hi> and the space <hi>AH</hi> de<g ref="char:EOLhyphen"/>scribe a circle <hi>HK,</hi> and from the point <hi>H</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> subtend vnto the circumference of the circle a right lyne equall to the right line <hi>FG,</hi> whiche let bee <hi>HK</hi>: and drawe a right lyne from <hi>A</hi> to <hi>K.</hi> Then I say that the angle <hi>HAK,</hi> is equall to the angle <hi>CDE.</hi> The proofe whereof I now omitte for that it dependeth of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>8 and 27 propositions of the third booke.</p>
                  <p>But now, as I sayd, by this you may very redily, and mechanically,<note place="margin">How to do this pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>position redi<g ref="char:EOLhyphen"/>ly and mechani<g ref="char:EOLhyphen"/>cally.</note> without demon<g ref="char:EOLunhyphen"/>stration, vpon a line geuen, and to a point in it geuen describe a rectiline angle equall to a rectiline angle geuen. For
<figure/>
in the rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g>, you neede onely to marke the two pointes, where the circumfe<g ref="char:EOLhyphen"/>rence of the circle cutteth the lines contayning the angle ge<g ref="char:EOLhyphen"/>uen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as the points <hi>F</hi> and <hi>G</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and likewise to marke in the line ge<g ref="char:EOLhyphen"/>uen as in <hi>AB,</hi> the point <hi>H,</hi> &amp; so making the centre the point <hi>A,</hi> according to the space <hi>AH</hi> (which is put to be equal to <hi>FD</hi>) describe a pe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e of a circumference on that side that you wil haue the angle to be, as for example the circumference <hi>HK,</hi> and opening your compasse to the wideth from
<pb facs="tcp:7062:64"/>
the point <hi>F,</hi> to the point <hi>G,</hi> set one foote thereof fixed in the point <hi>H,</hi> and marke the point where the other foote cutteth the sayde circumference, which point let be <hi>K.</hi> And from that point to the point. <hi>A,</hi> draw a right line: and so shall you haue described at the point <hi>A,</hi> an angle equal to the angle at the point <hi>D.</hi> As in the figures in the end of the other side put.</p>
                  <p>
                     <note place="margin">
                        <hi>Oenopides the first inuenter of this proposition.</hi>
                     </note>
                     <hi>Oenopides</hi> was the first inuenter of this proposition as witnesseth Eudemius.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>15.</hi> Theoreme The <hi>24.</hi> Proposition. If two triangles haue two sides of the one equall to two sides of the other, ech to his correspondent side, and if the angle co<g ref="char:cmbAbbrStroke">̄</g>tai<g ref="char:EOLhyphen"/>ned vnder the equall sides of the one, be greater then the an<g ref="char:EOLhyphen"/>gle contayned vnder the equall sides of the other: the base also of the same, shalbe greater then the base of the other.</head>
               <p>
                  <hi>SVppose that there be two triangles ABC, and DEF, hauing two sides of the one, that is, AB, and AC, e<g ref="char:EOLhyphen"/>quall to two sides of the other, that is, to DE, and DF, ech to his correspondent side: that is, the side AB, to the side DE, and the side AC to the side DF: and suppose that the angle BAC be greater then the angle EDF. Then I saye that the base BC, is greater then the base EF. For forasmuch as the angle BAC is greater then the angle EDF,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>make (by the</hi> 23. <hi>proposition) vp<g ref="char:EOLhyphen"/>on
<figure/>
the right line DE, and to the point in it geue<g ref="char:cmbAbbrStroke">̄</g> D, an angle EDG equall to the angle geuen BAC. And to one of these lines, that is, either to AC, or DF, put an equall line DG. And (by the first peticio<g ref="char:cmbAbbrStroke">̄</g>) draw a right line from the point G, to the point E, and an other from the point F,</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>to the point G. And forasmuch as the line AB is equall to the line DE, and the line AC to the line DG, the one to y<hi rend="sup">e</hi> other, and the angle BAC is (by construction) equall to the angle EDG, therefore (by the</hi> 4. <hi>proposition) the base BC, is equall to the base EG. Agayne for as much as the line DG is equall to the line DF, ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> (by the</hi> 5. <hi>proposition) the angle DGF, is equall to the angle DFG. VVhere<g ref="char:EOLhyphen"/>fore the angle DFG is greater then the angle EGF. VVherefore the angle EFG is much greater then the angle EGF. And forasmuch as EFG is a trian<g ref="char:EOLhyphen"/>gle, hauing the angle EFG greater then the angle EGF, and (by the</hi> 18. <hi>pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>position) vnder the greater angle is subtended the greater side, therefore the side EG is greater then the side EF. But the side EG is equall to the side BC: wherefore the side BC is greater then the side EF. If therefore two triangles
<pb n="34" facs="tcp:7062:64"/>
haue two sides of the one equall to two sides of the other, eche to his correspon<g ref="char:EOLhyphen"/>dent side, and if the angle contayned vnder the equall sides of the one, be grea<g ref="char:EOLhyphen"/>ter then the angle contayned vnder the equall sides of the other: the base also of the same shalbe greater then the base of the other: which was required to be proued.</hi>
               </p>
               <p>In this Theoreme may be three cases.<note place="margin">
                     <hi>Three cases in this pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>position.</hi>
                  </note> For the angle <hi>EDG,</hi> being put equall to the angle <hi>BAC,</hi> and the line <hi>DG,</hi> being put equall to the line <hi>AC,</hi> and a line being drawen from <hi>E</hi> to <hi>G,</hi> the line <hi>EG</hi> shall either fall aboue the line <hi>GF,</hi> or vpon it, or vnder it. <hi>Eu<g ref="char:EOLhyphen"/>clides</hi> demonstration serueth,<note place="margin">
                     <hi>The first cas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> 2.</note> when the line <hi>GE</hi> falleth aboue the line <hi>GF,</hi> as we haue, before manifestly seene.</p>
               <p>But if it fall vpon it,<note place="margin">
                     <hi>Second case.</hi>
                  </note> as in this figure here put.
<figure/>
Then forasmuch as the two lines <hi>AB</hi> and <hi>AC,</hi> are equal to the two lines <hi>DE</hi> and <hi>DG,</hi> the one to the other, and they contayne equall angles by con<g ref="char:EOLhyphen"/>struction: therefore (by the 4. proposition) the base <hi>BC,</hi> is equall to the base <hi>EG,</hi> but the base <hi>EG,</hi> is greater then the base <hi>EF</hi>: wherfore also the base <hi>BC,</hi> is greater then the base <hi>EF</hi>: which was required to be proued.</p>
               <p>But now let the line <hi>EG,</hi>
                  <note place="margin">
                     <hi>Third case.</hi>
                  </note> fall vnder the line <hi>E
<figure/>
F,</hi> as in the figure here put. And forasmuch as these two lines <hi>AB,</hi> and <hi>AC,</hi> are equall to these two lines <hi>DE</hi> and <hi>DG,</hi> the one to the other, and they contayne equall angles, therefore (by the 4. pro<g ref="char:EOLhyphen"/>position) the base <hi>BC,</hi> is equal to the base <hi>EG.</hi> And forasmuch as within the triangle <hi>DEG,</hi> the two linnes <hi>DF</hi> and <hi>FE,</hi> are set vpon the side <hi>DE</hi>: ther<g ref="char:EOLhyphen"/>fore (by the 21. proposition) the lines <hi>DF</hi> and <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> are lesse then the outward lines <hi>DG</hi> and <hi>GE</hi>: but the line <hi>DG</hi> is equal to the line <hi>DF.</hi> Wherfore the line <hi>GE</hi> is greater then the line <hi>FE.</hi> But <hi>GE</hi> is e<g ref="char:EOLhyphen"/>quall to <hi>BC.</hi> Wherefore the line <hi>BC</hi> is greater the the line <hi>EF.</hi> Which was required to be proued.</p>
               <p>This third case may also another way be demonstrated.<note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration of the third case.</hi>
                  </note> Produce the lines <hi>DF</hi> and <hi>DG,</hi> which are equal, vnto the points
<figure/>
                  <hi>K</hi> and <hi>H</hi>: and draw a line from <hi>F</hi> to <hi>G</hi>: wherefore (by the second part of the fifth proposition) the angles <hi>KFG</hi> and <hi>FGH,</hi> which are vnder the base <hi>FG</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are equall: therefore the angle <hi>EFG</hi> is greater then the angle <hi>FG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> Wherfore (by the 18 proposition) the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>EG</hi> is greater then the side <hi>EF</hi>: but the base <hi>BC</hi> is equall vnto the base <hi>EG</hi>: Where<g ref="char:EOLhyphen"/>fore the base <hi>BC,</hi> is greater then the base <hi>EF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Which was required to be proued.</p>
               <p>It may peraduenture seme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that Euclide should here in this proposition haue proued, that not onely the bases of the triangles are vnequall, but also that the a<g ref="char:EOLhyphen"/>reas of the same are vnequall: for so in the fourth proposition, after he had pro<g ref="char:EOLhyphen"/>ued
<pb facs="tcp:7062:65"/>
the base to be equall, he proued also the areas to be equall. But hereto may be answered,<note place="margin">
                     <hi>Why Euclide &amp;c.</hi>
                  </note> that in equall angles and bases, and vnequall angles and bases, the consideration is not like. For the angles and bases being equall, the triangles al<g ref="char:EOLhyphen"/>so shall of necessitie be equall, but the angles and bases being vnequall, the areas shall not of necessitie be equall. For the triangles may both be equall and vne<g ref="char:EOLhyphen"/>quall: and that may be the greater, whiche hathe the greater angle, and the grea<g ref="char:EOLhyphen"/>ter base, and it may also be the lesse. And for that cause Euclide made no menci<g ref="char:EOLhyphen"/>on of the comparison of the triangles. VVhereof this also mought be a cause, for that to the demonstration thereof are required certayne Propositions concer<g ref="char:EOLhyphen"/>ning parallel lines, which we are not as yet come vnto.<note place="margin">
                     <hi>After the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> &amp;c.</hi>
                  </note> Howbeit after the 37<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proposition of his booke you shal find the comparison of the areas of triangles, which haue their sides equall, and their bases and angles at the toppe vnequall.</p>
            </div>
            <div type="part">
               <head>The <hi>16.</hi> Theoreme. The <hi>25.</hi> Proposition. If two triangles haue two sides of the one equall to two sydes of the other, eche to his correspondent syde, and if the base of the one be greater then the base of the other: the angle also of the same co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the equall right lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shall be grea<g ref="char:EOLhyphen"/>ter then the angle of the other.</head>
               <p>
                  <hi>SVppose that there be two triangles, A, B, C,
<figure/>
and DEF, hauing two sides of tb'one, that is, AB, and AC, equall to two sides of the o<g ref="char:EOLhyphen"/>ther, that is to DE, and DF, ech to his correspon<g ref="char:EOLhyphen"/>dent side, namely, the side AB to the side DF, and the side AC to the syde DF. But let the base BC be greater then the base EF. The<g ref="char:cmbAbbrStroke">̄</g> I say, thay the an<g ref="char:EOLhyphen"/>gle BAC is greater then the angle EDF. For if not,</hi>
                  <note place="margin">
                     <hi>Demonstration leading to an absurdi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y.</hi>
                  </note> 
                  <hi>then is it either equall vnto it, or lesse then it. But the angle BAC is not equall to the angle EDF: for if it were equall, the base also BC should (by the</hi> 4. <hi>proposition) be equal to the base EF: but by supposition it is not. VVherfore the angle BAC is not e<g ref="char:EOLhyphen"/>quall to the angle EDF. Neither also is the angle BAC lesse then the angle EDF: for then should the base BC be lesse the<g ref="char:cmbAbbrStroke">̄</g> the base EF (by the former pro<g ref="char:EOLhyphen"/>position) But by supposition it is not. VVherfore y<hi rend="sup">•</hi> angle BAC is not lesse the<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> angle EDF. And it is already proued, that it is not equall vnto it: wherfore y<hi rend="sup">e</hi> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLunhyphen"/>gle BAC is greater then the angle EDF. If therfore two triangles haue two sides of the one equall to two sides of the other, eche to his correspondent side, &amp; if the base of the one be greater then the base of the other, the angle also of the same contayned vnder y<hi rend="sup">e</hi> equal right lines shall be greater the<g ref="char:cmbAbbrStroke">̄</g> the angle of the o<g ref="char:EOLhyphen"/>ther: which was required to be proued.</hi>
               </p>
               <p>
                  <pb n="35" facs="tcp:7062:65"/>This proposition is plaine opposite to the eight, &amp; is the co<g ref="char:cmbAbbrStroke">̄</g>uerse of the foure and twenty which went before.<note place="margin">
                     <hi>All converses are commonly indirectly de<g ref="char:EOLhyphen"/>monstrated.</hi>
                  </note> And it is proued (as commonly all conuerses are) by a reason leading to an absurditie. But it may after Menelaus Alexandrinus be demonstrated directly,<note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration after Menelaus Alexandrinus.</hi>
                  </note> after this maner.</p>
               <p>Suppose that there be two triangles <hi>ABC</hi> &amp; <hi>DEF</hi>: hauing the two sides <hi>AB</hi> and <hi>AC</hi> equal to the two sides <hi>DE</hi> and <hi>DF,</hi> the one to the other: and le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the base <hi>BC</hi> be greater then the base <hi>EF.</hi> Then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> say that the angle at the point <hi>A,</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the an<g ref="char:EOLhyphen"/>gle at the point <hi>D.</hi> For from the base <hi>BC</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of (by the thirde) a line eqnall to the base <hi>EF,</hi> and let the same be <hi>BG.</hi> And vpon the line <hi>GB</hi> and to the point <hi>B</hi> put (by the 23, proposition) an angle equal to the angle <hi>DEF</hi>: which let be <hi>GBH</hi>: and let the line <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>H</hi> be equall to the line <hi>DE.</hi> And drawe a lyne from <hi>H</hi> to <hi>G,</hi> and produce it beyond the point <hi>G:</hi> whiche being produce shall fal either vppon the angle <hi>A,</hi>
                  <note place="margin">
                     <hi>Diuers cases in this demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> or vpon the line <hi>AB,</hi> or vpon the line <hi>AC.</hi> First let it fall vpon the angle <hi>A.</hi>
                  <note place="margin">
                     <hi>First case.</hi>
                  </note> And forasmuch as these two lines <hi>BG</hi> and <hi>BH</hi> are equall to these two lines <hi>EF</hi> and <hi>ED,</hi> the one to the other, and they contayne e<g ref="char:EOLhyphen"/>quall angles (by construction) namely, the angles <hi>GBH</hi> and <hi>DEF</hi>: therfore (by the 4. proposition) the base <hi>GH</hi> is equall to the <hi>DF,</hi> and the angle <hi>BHG</hi> to the angle <hi>ED</hi>F. Agayne forasmuch as the line <hi>BH</hi> is equall to the line <hi>BA</hi> (for the line <hi>AB</hi> is supposed to be equal to the line <hi>DE,</hi> vnto which line the line <hi>BH</hi> is put equal) therfore (by the 5. proposition) the angle <hi>BHA</hi> is equall to the angle <hi>BAH</hi>: wherfore also the angle EDF is equal to the angle <hi>BAH.</hi> But the angle <hi>BAC</hi> is greater then the angle <hi>BAH</hi>: wherfore also the angle <hi>BAC</hi> is greater then the angle EDF.<note place="margin">
                     <hi>Second case.</hi>
                  </note>
               </p>
               <p>But now let it fall vpon the line <hi>AB</hi> in the
<figure/>
point K, and drawe a line from <hi>A</hi> to <hi>H.</hi> And for<g ref="char:EOLhyphen"/>asmuch as these two lines <hi>BG</hi> and BH are equall to these two lines EF and ED, the one to the other, &amp; they containe equal angles (by constuction) name<g ref="char:EOLhyphen"/>ly, the angles GBH and DEF: therfore (by the 4. proposition) the base GH is equall to the base DF, and the angle BHG to the angle EDF. Agayne for<g ref="char:EOLhyphen"/>asmuch as in the triangle BAH, the side BA is equal to the side BH, therfore (by the 5. proposition) the angle BAH is equal to the angle BHA. But the an<g ref="char:EOLhyphen"/>gle BHA is greater then the angle BHG. wherfore also the angle BAH is greater then the angle BHG. Wherfore the angle BAC is much greater then the angle BHG. But it is proued that the angle BHG is equall to the angle at the point D. Wherefore the angle BAC is greater then the angle at the point<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> D: Which was required to be proued.</p>
               <p>But now suppose that the line HG beyng produced doo fall vppon the line <hi>AC,</hi>
                  <note place="margin">
                     <hi>Third case.</hi>
                  </note> namely, in the point K. And agayne draw also a line from <hi>A</hi> to <hi>H.</hi> And forasmuch as <hi>BG</hi> is equall to <hi>EF,</hi> and <hi>BH,</hi> to <hi>ED,</hi> therefore these two lines <hi>BG</hi> and <hi>BH</hi> are equall to these two lines <hi>EF,</hi> and <hi>ED,</hi> the one to the other, and (by construction) they contayne equall angles, namely, the angles GBH and <hi>FED.</hi> Wherfore (by the fourth propositio<g ref="char:cmbAbbrStroke">̄</g>)
<pb facs="tcp:7062:66"/>
the base GH is equal to the base <hi>DF:</hi> &amp; th'an<g ref="char:EOLunhyphen"/>gle
<figure/>
BHG is equall to th'angle <hi>EDF.</hi> And for<g ref="char:EOLhyphen"/>asmuch as GH is equall to <hi>DF,</hi> and <hi>DF</hi> is e<g ref="char:EOLhyphen"/>quall to <hi>AC</hi>: therfore GH also is equall to <hi>AC.</hi> Wherfore HK is greater then <hi>AC,</hi> where<g ref="char:EOLhyphen"/>fore HK is much greater then AK. Wherfore (by the 18. proposition) the angle KAH is greater then the angle KHA. Agayne foras<g ref="char:EOLhyphen"/>much as <hi>BH</hi> is equal to <hi>AB</hi> (for BH is put e<g ref="char:EOLhyphen"/>quall to <hi>DE,</hi> which is by supposition equal to <hi>AB</hi>) therfore (by the 5. proposition) the an<g ref="char:EOLhyphen"/>gle BHA is equall to the angle BAH. Wher<g ref="char:EOLhyphen"/>fore the whole angle BHK is lesse then the whole angle BAK. But it hath bene pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ed, that the angle BHK is equall to the angle at the point D, wherfore the angle <hi>BAC</hi> is grea<g ref="char:EOLhyphen"/>ter then the angle at the point <hi>D,</hi> which was required to be proued.</p>
               <p>Hero Mechanicus also demonstrateth it an other way, and that by a direct demonstration.</p>
               <p>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Hero Me<g ref="char:EOLhyphen"/>chanicus.</hi>
                  </note>Suppose that there be two triangles <hi>ABC</hi>
                  <figure/>
and <hi>DEF,</hi> hauyng the two sides <hi>AB,</hi> and <hi>AC,</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall to the two sides <hi>DE,</hi> &amp; <hi>DF,</hi> the one to the other, and let the base <hi>BC,</hi> be greater then the base <hi>EF.</hi> Then I say, that the angle at the point <hi>A,</hi> is greater then the angle at the point <hi>D.</hi> For forasmuch as <hi>BC,</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>EF,</hi> produce <hi>EF</hi> to the point <hi>G,</hi> and put the line <hi>EG,</hi> equall to the line <hi>BC.</hi> Likewise pro<g ref="char:EOLhyphen"/>duce the line <hi>DE</hi> to the point <hi>H,</hi> and put the line <hi>DH,</hi> equall to the line <hi>DF.</hi> Wherefore making the centre the point <hi>D,</hi> and the space <hi>DF,</hi> describe a circle, and it shall passe also by the point <hi>H.</hi> Let the same circle be <hi>FKH.</hi> And forasmuch as <hi>AC</hi> and <hi>AB</hi> are greater the <hi>BC</hi> (by the 20. propositio<g ref="char:cmbAbbrStroke">̄</g>) &amp; the lines <hi>AB</hi> &amp; <hi>AC,</hi> ar equal to the line <hi>EH,</hi> &amp; the line <hi>BC</hi> is equal to the line <hi>EG.</hi> Therefore the line <hi>EH</hi> is grea<g ref="char:EOLunhyphen"/>ter then the line <hi>EG.</hi> Wherefore making the centre the point <hi>E</hi> and the space <hi>EG</hi> describe a circle and it shall cut the line <hi>EH.</hi> Let the same circle be <hi>GK</hi>: and from the common sectio<g ref="char:cmbAbbrStroke">̄</g> of the circles, which let be the point <hi>K,</hi> draw these right lines <hi>KD</hi> and <hi>K</hi>E. And for<g ref="char:EOLhyphen"/>asmuch as the point <hi>D</hi> is the centre of the cir<g ref="char:EOLhyphen"/>cle <hi>HKF,</hi> therefore (by the 15. def<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nition) the line <hi>DK,</hi> is equall to the line <hi>DH,</hi> that is vnto the line <hi>AC.</hi> Agayne forasmuch as E is the centre of the circle <hi>GK,</hi> therefore the line E<hi>K</hi> is equal to the line EG, that is, to the line <hi>BC.</hi> And forasmuch as these two lines <hi>AB</hi> and <hi>AC,</hi> are equall to these two lines <hi>D</hi>E and <hi>DK,</hi> and the base <hi>BC</hi> is equal to the base E<hi>K</hi> (for E<hi>K</hi> is equall to E<hi>G</hi> (by the 15. definition) &amp; E<hi>G</hi> is put to be equal to <hi>BC</hi>). Wherefore (by the 4. proposition) the angle <hi>BAC</hi> is equal to the angle ED<hi>K.</hi> But the angle E<hi>DK</hi> is greater then the angle E<hi>DF</hi>: wherefore also the angle <hi>BAC,</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> greater then the angle <hi>EDF</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb n="36" facs="tcp:7062:66"/>
               <head>The <hi>17.</hi> Theoreme. The <hi>26.</hi> Proposition. If two triangles haue two angles of the one equall to two an<g ref="char:EOLhyphen"/>gles of the other, ech to his correspondent angle, and haue also one side of the one equall to one side of the other, either that side which lieth betwene the equall angles, or that which is subtended vnder one of the equall angles: the other sides also of the one, shalbe equall to the other sides of the other, eche to his correspondent side, and the other angle of the one shalbe equall to the other angle of the other.</head>
               <p>
                  <hi>SVppose that there be two triangles AB
<figure/>
C, and DEF, hauing two angles of the one, that is, the angles ABC, and BCA, equall to two angles of the other, that is, to the angles DEF, and EFD, ech to his correspo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLunhyphen"/>dent angle, that is, the angle ABC, to the angle DEF, and the angle BCA to the angle EFD, and one side of the one equall to one side of y<hi rend="sup">e</hi> other, first that side which lieth betwene the equall angles, that is, the side BC, to the EF. The<g ref="char:cmbAbbrStroke">̄</g> I say that the other sides also of the one shalbe equall to the other sides of the other, ech to his corre<g ref="char:EOLhyphen"/>spondent side, that is, that side AB, to the side DE, and the side AC, to the side DF, and the other angle of the one, to the other angle of the other, that is, the angle BAC to the angle EDF. For if the side AB be not equall to the side DE, the one of them is greater. Let the syde AB be greater: and (by the</hi> 3. <hi>propo<g ref="char:EOLunhyphen"/>sition) vnto the line DE, put an equall line GB, and draw a right line from the point G, to the point C. Now forasmuch as the line GB,</hi>
                  <note place="margin">
                     <hi>Demonstration leading to an absurditis.</hi>
                  </note> 
                  <hi>is equall to the line DE, and the line BC to the line EF, therefore these two lines GB and BC, are equall to these two lines DE and EF, the one to the other, and the angle GBC is (by supposition) equall to the angle DEF. VVherefore (by the</hi> 4. <hi>proposy<g ref="char:EOLhyphen"/>tion) the base GC is equall to the base DF, and the triangle GCB is equall to the triangle DEF, and the angles remay<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ing are equall to the angles remay<g ref="char:EOLhyphen"/>ning vnder which are subtended equall sydes. VVherefore the angle GCB is e<g ref="char:EOLhyphen"/>quall to the angle DFE. But the angle DFE is supposed to be equall to the an<g ref="char:EOLunhyphen"/>gle BCA. VVherefore (by the first common sentence) the angle BCG is equal to the angle BCA, the lesse angle to the greater: which is impossible. VVhere<g ref="char:EOLhyphen"/>fore the line AB is not vnequall to the line DE. VVherefore it is equall And the the line BC is equall to the line EF: now therefore there are two sydes AB and
<pb facs="tcp:7062:67"/>
BC equall to two sydes DE and EF, the one to the other, and the angle ABC, is equall to the angle DEF. VVherefore (by the</hi> 4. <hi>proposition) the base AC is equall to the base DF, and the angle remayning BAC is equall to the angle re<g ref="char:EOLunhyphen"/>mayning EDF.</hi>
               </p>
               <p>
                  <hi>Agayne suppose that the sydes subtending the equall angles be equall the one to the other, let the syde I say AB be equall to the syde DE. Then agayne I say, that the other sydes of the one are equall to the other sydes of the other, ech to his correspondent syde, that is the syde AC to the syde DF, and the syde BC to the syde EF: and moreouer the angle remayning, namely, BAC, is equall to the angle remayning, that is, to the angle EDF. For if the syde BC be not equall to the syde EF, the one of them is greater: let the syde BC, if it be possi<g ref="char:EOLhyphen"/>ble, be greater. And (by the third proposytion) vnto the line EF, put an equall line BH, and drawe a right line from the point A to the point H. And forasmuch as the line BH is equall to the line EF, and the line AB to the line DE, therefore these two sydes AB and BH, are equall to these two sydes DE and EF, the one to the other, and they containe
<figure/>
equall angles. VVherefore (by the</hi> 4. <hi>proposition) the base AH is equall to the base DF, and the triangle ABH, is equall to the triangle DEF, and the an<g ref="char:EOLhyphen"/>gles remayning are equall to the angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> remayning, vnder which ar subte<g ref="char:cmbAbbrStroke">̄</g>ded equal sydes. VVherfore the angle BHA is equall to the angle EFD. But the angle EFD is equall to the angle BCA. VVhere<g ref="char:EOLhyphen"/>fore the angle BHA is equal to the angle BCA. VVherefore the outward angle of y<hi rend="sup">e</hi> triangle AHC, namely, the angle BHA, is equall to the inward and opposite angle, namely, to the angle HCA, which (by the</hi> 16 <hi>proposition) is impossible. VVherfore the syde EF is not vnequall to the syde BC, wherefore it is equall. And the syde AB is equall to y<hi rend="sup">e</hi> syde DE: wherefore these two sydes AB and BC, are equall to these two sydes DE and EF, the one to the other, and they contayne equall angles: VVherfore (by the</hi> 4. <hi>proposition) the base AC is equall to the base DF: and the triangle ABC, is equall to the triangle DEF, and the angle remayning, name<g ref="char:EOLhyphen"/>ly, the angle BAC is equall to the angle remayning, that is, to the angle EDF. If therefore two triangles haue two angles of the one equall to two angles of the other, ech to his correspondent angle, and haue also one syde of the one equall to o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e syde of the other, either that syde which lieth betwene the equall angles, or that which is subtended vnder one of the equall angles: the other sydes also of the one shalbe equall to the other sydes of the other, eche to his correspondent side, and the other angle of the one shalbe equall to the other angle of the other: which was required to be proued.</hi>
               </p>
               <p>VVhereas in this proposition it is sayde, that triangles are equall, which hauing two angles of the one equall to two angles of the other, the one to the o<g ref="char:EOLhyphen"/>ther,
<pb n="37" facs="tcp:7062:67"/>
haue also one side of the one equall to one side of the other, either that side which lieth betwene the equall angles, or that side which subtendeth one of the equall angles: this is to be noted that without that caution touching the equall side, the proposition shall not alwayes be true. As for example.</p>
               <p>Suppose that there be a rectangle triangle <hi>ABC,</hi> whose right angle let be at the point <hi>B,</hi> &amp; let the side <hi>BC</hi> be greater the<g ref="char:cmbAbbrStroke">̄</g> the side <hi>BA</hi>: and produce the line <hi>AB,</hi> fro<g ref="char:cmbAbbrStroke">̄</g> the point <hi>B</hi> to the point <hi>D.</hi> And vpo<g ref="char:cmbAbbrStroke">̄</g> the right line
<figure/>
                  <hi>BC</hi> &amp; to the point in it <hi>C,</hi> make vnto the angle <hi>BAC</hi> an equal angle (by the 23. proposition), which let be <hi>BCD,</hi> &amp; let the lines <hi>BD</hi> &amp; <hi>CD,</hi> be<g ref="char:EOLunhyphen"/>ing produced co<g ref="char:cmbAbbrStroke">̄</g>curre in the point <hi>D.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> there are two triangles <hi>ABC,</hi> and <hi>BCD,</hi> which haue two angles of the one equall to two an<g ref="char:EOLhyphen"/>gles of the other, the one to the other, namely, the angle <hi>ABC</hi> to the angle <hi>DBC</hi> (for they are both right angles), &amp; the angle <hi>BAC,</hi> to the angle <hi>BCD</hi> (by construction) and haue al<g ref="char:EOLhyphen"/>so one side of the one equall to one side of the other, namely, the side <hi>BC,</hi> which is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon to them both. And yet notwithstanding the triangles are not equall: for the tri<g ref="char:EOLhyphen"/>angle <hi>BDC,</hi> is greater then the triangle <hi>ABC.</hi> For vpon the right line <hi>BC,</hi> and to the point in it <hi>C,</hi> describe an angle equall to the angle <hi>ACB</hi>: which let be <hi>FCB</hi> (by the 23. proposition). And forasmuch as the side <hi>BC</hi> was supposed to be greater then the side. <hi>AB,</hi> therefore (by the 18. proposition) the angle <hi>BAC</hi> is greater then the angle <hi>BCA,</hi> wherefore also the angle <hi>BCD</hi> is greater then the angle <hi>BCF.</hi> Wherefore the tri<g ref="char:EOLhyphen"/>angle <hi>BCD</hi> is greater then the triangle <hi>BCF.</hi> Agayne forasmuch as there are two tri<g ref="char:EOLhyphen"/>angles <hi>ABC</hi> and <hi>BCF</hi>; hauing two angles of the one equal to two angles of the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the one to the other, namely, the angle <hi>ABC</hi> to the angle <hi>FBC</hi> (for they are both right angles) and the angle <hi>ACB</hi> to the angle <hi>FCB</hi> (by construction), and one side of the one is equall to one side of the other, namely, that side which lieth betwene the equall an<g ref="char:EOLhyphen"/>gles, that is, the side <hi>BC</hi> which is common to both triangles. Wherefore (by this pro<g ref="char:EOLhyphen"/>position) the triangles <hi>ABC</hi> and F<hi>BC</hi> are equal. But the triangle <hi>DBC</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the triangle F<hi>BC.</hi> Wherefore also the triangle <hi>DBC</hi> is greater then the triangle <hi>ABC.</hi> Wherefore the triangles <hi>ABC</hi> and <hi>DBC,</hi> are not equall: notwithstanding they haue two angles of the one equall to two angles of the other, the one to the other, and one side of the one equall to one side of the other.</p>
               <p>The reason wherof is, for that the equal side in one triangle, subte<g ref="char:cmbAbbrStroke">̄</g>deth one of the equall angles, and in the other lieth betwene the equal angles. So that you see that it is of necessitie that the equall side do in both triangles, either subtend one of the equall angles, or lie betwene the equall angles.</p>
               <p>Of this proposition was Thales Milesius the inuentor,<note place="margin">
                     <hi>Thales Milesius the inuent<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r of this proposition.</hi>
                  </note> as witnesseth Eude<g ref="char:EOLhyphen"/>mus in his booke of Geometricall enarrations.</p>
            </div>
            <div type="part">
               <head>The <hi>18.</hi> Theoreme. The <hi>27.</hi> Proposition. If a right line falling vpon two right lines, do make the alter<g ref="char:EOLhyphen"/>nate angles equall the one to the other: those two right lines are parallels the one to the other.</head>
               <p>
                  <pb facs="tcp:7062:68"/>
                  <hi>SVppose that the right line EF falling vppon these two right lines AB and CD, do make the alternate angles, namely, the angles AEF &amp; EFD equall the one to the other. Then I say that AB is a parallel line to CD. For if not, then these lines produced shall
<figure/>
mete together, either on the side of B and D, or on the syde of A &amp; C.</hi>
                  <note place="margin">
                     <hi>Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>stration <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>eading to an absurditie.</hi>
                  </note> 
                  <hi>Let them be produced therfore, and let them mete if it be possible on the syde of B and D, in the point G. VVherfore in the triangle GEF, the outward angle AEF is equal to the in<g ref="char:EOLhyphen"/>ward and opposite angle EFG, which (by the</hi> 16. <hi>proposition) is impossible. VVherfore the lines AB and CD beyng produced on the side of B and D, shall not meete. In like sorte also may it be proued that they shall not mete on the syde of A and C. But lines whiche being produced on no syde meete together, are parrallell lines (by the last definition:) wherfore AB is a parrallel line to CD. If therfore a right line falling vpon two right lines, do make the alternate angles equall the one to the other: those two right lines are parrallels the one to the o<g ref="char:EOLhyphen"/>ther: which was required to be demonstrated.</hi>
               </p>
               <p>
                  <note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                        <desc>〈1 paragraph〉</desc>
                     </gap>
                  </note>This worde alternate is of Euclide in diuers places diuersly taken: somtimes for a kind of situation in place, and somtime for an order in proportion, in which signification he vseth it in the v. booke, and in his bokes of numbers. And in the first signification he vseth it here in this place, and generally in all hys other bokes,<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                        <desc>〈1 paragraph〉</desc>
                     </gap>
                  </note> h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uing to do with lines &amp; figures. And those two angles he calleth alter<g ref="char:EOLhyphen"/>nate, which beyng both contayned within two parallel or equidistant lynes are neither angles in order, nor are on the one and selfe same side, but are seperated the one from the other by the line which falleth on the two lines: the one angle beyng aboue, and the other beneath.</p>
            </div>
            <div type="part">
               <head>The <hi>19.</hi> Theoreme. The <hi>28.</hi> Proposition. If a right line falling vpon two right lines, make the outward angle equall to the inward and opposite angle on one and the same syde, or the inwarde angles on one and the same syde, e<g ref="char:EOLhyphen"/>quall to two right angles: those two right lines shall be paral<g ref="char:EOLhyphen"/>lels the one to the other.</head>
               <p>
                  <hi>SVppose that the right line EF, fallyng vppon these two right lines AB and CD, do make the outward angle EGB equall to the inward and opposite angle GHD, or do make the inward angles on one and
<pb n="38" facs="tcp:7062:68"/>
the same side, that is, the angles BGH and GHD
<figure/>
equall to two right angles. Then I say that the lyne AB is a parallel line to the lyne CD For forasmuch as the angle EGB is (by supposition) equall to the an<g ref="char:EOLhyphen"/>gle GHD,</hi>
                  <note place="margin">
                     <hi>Demonstration</hi>
                  </note> 
                  <hi>and the angle EGB is (by the</hi> 15. <hi>pro<g ref="char:EOLhyphen"/>position) equall to the angle AGH: therfore the an<g ref="char:EOLhyphen"/>gle AGH is equall to the angle GHD: and they are alternate angles. VVherfore (by the</hi> 27. <hi>proposi<g ref="char:EOLhyphen"/>tion) AB is a parallel line to CD.</hi>
               </p>
               <p>
                  <hi>Agayne forasmuch as the angles BGH and GHD are (by supposition) e<g ref="char:EOLhyphen"/>quall to two right angles, &amp; (by the</hi> 13. <hi>proposition) the angles AGH and BGH, are also equall to two right angles, wherefore the angles AGH and BGH, are equall to the angles BGH and GHD: take away the angle BGH which is common to them both VVherfore the angle remainyng, namely, AGH is equall to the angle remayning, namely, to GHD. And they are alternate an<g ref="char:EOLhyphen"/>gles. VVherfore (by the former proposition) AB is a parallell line to CD. If therfore a right line fallyng vpon two right lines, do make the outward angle e<g ref="char:EOLhyphen"/>quall to the inward and opposite angle on one and the same side, or the inwarde angles on one and the same side, equall to two right angles, those two right lines shall be parallels the one to the other: which was required to be proued.</hi>
               </p>
               <p>Ptolomeus demonstrateth the second part of this proposition, namely, that the two inward angles on one and the same side being equall, the right lines are parellels, after this manner.<note place="margin">
                     <hi>An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration of the second part of this proposition after Ptolome<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>
                     </hi>
                  </note>
               </p>
               <p>Suppose that there be two right lines <hi>AB</hi> and <hi>CD,</hi> and let a certayne right line <hi>EFGH</hi> cutte them in suche sorte, that
<figure/>
it make the angles <hi>BFG</hi> and <hi>FGD</hi> e<g ref="char:EOLhyphen"/>quall to two right angles. Then I say, that these right lines <hi>AB</hi> and <hi>CD</hi> are parallel lines, that is, they shall not con<g ref="char:EOLhyphen"/>curre. For if it be possible, let the lines <hi>BF</hi> and <hi>GD</hi> being produced concurre in the pointe K. Nowe forasmuche as the right line <hi>EF</hi> standeth vppon the right line <hi>AB,</hi> therfore (by the 13. propositi<g ref="char:EOLhyphen"/>on) it maketh the angles <hi>AFE,</hi> and <hi>BFE</hi> equall to two right angles: likewise forasmuch as the line <hi>FG</hi> standeth vpo<g ref="char:cmbAbbrStroke">̄</g> the line <hi>CD,</hi> therfore (by the same proposition) it maketh the angles <hi>CGF</hi> and <hi>DGF</hi> equall to two right angles. Wherfore the foure angles <hi>BFE, AFE, CGF,</hi> and <hi>DGF</hi> are equal to foure right angles: of which the two angles <hi>BFG</hi> and <hi>FGD</hi> are (by supposition) e<g ref="char:EOLhyphen"/>quall to two right angles, wherfore the angles remaining, namely, <hi>AFG</hi> and <hi>CGF</hi> are also equall to two right angles. If therfore the right lines FB and <hi>GD</hi> being produced (the inward angles being equall to two right angles) do concurre, then shall the lynes <hi>FA</hi> and <hi>GC</hi> being produced concurre. For the angles <hi>A</hi>F<hi>G</hi> and <hi>CG</hi>F are equall to two right angles. For either the right lines shall concurre on either side, or els on nei<g ref="char:EOLhyphen"/>ther side. For that on either side the angles are equall to two right angles. Wherefore let the right lines F<hi>A</hi> and <hi>GC</hi> concurre in the point L. Wherefore the two right lines LAFK and LCGK do comprehend a space, which (by the 6. peticion) is impossi<g ref="char:EOLhyphen"/>ble.
<pb facs="tcp:7062:69"/>
Wherfore it is not possible that the inward angles being equal to two right angles, the right lines should concurre. Wherefore they are parallels: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>20.</hi> Theoreme. The <hi>29.</hi> Proposition. A right line line falling vppon two parallel right lines: ma<g ref="char:EOLhyphen"/>keth the alternate angles equall the one to the other: and al<g ref="char:EOLhyphen"/>so the outwarde angle equall to the inwarde and opposite an<g ref="char:EOLhyphen"/>gle on one and the same side: and moreouer the inwarde an<g ref="char:EOLhyphen"/>gles on one and the same side equall to two right angles.</head>
               <p>SVppose that vpon these parallel lines AB and CD do fal the right line EF. Then I say that the alternate an<g ref="char:EOLhyphen"/>gles which it maketh, namely, the angles AGH and GHD, are equall the one
<figure/>
to the other and y<hi rend="sup">t</hi> the out<g ref="char:EOLhyphen"/>ward angle EGB is equal to the inwarde and oppo<g ref="char:EOLhyphen"/>site angle on the same side, namely, to y<hi rend="sup">e</hi> angle GHD: and y<hi rend="sup">t</hi> the inward an<g ref="char:EOLhyphen"/>gles on one and the selfe same side, that is, the an<g ref="char:EOLhyphen"/>gles BGH and GHD, are equall to two right angles. For if the angle AGH be not equal to the angle GHD,<note place="margin">Demonstration <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                  </note> the one of them is greater. Let the angle AGH be greater. And forasmuch as the angle AGH is greater then the angle GHD, put the angle BGH commo<g ref="char:cmbAbbrStroke">̄</g> to the<g ref="char:cmbAbbrStroke">̄</g> both. VVherfore y<hi rend="sup">e</hi> angles AGH and BGH, are greater the<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> angles BGH &amp; GHD. But by y<hi rend="sup">e</hi> 
                  <hi>13.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> angles AGH &amp; BGH are equall to two right angles, wherfore y<hi rend="sup">e</hi> angles BGH &amp; GHD are lesse the<g ref="char:cmbAbbrStroke">̄</g> two right angles. But (by y<hi rend="sup">e</hi> 
                  <hi>5.</hi> peticion) if vpo<g ref="char:cmbAbbrStroke">̄</g> two right lines do fall a right line, making y<hi rend="sup">e</hi> inward angles on one and y<hi rend="sup">e</hi> same side, lesse the<g ref="char:cmbAbbrStroke">̄</g> two right angles, those right lines being infinitly produced must needes at y<hi rend="sup">e</hi> length meete on the side wherin are the angles lesse the<g ref="char:cmbAbbrStroke">̄</g> two right angles. VVherfore the right lines AB and CD being infinitely produced will at y<hi rend="sup">e</hi> length meete. But they cannot meete, because they are parallels (by supposition): wherfore the angle AGH is not vnequall to the angle GHD: wherfore it is equall.</p>
               <p>And the angle AGH is (by the <hi>15.</hi> proposition) equall to the angle EGB.<note place="margin">Second part.</note> VVherfore (by the first common sentence) the angle EGB is equall to the an<g ref="char:EOLhyphen"/>gle GHD.</p>
               <p>Put the angle BGH common to them both: wherfore the angles EGB and BGH,<note place="margin">Third part.</note> are equall to the angles BGH and GHD. But the angles EGB
<pb n="39" facs="tcp:7062:69"/>
and BGH are (by the <hi>13.</hi> proposition) equall to two right angles. VVherefore the angles BGH and GHD are also equall to two right angles. If a lyne therfore do fall vpo<g ref="char:cmbAbbrStroke">̄</g> two parallel right lines: it maketh the alternate angles equal the one to the other: and also the outward angle equall to the inward and oppo<g ref="char:EOLhyphen"/>site angle on one and the same side: and moreouer the inward angles on one and the same syde equall to two right angles: whiche was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
               <p>
                  <hi>This proposition is the conuerse of the two propositions next going before.</hi>
                  <note place="margin">This proposition is the conuerse of the two for<g ref="char:EOLhyphen"/>mer propositio<g ref="char:cmbAbbrStroke">̄</g>s.</note> 
                  <hi>For, that which in either of them is the thing sought, or co<g ref="char:cmbAbbrStroke">̄</g>clusion, is in this the thing geuen, or supposition. And contrariwise the thinges which in them were geuen or suppositions, are in this proued, and are conclusions.</hi>
               </p>
               <div type="section">
                  <head>
                     <hi>Pelitarius</hi> after this proposition addeth this witty conclusion.</head>
                  <p>
                     <hi>If two right lines which cut two parallel lines,</hi>
                     <note place="margin">
                        <hi>An addition of Pelitarius.</hi>
                     </note> 
                     <hi>do betwene the sayde parallel lines concurre in a point, and make the alternate angles equall, or the outward angle equall to the inward and opposite angle on the same side, or finally the two inward angles on one and the selfe same side equall to two right angles: those two right lines are drawen directly and make one right line.</hi>
                  </p>
                  <p>Suppose that there be two right lines <hi>AB</hi> and <hi>CB,</hi> which let cut two parallel lines <hi>DE</hi> and FG: and let <hi>AB</hi> cut the line <hi>DE</hi> in the point <hi>H</hi>: and let <hi>CB</hi> cut the line FG in the point <hi>K:</hi> &amp; let the lines <hi>AB</hi> &amp; <hi>CB,</hi> concurre betwene the two parallel lines <hi>DE</hi> &amp; FG in the point <hi>B</hi>: and let the angle <hi>DHB</hi> be e<g ref="char:EOLhyphen"/>qual
<figure/>
to the angle BKG: or let the angle <hi>AHD</hi> be equall to the angle BKF: or finally let the angles <hi>BHD</hi> and BKF be equal to two right angles. The<g ref="char:cmbAbbrStroke">̄</g> I say that the two lines <hi>AB</hi> and <hi>BC</hi> are drawen di<g ref="char:EOLunhyphen"/>rectly, and do make one right line. For if they be not, then produce <hi>AB</hi> vntil it cut FG in the point L, and let AL be one right line,<note place="margin">
                        <hi>Demonstration leading to an absurditie.</hi>
                     </note> and so shal be made the triangle <hi>BLK.</hi> Now then (by the first part of this 29. proposition) the angle <hi>DHB</hi> shalbe equal to the alternate angle <hi>GLB</hi>:<note place="margin">
                        <hi>First part.</hi>
                     </note> but (by supposition) the angle <hi>DHB</hi> is equall to the angle <hi>BKG.</hi> Wherefore the angle <hi>BLG</hi> is equall to the angle <hi>BKL,</hi> namely, the outward angle to the inwarde and opposite angle: which (by the 16. proposition) is impossible.</p>
                  <p>Moreouer (by the seco<g ref="char:cmbAbbrStroke">̄</g>d part of this 29. propositio<g ref="char:cmbAbbrStroke">̄</g>) the angle AHD shalbe equal to the angle <hi>BLK,</hi>
                     <note place="margin">
                        <hi>Second part.</hi>
                     </note> namely, the outward angle to the inward and opposite angle on one and the same side. But the same angle <hi>AHD</hi> is supposed to be equall to the angle BKF: wherefore the angle BKF is equall to the angle <hi>BLK.</hi> Which (by the selfe same 16. proposition) is impossible.</p>
                  <p>Lastly forasmuch as the angles <hi>BHD</hi> and BKF are supposed to be equall to two right angles,<note place="margin">
                        <hi>Third part.</hi>
                     </note> &amp; the angles <hi>BHD</hi> &amp; BLK are also by the last part of this 29 propositi<g ref="char:EOLhyphen"/>on equal to two right angles, therefore the angle BKF shalbe equal to the angle BLK: which agayne by the selfe same 16. proposition is impossible.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>21.</hi> Theoreme The <hi>30.</hi> Proposition. Right lines which are parallels to one and the selfe same right line: are also parrallel lines the one to the other.</head>
               <p>
                  <pb facs="tcp:7062:70" rendition="simple:additions"/>
                  <hi>SVppose that these right lines
<figure/>
AB and CD, be parallel lines to the right line EF. Then I say, that the line AB is a parallel line to CD.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>Let there fall vpon these thre lines a right line GHK. And foras<g ref="char:EOLhyphen"/>much as the right line GHK falleth vppon these parallell right lines AB and EF, therfore (by the proposition going before) the angle AGH is e<g ref="char:EOLhyphen"/>quall to the angle GHF. Agayne for<g ref="char:EOLhyphen"/>asmuch as the right line GK falleth vppon these parallell right lines EF and CD, therefore (by the same) the angle GHF is equall to the angle GKD. Now then it proued that the angle AGH is equall to y<hi rend="sup">e</hi> angle GHF, and y<hi rend="sup">t</hi> the angle GKD is equall to the angle GHF. VVherfore the angle AGK is equall to the angle GKD. And they are alternate angles: wherfore AB is a parallel line to CD. Right lines therfore which are parallels to one and the selfe same right line, are also parallel lines the one to the other: which was required to be proued.</hi>
               </p>
               <p>Euclide in the demonstration of this proposition, setteth the two parallel lines which are compared to one, in the extremes, and the parallel to whome they are compared, he placeth in the middle, for the easier demonstration. It may also be proued euen by a principle onely. For if they shoulde concurre on any one side, they should concurre also with the middle line, and so should they not be parallels vnto it, which yet they are supposed to be.</p>
               <p>
                  <note place="margin">
                     <hi>An other case in <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> Problem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note>But if you will alter their position and placing, and set that line to which you will co<g ref="char:cmbAbbrStroke">̄</g>pare the other two lines, aboue, or beneath: you may vse the same demon<g ref="char:EOLunhyphen"/>stration which you had before. As for example.</p>
               <p>Suppose that the lines <hi>AB</hi> and <hi>CD</hi> be
<figure/>
parallels to the line <hi>EF</hi>: and let both the lines <hi>AB</hi> and <hi>CD,</hi> be aboue, and let the line <hi>EF</hi> be beneath, and not in the middest. Vpon which let the right line <hi>GHK</hi> fall. And forasmuch as either of the angles <hi>KHD</hi> and <hi>HGB</hi> is equall to the angle <hi>HKE,</hi> (for they ar alternate angles) therfore they are (by the first common sentence) equall the one to the other. Wherefore (by the 28 proposition) the right lines <hi>AB</hi> and <hi>EF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are paralles.</p>
               <p>
                  <note place="margin">
                     <hi>An obiection.</hi>
                  </note>But here if a man will obiect that the lines EK and KF, are parallels vnto the line CD,<note place="margin">
                     <hi>Answer.</hi>
                  </note> and therefore are parallels the one to the other. VVe will answere that the lines EK and KF are partes of one parallel line, and are not two parallel
<pb n="40" facs="tcp:7062:70"/>
lines.<note place="margin">
                     <hi>Parallel lines are vnderstan<g ref="char:EOLhyphen"/>ded to be produ<g ref="char:EOLhyphen"/>ced infinitely.</hi>
                  </note> For parallel lines ar vnderstanded to be produced infinitly But EK being produced falleth vpon KF. Wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> it is one and the selfe same with it, and not an other. Wherefore all the partes of a parrallel line are parallels, both to the right line vnto which the whole parallel line is a parallel, and also to al the parts of the same right line. As the line EK is a parallel vnto HD, and the line KE to the line CH. For if they be produced infinitly, they will neuer concurre.</p>
               <p>Howbeit there are some which like not, that two distinct parellel lines, should be taken and counted for one parallel line: for that the continuall quan<g ref="char:EOLhyphen"/>tity, namely, the line is cut asonder, and cesseth to be one. VVherefore they say, that there ought to be two distinct parallel lines compared to one. And therfore they adde to the proposition a correction, in this maner. <hi>Two lines being parallels to one line: are either parallels the one the other, or els the one is set directly againste the other, so that if they be produced they should make one right line. As for example.</hi>
               </p>
               <p>Suppose that the lines <hi>CD</hi> and <hi>EF</hi> be parallels to one and the selfe same line <hi>AB,</hi>
                  <figure/>
and let them not be parallels the one to the other. Then I say, that the two lines <hi>CD</hi> &amp; <hi>EF,</hi> are directly set the one to the other. For for as much as they are not parallel lines, let them concurre in the point <hi>G,</hi> And from the point <hi>G</hi> draw a line cutting the line <hi>AB</hi> in the point <hi>H.</hi> Now by the former proposition the angles <hi>AHG</hi> &amp; <hi>HGC</hi> are equall to two right angles, but by the same propositio<g ref="char:cmbAbbrStroke">̄</g>, the angle <hi>AHG,</hi> is equall to the alternate angle <hi>HGE.</hi> Wherefore the angles <hi>HGC</hi> and <hi>HGF</hi> are equal to two right angles. Wherefore (by the 14 proposition) the lines <hi>CG</hi> and <hi>FG</hi> are drawen directly and make one right line. Wherfore al<g ref="char:EOLhyphen"/>so the lines <hi>CD</hi> and <hi>EF</hi> are set directly the one to the other: and being produced they will make one right line.</p>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Probleme. The <hi>31.</hi> Proposition. By a point geuen, to draw vnto a right line geuen, a parallel line.</head>
               <p>SVppose that the point geuen be
<figure/>
A, and let the right line geuen be BC. It is required by the point geuen, namely A, to draw vuto the right line BC, a parallel line.<note place="margin">Construction.</note> Take in the line BC a point at all aduentures, and let the same be D. and (by the first peticio<g ref="char:cmbAbbrStroke">̄</g>) draw a right line from the point A, to the point D. And (by the <hi>23.</hi> proposition) vpon the right line geuen AD, and to the point in it geuen A, make an angle DAE, equall to the angle geuen ADC. And (by the <hi>14.</hi> proposition) put vnto the line AE the line AF directly, in such sorte that they
<pb facs="tcp:7062:71"/>
both make one right line. And forasmuch as the right line AD falling vpon the right lines BC and EF, doth make the alternate angles, namely, EAD, and ADC equall y<hi rend="sup">e</hi> one to the other, therfore (by the <hi>27.</hi> proposition) EF is a parallel line to BC. VVherfore by the point geuen, namely A, is drawne to the right line geuen BC a parallel line EAF: which was required to be done.</p>
               <p>
                  <hi>This proposition is to be vnderstanded of a point geuen without the line ge<g ref="char:EOLhyphen"/>uen, and in such sorte also, that the same line geuen being produced, doo not fall vppon the pointe geuen.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>22.</hi> Theoreme. The <hi>32,</hi> Proposition. If one of the sydes of any triangle be produced: the outwarde angle that it maketh, is equal to the two inward and opposite angles. And the three inwarde angles of a triangle are equall to two right angles.</head>
               <p>
                  <hi>SVppose y<hi rend="sup">•</hi> ABC be a triangle,
<figure/>
&amp; produce one of y<hi rend="sup">e</hi> sides ther<g ref="char:EOLhyphen"/>of namely, CB to the pointe D. Then I say, that the outwarde angle ACD is equall to the two inwarde and opposite angles CAB &amp; ABC: and y<hi rend="sup">t</hi> the three inwarde angles of the triangle, that is, the angles ABC, BCA, and CAB are equall to two right angles.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>For (by the proposition going before) rayse vp fro<g ref="char:cmbAbbrStroke">̄</g> the point C, a parallel to the right line AB, and let the same be CE.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>And forasmuch as AB is a parallel to CE, and vpon them falleth the right line AC: therefore the alternate angles BAC and ACE are equall the one to the other. Agayne forasmuch as AB is a parallel vnto CE, and vpon them falleth the right line BD, therfore the outward angle ECD is (by the</hi> 29. <hi>proposition) equall to the inward and opposite angle ABC. And it is proued that the angle ACE is equal to the angle BAC: wherfore the whole outwarde angle ACD is equall to the two inward and opposite angles, that is, to the angles BAC and ABC. Put the angle ACB common to them both. VVherfore the angles ACD and ACB, are equall to these three angles ABC, BCA, and BAC. But the angles ACD &amp; ACB are equall to two right angles (by the</hi> 13. <hi>proposition): wherfore the angles ACB, CBA, and CAB are equall to two right angles. If therfore one of the sides of any triangle be produced, the outward angle that it maketh, is equall to the two inward and opposite angles. And the three inward angles of a triangle
<pb n="41" facs="tcp:7062:71"/>
are equall to two right angles: which was required to be demonstrated.</hi>
               </p>
               <p>Euclide demonstrateth either part of this composed Theoreme, by drawyng fro<g ref="char:cmbAbbrStroke">̄</g> one angle of the triangle a parrallel line to one of the sides of the same triangle, without the triangle. Either part therof may also be proued without drawyng of a parallel line without the triangle, only chaunging the order of the thinges re<g ref="char:EOLhyphen"/>quired or conclusions. For Euclide first proueth that the outwarde angle of a tri<g ref="char:EOLhyphen"/>angle (one of his sides beyng produced) is equall to the two inwarde and opposite angles: and by that he proueth the second part: namely, that the 3. inward angles of a triangle are equall to two right angles. But here it is contrariwise. For first is proued that the three inward angles of a triangle are equall to two right angles, and by that is proued the other part of the Theoreme, namely, that one side of a tria<g ref="char:cmbAbbrStroke">̄</g>gle beyng produced, the outward angle is equal to the two inward and oppo<g ref="char:EOLhyphen"/>site angles. And that after this maner.</p>
               <p>Suppose that there be a triangle <hi>ABC</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration.</hi>
                  </note> and produce the side <hi>BC</hi> to the point <hi>E.</hi> And take in the line <hi>BC</hi> a point at al auentures which let be F:
<figure/>
&amp; draw a line from <hi>A</hi> to F. And by the point F drawe vnto the line <hi>AB</hi> a parallel line (by the former pro<g ref="char:EOLhyphen"/>position) which let be FD. Now forasmuch as FD is a parallell vnto <hi>AB,</hi> and vpon them falleth the right line AF, and also the right line <hi>BC,</hi> therfore the alter<g ref="char:EOLhyphen"/>nate angles are equall, and also the outward angle is equall to the inward angle. Wherefore the whole an<g ref="char:EOLhyphen"/>gle <hi>AFC</hi> is equall to the angles FAB and ABF. And by the same reason (if by the point F we draw a paral<g ref="char:EOLhyphen"/>lel line to the line <hi>AC</hi>) may we proue that the angle AFB is equall to the angles FAC, and <hi>ACF.</hi> Wherfore the two angles <hi>AFB</hi> &amp; <hi>AFC</hi> are equall to the three angles of the triangle <hi>ABC.</hi> But the two angles <hi>AFB</hi> &amp; <hi>AFC</hi> are (by the 13. proposition) equall to two right angles. Wherfore also the three angles of the triangle <hi>ABC</hi> are equall to two right angles.</p>
               <p>But the angles <hi>ACF</hi> and <hi>ACE</hi> are also (by the 13. proposition) equall to two right angles. Take away the angle <hi>ACF</hi> which is common, wherfore the angle remai<g ref="char:EOLhyphen"/>ning, namely, the outward angle <hi>ACE</hi> is equall to the two angles remaining, namely, to the two inwarde and opposite angles <hi>ABC</hi> and <hi>CAB</hi>: which was required to be proued.</p>
               <p>Eudemus affirmeth, that the latter part of this Theoreme, <hi>The three angles of a triangle are equall to two right angles,</hi> was first found out by Pithagoras,<note place="margin">
                     <p>
                        <hi>The latter part of this Theo<g ref="char:EOLunhyphen"/>reme first <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ound out by Pithago<g ref="char:EOLhyphen"/>ras.</hi>
                     </p>
                     <p>
                        <hi>The demon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tra<g ref="char:EOLhyphen"/>tion thereof af<g ref="char:EOLhyphen"/>ter him.</hi>
                     </p>
                  </note> whose demon<g ref="char:EOLhyphen"/>stration thereof is thus.</p>
               <p>Suppose that there be a triangle <hi>ABC</hi>: and by point <hi>A,</hi> draw (by the former proposition) vnto the line <hi>BC,</hi> a parallel line, which let be <hi>DE.</hi> And forasmuch as the right lines <hi>BC</hi> and <hi>DE</hi> are parallels, and vpon them falleth
<figure/>
the right lines <hi>AB</hi> and <hi>AC,</hi> therefore (by the 29. propositi<g ref="char:EOLhyphen"/>on) the alternate angles are equall. Wherefore the angle <hi>DAB</hi> is equall to the angle <hi>ABC,</hi> and the angle <hi>EAC</hi> to the angle <hi>ACB.</hi> Adde the angle <hi>BAC</hi> common. Wherefore the angles <hi>DAB, BAC, CAE,</hi> that is, the angles <hi>DAB</hi> and <hi>BAE,</hi> namely, two angles equal to two right angles, are equal to the thre angles of the triangle <hi>ABC.</hi> Wherfore the thre angles of a triangle are e<g ref="char:EOLhyphen"/>quall to two right angles: which was required to be proued.</p>
               <div type="section">
                  <head>The conuerse of this proposition is thus.</head>
                  <p>
                     <pb facs="tcp:7062:72"/>
                     <hi>If the outward angle of a triangle be equall to the two inward angles opposit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> against it: one of the sides of the triangles is produced,</hi>
                     <note place="margin">
                        <hi>The conuerse of this proposition.</hi>
                     </note> 
                     <hi>and the line without the triangle, is drawen directly to the side of the triangle, &amp; maketh one right line with it. And if the thre inward angles of a rectiline figure be equal to two right angles, the same rectiline figure is a triangle.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Demonstration of the first part of the conuerse.</hi>
                     </note>Suppose that there be a triangle <hi>ABC:</hi> and let the outward angle <hi>ACD</hi> be equal to the two inward &amp; opposite angles <hi>ABC</hi> and <hi>CAB.</hi> Then
<figure/>
I say that the side <hi>BC</hi> is produced to the poynt <hi>D.</hi> And that <hi>BCD</hi> is one right line. For forasmuch as the angle <hi>ACD</hi> is equal to the two inward &amp; opposite angles, adde the angle <hi>ACB</hi> common. Wherefore the angles <hi>ACD</hi> and <hi>ACB</hi> are equal to the three angles of the triangle <hi>ABC.</hi> But the three angles of the triangle <hi>ABC</hi> are equall to two right angles Wherefore also the two angles <hi>ACD</hi> and <hi>ACB</hi> are equall to two right angles. But if vnto a right line, and to a point in the same line be drawen two right lines, not both on one and the same side, making the side angles equal to two right angles: those two right lines shal be drawe<g ref="char:cmbAbbrStroke">̄</g> directly, and make one right line (by the 14. proposition.) Wherefore the right line <hi>BC</hi> is dra<g ref="char:EOLhyphen"/>wen directly to the line <hi>CD,</hi> and so is <hi>BCD</hi> one right line: which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>Demonstration of the second part of the conuerse.</hi>
                     </note>Agayne suppose that there be a certayne rectiline figure <hi>ABC,</hi> hauing onely three ang<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es, namely, at the pointes <hi>A, B, C</hi>: which angles let be equal to two right angles Then I say that <hi>ABC</hi> is a triangle. First <hi>AC</hi> is one right
<figure/>
line. For draw the line <hi>BD.</hi> And forasmuch as in either of the triangles <hi>ABD</hi> and <hi>DBC,</hi> the three angles are e<g ref="char:EOLhyphen"/>qual to two right angles, of which the angles at the points <hi>A, B, C,</hi> are equal to two right angles. Wherefore the an<g ref="char:EOLhyphen"/>gles remayning, namely, <hi>ADB</hi> and <hi>CDB</hi> are equall to two right angles. Wherefore (by the 14. proposition) the line <hi>DC</hi> is set directly to the line <hi>DA.</hi> Wherefore the side <hi>AC</hi> is one right line. And in like sort may we proue that the side <hi>AB</hi> is one right line, and also that the side <hi>BC</hi> is one right line. Wherefore the figure <hi>ABC</hi> is a triangle: which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>A Corrollary.</hi>
                     </note>By the seco<g ref="char:cmbAbbrStroke">̄</g>d part of this 29. propositio<g ref="char:cmbAbbrStroke">̄</g>, namely, <hi>three angles of a triangle are equall to two right angles,</hi> may easely be knowen, to how many right angles, the angles within any figure hauing right lines and many angles are equall. As are figures of fower angles, of fiue angles, of sixe angles, and so consequently: and infinitly, And this is to be noted, that euery right lined figure is resolued into triangle, For that a triangle is the first of all figures.<note place="margin">
                        <p>
                           <hi>Euery right li<g ref="char:EOLhyphen"/>ned figure is re<g ref="char:EOLhyphen"/>solued in tri<g ref="char:EOLhyphen"/>angles.</hi>
                        </p>
                        <p>
                           <hi>A triangle is the first of all fi<g ref="char:EOLhyphen"/>gures.</hi>
                        </p>
                        <p>
                           <hi>Into how many triangles a fi<g ref="char:EOLhyphen"/>gure may be re<g ref="char:EOLhyphen"/>solued.</hi>
                        </p>
                     </note> For two lines accomplish no figures VVher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore how many sides the figure hath, into so many triangles may it be re<g ref="char:EOLhyphen"/>solued, sauing two. As if the figure haue fower sides, it is resolued into two trian<g ref="char:EOLhyphen"/>gles, if it haue fiue sides, into 3. triangles: if 6 sides into 4. triangles, and so con<g ref="char:EOLhyphen"/>sequently, and infinitly. And it is proued that the three angles of euery triangle are equall to two right angles. VVherefore if you multiply the number of the triangles, into which the figure is resolued, by two, you shall haue the num<g ref="char:EOLhyphen"/>ber of right angles, to which the angles of the figure are equall. So the angles of euery qua<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rangled figure are equall to 4. right angles. For it is composed of two triangles. And the angles of a fiue angled figure are equal to 6. right angles, for it is composed of three triangles, and so forth in like order.</p>
                  <p>The rediest and aptest maner to reduce any rectiline figure into triangles, is
<pb n="42" facs="tcp:7062:72"/>
thus. From any one angle of the figure to euery other angle (of the same) beyng opposite vnto it, drawe a right line, so shall you haue all the triangles of that fi<g ref="char:EOLhyphen"/>gure described.
<figure/>
In a quadra<g ref="char:cmbAbbrStroke">̄</g>gle, from one angle you can drawe but one lyne to the opposite an<g ref="char:EOLunhyphen"/>gle, by which it is deuided into two triangles only. In a pen<g ref="char:EOLhyphen"/>tagon figure, from one angle you may draw lines to two op<g ref="char:EOLunhyphen"/>posite angles, and so you shal haue three triangles. In an Hexagon, you may from one angle draw lines to thre opposite angles, and so shall you haue 4. triangles. In an heptagon, from one an<g ref="char:EOLhyphen"/>gle may be drawne lines to foure opposite angles, and so shal there be fiue trian<g ref="char:EOLhyphen"/>gle. And so consequently of the rest. As you see in the figures here set.</p>
                  <p>This thing may also be thus expressed. In any figure of many sides,<note place="margin">
                        <hi>An other way to know the nu<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>ber of right angles vnto which the angles of a<g ref="char:EOLhyphen"/>ny figure are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>quall.</hi>
                     </note> the num<g ref="char:EOLhyphen"/>ber of the angles of the figure doubled, is the nu<g ref="char:cmbAbbrStroke">̄</g>ber of the right angles to which the angles of the figure are equall, sauing foure, As for example.</p>
                  <p>Let there be an hexagon figure ABCDEF, and within it take a point at all
<figure/>
auentures, namely, G. And draw from the same point to euery one of the angles a right line, &amp; so shal there be comprehended in the figure so many triangles, as there are angles in the same. VVherefore by this 32. proposition all the angles of these triangles taken to<g ref="char:EOLhyphen"/>gether, are equall to double so many right angles, as there be angles in the figure. VVherefore forasmuch as there are sixe triangles, there are twelue right an<g ref="char:EOLhyphen"/>gles. But all the angles at the point G are equall to 4. right angles by the 13. proposition. VVherefore take away foure out of twelue, and there rest eight. VVher<g ref="char:EOLunhyphen"/>fore the sixe angles in the Hexagon figure are equall to eight right angles.</p>
                  <p>By that which hath now bene declared,<note place="margin">
                        <hi>Another expres<g ref="char:EOLhyphen"/>sion of the for<g ref="char:EOLhyphen"/>mer Corollary.</hi>
                     </note> it foloweth that all the angles of any fi<g ref="char:EOLhyphen"/>gure hauing many sides, take<g ref="char:cmbAbbrStroke">̄</g> together, are equal to twise so many right angles, as the figure is in the reaw or order of figures. A tria<g ref="char:cmbAbbrStroke">̄</g>gle is the first figure in order,<note place="margin">
                        <hi>A triangle the first figure in order.</hi>
                     </note> &amp;<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> his angles are equal to two right angles, which are twise one. A quadrangle is the second figure in order.<note place="margin">
                        <hi>A quadrangle the second, and so consequently. How the order of figures is ga<g ref="char:EOLhyphen"/>thered.</hi>
                     </note> VVherfore his angles are equal to fower right angles, which are twise two. The order of figures is gathered of the sides. For if you take two from the number of the sides of a figure, the number of the sides remayning, is the number of the order of the figure. As if you will know, how many in order is a figure of sixe sides: from six (which is the number of this sides) take a way two, and there will remaine foure. VVherfore a figure of sixe sides is the fourth figure
<pb facs="tcp:7062:73"/>
in the order of figures. Then double foure, so shall you haue eight. VVherefore the angles therof are equall to eight right angles. And so of all other figures.</p>
                  <p>
                     <note place="margin">
                        <hi>An other Cor<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note>Hereby also it is manifest, that the outward angles of any figure of many sides taken together, are equall to foure right angles. For the inwarde angles together with the outward angles, are equall to twise so many right angles, as there be an<g ref="char:EOLhyphen"/>gles in the figure (by the 13. proposition) But the inward angles are equal to twise so many right angles, as there be angles in the figure, sauyng foure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as it was be<g ref="char:EOLhyphen"/>fore declared. VVherfore the outward angles are always equal to foure right an<g ref="char:EOLhyphen"/>gles. As for example.</p>
                  <p>Suppose that there be a pentagon, ABCDE. And produce the fiue sides ther<g ref="char:EOLhyphen"/>of to the points F, G, H, K, L. Now (by the 13.
<figure/>
propositio<g ref="char:cmbAbbrStroke">̄</g>) the two angles at the point A shall be equal to two rightangles. And (by the same) the two angles at the pointe B shall be also e<g ref="char:EOLhyphen"/>quall to two right angles. And so taking euery two angles, they shall be in all equall to tenne right angles. VVherfore takyng away the in<g ref="char:EOLhyphen"/>ward angles, whiche (as hath before bene pro<g ref="char:EOLhyphen"/>ued) are equall to sixe righte angles, the out<g ref="char:EOLhyphen"/>ward angles shall be equall to fower right an<g ref="char:EOLhyphen"/>gles. And so of all other figures<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>An other Cor<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note>It is also manifest, that euery pentago<g ref="char:cmbAbbrStroke">̄</g>, which is so described, that ech side therof deuideth two of the other sides, hath his fiue angles equall to two right angles.</p>
                  <p>For suppose that <hi>ABCDE</hi>: be such a pentagon as is there required so that let the side <hi>AC</hi> cut the side <hi>BE</hi> in the point <hi>G</hi>: &amp; let the side <hi>AD,</hi> cut
<figure/>
the same side <hi>BE</hi> in the point <hi>F.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> by this propo<g ref="char:EOLhyphen"/>sition the angle <hi>AFG</hi> shal be equall to the two angles at the point <hi>B</hi> and <hi>D</hi>: namely, the outward angle to the two inward and opposite angles. And by the same reason the angle <hi>FGA</hi> is equal to the angles at the points <hi>C</hi> and <hi>E</hi> which are in the triangle <hi>CEG.</hi> But the two angles <hi>AFG</hi> and <hi>FGA,</hi> together with the angle at the point <hi>A,</hi> are e<g ref="char:EOLhyphen"/>quall to two right angles (by this proposition). Where<g ref="char:EOLhyphen"/>fore the fower angles at the pointes, <hi>B, C, D, E,</hi> together with the angle at the point <hi>A,</hi> are equal to two right an<g ref="char:EOLhyphen"/>gles: which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>An other Cor<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note>By this proposition also it is manifest, that euery angle of an equilate trian<g ref="char:EOLhyphen"/>gle is two third partes of a right angle. And that in a triangle of two equall sides hauing a right angle at the toppe, either of the the two angles at the base is the halfe of a right angle. And in a triangle called Sealenum, such a Scalenu<g ref="char:cmbAbbrStroke">̄</g> (I say) which is made by the drawght of a perpendicular line from any one of the angles of an equilater tria<g ref="char:cmbAbbrStroke">̄</g>gle to the opposite side therof, one angle is a right angle, an other is two third parts of a right angle, namely, that angle which was also an angle of the equilater triangle, wherfore of necessity the angle remaining is one third part of a right angle. For the three angles of a tria<g ref="char:cmbAbbrStroke">̄</g>gle must be equall to two right angles.</p>
                  <p>
                     <note place="margin">
                        <hi>An other Cor<g ref="char:EOLhyphen"/>rallory.</hi>
                     </note>Moreouer by this proposition it is manifest, that if there be two triangles, and if two angles of the one be equal to two angles of the other: the angle remai<g ref="char:EOLhyphen"/>ning
<pb n="43" facs="tcp:7062:73"/>
shall also be equall to the angle remayning. For forasmuch as three angles of any triangle are equal to three angles of any other triangle (for that in ech the three angles are equal to two right angles)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> If from ech triangle be taken away the two equall angles, the angle remayning shall (by the 3. common sentence) be equall to the angle remayning.</p>
                  <p>And here I thinke it good to shew how to deuide a right angle into three e<g ref="char:EOLhyphen"/>quall partes, for that the demonstration thereof dependeth of this proposition.</p>
                  <p>Suppose that there be a right angle <hi>ABC,</hi> contayned of the right lines <hi>AB</hi> and <hi>BC</hi>:<note place="margin">
                        <hi>How to deuide <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> right angle into three equall partes.</hi>
                     </note> &amp;
<figure/>
in the line <hi>BC,</hi> take a point at all aduentures, which let be <hi>D.</hi> An<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> vpon the line <hi>BD</hi> describe (by the first) an equila<g ref="char:EOLhyphen"/>ter triangle <hi>BDE.</hi> And (by the 9. proposition) deuide the angle <hi>DBE</hi> into two equall partes by the right line <hi>BF.</hi> Then <hi>I</hi> say that the right angle <hi>ABC</hi> is deuided into thre equal parts by the right lines <hi>BE</hi> and <hi>BF.</hi> For forasmuch as <hi>EBD</hi> is an equilater triangle, therfore as hath<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> before bene declared, the angle <hi>EBD</hi> is two thirde partes of a right angle. But the whole angle <hi>ABC,</hi> is a right angle. Wherfore the angle remaining, namely, <hi>ABE</hi> is one third part of a right angle. Again forasmuch as the angle <hi>EBD,</hi> is two third partes of a right angle, and it is deuided into two equall parts by the right line <hi>BF,</hi> therefore either of these two angles <hi>EBF</hi> &amp; <hi>FBD</hi> is one third part of a right angle. Wherefore the three angles <hi>ABE, EBF</hi> and <hi>FBD</hi> are equall the one to the o<g ref="char:EOLhyphen"/>ther. Wherefore the right angle <hi>ABC</hi> is deuided into three equall partes by the right lines <hi>BE</hi> and <hi>BF</hi>: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>23.</hi> Theoreme. The <hi>33.</hi> Proposition. Two right lines ioyning together on one and the same side, two equall parallel lines: are also them selues equall the one to the other, and also parallels.</head>
               <p>SVppose that AB and CD be
<figure/>
right lines equal, and parallels: and let these two right lines AC and BD ioyne the<g ref="char:cmbAbbrStroke">̄</g> together, the one on the one side, and the other on y<hi rend="sup">e</hi> other side. Then I say that the lines AC &amp; BD are both equall, &amp; also parallels. Draw (by the first peticion) a right line from the point B to the point C.<note place="margin">D<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>on<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ratio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> And for<g ref="char:EOLhyphen"/>asmuch as AB is a parallel to CD, and vpo<g ref="char:cmbAbbrStroke">̄</g> them falleth the right line BC, ther<g ref="char:EOLhyphen"/>fore the alternate angles ABC and BCD, are equall the one to the other (by the 29. proposition). And forasmuch as the line AB is equall to the line CD, and the line BC is common to them both,
<pb facs="tcp:7062:74"/>
therefore these two lines AB and BC, are equall to these two lines BC and CD, and the angle ABC is equall to the angle BCD. VVherfore (by the <hi>4.</hi> pro<g ref="char:EOLhyphen"/>proposition) the base BD is equall to the base AC, and the triangle ABC, is e<g ref="char:EOLhyphen"/>quall to the triangle BCD, and the angles remayning are equall to the angles remayning the one to the other, vnder which are subtended equall sides, where<g ref="char:EOLhyphen"/>fore the angle ACB is equall to the angle CBD, and the angle BAC to the angle BDC. And forasmuch as vpon these right lines AC and BD falleth the right line BC making the alternate angles, that is the angles ACB and CBD, equall the one to the other, therefore (by the <hi>27.</hi> proposition) the line AC is a parallel to the line BD. And it is proued that it is equall vnto it. VVherefore two right lines ioyning together on one and the same side two equall lines which are parallels, are also themselues equall the one to the other, and also parallels: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>24.</hi> Theoreme. The <hi>34</hi> Proposition. In parallelogrammes, the sides and angles which are opposite the one to the other, are equall the one to the other, and their diameter deuideth them into two equall partes.</head>
               <p>
                  <hi>SVppose that ABCD be a parallelogramme and let the diameter ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of be BC. Then I say that the opposite sides and angles of the parallelo<g ref="char:EOLhyphen"/>gramme ACDB are equall the one to the other, and y<hi rend="sup">t</hi> the diameter BC deuideth it into two equall partes.</hi>
                  <note place="margin">
                     <hi>Demonstration</hi>
                  </note> 
                  <hi>For foras<g ref="char:EOLhyphen"/>much
<figure/>
as AB is a parallel line vnto CD, and v<g ref="char:EOLhyphen"/>pon them falleth a right line BC: therfore (by the</hi> 29. <hi>proposition) the alternate angles ABC and BCD are equall the one to the other. A<g ref="char:EOLhyphen"/>gayne forasmuch as AC is a parallel line to BD, and vppon them falleth the right lyne BC: therfore (by the same) the alternate angles, that is, the angles ACB and CBD are equall the one to the other. Now therfore there are two triangles ABC and BCD, hauing two an<g ref="char:EOLhyphen"/>gles of the one, namely, the angles ABC and ACB equall to two angles of the other, that is, to the angles BCD and CBD, the one to the other, and one side of the one equal to one side of the other, namely, that syde that lieth betwene the equall angles, which syde is common to them both, namely, the side BC. VVher<g ref="char:EOLhyphen"/>fore (by the</hi> 26. <hi>proposition) the other sides remaining are equall to the other sides remaining, the one to the other, and the angle remaining is equal to the an<g ref="char:EOLhyphen"/>gle remayning. VVherfore the side AB is equall to the side CD, and the side AC
<pb n="44" facs="tcp:7062:74"/>
to the side BD, &amp; the angle BAC is equal to the angle BDC. And for asmuch as the angle ABC is equal to the angle BCD, and the angle CBD to the an<g ref="char:EOLhyphen"/>gle ACB: therfore (by the second common sentence) the whole angle ABD is equall to the whole angle ACD. And it is proued that the angle BAC is e<g ref="char:EOLhyphen"/>quall to the angle CDB. VVherfore in parallelogrammes, the sides and angles which are opposite are equall the one to the other. I say also that the diameter therof deuideth it into two equall partes. For forasmuch as AB is equall to CD, and BC is common to them both, therfore these two AB and BC are equall to these two BC and CD, and the angle ABC is equal to the angle BCD. VVher<g ref="char:EOLunhyphen"/>fore (by the</hi> 4. <hi>proposition) the base AC is equall to the base BD, and the tri<g ref="char:EOLhyphen"/>angle ABC is equall to the triangle BCD. VVherfore the diameter BC de<g ref="char:EOLhyphen"/>uideth the parallelogramme ABCD into two equall partes: which is all that was required to be proued.</hi>
               </p>
               <p>In this Theoreme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are demonstrated three passions or properties of paralle<g ref="char:EOLhyphen"/>logrammes.<note place="margin">
                     <hi>Thre passions of parallolegrames demo<g ref="char:cmbAbbrStroke">̄</g>strated in this Thoreme.</hi>
                  </note> Namely, that thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> opposite sides are equall: that their opposite an<g ref="char:EOLhyphen"/>gles are equall: and that the diameter deuideth the parallelogramme into two e<g ref="char:EOLhyphen"/>quall partes. VVhich is true in all kindes of parallelogrammes. There are fower kindes of parallelogrammes,<note place="margin">
                     <hi>Fower kindes of parallelo<g ref="char:EOLhyphen"/>grammes.</hi>
                  </note> a square, a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure of one side longer then the other, a Rhombus, or diamond figure, and a Rhom<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oides or diamondlike figure. And here is to be noted, that in those parallelogrammes, all whose angles ar right an<g ref="char:EOLhyphen"/>gles (as is a square, and a figure on the one side longer) the diameters do not only deuide the figure into two equall partes, but also they are equal the one to the o<g ref="char:EOLhyphen"/>ther. As for example.</p>
               <p>Suppose that <hi>ABCD</hi> be a square,
<figure/>
or a figure on the one side longer, and draw in it these diametres <hi>AD</hi> and <hi>BC.</hi> And forasmuch as the line <hi>AB</hi> is e<g ref="char:EOLhyphen"/>quall to the line <hi>CD</hi> (by the definitio<g ref="char:cmbAbbrStroke">̄</g> of a square, and of a figure on the also one side lo<g ref="char:cmbAbbrStroke">̄</g>ger) &amp; the line <hi>AC</hi> is com<g ref="char:EOLhyphen"/>mon to the both: therfore two sides of the triangle <hi>ABC</hi> are equal to two sides of the triangle <hi>ACD,</hi> the one to the other, and the angles which they contayne are equall, namely, the an<g ref="char:EOLhyphen"/>gles <hi>BAC</hi> &amp; <hi>ACD,</hi> for they are rightangles. Wherefore the bases namely, the diame<g ref="char:EOLhyphen"/>ters <hi>AD</hi> and <hi>BC,</hi> are (by the 4. proposition) equal.</p>
               <p>But in those parallelogrames whose angles are not right angles, as is a Rhom<g ref="char:EOLhyphen"/>bus, and a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>homboides, the diameters be euer vnequall. As for example.</p>
               <p>Suppose that <hi>ABCD</hi> he
<figure/>
a Rhombus, or a Rhombaides and drawe in it these diame<g ref="char:EOLhyphen"/>ters <hi>AC</hi> and <hi>BD.</hi> And foras<g ref="char:EOLhyphen"/>much as <hi>AB</hi> is equall to <hi>CD</hi> and <hi>BC</hi> is common to them both, &amp; the angle <hi>ABC</hi> is not equall to the angle <hi>BCD</hi> (by the definition of a Rhombus and also of a Rhombaides)
<pb facs="tcp:7062:75"/>
therefore (by the 24. proposition) the bases also are vnequall, namely, the diameters <hi>AC</hi> and <hi>BD.</hi>
               </p>
               <p>Agayne. In parallelogrammes of equall sides, as are a square, and a Rhom<g ref="char:EOLhyphen"/>bus, the diameters do not onely deuide the figures into two equall partes, but also they deuide the angles into two equall partes.</p>
               <p>For suppose that there be a square or Rhombus <hi>ABCD,</hi> and draw the diameter <hi>AD.</hi> And forasmuch as the sides <hi>AB</hi> and <hi>BD</hi> are e<g ref="char:EOLhyphen"/>quall
<figure/>
to the sides <hi>AC</hi> and <hi>CD</hi> (for the figures are equilater) and the angles <hi>ABD</hi> and <hi>ACD</hi> are equall (for they are opposite angles) and the base AD is common to both triangles. Therefore (by the fourth proposition) the angles <hi>BAD</hi> &amp; <hi>CAD</hi> are equall, and so also are the angles <hi>BDA</hi> and <hi>CDA</hi> equall. Wh<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>fore the angles <hi>BAC</hi> and <hi>CDB</hi> are deuided into two equall partes.</p>
               <p>But in parallogrammes whose sides are not equall, such as are a figure on the one side longer, and a Rhomboides it is not so.</p>
               <p>For suppose <hi>ABCD</hi> to be a figure on the one
<figure/>
side longer or a Romboides. And draw the dia<g ref="char:EOLhyphen"/>meter <hi>AD.</hi> And now if the angles <hi>BAC</hi> and <hi>CDB,</hi> be deuided into two equall partes by the dia<g ref="char:EOLhyphen"/>meter <hi>AD,</hi> then forasmuch as the angle. <hi>CAD</hi> is (by the 29. proposition) equall to the angle <hi>ADB,</hi> the angle also <hi>BAD</hi> shal be equal to the an<g ref="char:EOLhyphen"/>gle <hi>ADB</hi> (by the first common sentence). Wher<g ref="char:EOLhyphen"/>fore also the side <hi>AB</hi> is equall to the side <hi>BD</hi> (by the 6. propositio<g ref="char:cmbAbbrStroke">̄</g>). But the sayd sides are vnequal: which is impo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ible. Wherefore the angles <hi>BAC</hi> and <hi>CDB</hi> are not deuided into two equall partes.</p>
               <p>
                  <note place="margin">
                     <hi>The conuerse of th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> proposition after Pr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>lus.</hi>
                  </note>The conuerse of the first and second part of this proposition after Proclus. <hi>Is <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> figure whatsoeuer haue his opposite sides and angles equall: then is a parallelograms.</hi>
               </p>
               <p>For suppose that <hi>ABCD</hi> be such a figure, namely, which hath his opposite sides and angles equall. And let the diameter thereof be <hi>AD.</hi>
                  <figure/>
Now forasmuch as the sides <hi>AB</hi> and <hi>BD</hi> are equall to to the sides <hi>DC</hi> and <hi>AC,</hi> and the angles which they co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayne are equall, and the base <hi>AD</hi> is common to ech tri<g ref="char:EOLhyphen"/>angle, therefore (by the 4. proposition) the angles remay<g ref="char:EOLhyphen"/>ning are equall to the angles remayning, vnder which are subtended equal sides. Wherfore the angle <hi>BAD</hi> is equal to the angle <hi>ADC,</hi> and the angle <hi>ADB</hi> to the angle <hi>CAD.</hi> Wherefore (by the 27. proposition) the line <hi>AB</hi> is a parallel to the line <hi>CD,</hi> and the line <hi>AC</hi> to the line <hi>BD.</hi> Wherefore the figure <hi>ABCD</hi> is a parallogramme: which was required to be proued.</p>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>A Corollary ta<g ref="char:EOLhyphen"/>ken out of Flus<g ref="char:EOLhyphen"/>sates.</hi>
                     </note>A Corrollary taken out of Flussates.</head>
                  <p>
                     <hi>A right line cutting a parallelogramme which way soeuer into two equall partes, shall also de<g ref="char:EOLhyphen"/>uide the diameter thereof into two equall partes.</hi>
                  </p>
                  <p>
                     <pb n="45" facs="tcp:7062:75"/>For if it be possible let the right line <hi>GC</hi> deuide the parallelogramme <hi>AEBD</hi> into two equal partes, but let it deuide the diameter <hi>DE</hi> into two vnequall partes in the point <hi>I.</hi> And l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the part <hi>IE</hi> be greater then the part <hi>ID.</hi> And vnto the line <hi>ID</hi> put the line <hi>IO</hi> equall (by th3. propositio<g ref="char:cmbAbbrStroke">̄</g>).<note place="margin">
                        <hi>Demonstration leading to <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> absurdi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>e.</hi>
                     </note> And by the
<figure/>
point <hi>O,</hi> draw vnto the lines <hi>AD</hi> and <hi>BE</hi> a pa<g ref="char:EOLhyphen"/>rallel line <hi>OF</hi> (by the 31. proposition.) Where<g ref="char:EOLhyphen"/>fore in the triangles <hi>FOI</hi> and <hi>CDI,</hi> two angles of the one are equal to two angles of the other, namely, the angles <hi>IOF</hi> and <hi>IDC</hi> (by the 29. proposition), &amp; the angles <hi>FIO</hi> &amp; <hi>CID</hi> (by the 15. propositio<g ref="char:cmbAbbrStroke">̄</g>), &amp; the side <hi>ID</hi> is equal to the side <hi>IO.</hi> Wherefore (by the 26. proposition) the tri<g ref="char:EOLhyphen"/>angles are equall. Wherefore the whole triangle <hi>EIG</hi> is greater then the triangle <hi>DIC.</hi> And forasmuch as the trapesium <hi>GBDC</hi> is sup<g ref="char:EOLhyphen"/>posed to be the halfe of the parrallelograme, and the halfe of the same parallelograme is the triangle <hi>EBD</hi> (by this proposition). From the trapesium <hi>GBDC</hi> and the trian<g ref="char:EOLhyphen"/>gle <hi>EBD</hi> which are equall, take away the trapesium <hi>GBDI</hi> which is common to them both, and the residue namely, the triangle <hi>DIC</hi> shalbe equall to the residue, namely, to the tria<g ref="char:cmbAbbrStroke">̄</g>gle <hi>EIG</hi>: but it is also lesse (as hath before ben proued): which is impossible. Wherefore a right line deuiding a parallelogramme into two equall partes, shall not deuide the diameter thereof vnequally. Wherefore it shall deuide it equally: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>An addition of Pelitarius.</head>
                  <p>
                     <hi>Betwene two right lines being infinite and making an angle geuen:</hi>
                     <note place="margin">
                        <hi>An addition o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> Pelitarius.</hi>
                     </note> 
                     <hi>to place a line equall to a line geuen, in such sorte, that it shall make with one of those lines an angle equall to an other angle ge<g ref="char:EOLhyphen"/>uen. Now it behoueth that the two angles geuen be lesse then two right angles.</hi>
                  </p>
                  <p>Suppose that there be two lines <hi>AB</hi> and <hi>AC,</hi> making an angle geuen <hi>BAC</hi>: and let them be infinite on that side where they open one from the other. And let the line geuen be D, and let the other angle geuen be <hi>E.</hi>
                     <figure/>
And let the two angles <hi>A</hi> and <hi>E</hi> be lesse then two right angles (otherwise there coulde not be made a triangle, as it is manifest by the 17 proposition). It is required betwene the lines <hi>AB</hi> and <hi>AC</hi> to place a line equall to the line geuen <hi>D,</hi> which with one of them as for exa<g ref="char:cmbAbbrStroke">̄</g>ple with the line <hi>AC,</hi>
                     <note place="margin">
                        <hi>Construction.</hi>
                     </note> may make an angle equal to the angle ge<g ref="char:EOLunhyphen"/>uen <hi>E.</hi> Now then vpon the line <hi>AC</hi> and to the point in it <hi>A,</hi> make an angle equall to the angle geue<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>E</hi> (by the 23. proposition), which let be <hi>CAF.</hi> And produce the line <hi>FA</hi> on the other side of the point <hi>A</hi> to the point <hi>G</hi>: and let <hi>AG</hi> be equall to the line geuen <hi>D</hi> (by the 3. proposition). And by the point <hi>G,</hi> draw (by the 31. proposition) a parallel line to the line <hi>AC,</hi> which let be <hi>GH,</hi> and produce it vntil it concurre with the line <hi>AB</hi>: which concurse let be in the point <hi>H.</hi> And agayne by the point <hi>H</hi> draw the line <hi>HK</hi> parallel vnto the line <hi>GF:</hi> which let cut the line <hi>AC</hi> in the point <hi>K.</hi> Then I say that the line <hi>HK</hi> is placed betwene the lines <hi>AB</hi> &amp; <hi>AC</hi> &amp; is equall to the line <hi>D.</hi> And that the angle at the point <hi>K</hi> is equall to the angle geuen <hi>E.</hi>
                     <note place="margin">
                        <hi>Demonstration</hi>
                     </note> For forasmuch as (by construction) <hi>AGHK</hi> is a parallelogramme the line <hi>KH</hi> is equall to the line <hi>AG</hi> (by this proposition). Wherefore also it is equall to the line <hi>D.</hi> And forasmuch as the line <hi>AK</hi> fall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th vppon the two parallel lines, <hi>FG</hi> and <hi>KH,</hi> therfore the angle <hi>AKH</hi> is equal to the angle <hi>FAK</hi> (by the 29. propositio<g ref="char:cmbAbbrStroke">̄</g>) for that they are alternate angles. Wherfore also the same angle at the point <hi>K</hi> is equal to the angle geuen <hi>E.</hi> Wherefore the line <hi>HK</hi> being placed betwene the two lines <hi>AB</hi> and <hi>AC</hi> and being equall to the line <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> maketh the angle at the point K equall to the angle geuen <hi>E</hi>: which was required to be done.</p>
                  <p>Though this addition of Pelitarius be not so muche pertayning to the
<pb facs="tcp:7062:76"/>
proposition: yet because it is witty and semeth somewhat difficult, I thought it good here to anexe it.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>25.</hi> Theoreme. The <hi>35.</hi> Proposition. Parallelogrammes consisting vppon one and the same base, and in the selfe same parallel lines, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that these parallelo<g ref="char:EOLhyphen"/>grammes ABCD and EB
<figure/>
CF do consist vpon one and the same base, that is, vppon BC, and in the selfe same parallel lines, that is AF, and BC. Then I say, that the parallelograme ABCD is equal to the parallelograme EBCF.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>For forasmuch as ABCD is a parallelogramme, ther<g ref="char:EOLhyphen"/>fore (by the</hi> 34. <hi>proposition) the side AD, is equall to the side BC, and by the same reason also the side EF is equall to the side BC, wherfore AD is equall to EF: and DE is common to them both. VVherfore the whole line AE is equall to the whole line DF. And the side AB is equall to y<hi rend="sup">e</hi> side DC: wherfore these two EA and AB are equall to these two FD and DC, the one to the other: and y<hi rend="sup">e</hi> angle FDC is equall to the angle EAB, namely, the outward angle to y<hi rend="sup">e</hi> inward angle (by y<hi rend="sup">e</hi>
                  </hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>9. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>): wherfore (by y<hi rend="sup">e</hi>
                  </hi> 4 <hi>proposition) the base EB is equall to the base FC, and the triangle EAB is equall to the triangle FDC. Take away the triangle DGE, which is common to them both. VVherefore the residue, namely, the trapesium ABGD is equall to the residue, that is, to the trapesium EGCF. Put the triangle GBC commo<g ref="char:cmbAbbrStroke">̄</g> to them both. VVherefore the whole parallelogramme ABCD is equall to the whole parallelogramme EBCF. VVherefore parallelogrammes consisting vp<g ref="char:EOLhyphen"/>on one and the same base, and in the selfe same parallel lines, are equal the one to the other: which was required to be demonstrated.</hi>
               </p>
               <p>Parallelogrammes are sayde to be in the selfe same parallel lines, when their bases, and the opposite sides vnto them, are one and the selfe same lynes wyth the parallels.</p>
               <p>
                  <note place="margin">
                     <p>
                        <hi>Three cases in this proposition.</hi>
                     </p>
                     <p>
                        <hi>The first case.</hi>
                     </p>
                  </note>In this proposition are three cases. For the line BE may cutte the line AF, either beyond the point D, or in the point D, or on this side the point D. When
<pb n="46" facs="tcp:7062:76"/>
it cutteth the line AF beyond the point D the demonstration before put ser<g ref="char:EOLhyphen"/>ueth.</p>
               <p>But if the line <hi>BE</hi> do cutte the line <hi>AF</hi> in
<figure/>
the point D,<note place="margin">
                     <hi>The second case.</hi>
                  </note> then forasmuch as (by the former proposition) the triangle <hi>BCD</hi> or <hi>BCE</hi> is the halfe of either of these para<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lelogrammes <hi>ABCD</hi> and <hi>EBCF</hi> (for in the parallelogramme <hi>ABCD</hi> the diameter <hi>BD</hi> maketh the triangle <hi>B</hi>D<hi>C</hi> the halfe of the same parallelogramme, and in the parallelogramme <hi>EBCF</hi> the diameter <hi>EC</hi> or DC maketh the selfe same triangle <hi>BDC</hi> the halfe of the parallelograme <hi>EBCF</hi>) ther<g ref="char:EOLhyphen"/>fore (by the 7. common sentence) the parallelo<g ref="char:EOLhyphen"/>grammes <hi>ABC</hi>D and <hi>EBCF</hi> are equall.</p>
               <p>But if the liue <hi>BE</hi> do cutte the lyne <hi>AF</hi>
                  <figure/>
on this side the point D,<note place="margin">
                     <hi>The third case.</hi>
                  </note> then forasmuch as ey<g ref="char:EOLhyphen"/>ther of the lines AD and <hi>EF</hi> is equall to the line <hi>BC,</hi> therefore by the first common sentence they are equall the one to the other. Wherefore taking away ED, which is common to both, the residue <hi>AE</hi> shalbe equall to the residue DF. Agayne forasmuch as (by the 34. propositio<g ref="char:cmbAbbrStroke">̄</g>) the line <hi>AB</hi> is equall to the line CD, and (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7. proposition) the angle <hi>EAB</hi> is equal to the angle FDC: therfore (by the 4. proposi<g ref="char:EOLhyphen"/>tion) the triangles <hi>EBA</hi> and FCD are equal. Adde the trapesium CDEB common to them both: and so (by the seconde common sen<g ref="char:EOLhyphen"/>tence) the two parallelogrammes <hi>ABCD</hi> and <hi>EBCF</hi> shalbe equall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>26.</hi> Theoreme. The <hi>36.</hi> Proposition. Parallelogrammes consisting vpon equall bases, and in the selfe same parallel lines, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that these parallelogrammes ABCD and EFGH do consist vpon equall bases, that is, vpon BC and FG, and in the selfe same paral<g ref="char:EOLhyphen"/>lel lines, that is, AH and BG. Then I say, that the parallelogramme ABCD is equall to the parallelogramme EFGH.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Draw a right line from the point B to the point E, and an other from the point C to the point H.</hi>
                  <note place="margin">
                     <hi>Demonstration</hi>
                  </note> 
                  <hi>And for<g ref="char:EOLhyphen"/>asmuch as BC is equall to FG, but FG is equall to EH, therfore BC also is e<g ref="char:EOLhyphen"/>quall to EH, and they are parallel lines, and the lines BE and CH ioyne them together: but two right lynes ioynyng together two equall right
<pb facs="tcp:7062:77"/>
lines being parallels, are themselues
<figure/>
also (by the</hi> 33 <hi>proposition) equall the one to the other, and parallels. VVherfore EBCH is a parallelo<g ref="char:EOLhyphen"/>gramme, and is equall to the paralle<g ref="char:EOLhyphen"/>lograme ABCD for they haue both one and y<hi rend="sup">e</hi> same base, that is, BC. And are in y<hi rend="sup">e</hi> selfe same parallel lines, that is, BC &amp; EH. And by y<hi rend="sup">e</hi> same reason also the parallelograme EFGH is e<g ref="char:EOLhyphen"/>qual to the parallelograme EBCH. VVherfore the parallelograme ABCD is equal to the parallelograme EFGH. VVherfore parallelogrames consisting vppon equall bases, and in the selfe same parrallel lines, are equall the one to the oth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r: which was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Three cases in this proposition.</hi>
                  </note>In this proposition also are three cases. For the equall bases may either be vt<g ref="char:EOLhyphen"/>terly seperated a sonder: or they may touche a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> one of the endes: or they may haue one part common to them both.</p>
               <p>
                  <note place="margin">
                     <hi>The first case.</hi>
                  </note>Euclides demonstration serueth when the bases be vtterly seperated a sonder. VVhich yet may happen seuen diuers wayes.<note place="margin">
                     <hi>Euery case may happen seuen diuers wayes.</hi>
                  </note> For the bases being seperated a so<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>der, their opposite sides also may be vtterly seperated a sonder beyond the point D, as the sides AD and EH in the first figure.</p>
               <p>
                  <milestone type="tcpmilestone" unit="unspecified" n="2"/> Or they may touche together in one of the endes, and the whole side may be beyond the point D, as the sides AD and EH do in the second figure.</p>
               <p>
                  <milestone type="tcpmilestone" unit="unspecified" n="3"/> Or one part may be beyond the point D, and an other part common to them
<figure/>
                  <pb n="47" facs="tcp:7062:77"/>
both, as in the third figure, the sides AD and EH haue the part ED common to them both.</p>
               <p>Or they may iustly agree the one with the other, that is, the pointes A and D <milestone type="tcpmilestone" unit="unspecified" n="4"/> may fall vpon the pointes E and H: as in the fourth figure.</p>
               <p>Or the side AD being produced on this side the point A, part of the opposite <milestone type="tcpmilestone" unit="unspecified" n="5"/> side vnto the base FG may be on this side the point A, and an other part may be common with the line AD, as in the fifth figure.</p>
               <p>Or one ende of the side EH may light vpon the pointe A, and the whole side <milestone type="tcpmilestone" unit="unspecified" n="6"/> on this side of it: As in the sixt figure.</p>
               <p>Or the said side EH may vtterly be seperated a sonder on this side the pointe <milestone type="tcpmilestone" unit="unspecified" n="7"/> A, as in the seuenth figure.</p>
               <p>And the two other cases also may in like maner haue seuen variet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es:<note place="margin">
                     <hi>The like varie<g ref="char:EOLhyphen"/>ty in ech of the other two cases.</hi>
                  </note> as in the figures here vnderneth and on the other side of this leafe set it is manifest. And here is to be noted, that in these three cases and in all their varieties also,<note place="margin">
                     <hi>Euclides con<g ref="char:EOLhyphen"/>struction and demostration serueth in all these cases, and in their varities also.</hi>
                  </note> the construction &amp; demonstration put by Euclide (namely, the drawing of lines fro<g ref="char:cmbAbbrStroke">̄</g> the point B to the point E &amp; from the pointe C to the point H, and so prouing it by the former proposition) will serue onely in the fourth varietie of ech case, there nedeth no farther construction: for that the conclusion straight way follo<g ref="char:EOLhyphen"/>weth by the former proposition.</p>
               <p>
                  <figure/>
                  <pb facs="tcp:7062:78"/>
                  <figure/>
               </p>
            </div>
            <div type="part">
               <head>The <hi>27.</hi> Theoreme. The <hi>37.</hi> Proposition. Triangles consisting vpon one and the selfe same base, and in the selfe same paralles: are equall the one to the other.</head>
               <p>
                  <hi>SVppose that these triangles ABC and DBC do consist vpon one and the same base, namely, BC, and in the selfe same parallel lines, that is, AD and BC. Then I say, that the triangle ABC is equall to the triangle BDC.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Produce (by the 2. peticion) the line AD on ech side to the pointes E and F. And (by the</hi> 31. <hi>proposition) by the point B, draw vnto the line CA a parallel line BE and (by y<hi rend="sup">e</hi> same) by the point C, draw vnto the line BD a pa<g ref="char:EOLhyphen"/>rallel line CF.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>VVherefore EBCA.
<figure/>
and DBCF are parallelogrammes. And the parallelogramme EBCA, is (by the</hi> 35. <hi>proposition) equall to the parallelogramme DBCF. For they consist vppon one and the selfe same base, namely, BC, and are in the selfe same parallel lines, that is, BC and EF. But the triangle ABC is (by the</hi> 34. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>) the halfe of the parallelogramme EBCA, for the diameter AB deuideth it into two equall parts: &amp; (by the same) the tri<g ref="char:EOLhyphen"/>angle DBC is the halfe of the pa<g ref="char:EOLhyphen"/>rallelogramme DBCF, for the diameter DC deuideth it into two equall parts: but the halues of thinges equall are also equall the one to the other (by the</hi> 7. <hi>common sentence), wherefore ehe triangle ABC is equall to the triangle DBC VVherefore triangles consisting vpon one and the selfe same base, and in the selfe same parallels: are equall the one to the other: which was required to be demonstrated.</hi>
               </p>
               <p>
                  <pb n="48" facs="tcp:7062:78" rendition="simple:additions"/>Those triangles are saide to be contayned within the selfe same parallel lines,<note place="margin">
                     <hi>How triangles are sayde to be in the selfe same parallel lines.</hi>
                  </note> which hauing their bases in one of the parallel lines, haue their <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oppes in the other.</p>
               <p>Here as I promised will I shew out of Proclus the comparison of two trian<g ref="char:EOLhyphen"/>gles, which hauing their sides equall, haue the bases and angles at the toppe v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e<g ref="char:EOLhyphen"/>quall.<note place="margin">
                     <hi>Comparison of two triangles whose sides be<g ref="char:EOLhyphen"/>ing equal, their bases and angles at the toppe are vnequall.</hi>
                  </note> And first I say that the vnequall angles at the toppe being equall to two right angles, the triangles shalbe equall. As for example.</p>
               <p>Suppose that these two triangles <hi>ABC</hi> and <hi>DEF</hi> haue two sides of the one, name<g ref="char:EOLhyphen"/>ly, <hi>AB</hi> and <hi>AC,</hi> equall to two sides of the other, namely, to <hi>DE</hi> and <hi>DF,</hi> eche to his correspondent side,<note place="margin">
                     <hi>When the two angles at the toppes are equall to two right an<g ref="char:EOLhyphen"/>gles.</hi>
                  </note> that is, <hi>AB</hi> to <hi>DE,</hi> and <hi>AC</hi> to <hi>DF,</hi> and let the base <hi>BC</hi> be greater then the base <hi>EF</hi>: and let the angle at the point <hi>A</hi> be greater then the angle at the point D. But let the sayde angles at the pointes <hi>A</hi> and D,
<figure/>
be equall to two right angles. Then I say that the triangles <hi>ABC</hi> and D<hi>EF</hi> are equall. For forasmuch as the angle <hi>BAC</hi> is greater then the angle <hi>E</hi>D<hi>F,</hi> vpon the line <hi>E</hi>D, and to the point D describe (by the 23. proposition) an angle equall to the angle <hi>BAC,</hi> which let be EDG: and put the line DG equall to the line <hi>AC</hi>: and draw a line from E to G, and an other from <hi>F</hi> to G: and produce the lines ED &amp; FD beyond the poynt D to the pointes <hi>H</hi> and <hi>K.</hi> Now for<g ref="char:EOLunhyphen"/>asmuch as the angle <hi>BAC</hi> is equall to the angle EDG, and the angles <hi>BAC</hi> and EDF are equall to two right an<g ref="char:EOLhyphen"/>gles, therefore the angles EDG and EDF are equall to two right angles. But the angles EDG and KDG, are also equal to two right angles: take away the angle FDG com<g ref="char:EOLhyphen"/>mon to them both: wherefore the angle remayning EDF is equall to the angle remayning GDK. But the angle EDF is equall to the angle HDK (by the 15. proposition) for they are hed angles. Wherefore the angle GDK is equall to the angle HDK. And forasmuch as in the triangle GDF the outward angle GDH is (by the 32. proposition) equal to the two inward and opposite angles at the points <hi>G</hi> and <hi>F</hi>: which two angles also are (by the 5. proposition) equall the one to the other: for the line DG is by construction equall to the line <hi>AC,</hi> namely, to the line DF. Wherefore the angle GDH is double both to the angle at the point <hi>G,</hi> and to the an<g ref="char:EOLhyphen"/>gle at the point <hi>F.</hi> But the angle <hi>GDH</hi> is also double to the angle GDK (for the an<g ref="char:EOLhyphen"/>gle GDK is proued to be equall to the angle KDH) wherefore the angle at DGF is equall to the angle GDK: and they are alternate angles. Wherefore (by the 27. proposition) the line D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is a parallel to the line <hi>FG.</hi> Wherefore the triangles GDE and FDE are vppon one and the selfe same base, namely, <hi>DE,</hi> and in the selfe same parallel lines: DE and <hi>GF.</hi> Wherefore by this proposition they are equall. But the tri<g ref="char:EOLhyphen"/>angle GDE is by construction equall to the triangle <hi>ABC.</hi> Wherefore also the tri<g ref="char:EOLhyphen"/>angle DEF is equall to the triangle <hi>ABC</hi>: which was required to be proued.</p>
               <p>But now let the angles <hi>BAC</hi> and <hi>EDF</hi> be greater then two right angles:<note place="margin">
                     <hi>When they are greater the<g ref="char:cmbAbbrStroke">̄</g> two right angles.</hi>
                  </note> &amp; let the angle at the point <hi>A</hi> be greater then the angle at the point <hi>D,</hi> as it was before. The<g ref="char:cmbAbbrStroke">̄</g> I say that the triangle <hi>ABC</hi> is le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se then the triangle <hi>DEF.</hi> Let the same construction be here that was in the former. And forasmuch as the angles <hi>BAC</hi> and <hi>EDF,</hi> that is, the angles <hi>EDG</hi> and <hi>EDF</hi> are greater then two right angles, but the angles <hi>EDG</hi> and <hi>GDK</hi> are equall to two right angles: take away the angle <hi>FDG</hi> which is common to them both. Wherefore the angle remayning, namely, <hi>EDF</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> the angle remayning, namely, then <hi>GDK</hi>: that is, the angle <hi>KDH</hi> (which by the 15. proposition is equall to the angle <hi>EDF</hi>) is greater then the angle <hi>GDK,</hi> wherefore the angle <hi>GDH</hi> is more then double to the angle <hi>GDK</hi>: but the angle <hi>GDH</hi> is double to the an<g ref="char:EOLhyphen"/>gle
<pb facs="tcp:7062:79"/>
                  <hi>DGF,</hi> as was before proued. Wherefore the
<figure/>
angle <hi>GDK</hi> is lesse then the angle <hi>DGF.</hi> Vnto the angle <hi>GDK</hi> put (by the 23. proposition) the an<g ref="char:EOLhyphen"/>gle <hi>DGL</hi> equall: and produce the line <hi>GL</hi> till it concurre with the line <hi>EF</hi> in the pointe <hi>L.</hi> And draw a line from <hi>D</hi> to <hi>L.</hi> Wherefore (by the 27. proposition) <hi>GL</hi> is a parallel line to <hi>DE,</hi> for that the alternate angles <hi>DGL</hi> and <hi>GDK</hi> are equal. Wherfore the triangles <hi>GDE</hi> and <hi>LDE</hi> are (by this proposition) equal (for they consist vpon one and the self same base, namely, <hi>DE,</hi> and are in the selfe same parallel lines, namely, <hi>ED</hi> and <hi>GL</hi>) But the triangle <hi>LDE</hi> is lesse then the triangle <hi>FDE,</hi> Wherfore also the triangle <hi>GDE</hi> is lesse the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the triangle <hi>FDE.</hi> But the triangle <hi>GDE</hi> is equal to the triangle <hi>ABC.</hi> Wherfore the triangle <hi>ABC</hi> is lesse then the triangle <hi>DEF</hi>: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               <p>
                  <note place="margin">
                     <hi>When they are lesse then two right angles.</hi>
                  </note>But now let the angles <hi>BAC</hi> and <hi>EDF</hi> be lesse then two right angles: and agayne let the angle at the pointe <hi>A</hi> be greater then the angle at the point <hi>D.</hi> Then <hi>I</hi> say that the tria<g ref="char:cmbAbbrStroke">̄</g>gle <hi>ABC</hi> is grea<g ref="char:EOLhyphen"/>ter then the triangle <hi>DEF.</hi> Let the same constru<g ref="char:EOLhyphen"/>ctio<g ref="char:cmbAbbrStroke">̄</g> be also here that was in the two former. And forasmuch as the angles <hi>BAC</hi> and <hi>EDF,</hi> that is, the angles <hi>EDG</hi> &amp; <hi>EDF,</hi> are lesse then two right angles, but the angles <hi>EDG</hi> and <hi>GDK</hi> are equal to two right angles, take away the angle <hi>FDG</hi> which is common to them both, wherefore the angle remayning, namely, <hi>EDF</hi> is lesse then the angle remayning, namely, then <hi>GDK</hi>: that is, the angle <hi>HDK</hi> (which by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. proposition is e<g ref="char:EOLhyphen"/>quall to the angle <hi>E</hi>D<hi>F</hi>) is lesse then the angle <hi>GDK.</hi> Wherfore the whole angle <hi>G</hi>D<hi>H</hi> is lesse then double to the angle <hi>G</hi>D<hi>K.</hi> But it is double to the angle DGF (as before it was proued): wherfore the angle <hi>G</hi>D<hi>K</hi> is grea<g ref="char:EOLhyphen"/>ter then the angle DGF. Put the angle DGL equall to the angle <hi>G</hi>D<hi>K</hi> (by the 23. proposition) and produce the line <hi>GL</hi> till it concurre with the line <hi>EF</hi> also produced<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp; let the concurse be in the point <hi>L.</hi> And draw a line from D to <hi>L.</hi> And for as much as the angle D<hi>GL</hi> is equall to the angle <hi>G</hi>D<hi>K,</hi> and they are alternate angles, therefore the line <hi>GL</hi> is a parallel to D<hi>E</hi> (by the 27. proposition). Wherefore (by this propositi<g ref="char:EOLhyphen"/>on) the triangles <hi>G</hi>D<hi>E</hi> and <hi>L</hi>D<hi>E</hi> are equal: but the triangle LDE is greater then the triangle <hi>F</hi>D<hi>E,</hi> and the triangle <hi>G</hi>D<hi>E</hi> is equall to the triangle <hi>ABC.</hi> Wherefore the triangle <hi>ABC</hi> is greater then the triangle D<hi>EF</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>28.</hi> Theoreme. The <hi>38.</hi> Proposition. Triangles which consist vppon equall bases, and in the selfe same parallel lines, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that these triangles ABC and DEF do consist vpon equal ba<g ref="char:EOLhyphen"/>ses, that is, vpon BC and EF, and in the selfe same parallel lines, that is BF and AD. Then I say that the triangle ABC is equall to the trian<g ref="char:EOLhyphen"/>gle
<pb n="49" facs="tcp:7062:79"/>
ABC is equall to the triangle DEF.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Produce (by the second peticion) the line AD on eche side to the pointes G and H. And (by the</hi> 31. <hi>proposition) by the point B drawe vnto CA a paral<g ref="char:EOLhyphen"/>lel
<figure/>
line BG, and (by the same) by the pointe F drawe vnto DE a parallel line FH.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>VVherfore GBCA and DEFH are parallelogrammes. But the parallelograme GBCA is (by the</hi> 36 <hi>proposition) equal to the paralle<g ref="char:EOLhyphen"/>logramme DEFH, for they consist vpon equall bases, that is, BC and EF, and are in the selfe same parallel lines, that is, BF and GH. But (by the</hi> 34. <hi>proposition) the triangle ABC is the halfe of the parallelogramme GBCA, for the diameter AB deui<g ref="char:EOLhyphen"/>deth it into two equall partes: and the triangle DEF is (by the same) the halfe of the parallelogramme DEFH, for the diameter FD deuideth it into two e<g ref="char:EOLhyphen"/>quall partes. But the halues of thinges equall are (by the</hi> 7. <hi>common sentence) e<g ref="char:EOLhyphen"/>quall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he one to the other. VVherfore the triangle ABC is equall to the trian<g ref="char:EOLhyphen"/>gle DEF. VVherefore triangles which consist vppon equall bases, and in the selfe same parallel lines, are equall the one to the other: which was required to be proued.</hi>
               </p>
               <p>In this proposition are three cases.<note place="margin">
                     <hi>Thre cases in this proposition.</hi>
                  </note> For the bases of the triangles either haue one part common to them both or the base of the one toucheth the base of the other onely in a point: or their bases are vtterly seuered a sunder. And ech of these cases may also be diuersly,<note place="margin">
                     <hi>Ech of these ca<g ref="char:EOLhyphen"/>ses also may be diuersly.</hi>
                  </note> as we before haue sene in parallelogrammes con<g ref="char:EOLunhyphen"/>sisting on equall bases, and being in the selfe same parallel lines. So that he which diligently noteth the variety that was there put touching them, may also easely frame the same varietie to ech case in this proposition. VVherefore I thinke it nedeles here to repeate the same agayne: for how soeuer the bases be put, or the toppes, the manner of construction and demonstration here put by Euclide will serue: namely, to draw parallel lines to the sides.</p>
               <div type="section">
                  <head>An addition of Pelitarius.</head>
                  <p>
                     <hi>To deuide a triangle geuen into two equall partes.</hi>
                  </p>
                  <p>Suppose that the triangle geuen to be deuided in to two
<figure/>
equall partes be <hi>ABC.</hi>
                     <note place="margin">
                        <hi>An addition of Pelitarius, to deuide a trian<g ref="char:EOLhyphen"/>gle into two e<g ref="char:EOLhyphen"/>quall partes.</hi>
                     </note> Deuide one of the sides therof, namely, <hi>BC</hi> into two equall partes (by the 10. propo<g ref="char:EOLhyphen"/>sition) in the point D. And draw a line from the point D to the point <hi>A.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say that the two triangles <hi>ABD</hi> &amp; <hi>ACD,</hi> are equal, which is easy to proue (by the 38. pro<g ref="char:EOLhyphen"/>position) if by the point <hi>A</hi> we drawe vnto the line <hi>BC</hi> a paral<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>el line (by the 31. proposition), which let by <hi>HK</hi>: for so the triangles <hi>AB</hi> D and <hi>ADC,</hi> consisting vppon equal bases <hi>BD</hi> &amp; <hi>DC,</hi> and being in the selfe same paral<g ref="char:EOLhyphen"/>lel lines <hi>HK</hi> and <hi>BC</hi> are of necessitie equall. The selfe
<pb facs="tcp:7062:80"/>
same thing also wil happen if the side <hi>BA</hi> be deuided into two equall parts in the point <hi>E,</hi> and so be drawen a right line from the point <hi>E,</hi> to the point <hi>C.</hi> Or if the side <hi>AC</hi> be deuided into two equall partes in the point <hi>F,</hi> and so be drawen a right line from the point <hi>F</hi> to the point <hi>B</hi>: which is in like manner proued by drawing parallel lines by the pointes <hi>B, and C,</hi> to the lines <hi>BA</hi> and <hi>AC,</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Note.</hi>
                     </note>And so by this you may deuide any triangle into so many partes as are sig<g ref="char:EOLhyphen"/>nified by any number that is euenly euen: as into 14.16.32.64. &amp;c.</p>
               </div>
               <div type="section">
                  <head>An other addition of Pelitarius.<note place="margin">
                        <hi>An other addi<g ref="char:EOLhyphen"/>tion of Pelita<g ref="char:EOLhyphen"/>rius.</hi>
                     </note>
                  </head>
                  <p>
                     <hi>From any point geuen in one of the sides of a triangle, to draw a line which shal deuide the trian<g ref="char:EOLhyphen"/>gle into two equall partes.</hi>
                  </p>
                  <p>Let the triangle geuen be <hi>BC</hi>D: and let the point geuen in the side <hi>BC</hi> be <hi>A.</hi> It is required from the point <hi>A</hi> to draw a line which shal deuide the triangle <hi>BCD</hi> into two equall partes.<note place="margin">
                        <hi>Construction.</hi>
                     </note> Deuide the side <hi>BC</hi> into two equall partes in the point <hi>E.</hi> And drawe a right line from the point <hi>A</hi> to the point <hi>D.</hi> And (by the
<figure/>
31. proposition) by the point <hi>E</hi> draw vnto the line <hi>AD</hi> a parallel line <hi>EF</hi>: which let cutte the side <hi>DC</hi> in the point <hi>F.</hi> And draw a line from the point <hi>A</hi> to the point <hi>F.</hi> Then I say that the line <hi>AF</hi> deuideth the triangle <hi>BCD</hi> into two equall partes: namely, the trapesium <hi>ABDF</hi> is equall to the triangle <hi>ACF.</hi> For draw a line from <hi>E</hi> to <hi>D,</hi>
                     <note place="margin">
                        <hi>Demonstration</hi>
                     </note> cutting the line <hi>AF</hi> in the point <hi>G.</hi> Now then it is manifest (by the 38. proposition) that the two trian<g ref="char:EOLhyphen"/>gles <hi>BED</hi> and <hi>CED</hi> are equall (if we vnderstand a line to be drawen by the point D parallel to the line <hi>BC,</hi> for the bases <hi>BE</hi> and <hi>EC</hi> are equal). The two triangles also D<hi>EF</hi> and <hi>AEF</hi> are (by the 37. proposition) equall: for they consist vpon one and the selfe same base <hi>EF,</hi> and are in the selfe same parallel lines <hi>A</hi>D and <hi>EF.</hi> Wherefore taking away the triangle <hi>EFG</hi> which is co<g ref="char:cmbAbbrStroke">̄</g>mo<g ref="char:cmbAbbrStroke">̄</g> to the<g ref="char:cmbAbbrStroke">̄</g> both, the triangle <hi>AEG</hi> shalbe equall to the triangle D<hi>FG</hi>: wher<g ref="char:EOLunhyphen"/>fore vnto either of the<g ref="char:cmbAbbrStroke">̄</g> adde the trapesiu<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>CFGE,</hi> and the triangle <hi>ACF</hi> shalbe equal to the triangle <hi>DEC.</hi> But the triangle <hi>DEC</hi> is the halfe part of the whole triangle <hi>BCD</hi> wherefore the triangle <hi>ACF</hi> is the halfe part of the same triangle <hi>BCD.</hi> Wherfore the residue, namely, the trapesium <hi>ABFD</hi> is the other halfe of the same triangle. Where<g ref="char:EOLhyphen"/>fore the line <hi>AF</hi> deuideth the whole triangle <hi>BCD</hi> into two equall partes: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>29.</hi> Theoreme. The <hi>39.</hi> Proposition. Equall triangles consisting vpon one and the same base, and on one and the same side: are also in the selfe same parallel lines.</head>
               <p>
                  <hi>SVppose that these two equall triangles ABC and DBC do consist vp<g ref="char:EOLhyphen"/>pon one and the same base, namely, BC and on one and the same side. The<g ref="char:cmbAbbrStroke">̄</g> I say that they are in the selfe same parallel lines. Drawe a right line from the point A to the point D. Now I say that AD is a parallel line to BC. For if not, then (by the</hi> 31. <hi>proposition) by the point A drawe vnto the right line BC a parallel line AE, and draw a right line from the point E to the
<pb n="50" facs="tcp:7062:80"/>
point C. VVherfore y<hi rend="sup">e</hi> triangle EBC is
<figure/>
(by y<hi rend="sup">e</hi>
                  </hi> 37. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>) equal to the triangle ABC, for they consist vpon one and the selfe same base, namely, BC, and are in y<hi rend="sup">e</hi> selfe same parallels, that is, AE and BC. But the triangle DBC is (by supposition) equall to the triangle ABC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> VVherfore the triangle DBC is equal to the triangle EBC, the greater vnto the lesse: which is impossible. VVhere<g ref="char:EOLhyphen"/>fore the line AE, is not a parallel to the line BC. And in like sorte may it be proued that no other line besides AD is a parallel line to BC, wherefore AD is a parallel line to BC. VVherfore equall triangles consisting vpon one and the same base, and on one and the same side, are also in the selfe same parallel lines: which was required to be proued.</hi>
               </p>
               <p>This proposition is the conuerse of the 37. proposition.<note place="margin">
                     <hi>This Theoreme the conuerse of the</hi> 37. <hi>proposi<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> And here is to be noted that if by the point A, you draw vnto the line BC a parallel line, the same shal of necessitie either light vpo<g ref="char:cmbAbbrStroke">̄</g> the point D, or vnder it, or aboue it. If it light vpo<g ref="char:cmbAbbrStroke">̄</g> it, then is that manifest which is required: but if it light vnder it, then foloweth that absurditie which Euclide putteth, namely, that the greater triangle is equall to the lesse: which selfe same absurditie also will follow, if it fall aboue the point D. As for example.</p>
               <p>Suppose that these equall triangles <hi>ABC</hi> and D<hi>BC</hi> do consist vppon one and the selfe same base <hi>BC,</hi> and on one and the same side. Then <hi>I</hi>
                  <figure/>
say, that they are in the selfe same parallel lines, and that a right line ioyning together their toppes is a parallel to the base <hi>BC.</hi> Draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>A</hi> to <hi>D.</hi> Now if this be not a parallel to the base <hi>BC,</hi> let <hi>AE</hi> be a parallel vnto it, and let <hi>AE</hi> fall without the line <hi>AD.</hi> And produce the line <hi>BD</hi> till it concurre with the line <hi>AE</hi> in the pointe <hi>E</hi> and draw a line from <hi>E</hi> to <hi>C.</hi> Wherfore the triangle <hi>ABC</hi> is equal to the triangle <hi>EBC</hi>: but the triangle <hi>ABC</hi> is equall to the triangle <hi>DBC</hi>: Wherfore the triangle <hi>EBC</hi> is equall to the triangle D<hi>BC.</hi> Namely, the whole to the part: which is impossible. Wherfore the parallel line fal<g ref="char:EOLhyphen"/>leth not without the line <hi>AD.</hi> And Euclide hath proued that it falleth not within. Wherfore the line <hi>AD</hi> is a pa<g ref="char:EOLhyphen"/>rallel vnto the line <hi>BC.</hi> Wherfore equall triangles which are on the selfe same side, and on one and the selfe same base, are also in the selfe same parallel lines: which was required to be proued.</p>
               <div type="section">
                  <head>An addition of Flussates.</head>
                  <p>The selfe same also followeth in parallelogrames.<note place="margin">
                        <hi>An addition of Fl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ssases.</hi>
                     </note> For if vpon the base AB be set on one &amp; the same side these equal parallelogrames ABCD &amp; ABGE, they shall of necessitie be in the selfe same parallel lines. For if not, but one of them is
<pb facs="tcp:7062:81"/>
set eyther within or without, let the parallelo<g ref="char:EOLhyphen"/>grame
<figure/>
B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> being equall to the parallelograme ABCD be set within the same parallel lines: wherefore the same parallelograme BF beyng e<g ref="char:EOLhyphen"/>quall to the parallelograme ABCD (by the 35. proposition) shall also be equall to the other pa<g ref="char:EOLhyphen"/>rallelograme ABGE (by the first common sen<g ref="char:EOLhyphen"/>tence) For the parallelograme ABGE is by sup<g ref="char:EOLhyphen"/>position equall to the parallelogramme ABCD: whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore the parallelograme BF being equall to the parallelograme ABGE, the parte shall bee e<g ref="char:EOLhyphen"/>qual to the whole, which is absurde. The same in<g ref="char:EOLhyphen"/>conuenience also will followe, if it fall without. VVherefore it can neither fall within nor with<g ref="char:EOLhyphen"/>out. VVherfore equall parallelogrames beyng vpon one and the selfe same base and on one and the same side, are also in the selfe same parallel lines.</p>
               </div>
               <div type="section">
                  <head>An addition of Campanus.</head>
                  <p>
                     <note place="margin">
                        <hi>An addition of Campanus.</hi>
                     </note>
                     <hi>If a right line deuide two sides of a triangle into two equall partes: it shall be equidistant vnto the third side.</hi>
                  </p>
                  <p>Suppose that there be a triangle ABC: and let there bee a right lyne DE, which let deuide the two sides AB and BC into two equall partes in the pointes D and E Then <hi>I</hi> say, that the line DE is a parallel to the line A
<figure/>
C. Drawe these two lines AE and DC. Now then imagining a line to be drawne by the point E parallel to the line AB, the triangle BDE shall (by the 38. proposition) bee equall to the triangle DAE (for their two bases AD and DB are put to be equall) And by the same reason the triangle BDE is equall to the triangle CED. VVherfore (by the first common sente<g ref="char:cmbAbbrStroke">̄</g>ce) the triangles EAD and ECD are equall, and they are erected on one and the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e same base, namely, DE, and on one and the same side. VVherefore (by the 39. proposition) they are in the selfe same parallel lines, and the line which ioyneth together their toppes is a parallel to their base. VVherfore the lynes DE and AC are pa<g ref="char:EOLhyphen"/>ralles: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>30.</hi> Theoreme. The <hi>40.</hi> Proposition. Equall triangles consisting vpon equall bases, and in one and the same side: are also in the selfe same parallel lines.</head>
               <p>SVppose that these equall triangles ABC and CDE do consist vppon e<g ref="char:EOLhyphen"/>quall bases, that is, vppon BC and CE, and on one and the same syde, namely, on the side of A. Then I say that they are in the selfe same paral<g ref="char:EOLhyphen"/>lell
<pb n="51" facs="tcp:7062:81"/>
lines.<note place="margin">Construction.</note> Drawe by the first peticion a
<figure/>
right line from the point A to the pointe D. Now I say that AD is a parallel line to BE. For if not, then (by the <hi>31.</hi> propo<g ref="char:EOLhyphen"/>sition) by the point A draw vnto the line BE a parallel line AF. And drawe a right line from the point F to the pointe E.<note place="margin">Demonstration leading to an absurditie.</note> VVherfore (by the <hi>38.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g>) the triangle BAC is equall to the triangle CFE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for they consist vpon equal bases, that is BC and CE, and are in the selfe same parallel lines, namely, BE and AF. But by supposition the triangle ABC is equal to the triangle CDE. VVher<g ref="char:EOLhyphen"/>fore the triangle DCE is equall to the triangle FCE, namely, the greater vnto the lesse, which is impossible. VVherfore AF is not a parallel line to BE. And in like sorte may we proue that no other line besides AD is a parallell line to BE. VVherfore AD is a parallel lyne to BE Equall triangles therfore consisting vppon equall bases, and in one and the same side: are also in the selfe same pa<g ref="char:EOLhyphen"/>rallel lines: which was required to be proued.</p>
               <p>
                  <hi>This proposition is the conuerse of the 38. proposition.</hi>
                  <note place="margin">This proposition is the conuerse of the <hi>38.</hi> pro<g ref="char:EOLhyphen"/>positio<g ref="char:cmbAbbrStroke">̄</g>s.</note> 
                  <hi>And in this as in the former proposition, if the parallel line drawen by the point A, should not passe by the point D, it must passe eyther beneath it, or aboue it. Euclide setteth forth onely the absurdity which should follow if it passe beneath it: but the selfe same absurditie also wil follow if it should passe aboue it: as it is not hard to see by the gathering thereof in the former proposition. And therefore here I omitte it.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>31.</hi> Theoreme. The <hi>41.</hi> Proposition. If a parallelograme &amp; a triangle haue one &amp; the selfe same base, and be in the selfe same parallel lines: the parallelo<g ref="char:EOLhyphen"/>grame shalbe double to the triangle.</head>
               <p>
                  <hi>SVppose that the parallelo<g ref="char:EOLhyphen"/>grame
<figure/>
ABCD and the tri<g ref="char:EOLhyphen"/>angle EBC haue one &amp; the same base, namely, BC, and let the<g ref="char:cmbAbbrStroke">̄</g> be in the selfe same parallel lines, namely, BC &amp; AE. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the parallelograme ABCD is double to the triangle BEC. Draw (by the first peti<g ref="char:EOLhyphen"/>cion) a ryght line from the pointe A to y<hi rend="sup">e</hi> pointe C.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>VVherfore (by the</hi> 37. <hi>propo<g ref="char:EOLhyphen"/>sition)
<pb facs="tcp:7062:82"/>
the triangle ABC is equall to
<figure/>
the triangle EBC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for they are vppon one and the selfe same base BC, and in the selfe same parallell lines BC and EA: but the parallelograme ABCD is double to the triangle ABC (by the</hi> 34. <hi>proposition) for the diameter thereof AC deuideth it into two equal parts: wher<g ref="char:EOLunhyphen"/>fore the parallelogramme ABCD is double to the triangle EBC. If therfore a parallelogramme and a triangle haue one and the selfe same base, and be in the selfe same parallels, the parallelo<g ref="char:EOLhyphen"/>grame shall be double to the triangle: which was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Two cases in this proposition.</hi>
                  </note>This proposition hath two cases. For the base beyng one, the triangle may haue his toppe without the parallelograme, or within. The first case is demon<g ref="char:EOLhyphen"/>strated of the author. The second case is thus.</p>
               <p>Suppose that there be a parallelograme <hi>ABCD,</hi> and let
<figure/>
the triangle be <hi>ECD,</hi> either of which let haue one and the selfe same base, namely, <hi>CD,</hi> and let them be in the selfe same parallel lines <hi>CD</hi> and <hi>AB,</hi> and let the toppe of the triangle <hi>ECD,</hi> namely, the point <hi>E,</hi> be within the paralle<g ref="char:EOLhyphen"/>lograme <hi>ABCD.</hi> Then I say that the parallelograme <hi>ABCD</hi> is double to the triangle <hi>ECD.</hi> Draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point <hi>A</hi> to the point <hi>D.</hi> Now forasmuch as the paral<g ref="char:EOLhyphen"/>lelograme <hi>ABCD</hi> is double to the triangle <hi>AD</hi> (by the 34. proposition): and the triangle <hi>ADC</hi> is equall to the triangle <hi>EDC</hi> (by the 37. proposition). Therfore the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>ABCD</hi> is double to the triangle <hi>ECD:</hi> which was required to be proued.</p>
               <p>
                  <note place="margin">
                     <hi>A corollary.</hi>
                  </note>By this proposition it is manifest that if the base be doubled, the triangle e<g ref="char:EOLhyphen"/>rected vppon it shalbe equall to the parallelogramme.</p>
               <p>
                  <note place="margin">
                     <hi>The selfe same demonstration will serue if the triangle &amp; the parallelogramme be vpon equall bases.</hi>
                  </note>And if the bases of the triangle and of the parallelogramme be equall, the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e same demonstration will serue if you drawe the diameter of the parallelo<g ref="char:EOLhyphen"/>grame. For the triangles being equal, the parallelogramme which is double to the one, shal also be double to the other. And the triangles must nedes be equall (by the 38. proposition) for that their bases are equal, and for that they are in the selfe same parallel lines.</p>
               <div type="section">
                  <head>The conuerse of this proposition is thus.</head>
                  <p>
                     <hi>If a parallelogramme and a triangle haue one and the selfe same base, or equall bases the one to the other, and be described on one and the same side of the base: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>f the parallelogramme be double to the triangle, they shalbe in the selfe same parallel lines.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>The conuerse of this proposition.</hi>
                     </note>For if they be not, the whole shalbe equall to his parte. For then the toppe of the triangle must nedes fall either within the parallel lines or without. And
<pb n="52" facs="tcp:7062:82"/>
whether of both soeuer it do, one and the selfe same impossibilitie will follow, if by the toppe of the triangle be drawen vnto the base a parallel line.</p>
               </div>
               <div type="section">
                  <head>An other conuerse of the same proposition.</head>
                  <p>
                     <hi>If a parallelogramme be the double of a triangle, being both within the selfe same parallel lines:</hi>
                     <note place="margin">
                        <hi>An other con<g ref="char:EOLhyphen"/>uerse of the same propositio<g ref="char:cmbAbbrStroke">̄</g>.</hi>
                     </note> 
                     <hi>then are they vpon one and the selfe same base, or vpon equall bases.</hi> For if in that case their ba<g ref="char:EOLhyphen"/>ses should be vnequal, then might straight way be proued, that the whole is e<g ref="char:EOLhyphen"/>quall to his part: which is impossible.</p>
                  <p>A trapesium hauing two sides onely parallel lines,<note place="margin">
                        <hi>Comparison of a triangle and a trapesium being vpon one &amp; the selfe same base, and in the selfe same parallel lines.</hi>
                     </note> is eyther more then dou<g ref="char:EOLhyphen"/>ble, or lesse then double to a triangle contayned within the selfe same parallel lines, and hauing one and the selfe same base with the trapesium, or table. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ust the double it cannot be, for then it should be a parallelogramme. A trapesium ha<g ref="char:EOLhyphen"/>uing two sides parallels hath of necessitie the one of them longer then the other: for if they were equall then should the other two sides enclosing them be also e<g ref="char:EOLhyphen"/>quall (by the 33. proposition.) If the greater side of the trapesium be the base of the triangle, then shal the trapesium be lesse then the double of the triangle And if the lesse side of the trapesium be the base of the triangle then shall the trapesi<g ref="char:EOLhyphen"/>um be greater then the triangle.</p>
                  <p>For suppose that <hi>ABCD</hi> be a trapesium,<note place="margin">
                        <hi>When the grea<g ref="char:EOLhyphen"/>ter side of the trapesium is the base of the tri<g ref="char:EOLhyphen"/>angle.</hi>
                     </note> and let
<figure/>
two sides thereof, namely, <hi>AB</hi> and <hi>CD</hi> be parallel lines, and let the side <hi>AB</hi> be lesse then the side <hi>CD,</hi> &amp; produce the side <hi>AB</hi> infinitlye on the side <hi>B</hi> to the point <hi>F.</hi> And let the triangle <hi>ECD</hi> haue one and the selfe same base with the trapesium, namely, the line <hi>CD.</hi> Then I say that the trapesium <hi>ABCD</hi> is lesse the<g ref="char:cmbAbbrStroke">̄</g> the double of the triangle <hi>ECD.</hi> For put the line <hi>AF</hi> equall to the line CD (by the 3. propositio<g ref="char:cmbAbbrStroke">̄</g>) and draw a line from D to <hi>F.</hi> Wherefore <hi>AC</hi>D<hi>F</hi> is a parallelo<g ref="char:EOLhyphen"/>gramme (by the 33. proposition). Wherefore (by the 34 proposition) it is double to the triangle <hi>EC</hi>D. But the trapesium <hi>ABC</hi>D is a part of the parallelo<g ref="char:EOLhyphen"/>gramme <hi>AC</hi>D<hi>F.</hi> Wherefore the trapesium <hi>ABC</hi>D is lesse then the double of the triangle <hi>EC</hi>D: which was requi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed to be proued.</p>
                  <p>Agayne let the triangle haue to his base the side
<figure/>
                     <hi>AB.</hi>
                     <note place="margin">
                        <hi>When the lesse side is the base.</hi>
                     </note> Then I say that the trapesium <hi>ABCD</hi> is grea<g ref="char:EOLhyphen"/>ter then the double of the triangle <hi>AEB.</hi> For from the side <hi>CD</hi> cut of the line <hi>CF</hi> equall to the line <hi>AB</hi> (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. proposition). And draw a line from <hi>B</hi> to <hi>F.</hi> Wherfore (by the 33. proposition) <hi>ABCF</hi> is a parallelogramme: and therefore is (by the 34. pro<g ref="char:EOLhyphen"/>position) double to the triangle <hi>AEB.</hi> Where<g ref="char:EOLhyphen"/>fore the trapesium <hi>ABCD</hi> is more then the dou<g ref="char:EOLhyphen"/>ble of the triangle <hi>AEB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:83"/>
               <head>The <hi>11.</hi> Probleme. The <hi>42.</hi> proposition. Vnto a triangle geuen, to make a parallelograme equal, whose angle shall be equall to a rectiline angle geuen.</head>
               <p>SVppose that the triangle geuen be ABC, and let the rectiline angle geuen be D. It is required that vnto the triangle ABC there be made a parallelograme equall, whose angle shal be equall to the rectiline angle geuen, namely,<note place="margin">Construction.</note> to the angle D. Deuide (by the <hi>10.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g>) the line BC into two equall partes in the pointe E. And (by the first peticion) draw a right line from the point A to the point E. And (by the <hi>23.</hi> proposition) vpon the right line geuen EC, and to the point in it
<figure/>
geuen E, make the angle CEF equal to the angle D. And (by the <hi>31.</hi> pro<g ref="char:EOLhyphen"/>position) by the point A draw vnto the line EC a parallel line AH: and let the line EF cut the line AH in the point F. and (by the same) by the point C, drawe vnto the line EF a parallel line CG. VVherfore FECG is a parallelograme.<note place="margin">Demonstration</note> And forasmuche as BE is equall to EC, therfore (by the <hi>38.</hi> proposition) the triangle ABE is equall to the triangle AEC, for they consist vpo<g ref="char:cmbAbbrStroke">̄</g> equall bases that is BE and EC, and are in the selfe same parallel lines, namely, BC and AH. VVherfore the triangle ABC is double to the triangle AEC. And the paralle<g ref="char:EOLhyphen"/>lograme CEFG is also double to the triangle AEC: for they haue one &amp; the selfe same base, namely, EC: and are in the selfe same parallel lines, that is, EC and AH. VVherfore the parallelograme FECG is equall to the triangle ABC, and hath the angle CEF equall to the angle geuen D. VVherefore vnto the triangle geuen ABC is made an equall parallelograme, namely, FECG, whose angle CEF is equall to the angle geuen D: which was required to be done.</p>
               <div type="section">
                  <head>The conuerse of this proposition after Pelitarius.</head>
                  <p>
                     <hi>Vnto a parallelogramme geuen,</hi>
                     <note place="margin">
                        <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                           <desc>〈1 paragraph〉</desc>
                        </gap>
                     </note> 
                     <hi>to make a triangle equall, hauy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>g an angle equall to a rectiline angle geuen.</hi>
                  </p>
                  <p>Suppose that the parallelograme geuen be <hi>ABCD,</hi> and let the angle geuen be <hi>E.</hi> It is required vnto the parallelograme <hi>ABCD</hi> to make a triangle equall hauyng an
<pb n="53" facs="tcp:7062:83"/>
angle equal to the angle <hi>E.</hi> Vpon
<figure/>
the line <hi>CD</hi> and to the pointe in it <hi>C,</hi> describe (by the 23. propo<g ref="char:EOLhyphen"/>position) an angle equall to the angle <hi>E</hi>: which let be <hi>DCF</hi>: a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d let the line <hi>CF</hi> cut the line <hi>AB</hi> being produced, in the point <hi>F</hi>: and produce the line <hi>CD</hi> (which is parallell to the line <hi>AF</hi>) to the point <hi>G.</hi> And put the line <hi>DG</hi> e<g ref="char:EOLhyphen"/>quall to the line <hi>CD</hi> and draw a line from <hi>F</hi> to <hi>G.</hi> Then <hi>I</hi> say that the triangle <hi>CFG</hi> is equal to the parallelograme <hi>ABCD.</hi> For for<g ref="char:EOLhyphen"/>asmuch as (by the 38. proposition) the whole triangle <hi>CFG</hi> is double to the triangle <hi>CDF.</hi> A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d (by the 41. proposition) the parallelograme <hi>ABCD</hi> is double to the same triangle <hi>CDF</hi>: therfore the parallelograme <hi>ABCD</hi> and the triangle <hi>CFG</hi> are equall the one to the other: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>32.</hi> Theoreme. The <hi>43.</hi> Proposition. In euery parallelograme, the supplementes of those parallelo<g ref="char:EOLhyphen"/>grammes which are about the diameter, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that ABCD be a parallelograme, and let the diameter ther<g ref="char:EOLhyphen"/>of be AC: and about the diameter AC let these parallelogrames EH and GF consist: and let the supplementes be BK and KD. Then I say that the supplement BK is equall to the supplement KD. For forasmuch as ABCD is a parallelograme
<figure/>
and the diameter therof is AC, therfore (by the</hi> 34. <hi>proposition) the triangle ABC is equall to the triangle ADC. A<g ref="char:EOLhyphen"/>gayne forasmuch as AEKH is a pa<g ref="char:EOLhyphen"/>rallelograme, and the diameter therof is AK, therfore (by the same) the trian<g ref="char:EOLhyphen"/>gle AEK is equall to the triangle AHK. And by the same reason also the tri<g ref="char:EOLhyphen"/>angle KFC is equall to the triangle KGC. And forasmuch as the triangle AEK is equall to the triangle AHK, and the triangle KFC to the triangle NGC, therfore the triangles AEK and KGC are equall to the triangles AHK and KFC: and the whole triangle ABC is equall to the whole triangle AD
<pb facs="tcp:7062:84"/>
C: wherfore the residue, namely, the supplement BK is (by the</hi> 3. <hi>common sen<g ref="char:EOLhyphen"/>tence) equall to the residue, namely, to the supplement KD. VVherefore in e<g ref="char:EOLhyphen"/>uery parallelogramme, the supplementes of those parallelogrammes whiche are about the diameter, are equall the one to the other: whiche was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>How parallelo<g ref="char:EOLhyphen"/>grammes are sayde to consiste about a dia<g ref="char:EOLhyphen"/>meter.</hi>
                  </note>Those parallelogrames are sayde to consist about a diameter which haue to their diameters part of the diameter of the whole and great parallelograme, as in the example of Euclide. And such parallelo<g ref="char:EOLhyphen"/>grames
<figure/>
which haue not to their diameters part of the diameter of the greater parallelograme, are sayde not to consist about the diameter. For the<g ref="char:cmbAbbrStroke">̄</g> the diameter of the greater parallelograme cutteth the side of the lesse co<g ref="char:cmbAbbrStroke">̄</g>tayned wythin it. As in the parallelogramme <hi>AB,</hi> the diameter <hi>CD,</hi> cutteth the side <hi>EH</hi> of the parallelogramme <hi>CE.</hi> Wherefore the parallelogramme <hi>CE</hi> is not about one and the selfe same diameter with the parallelo<g ref="char:EOLhyphen"/>gramme <hi>CD.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Supplements &amp; Complementes.</hi>
                  </note>Supplementes or Complementes are those figures which with the two pa<g ref="char:EOLhyphen"/>rallelogrammes accomplish the whole parallelogramme. Although Pel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tarius for distinction sake putteth a difference betwene Supplementes and Comple<g ref="char:EOLhyphen"/>mente<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. The parallelogrammes about the diameter he calleth Complementes, the other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wo figures he calleth Supplementes.</p>
               <p>
                  <note place="margin">
                     <hi>Three cases in this Theoreme.</hi>
                  </note>This theoreme hath three cases onely. For the parallelogrammes which consist about the diameter, eyther touch the one the other in a point: or by a cer<g ref="char:EOLunhyphen"/>tayne parte of the diameter are seuered the one from the other: or els they cutte the one the other.<note place="margin">
                     <hi>The first case.</hi>
                  </note> For the first case is the example of Euclide before set. The se<g ref="char:EOLhyphen"/>cond case is thus.</p>
               <p>
                  <note place="margin">
                     <hi>The second case.</hi>
                  </note>Suppose that <hi>AB</hi> be a parallelograme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
whose diameter let be <hi>CD</hi>: and aboute the same diameter let these parallelogrammes <hi>CK</hi> and <hi>DL</hi> consist: and betwene the<g ref="char:cmbAbbrStroke">̄</g> let there be a certayne part of the diameter, namely, <hi>LK.</hi> Then I say that the two supplementes <hi>AGLKE</hi> &amp; <hi>BFKLH</hi> are equall. For we may as before (by the 34. proposition) proue that the triangle <hi>ACD,</hi> is equall to the triangle <hi>BCD,</hi> and the triangle <hi>ECK</hi> to the triangle K<hi>CF,</hi> and also the triangle <hi>DGL</hi> to the tri<g ref="char:EOLhyphen"/>angle <hi>DHL.</hi> Wherfore the residue, namely, the fiue sided figure <hi>AGLKE</hi> is equall to the residue, namely, to the fiue sided figure, <hi>BFKLH</hi>: that is, the one supplement to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
               <p>
                  <note place="margin">
                     <hi>The third case.</hi>
                  </note>But now suppose <hi>AB</hi> to be a parallelogramme, and let the diameter thereof be <hi>CD</hi>: and let the one of the parallelogrammes about it be <hi>ECFL,</hi> and let the other be
<pb n="54" facs="tcp:7062:84"/>
                  <hi>DGKH,</hi> of which let the one cut the other.
<figure/>
Then I say that the supplementes <hi>EG</hi> and <hi>EH</hi> are equall. For forasmuch as the whole trian<g ref="char:EOLhyphen"/>gle <hi>DG</hi>K is equal to the whole triangle <hi>DH</hi>K (by the 34. proposition), and part also of the one, namely, the triangle K<hi>LM</hi> is equall to part of the other, namely, to the triangle K<hi>LN</hi> (by the same), for LK is a parallelograme: therefore the residue, namely, the Trapesium <hi>DLNH</hi> is equall to the residue, namely, to the trapesiu<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>DLMG</hi>: but the triangle <hi>ADC</hi> is equal to the triangle <hi>BCD,</hi> and in the pa<g ref="char:EOLhyphen"/>rallelograme <hi>EF,</hi> the triangle <hi>FCL,</hi> is equall to the triangle <hi>ECL,</hi> and the trapesium <hi>DG</hi>ML is (as it hath bene proued) equall to the trapesium <hi>DHN</hi>L. Wherefore the residue, namely, the quadrilater figure <hi>GF</hi> is equall to the residue, namely, to the quadrilater figure <hi>EH,</hi> that is, the one supplement to the other: which was required to be proued.</p>
               <p>This is to be noted that in <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>h of those three cases it may so happen, that the parallelogrammes aboute the diameter shall not haue one angle common wyth the whole parallelogramme, as they haue in the former figures. But yet though they haue not, the selfe same demonstration wil serue, as it is playne to see in the figures here vnderneath put. For alwayes, if from thinges equall be taken away thinges equall, the residue shalbe equall.</p>
               <figure/>
               <p>This proposition P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>litarius calleth Gnomicall, and misticall,<note place="margin">
                     <hi>This proposition called Gnomical and mistical.</hi>
                  </note> for that of it (sayth he) spring infinite demonstrations, and vses in geometry. And he putteth the conuerse thereof after this manner.</p>
               <p>
                  <hi>If a parallelogramme be deuided into two equall supplementes,</hi>
                  <note place="margin">
                     <hi>The conuerse of this proposition.</hi>
                  </note> 
                  <hi>and into two complements what<g ref="char:EOLhyphen"/>soeuer: the diameter of the two complementes shall be set directly, and make one diameter of the whole parallelogramme.</hi>
               </p>
               <p>Here is to be noted as I before admonished that Pelitarius for distinction sake putteth a difference betwene supplementes and complementes, which diffe<g ref="char:EOLhyphen"/>rence, for that I haue before declared, I shall not neede here to repete agayne.</p>
               <p>Suppose that there be a parallelogramme <hi>ABCD,</hi> whose two equall supplements let be <hi>AEFG</hi> and F<hi>HD</hi>K, and let the two complementes thereof be GFCK and <hi>EBFH</hi>: whose diameters let be <hi>CF</hi> and <hi>FB.</hi> Then I say that <hi>CFB</hi> is one right line, and is the diameter of the whole parallelogramme <hi>ABCD</hi>: for if it be not, then is there an
<pb facs="tcp:7062:85"/>
other diameter of the whole parrallelogramme, which let
<figure/>
be CLB being drawen vnder the diameters CF and FB, and cutting the line <hi>GH</hi> in the in the point L. And (by the 31. proposition) by the point L, draw vnto the line <hi>AC</hi> a parallel line MLN. And so are there in the whole paral<g ref="char:EOLhyphen"/>lelogramme <hi>ABCD</hi> two supplements AMGL and LHND, which by this proposition shalbe equall the one to the other. For that they are about the diameter CLB. But the supplement <hi>AEFG</hi> is (by supposition) equall to the supplement FHDK: and forasmuch as FHDK is greater then LH<hi>D</hi>N, AEFG also shalbe greater then AMGL, namely, the part greater then the whole: which is impossible. And by the same reaso<g ref="char:cmbAbbrStroke">̄</g> may it be proued, that the diameter cannot be drawen aboue the diameters CF and FB. Wherefore CFB is one diameter of the whole parallelogramme <hi>ABCD</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Probleme. The <hi>44.</hi> Proposition. Vppon a right line geuen, to applye a parallelograme equall to a triangle geuen, and contayning an angle equall to a rec<g ref="char:EOLhyphen"/>tiline angle geuen.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB, and let the triangle geuen be C, and let the rectiline angle geuen be D. It is required vpon the right line geuen AB, to applye a parallelogramme equal to the trian<g ref="char:EOLhyphen"/>gle geuen C,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>and contayning an angle equall to the rectiline angle ge<g ref="char:EOLhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g> 
                     <milestone type="tcpmilestone" unit="unspecified" n="42"/> D. Describe (by the</hi> 44. <hi>proposition) y<hi rend="sup">e</hi>
                     <figure/>
parallelograme BGEF equall to the tri<g ref="char:EOLhyphen"/>angle C, and hauing the angle BGF e<g ref="char:EOLhyphen"/>quall to the angle D. And vnto the line EB ioyne the line AB in such sort that they make both one right line. And extend the line FG beyond the point G to the poynte H. And (by the</hi> 31. <hi>proposition) by the point A drawe to either of these lines BG and EF a parallel line AH. And (by the first peticion) draw a right line from the point H to the point B. And forasmuch as vpon the parallel lines AH and EF falleth a certayne right line HF, therefore (by the</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>9 <hi>proposition) the angles AHF and HFE are equall to two right angles: wherefore the angles BHG and GFE are lesse then two right angles: but if vpon two right lines fall a right line making the inward angles on one and the
<pb n="55" facs="tcp:7062:85"/>
same side lesse then two right angles, those right lines being infinitly produced shall at the length mete on that side in which are the angles lesse then two right angles (by the</hi> 5. <hi>peticion). VVherfore the lines HB and FE being infinitly produced will at the length mete. Let them be produced, &amp; let them mete in the point K. And (by the</hi> 31 <hi>proposition) by the point K draw to either of these lines EA and FH a parallel line KL. And (by the</hi> 2. <hi>peticion) extend the lines HA and GB till they co<g ref="char:cmbAbbrStroke">̄</g>curre with the line KL in the pointes L and M.</hi>
                  <note place="margin">
                     <hi>Demonstration</hi>
                  </note> 
                  <hi>VVhere<g ref="char:EOLhyphen"/>fore HLKF is a parallelogramme, and the diameter thereof is HK: and a<g ref="char:EOLhyphen"/>bout the diameter HK are the parallelogrammes AG and ME, and the sup<g ref="char:EOLhyphen"/>plementes are LB and BF: wherefore (by the</hi> 43. <hi>prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sition) the supplement LB is equall to the supplement BF: but by construction the parallelograme BF is equall to the triangle C: wherefore also the parallelogramme LB is equall to the triangle C. And forasmuch as the line FH is a parallel to the line KL, and vpon them lighteth the line GM, therefore (by the</hi> 27. <hi>proposition) the angle FGB is equall to the angle BML. But the angle FGB is equall to the angle D, therfore the angle BML is equal to the angle D. VVherfore vpo<g ref="char:cmbAbbrStroke">̄</g> the right line geuen AB is applied the parallelograme LB, equal to the triangle geuen C, and contayning the angle BML equal to the rectiline angle geuen D: which was required to be done.</hi>
               </p>
               <p>Applications of spaces or figures to lines with excesses or wantes is (sayth Eudemus) an auncient inuention of Pithagoras.<note place="margin">
                     <hi>Applications of spaces with ex<g ref="char:EOLhyphen"/>cesses or wants an auncient in<g ref="char:EOLhyphen"/>uention of Pi<g ref="char:EOLhyphen"/>thagoras.</hi>
                  </note>
               </p>
               <p>VVhen the space or figure is ioyned to the whole line,<note place="margin">
                     <hi>How a figure is sayde to be ap<g ref="char:EOLhyphen"/>plied to a line.</hi>
                  </note> the<g ref="char:cmbAbbrStroke">̄</g> is the figure sayd to be applied to the line. But if the length of the space be longer then the line, the<g ref="char:cmbAbbrStroke">̄</g> it is sayde to exceede: and if the length of the figure be shorter then the line, so that part of the line remayneth without the figure described, then is it sayde to want.</p>
               <p>In this probleme are three thinges geuen.<note place="margin">
                     <hi>Three thinges geuen in this proposition.</hi>
                  </note> A right line to which the applica<g ref="char:EOLhyphen"/>tion is made, which here must be the one side of the parallelogramme applied. A triangle whereunto the parallelogramme applied must bee equall: and an angle wherunto the angle of the parallelograme applied must be equall. And if the an<g ref="char:EOLhyphen"/>gle geuen be a right angle, the<g ref="char:cmbAbbrStroke">̄</g> shal the parallelograme applied be either a square, or a figure on the one side longer. But if the angle geuen be an obtuse or an acute angle, then shall the parallelograme applied be a Rhombus or diamond fi<g ref="char:EOLhyphen"/>gure, or els a Rhomboides or diamondlike figure.</p>
               <div type="section">
                  <head>The conuerse of this proposition after Politarius.</head>
                  <p>
                     <hi>Vpon a right line geuen, to applie vnto a parallelograme geuen an equall triangle hauyng an an<g ref="char:EOLhyphen"/>gle equall to an angle geuen.<note place="margin">
                           <hi>The conuerse of this proposition.</hi>
                        </note>
                     </hi>
                  </p>
                  <p>Suppose that the right line geuen be <hi>AB,</hi> and let the parallelograme geuen be <hi>CDEF,</hi> and let the angle geuen be <hi>G.</hi> It is required vpon the line <hi>AB</hi> to describe a tria<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>gle equall to the parallelograme <hi>CDEF,</hi> hauing an angle equall to the angle <hi>G.</hi> Drawe the diameter <hi>CF</hi> &amp; produce <hi>CD</hi> beyond the point D to the point <hi>H.</hi> And put the line
<pb facs="tcp:7062:86"/>
                     <hi>DH</hi> equall to the line <hi>CD.</hi> And
<figure/>
draw a line from <hi>F</hi> to <hi>H.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> (by the 41. proposition) the tria<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>gle <hi>CHF</hi> is equall to the paralle<g ref="char:EOLhyphen"/>lograme <hi>C</hi>D<hi>E</hi>F. And (by this proposition) vppon the line <hi>AB</hi> describe a parallelograme <hi>ABKL</hi> equall to the triangle <hi>CHF,</hi> ha<g ref="char:EOLhyphen"/>uing the angle <hi>ABL</hi> equal to the angle geuen <hi>G:</hi> and produce the line <hi>BL</hi> beyonde the pointe L to the point M. And put the line LM equall to the line BL, and draw a line from <hi>A</hi> to M. Then I say that vpon the line <hi>AB</hi> is described the triangle <hi>AB</hi>M, which is such a trian<g ref="char:EOLhyphen"/>gle as is required. For (by the 41. proposition) the triangle <hi>AB</hi>M is equal to the paral<g ref="char:EOLhyphen"/>lelogramme <hi>ABKL</hi> (for that they are betwene two parallel lines <hi>B</hi>M and <hi>AK,</hi> &amp; the base of the triangle is double to the base of the parallelogramme): but <hi>ABKL</hi> is by construction equall to the triangle <hi>CHF</hi>: and the triangle <hi>CHF</hi> is equall to the pa<g ref="char:EOLhyphen"/>rallelograme <hi>CDEF.</hi> Wherfore (by the first common sentence) the triangle <hi>AB</hi>M is equall to the parallelograme geuen <hi>C</hi>D<hi>EF,</hi> and hath his angle <hi>A</hi>BM equal to the an<g ref="char:EOLhyphen"/>gle geuen <hi>G</hi>: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Probleme. The <hi>45</hi> Proposition. To describe a parallelograme equal to any rectiline figure ge<g ref="char:EOLhyphen"/>uen, and contayning an angle equall to a rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g>.</head>
               <p>
                  <hi>SVppose that the rectiline figure geuen be ABCD, and let the rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g> be E. It is required to describe a parallelograme equall to the rectiline figure geuen ABCD, and contayning an angle equal to the re<g ref="char:EOLhyphen"/>ctiline angle geuen E.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Draw (by the first peticion) a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point D to the point B. And (by the</hi> 42. <hi>proposition) vnto the triangle ABD describe an e<g ref="char:EOLhyphen"/>quall parallelograme FH, hauing his angle FKH equall to the angle E. And (by the</hi> 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. <hi>of the first) vpo<g ref="char:cmbAbbrStroke">̄</g> the right
<figure/>
line GH apply the parallelogramme GM equal to the triangle DBC, ha<g ref="char:EOLhyphen"/>uing his angle GHM equall to the angle E.</hi>
                  <note place="margin">
                     <hi>Demonstration.</hi>
                  </note> 
                  <hi>And forasmuch as eyther of those angles HKF and GHM is equall to the angle E: therefore the angle HKF is equall to the angle GHM: put the angle KHG com<g ref="char:EOLhyphen"/>mon to them both, wherfore the an<g ref="char:EOLhyphen"/>gles FKH and KHG are equall to the angles KHG and GHM. but the angles FKH and KHG are (by the</hi> 29. <hi>proposition) equall to two right angles. VVherfore the angles KHG and GHM are equall to two
<pb n="56" facs="tcp:7062:86"/>
right angles. Now then vnto a right line GH, and to a point in the same H, are drawen two right lines KH and HM not both on one and the same side, ma<g ref="char:EOLhyphen"/>king the side angles equall to two right angles. VVherfore (by the</hi> 14. <hi>propositi<g ref="char:EOLhyphen"/>on) the lines KH and HM make directly one right line. And forasmuch as v<g ref="char:EOLhyphen"/>pon the parallel lines. KM and FG falleth the right line HG, therefore the alternate aagles MHG and HGF are (by the</hi> 29 <hi>proposition) equall the one to the other: put the angle HGL common to them both. VVherfore the angles MHG and HGL are equall to the angles HGF and HGL. But the angles MHG &amp; HGL are equall to two right angles (by y<hi rend="sup">e</hi>
                  </hi> 29. <hi>proposition). VVher<g ref="char:EOLhyphen"/>fore also the angles HGF and HGL are equall to two right angles. VVher<g ref="char:EOLhyphen"/>fore (by the</hi> 14. <hi>proposition) the lines FG and GL make directly one right line. And forasmuch as the line KF is (by the</hi> 34. <hi>proposition) equal to the lyne HG, and it is also parallel vnto it: and the line HG is (by the same) equall to the line ML, therfore (by the first common sentence) the line FK is equall to the lyne ML, and also a parallel vnto it (by the</hi> 30. <hi>proposition). But the right lynes KM and FL ioyne them together. VVherfore (by the</hi> 33. <hi>proposition) the lines KM and FL are equall the on to the other and parallel lines. VVherfore KFLM is a parallelograme. And forasmuch as the triangle ABD is eqnal to the paral<g ref="char:EOLhyphen"/>lelograme FH, and the triangle DBC to the parallelogramme GM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therfore the whole rectiline figure ABCD is equall to the whole parallelograme KFLM. VVherfore to the rectiline figure geuen ABCD is made an equall paralle<g ref="char:EOLhyphen"/>grame KFLM, whose angle FKM is equal to the angle geuen, namely, to E: which was required to be done.</hi>
               </p>
               <p>The rectiline figure geue<g ref="char:cmbAbbrStroke">̄</g> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> in the example of Euclide is a parallelograme. But if the rectiline figure be of many sides, as of 5.6. or mo, the<g ref="char:cmbAbbrStroke">̄</g> must you resolue the figure into his triangles, as hath bene before taught in the 32. proposition. And the<g ref="char:cmbAbbrStroke">̄</g> apply a parallelograme equal to euery triangle vpon a line geue<g ref="char:cmbAbbrStroke">̄</g>, as before in the example of the author. And the same kind of reasoning wil serue that was be<g ref="char:EOLunhyphen"/>fore, only by reaso<g ref="char:cmbAbbrStroke">̄</g> of the multitude of triangles, you shall haue neede of oftener repeticio<g ref="char:cmbAbbrStroke">̄</g> of the 29. and 14. propositio<g ref="char:cmbAbbrStroke">̄</g>s to proue that the bases of al the parallelo<g ref="char:EOLhyphen"/>grames made equall to all the triangles make one right line, and so also of the toppes of the said parallelogrames. Pelitarius addeth vnto this proposition this Probleme following.</p>
               <p>
                  <hi>Two vnequall rectiline superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s beyng geuen, to find out the exc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sse of the greater aboue the lesse.</hi>
                  <note place="margin">
                     <hi>An addition of Pelitarius.</hi>
                  </note>
               </p>
               <p>Suppose that
<figure/>
there be two vne<g ref="char:EOLhyphen"/>quall rectiline su<g ref="char:EOLhyphen"/>perficieces <hi>A</hi> &amp; <hi>B</hi> of which let <hi>A</hi> be the greater. It is required to finde out the excesse of the superficies <hi>A</hi> aboue the super<g ref="char:EOLhyphen"/>ficieces <hi>B.</hi> De<g ref="char:EOLhyphen"/>scribe (by the 44.
<pb facs="tcp:7062:87"/>
proposition) the parallelograme <hi>CDEF</hi> equall to the rectiline figure <hi>A,</hi> contayning a right angle. And produce the line <hi>CD</hi> beyond the point <hi>D</hi> to the point <hi>G</hi>: &amp; put the line <hi>DG</hi> equall to the line <hi>CD.</hi> And againe (by the 44. proposition) vpon the line <hi>DG</hi> describe the parallelograme <hi>DGHK</hi> equall to the rectiline figure <hi>B,</hi> and hauyng the angle <hi>DGK</hi> a right angle. And produce the line KH beyond the point H, vntill it cutte the line <hi>CE</hi> in the point L. Then I say that HLEF, is the excesse of the rectiline figure <hi>A</hi> aboue the rectiline figure <hi>B.</hi> For first that CGKL is a parallelogramme it is mani<g ref="char:EOLhyphen"/>fest, neither nedeth it to be demonstrated. And forasmuch as the lines <hi>CD</hi> and <hi>DG</hi> are by supposition equal and either of them is a parallel to KL, therfore (by the 36, propo<g ref="char:EOLhyphen"/>sition) the two parallelogrames <hi>CH</hi> and <hi>DK</hi> are equall. And forasmuch as DK is sup<g ref="char:EOLhyphen"/>posed to be equall to the rectiline figure <hi>B, CH</hi> also shall be equall to the same rectiline figure <hi>B.</hi> Wherfore forasmuch as the whole parallelograme <hi>CF</hi> is equall to the rectiline figure <hi>A,</hi> and LF is the excesse of <hi>CF</hi> aboue DL or DK, it followeth that LF is the ex<g ref="char:EOLhyphen"/>cesse of the rectiline figure <hi>A</hi> aboue the rectiline figure <hi>B</hi>: whiche was required to be done.</p>
               <div type="section">
                  <head>An other more redy way.</head>
                  <p>Let the parallelograme <hi>CDEF</hi> remayne equall to the rectiline figure <hi>A,</hi> &amp; produce the line <hi>CD</hi> beyond the point <hi>D</hi> to the pointe G. And vpon the line <hi>DG</hi> describe the parallelograme DGHK equall to the rectiline figure B. And produce the lines EC &amp; HK beyond the points C and K till they concurre in the point L. And by the pointe D draw the diame<g ref="char:EOLhyphen"/>ter
<figure/>
LDM, which let cutte the line HG beyng pro<g ref="char:EOLhyphen"/>duced beyonde the pointe G in the point M, &amp; by the pointe M drawe vnto the line HL a paral<g ref="char:EOLhyphen"/>lel MN cuttyng the line EL in the pointe N: and by that meanes is HLMN a paralle<g ref="char:EOLhyphen"/>lograme. Then <hi>I</hi> say that NF is the evcesse of the rectiline figure <hi>A</hi> aboue the rectiline figure <hi>B.</hi> For forasmuch as the parallelograme HD is equall to the rectiline figure <hi>B,</hi> &amp; the supplementes HD and DN ar (by the 43. proposition) equall: therfore DN also is equall to the rectiline figure B, which rectiline figure DN being taken away fro<g ref="char:cmbAbbrStroke">̄</g> the parallelograme CF (which is supposed to be equall to the rectiline figure <hi>A</hi>) the resi<g ref="char:EOLhyphen"/>due NF shall be the ezcesse of the rectiline figure <hi>A</hi> aboue the rectiline figure <hi>B</hi>: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Probleme. The <hi>46.</hi> Proposition. Vppon a right line geuen, to describe a square.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB. It is required vpo<g ref="char:cmbAbbrStroke">̄</g> the right line AB, to describe a square. Vpon the right line AB, and from a point in it geuen, namely, A, rayse vp (by the</hi> 11 <hi>proposition) a per<g ref="char:EOLhyphen"/>pendiculer line AC. And (by the</hi> 3. <hi>proposition) vnto AB put an e<g ref="char:EOLhyphen"/>quall line AD. And (by the</hi> 31. <hi>proposition) by the point D drawe vnto AB a parallel line DE. And (by the same) by the point B drawe vnto AD a parallel
<pb n="57" facs="tcp:7062:87" rendition="simple:additions"/>
line BE. VVherefore ADEB is a pa<g ref="char:EOLhyphen"/>rallelogramme.
<figure/>
VVherefore the line AB is equall to the line DE, and the line AD to the line BE: but the line AB is e<g ref="char:EOLhyphen"/>quall to the line AD wherefore these fo<g ref="char:EOLhyphen"/>wer lines BA, AD, DE, EB, are equall the one to the other. VVherefore the pa<g ref="char:EOLhyphen"/>rallelogramme ADEB consisteth of e<g ref="char:EOLhyphen"/>quall sides. I say also that it is rectangle. For forasmuch as vpon the parallel lines AB and DE falleth a right line AD: there<g ref="char:EOLhyphen"/>fore (by the</hi> 29. <hi>proposition) the angle: BAD and ADE are equal to two right angles: but the angle BAD is a right angle. VVherfore the angle ADE also is a right angle. But in parallelogrames the sides and angles which are opposite are equall the one to the other (by the</hi> 34 <hi>proposition). VVherefore the two opposite angles ABE and BED are ech of them a right angle. VVherefore the parallelograme ABED is rectangle: &amp; it is also proued that it is equilater. VVherfore it is a square, &amp; it is described vp<g ref="char:EOLhyphen"/>on the right line geuen AB: which was required to be done.</hi>
               </p>
               <p>This is to be noted that if you will mechanically and redily,<note place="margin">
                     <hi>To describe a square mecha<g ref="char:EOLhyphen"/>nically.</hi>
                  </note> not regarding demo<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>stration describe a square vpon a line geuen, as vpon the line <hi>AB,</hi> after that you haue erected the perpendiculer line <hi>CA</hi> vpon the line <hi>AB,</hi> and
<figure/>
put the line <hi>AE</hi> equall to the line <hi>AB</hi>: then open your compasse to the wydth of the line <hi>AB</hi> or <hi>AE,</hi> &amp; set one foote thereof in the point <hi>E,</hi> and describe a peece of the circumference of a circle: and againe make the centre the point <hi>B,</hi> and describe also a piece of the circumference of a circle, namely, in such sort that the peece of the circum<g ref="char:EOLhyphen"/>fere<g ref="char:cmbAbbrStroke">̄</g>ce of the one may cut the peece of the circumference of the other, as in the point <hi>D</hi>: and from the point of the intersection, draw vnto the points <hi>E</hi> &amp; <hi>B</hi> right lines: &amp; so shalbe described a square. As in this figure here put, wher<g ref="char:EOLhyphen"/>in <hi>I</hi> haue not drawen the lines <hi>ED</hi> and <hi>DB,</hi> that the pee<g ref="char:EOLhyphen"/>ces of the circumference cutting the one the other might the plainlier be sene.</p>
               <div type="section">
                  <head>An addition of Proc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us.</head>
                  <p>
                     <hi>If the lines vpon which the squares be described be equall, the squares also are equall.</hi>
                  </p>
                  <p>Suppose that these right lines
<figure/>
                     <hi>AB</hi> and <hi>CD</hi> be equall,<note place="margin">
                        <hi>An addition of Proc<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>.</hi>
                     </note> &amp; vpon the line <hi>AB</hi> describe a square <hi>ABEG</hi>: and vpon the line <hi>C</hi>D de<g ref="char:EOLhyphen"/>scribe a square <hi>CDHF.</hi> Then <hi>I</hi> say that the two squares <hi>ABEG</hi> &amp; <hi>CDHF</hi> are equal. For draw these right lines <hi>GB</hi> and <hi>HD.</hi> And for as much as the right lines <hi>AB</hi> and <hi>CD</hi> are equall, &amp; the lines <hi>AG</hi> and <hi>HC</hi> are also equall, and they contayne eqaul
<pb facs="tcp:7062:88"/>
angles, namely, right angles (by the definition of a square) therefore (by the 4. propo<g ref="char:EOLhyphen"/>sition) the base <hi>BG</hi> is equall to the base <hi>HD.</hi> And the triangle <hi>ABG</hi> is equall to the triangle <hi>CDH.</hi> Wherefore the doubles of the saide triangles are equall. Wherefore the square <hi>AE</hi> is equall to the square <hi>CF</hi>: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The conuerse thereof is thus.</head>
                  <p>
                     <hi>If the squares be equall:<note place="margin">
                           <hi>The conuerse thereof.</hi>
                        </note> the lines also vppon which they are described are equall.</hi>
                  </p>
                  <p>Suppose that there be two equall squares <hi>AF</hi> and <hi>CG</hi> described vpon the lines <hi>AB</hi> &amp; <hi>BC.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say, that the lines <hi>AB</hi> and <hi>BC</hi> are equall. Put the line <hi>AB</hi> directly to the line <hi>BC,</hi> that they both make on right line. And
<figure/>
forasmuch as the angles are right angles, ther<g ref="char:EOLhyphen"/>fore also (by the 14. proposition) the right line <hi>FB</hi> is set directly to the right line <hi>BG.</hi> Drawe these right lines <hi>FC, AG, AF,</hi> and <hi>CG.</hi> Now for asmuch as the square <hi>AF</hi> is equal to the square <hi>CG,</hi> the triangle also <hi>AFB</hi> shalbe equall to the triangle <hi>CBG</hi>: put the triangle <hi>BCF</hi> co<g ref="char:cmbAbbrStroke">̄</g>mon to them both. Wherfore the whole triangle <hi>ACF</hi> is equall to the whole triangle <hi>CFG.</hi> Where<g ref="char:EOLhyphen"/>fore the line <hi>AG</hi> is a parallel vnto the line <hi>CF</hi> (by the 38. propositio<g ref="char:cmbAbbrStroke">̄</g>): for the triangles consist vpon one and the selfe same base, namely <hi>CF.</hi> Againe forasmuch as either of these angles <hi>AFG</hi> &amp; <hi>CBG</hi> is the halfe of a right angle, therfore (by the 17. proposition) the line <hi>AF</hi> is a paral<g ref="char:EOLhyphen"/>lel to the line <hi>CG.</hi> Wherfore the right line <hi>AF</hi> is equal to the right line <hi>CG</hi> (for the op<g ref="char:EOLhyphen"/>posite sides of a parallelograme are equall). And forasmuch as there are two triangles <hi>ABF</hi> and <hi>BCG,</hi> whose alternate angles are equall, namely, the angle <hi>AFB</hi> to the an<g ref="char:EOLhyphen"/>gle <hi>BGC,</hi> and the angle <hi>BAF</hi> to the angle <hi>BCG,</hi> and one side of the one is equall to one side of the other, namely, the side which lieth betwene the equal angles, that is, the side <hi>AF</hi> to the side <hi>CG,</hi> therefore (by the 26. proposition) the side <hi>AB</hi> is equal to the side <hi>BC,</hi> and the side <hi>BF</hi> to the side <hi>BG.</hi> Wherefore it is proued that the squares of the lines <hi>AF</hi> and <hi>CG</hi> being equall, their sides also shalbe equall: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>33.</hi> Theoreme. The <hi>47.</hi> Proposition. In rectangle triangles, the square whiche is made of the side that subtendeth the right angle, is equal to the squares which are made of the sides contayning the right angle.</head>
               <p>SVppose that ABC be a rectangle triangle, hauyng the angle BAC a right angle. The<g ref="char:cmbAbbrStroke">̄</g> I say y<hi rend="sup">t</hi> the square which is made of the line BC is equall to the squares which are made of y<hi rend="sup">e</hi> lines AB and AC.<note place="margin">Construction.</note> Describe (by y<hi rend="sup">e</hi> 
                  <hi>46.</hi> proposi<g ref="char:EOLhyphen"/>tion) vpon y<hi rend="sup">e</hi> line BC a square BDCE, and (by y<hi rend="sup">e</hi> same) vpon the lines BA and AC describe the squares ABFG and ACKH. And by the point A draw (by the 31. pro<g ref="char:EOLhyphen"/>position) to either of these lynes BD and CE a parallel
<pb n="58" facs="tcp:7062:88" rendition="simple:additions"/>
line AL. And (by the first peticion) draw a right lyne from the point A to the point D, and an other from the point C to the point E.<note place="margin">Demonstration.</note> And forasmuch as the an<g ref="char:EOLhyphen"/>gles BAC and BAG are right angles, therfore vnto a right line BA, and to a point in it geuen A, are drawen two
<figure/>
right lines AC and AG, not both on one and the same side, makyng the two side angles equall to two right angles: wherfore (by the <hi>14.</hi> proposi<g ref="char:EOLhyphen"/>tion) the lines AC and AG make di<g ref="char:EOLhyphen"/>rectly one right line. And by the same reason the lines BA and AH make also directly one right line. And for<g ref="char:EOLhyphen"/>asmuch as the angle DBC is equall to the angle FBA (for either of the<g ref="char:cmbAbbrStroke">̄</g> is a right angle) put the angle ABC common to them both: wherfore the whole angle DBA is equall to the whole angle FBC. And forasmuch as these two lines AB and BD are equal to these two lines BF and BC, the one to the other, and the angle DBA is equal to the angle FBC: therfore (by the <hi>4.</hi> proposition) the base AD is equall to the base FC, and the triangle ABD is equall to the triangle FBC. But (by the <hi>31.</hi> proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) the parallelogramme BL is double to the triangle ABD, for they haue both one aud the same base, namely, BD, and are in the selfe same paral<g ref="char:EOLhyphen"/>lel lynes, that is, BD and AL and (by the same) the square GB is double to the triangle FBC, for they haue both one and the selfe same base, that is, BF, and are in the selfe same parallel lynes, that is, FB and GC. But the dou<g ref="char:EOLhyphen"/>bles of thinges equall, are (by the sixte common sentence) equall the one to the other. VVherfore the parallelograme BL is equall to the square GB. And in like sorte if (by the first peticion) there be drawen a right line from the point A to the point E, and an other from the point B to the point K, we may proue y<hi rend="sup">t</hi> the parallelograme CL is equal to the square HC. VVherfore the whole square BDEC is equall to the two squares GB and HC. But the square BDEC is described vpon the line BC, and the squares GB and HC are described vppon the lines BA &amp; AC: wherfore the square of the side BC, is equal to the squares of the sides BA and AC. VVherefore in rectangle triangles, the square whiche is made of the side that subtendeth the right angle, is equal to the squares which are made of the sides contayning the right angle: which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
               <p>
                  <hi>This most excellent and notable Theoreme was first inuented of the greate philosopher Pithagoras,</hi>
                  <note place="margin">Pithagoras the first inuenter of this proposition.</note> 
                  <hi>who for the exceeding ioy conceiued of the inuention therof, offered in sacrifice an Oxe, as recorde Hi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rone, Proclus, Lycius, &amp; Vi<g ref="char:EOLhyphen"/>truutus. And it hath bene commo<g ref="char:cmbAbbrStroke">̄</g>ly called of barbarous writers of the latter time Dulcarnon.</hi>
               </p>
               <div type="section">
                  <pb facs="tcp:7062:89"/>
                  <head>
                     <note place="margin">
                        <hi>An addition of P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tari<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>.</hi>
                     </note>An addition of Pelitarius.</head>
                  <p>To reduce two vnequall squares to two equall squares.</p>
                  <p>Suppose that the squares of the lines <hi>AB</hi> and <hi>AC</hi> be vnequall. It is required to re<g ref="char:EOLhyphen"/>duce them to two equall squares. Ioyne the two lines <hi>AB</hi> and <hi>AC</hi> at their endes in such sort that they make a right angle <hi>BAC.</hi> And draw a line from <hi>B</hi> to <hi>C.</hi> Then vppon the two endes <hi>B</hi> and <hi>C</hi> make two angles eche of which may be equal to halfe a right an<g ref="char:EOLhyphen"/>gle (This is done by erecting vpon the line <hi>BC</hi> perpe<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>diculer
<figure/>
lines, from the pointes <hi>B</hi> and <hi>C:</hi> and so (by the 9. proposition) deuiding <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>che of the right angles into two equall partes): and let the angles <hi>BCD</hi> and <hi>CBD</hi> be either of the<g ref="char:cmbAbbrStroke">̄</g> halfe of a right angle. And let the lines <hi>BD</hi> and <hi>CD</hi> concurre in the point <hi>D.</hi> Then <hi>I</hi> say that the two squares of the sides <hi>BD</hi> and <hi>CD,</hi> are equall to the two squares of the sides <hi>AB</hi> and <hi>AC.</hi> For (by the 6 proposition) the two sides <hi>DB</hi> and <hi>DC</hi> are equall, and the angle at the pointe D is (by the 32. proposition) a right angle. Wherefore the square of the side <hi>BC</hi> is e<g ref="char:EOLhyphen"/>qual to the squares of the two sides <hi>DB</hi> and <hi>DC</hi> (by the 47. proposition): but it is also equall to the squares of the two sides <hi>AB</hi> and <hi>AC</hi> (by the self same proposition) wher<g ref="char:EOLhyphen"/>fore (by the common sentence) the squares of the two sides <hi>BD</hi> and <hi>DC</hi> are equall to the squares of the two sides <hi>AB</hi> and <hi>AC</hi>: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>An other addition of Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>An other aditio<g ref="char:cmbAbbrStroke">̄</g> of Pelitarius.</hi>
                     </note>
                     <hi>If two right angled triangles haue equall bases. the squares of the two sides of the one are equall to the squares of the two sides of the other.</hi>
                  </p>
                  <p>This is manifest by the former construction and demonstration.</p>
               </div>
               <div type="section">
                  <head>An other addition of Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>An other additi<g ref="char:EOLunhyphen"/>on of Pelitarius.</hi>
                     </note>
                     <hi>Two vnequall lines being geuen, to know how much the square of the one is greater then the square of the other.</hi>
                  </p>
                  <p>Suppose that there be two vnequal lines <hi>AB</hi> and <hi>BC</hi>: of which let <hi>AB</hi> be the grea<g ref="char:EOLhyphen"/>ter. It is required to search out how much the square of <hi>AB</hi> excedeth the square of <hi>BC.</hi> That is <hi>I</hi> wil finde out the square, which with the square of the line <hi>BC</hi> shalbe equal to the square of the line <hi>AB.</hi> Put the lines <hi>A
<figure/>
B</hi> and <hi>BC</hi> directly, that they make both one right line: and making the centre the point <hi>B,</hi> and the space <hi>BA</hi> describe a circle <hi>ADE.</hi> And produce the line <hi>AC</hi> to the circumference, and let it concurre with it in the point <hi>E.</hi> And vpon the lyne <hi>AE</hi> and fro<g ref="char:cmbAbbrStroke">̄</g> the point <hi>C</hi> erect (by the 11. proposition) a perpendiculer line <hi>CD,</hi> which produce till it concurre with the circumference in the point <hi>D</hi>: &amp; draw a line from <hi>B</hi> to <hi>D.</hi> Then <hi>I</hi> say, that the square of the line <hi>CD,</hi> is the excesse of the square of the line <hi>AB</hi> aboue the square of the line <hi>BC.</hi> For for<g ref="char:EOLhyphen"/>asmuch as in the triangle <hi>BCD,</hi> the angle at the point <hi>C</hi> is a right angle, the square of the base <hi>BD</hi> is equall to the squares of the two sides <hi>BC</hi> and <hi>C</hi>D (by this 47. proposition). Wherefore also the square of the line <hi>AB</hi> is equall to the selfe same squares of the lines <hi>BC</hi> and <hi>CD.</hi> Wherefore the square of the line <hi>BC</hi> is so much lesse then the square of the line <hi>AB,</hi> as is the square of the line <hi>CD</hi>: which was required to search out.</p>
               </div>
               <div type="section">
                  <pb n="59" facs="tcp:7062:89"/>
                  <head>An other addition of Pelitarius.<note place="margin">
                        <hi>An other aditio<g ref="char:cmbAbbrStroke">̄</g> of Pelitarius.</hi>
                     </note> The diameter of a square being geuen, to geue the square thereof.</head>
                  <p>This is easie to be done. For if vpon the two endes of the line be drawen two halfe right angles, and so be made perfect the triangle then shalbe described half of the square: the other halfe whereof also is after the same manner easie to be de<g ref="char:EOLhyphen"/>scribed.</p>
                  <p>
                     <hi>Hereby it is manifest, that the square of the diameter is double to that square whose diameter it is.</hi>
                     <note place="margin">
                        <hi>A Corrollary.</hi>
                     </note>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>34.</hi> Theoreme. The <hi>48.</hi> Proposition. If the square which is made of one of the sides of a triangle, be equall to the squares which are made of the two other sides of the same triangle: the angle comprehended vnder those two other sides is a right angle.</head>
               <p>
                  <hi>SVppose that ABC be a triangle, and let the square which is made of one of the sides there, namely, of the side BC, be equall to the squares which are made of the sides BA and AC. Then I say that the angle BAC is a right angle. Rayse vp (by the</hi> 11. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>) from the point A vnto the right line AC a perpendicular line AD. And (by the thirde proposition) vnto the line AB put an equall line AD. And by the first peticion draw a right line from the point D to the poin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> C. And forasmuch as
<figure/>
the line DA is equall to the line AB, the square which is made of the line DA is e<g ref="char:EOLhyphen"/>quall to the square whiche is made of the line AB Put the square of the line AC, common to them both. VVherefore the squares of the lines DA and AC are equal to the squares of the lines BA and AC. But (by the proposition going before) the square of the line DC is equal to y<hi rend="sup">e</hi> squares of the lines. AD and AC. (For the angle DAC is a right angle) and the square of BC is (by supposition) equall to the squares of AB and AC. VVherefore the square of DC is equall to the square of BC: wherefore the side DC is equall to the side BC. And forasmuch as AB is equall to AD <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd AC is common to them both, therefore these two sides DA and AC are equall to these two sides BA and AC, the one to the other, and the base DC is equall to the base BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wher<g ref="char:EOLhyphen"/>fore (by the</hi> 8. <hi>proposition) the angle DAC is equall to the angle BAC. But the angle DAC is a right angle, wherefore also the angle BAC is a right angle. If
<pb facs="tcp:7062:90"/>
therefore the square which is made of one of the sides of a triangle, be equall to the squares which are made of the two other sides of the same triangle, the an<g ref="char:EOLhyphen"/>gle comprehended vnder those two other sides is a right angle. which was re<g ref="char:EOLhyphen"/>quired to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>This proposition is the conue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>se of the former.</hi>
                  </note>This proposition is the conuerse of the former, and is of Pelitarius demon<g ref="char:EOLhyphen"/>strated by an argument leading to an impossibilitie after this maner.</p>
               <p>
                  <note place="margin">
                     <hi>An other De<g ref="char:EOLhyphen"/>monstration af<g ref="char:EOLhyphen"/>ter Pelitarius.</hi>
                  </note>Suppose that <hi>ABC</hi> be a triangle: &amp; let the square of the side <hi>AC,</hi> be equal to the squares of the two sides <hi>AB</hi> and <hi>BC.</hi> Then <hi>I</hi> say that the angle at the point <hi>B,</hi> which is opposite to the side <hi>AC,</hi> is a right angle. For if the angle at the point
<figure/>
                  <hi>B</hi> be not a right angle, then shal it be eyther greater or lesse the<g ref="char:cmbAbbrStroke">̄</g> a right angle. First let it be is greater. And let the angle <hi>DBC</hi> be a right angle, by erecting from the point <hi>B</hi> a per<g ref="char:EOLhyphen"/>pendicular line vnto the line <hi>BC</hi> (by the 11. proposition) which let be <hi>BD:</hi> and put the line <hi>BD</hi> equall to the lyne <hi>AB</hi> (by the thirde proposition). And drawe a line from <hi>C</hi> to D. Now (by the former proposition) the square of the side <hi>CD</hi> shalbe equall to the squares of the two sides <hi>BD</hi> and <hi>BC</hi>: wherefore also to the squares of the two sides <hi>BA</hi> and <hi>BC.</hi> Wherefore the base <hi>C</hi>D shalbe equall to the base <hi>CA,</hi> when as their squares are equall: which is con<g ref="char:EOLhyphen"/>trary to the 24. proposition. For forasmuch as the angle <hi>ABC</hi> is greater then the angle D<hi>BC,</hi> and the two sides <hi>AB</hi> and <hi>BC</hi> are equall to the two sides <hi>DB</hi> and <hi>BC,</hi> the one to the other, the base <hi>CA</hi> shall be greater then the base <hi>CD.</hi> It is also contrary to the 7. proposition, for from the two endes of one &amp; the same line, namely, fro<g ref="char:cmbAbbrStroke">̄</g> the points <hi>B</hi> &amp; <hi>C</hi> should be drawn on one and the same side two lines <hi>BD</hi> and <hi>DC</hi> ending at the pointe <hi>D,</hi> e<g ref="char:EOLhyphen"/>quall to two other lines <hi>BA</hi> and <hi>AC</hi> drawen from the same endes and ending at an other point, namely, at <hi>A,</hi> which is impos<g ref="char:EOLhyphen"/>sible. By the same reason also may we proue that the whole angle at the pointe <hi>B</hi> is not lesse then a right angle. Wherfore it is a right angle: which was required to be proued.</p>
            </div>
            <trailer>The ende of the first booke of Euclides Elementes.</trailer>
         </div>
         <div n="2" type="book">
            <pb n="60" facs="tcp:7062:90"/>
            <head>¶The second booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>IN this second booke Euclide sheweth,<note place="margin">
                  <hi>The argument of the second booke.</hi>
               </note> what is a Gnomo<g ref="char:cmbAbbrStroke">̄</g>, and a right angled parallelogramme. Also in this booke are set forth the powers of lines, deui<g ref="char:EOLhyphen"/>ded euenly and vneuenly, and of lines added one to an other. The power of a line,<note place="margin">
                  <hi>What is the power of a line.</hi>
               </note> is the square of the same line: that is, a square, euery side of which is e<g ref="char:EOLhyphen"/>quall to the line. So that here are set forth the quali<g ref="char:EOLhyphen"/>ties and proprieties of the squares and right lined fi<g ref="char:EOLhyphen"/>gures, which are made of lines &amp; of their parts.<note place="margin">Many compe<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>dious rules of reckoning ga<g ref="char:EOLhyphen"/>thered one of this booke, and also many rules of Alge<g ref="char:EOLhyphen"/>bra.</note> The Arithmetician also our of this booke gathereth ma<g ref="char:EOLhyphen"/>ny compendious rules of reckoning, and many rules also of Algebra, with the equatio<g ref="char:cmbAbbrStroke">̄</g>s therein vsed. The groundes also of those rules are for the most part by this second booke demon<g ref="char:EOLhyphen"/>strated.<note place="margin">
                  <hi>Two wonder<g ref="char:EOLhyphen"/>full propositi<g ref="char:EOLhyphen"/>ons in this booke.</hi>
               </note> This booke moreouer contayneth two wonderfull propositions, one of an obtuse angled triangle, and the other of an acute: which with the ayde of the 47. proposition of the first booke of Euclide, which is of a rectangle triangle, of how great force and profite they are in matters of astronomy, they knowe which haue trauayled in that arte. VVherefore if this booke had none other profite be side, onely for these 2. propositions sake it were diligently to be embraced and studied.</p>
            <div type="part">
               <head>The definitions.</head>
               <div type="section">
                  <head>
                     <hi>1.</hi> Euery rectangled parallelogramme,<note place="margin">
                        <hi>First definition.</hi>
                     </note> is sayde to be contayned vnder two right lines comprehending a right angle.</head>
                  <p>A parallelogramme is a figure of fower sides,<note place="margin">
                        <hi>What a pa<g ref="char:EOLhyphen"/>rallelogramme is.</hi>
                     </note> whose two opposite or contra<g ref="char:EOLhyphen"/>ry sides are equall the one to the other. There are of parallelogrammes fower kyndes,<note place="margin">
                        <hi>Fower kindes of parallelo<g ref="char:EOLhyphen"/>grammes.</hi>
                     </note> a square, a figure of one side longer, a Rombus or diamond, and a Rom<g ref="char:EOLhyphen"/>boides or diamond like figure, as before was sayde in the 33. definition of the first booke. Of these fower sortes, the square and the figure of one side longer are onely right angled Parallelogrammes: for that all their angles are right angles. And either of them is contayned (according to this definition) vnder two right lynes whi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h concurre together, and cause the right angle, and containe the same. Of which two lines the one is the length of the figure, &amp; the other the breadth. The parallelogramme is imagined to be made by the draught or motion of one of the lines into the length of the other. As if two numbers shoulde be multipli<g ref="char:EOLhyphen"/>ed the one into the other. As the figure ABCD is a parallelograme, and is sayde to be contayned vnder the two right lines AB and AC, which contayne the right angle BAC, or vnder the two right lines AC and
<figure/>
CD, for they likewise contayne the right angle ACD: of which 2. lines the one, namely, AB is the length, and the o<g ref="char:EOLhyphen"/>ther, namely, AC is the breadth. And if we imagine the line AC to be drawen or moued directly according to the le<g ref="char:cmbAbbrStroke">̄</g>gth
<pb facs="tcp:7062:91"/>
of the line AB, or contrary wise the line AB to be moued directly according to the length of the line AC, you shall produce the whole rectangle parallelo<g ref="char:EOLhyphen"/>gramme ABCD which is sayde to be contayned of them: euen as one number multiplied by an other produceth a plaine and righte angled superficiall num<g ref="char:EOLhyphen"/>ber, as ye see in the figure here set, where the number of sixe
<figure/>
or sixe vnities, is multiplied by the number of fiue or by fiue vnities: of which multiplication are produced 30. which number being set downe and described by his vnities repre<g ref="char:EOLhyphen"/>senteth a playne and a right angled number. VVherefore e<g ref="char:EOLhyphen"/>uen as equall numbers multipled by equal numbers produce numbers equall the one to the other: so rectangle parallelo<g ref="char:EOLhyphen"/>grames which are comprehended vnder equal lines are equal the one to the other.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Second de<g ref="char:EOLhyphen"/>fini<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion.</hi>
                     </note>
                     <hi>2.</hi> In euery parallelogramme, one of those parallelogrammes, which soeuer it be, which are about the diameter, together with the two supplementes, is called a Gnomon.</head>
                  <p>Those perticuler parallelogrames are sayde to be about the diameter of the parallelograme, which haue the same diameter which the whole parallelograme hath. And supplementes are such, which are without the diameter of the whole parallelograme. As of the parallelograme ABCD the partial or perticuler paral<g ref="char:EOLhyphen"/>lelogrames GKCF and EBKH are parallelogrames about the diameter, for that ech of them hath for his diameter a part of the diameter of the whole paral<g ref="char:EOLhyphen"/>lelogramme.
<figure/>
As CK and KB the perticuler diameters, are partes of the line CB, which is the diameter of the whole parallelogramme. And the two paralle<g ref="char:EOLhyphen"/>logrammes AEGK and KHFD, are supplementes, because they are wythout the diameter of the whole parallelogramme, namely, CB. Now any one of those partiall parallelogrammes about the diameter together with the two supple<g ref="char:EOLhyphen"/>mentes make a gnomon. As the parallelograme EBKH, with the two supple<g ref="char:EOLhyphen"/>mentes AEGK and KHFD make the gnomon FGEH. Likewise the paral<g ref="char:EOLhyphen"/>lelogramme GKCF with the same two supplementes make the gnomon EHFG. And this diffinition of a gnomon extendeth it selfe, and is generall to all kyndes of parallelogrammes, whether they be squares or figures of one side lon<g ref="char:EOLhyphen"/>ger or Rhombus or Romboides. To be shorte, if you take away from the whole parallelogramme one of the partiall parallelogrammes which are about the di<g ref="char:EOLhyphen"/>ameter
<pb n="61" facs="tcp:7062:91"/>
whether ye will, the rest of the figure is a gnomon.</p>
                  <p>Campa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e after the last proposition of the first booke addeth this propositio<g ref="char:cmbAbbrStroke">̄</g>.<note place="margin">
                        <hi>A proposition added by Cam<g ref="char:EOLhyphen"/>pane after the last propositi<g ref="char:EOLhyphen"/>on of the first booke.</hi>
                     </note> 
                     <hi>Two squares being geuen, to adioyne to one of them a Gnomon equall to the other square:</hi> which, for that as then it was not taught what a Gnomon is, I there omitted, thinking that it might more aptly be placed here. The doing and demonstration whereof, is thus.</p>
                  <p>Suppose that there be two squares <hi>AB</hi> and <hi>CD</hi>: vnto one of which, namely, vnto <hi>AB,</hi> it is required to adde a Gnomon equall to the other square, namely, to <hi>CD.</hi> Pro<g ref="char:EOLhyphen"/>duce the side <hi>BF</hi> of the square <hi>AB</hi> di<g ref="char:EOLhyphen"/>rectly
<figure/>
to the point <hi>E.</hi> and put the line <hi>FE</hi> equall to the side of the square <hi>CD.</hi> And draw a line from <hi>E</hi> to <hi>A.</hi> Now then forasmuch as <hi>EFA</hi> is a rectangle trian<g ref="char:EOLhyphen"/>gle, therefore (by the 47. of the first) the square of the line <hi>EA</hi> is equall to the squares of the lines <hi>EF</hi> &amp; <hi>FA.</hi> But the square of the line <hi>EF</hi> is equall to the square <hi>CD,</hi> &amp; the square of the side <hi>FA</hi> is the square <hi>AB.</hi> Wherefore the square of the line <hi>AE</hi> is equall to the two squares <hi>CD</hi> and <hi>AB.</hi> But the sides <hi>EF</hi> and <hi>FA</hi> are (by the 21. of the first) longer then the side <hi>AE,</hi> and the side <hi>FA</hi> is equall to the side <hi>FB.</hi> Wherfore the sides <hi>EF</hi> and <hi>FB</hi> are longer the<g ref="char:cmbAbbrStroke">̄</g> the side <hi>AE.</hi> Wherefore the whole line <hi>BE</hi> is longer then the line <hi>AE.</hi> From the line <hi>BE</hi> cut of a line equall to the line <hi>AE,</hi> which let be <hi>BC.</hi> And (by the 46. proposition) vpon the line <hi>BC</hi> describe a square, which let be <hi>BCGH</hi>: which shalbe equal to the square of the line <hi>AE,</hi> but the square of the line <hi>AE</hi> is equal to the two squares <hi>AB</hi> and <hi>DC.</hi> Wherefore the square <hi>BCGH</hi> is equal to the same squares. Wherfore forasmuch as the square <hi>BCGH</hi> is composed of the square <hi>AB</hi> and of the gnomon <hi>FGAH,</hi> the sayde gnomon shalbe equall vnto the square <hi>CD</hi>: which was required to be done.</p>
                  <div type="subsection">
                     <head>An other more redy way after Pelitarius.</head>
                     <p>Suppose that there be two squares, whose sides let be <hi>AB</hi> and <hi>BC.</hi> It is required vnto the square of the line <hi>AB,</hi> to adde a gnomon equall to the square of the line <hi>BC.</hi> Set the lines <hi>AB</hi>
                        <figure/>
and <hi>BC</hi> in such sort that they make a right angle <hi>ABC.</hi> And draw a line fro<g ref="char:cmbAbbrStroke">̄</g> 
                        <hi>A</hi> to <hi>C.</hi> And vpo<g ref="char:cmbAbbrStroke">̄</g> the line <hi>AB</hi> describe a square which let be <hi>ABDE.</hi> And produce the line <hi>BA</hi> to the point <hi>F,</hi> and put the line <hi>BF</hi> equall to the line <hi>AC.</hi> And vpon the line BF describe a square which let be BFGH: which shalbe equal to the square of the line <hi>AC,</hi> whe<g ref="char:cmbAbbrStroke">̄</g> as the lines <hi>BF</hi> and <hi>AC</hi> are equal: and therefore it is equal to the squares of the two lines <hi>AB</hi> and <hi>BC.</hi> Now forasmuch as the square <hi>BFGH</hi> is made complete by the square <hi>ABDE</hi> and by the gnomon <hi>FEGD,</hi> the gnomon <hi>FEGD</hi> shalbe equal to the square of the line <hi>BC</hi>: which was required to be done.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:92"/>
               <head>The <hi>1.</hi> Theoreme. The <hi>1.</hi> Proposition. If there be two right lines, and if the one of them be deuided into partes howe many soeuer: the rectangle figure compre<g ref="char:EOLhyphen"/>hended vnder the two right lines, is equall to the rectangle fi<g ref="char:EOLhyphen"/>gures whiche are comprehended vnder the line vndeuided, and vnder euery one of the partes of the other line.</head>
               <p>
                  <hi>SVppose that there be two right lynes A and BC and let one of them, namely, BC be deui<g ref="char:EOLhyphen"/>ded at all aduentures in the pointes D and E. Then I say that the rectangle figure comprehended vn<g ref="char:EOLhyphen"/>der the lines A, and BC, is equall vnto the rectan<g ref="char:EOLhyphen"/>gle figure comprehended vnder the lines A, and BD, &amp; vnto the rectangle figure which is co<g ref="char:cmbAbbrStroke">̄</g>prehen<g ref="char:EOLhyphen"/>ded vnder the lines A and DE, and also vnto the rectangle figure which is comprehended vnder the lines A and EC.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>For from the pointe Brayse vp (by the</hi> 11. <hi>of the first) vnto the right line BC a perpendiculer line BF, &amp; vnto the
<figure/>
line A (by the third of the first) put the line BG e<g ref="char:EOLhyphen"/>quall, and by the point G (by the</hi> 31. <hi>of the first) draw a parallel line vnto the right line BC and let the same be GM, and (by the selfe same) by y<hi rend="sup">e</hi> points D, E, and C, draw vnto the line BG these parallel lines DK,</hi>
                  <note place="margin">
                     <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                     </hi>
                  </note> 
                  <hi>EL and CH. Now then the parallelo<g ref="char:EOLhyphen"/>grame BH is equall to these parallelogrammes BK, DL, and EH. But the parallelograme BH is equall vnto that which is contayned vnder the lines A and BC. (For it is com<g ref="char:EOLhyphen"/>prehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder the lines GB &amp; BC, and the line GB is equall vnto the line A) And the parallelograme BK is equall to that which is contayned vnder the lines A and BD: (for it is comprehended vnder the line GB and BD, and BG is e<g ref="char:EOLhyphen"/>quall vnto A) And the parallelograme DL is equall to that which is contayned vnder the lines A and DE (for the line DK, that is, BG is equal vnto A) And moreouer likewise the parallelograme EH is equall to that which is contained vnder the lines A &amp; EC. VVherfore that which is comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder y<hi rend="sup">e</hi> lines A &amp; BC is equall to that which is comprehended vnder the lines A &amp; BD, &amp; vn<g ref="char:EOLhyphen"/>to y<hi rend="sup">t</hi> which is comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder the lines A and DE, and moreouer vnto that which is comprehended vnder the lines A and EC. If therfore there be two right lines, and if the one of them be deuided into partes how many soeuer, the rectan<g ref="char:EOLhyphen"/>gle
<pb n="62" facs="tcp:7062:92"/>
figure comprehended vnder the two right lines, is equall to the rectangle fi<g ref="char:EOLhyphen"/>gures which are comprehended vnder the line vndeuided and vnder euery one of the partes of the other line: which was required to be demonstrated.</hi>
               </p>
               <p>Because that all the Propositions of this second booke for the most part are true both in lines and in numbers, and may be declared by both: therefore haue I haue added to euery Proposition conuenient numbers for the manifestation of the same. And to the end the studious and diligent reader may the more fully perceaue and vnderstand the agrement of this art of Geometry with the science of Arithmetique, and how nere &amp; deare sisters they are together, so that the one cannot without great blemish be without the other, I haue here also ioyned a little booke of Arithmetique written by one <hi>Barlaam,</hi> a Greeke authour a man of greate knowledge. In whiche booke are by the authour demonstrated many of the selfe same proprieties and passions in number, which <hi>Euclide</hi> in this his second boke hath demonstrated in magnitude, namely, the first ten pro<g ref="char:EOLhyphen"/>positions as they follow in order. VVhich is vndoubtedly great pleasure to co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>sider, also great increase &amp; furniture of knowledge. VVhose Propositio<g ref="char:cmbAbbrStroke">̄</g>s are set orderly after the propositio<g ref="char:cmbAbbrStroke">̄</g>s of <hi>Euclide,</hi> euery one of <hi>Barlaam</hi> correspo<g ref="char:cmbAbbrStroke">̄</g>dent to the same of Euclide. And doubtles it is wonderful to see how these two co<g ref="char:cmbAbbrStroke">̄</g>trary kynds of quantity, quantity discrete or number, and quantity continual or magnitude (which are the subiectes or matter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of Arithmitique and Geometry) shoulde haue in them one and the same proprieties common to them both in very ma<g ref="char:EOLhyphen"/>ny points, and affections, although not in all. For a line may in such sort be de<g ref="char:EOLhyphen"/>uided, that what proportion the whole hath to the greater parte the same shall the greater part haue to the lesse. But that can not be in number. For a number can not so be deuided, that the whole number to the greater part thereof, shall haue that proportion which the greater part hath to the lesse, as Iordane very playnely proueth in his booke of Arithmetike, which thynge Campane also (as we shall afterward in the 9. booke after the 15. proposition see) proueth. And as touching these tenne firste propositions of the seconde booke of Eu<g ref="char:EOLhyphen"/>clide, demonstrated by Barlaam in numbers, they are also demo<g ref="char:cmbAbbrStroke">̄</g>strated of Cam<g ref="char:EOLhyphen"/>pane after the 15. proposition of the 9. booke, whose demonstrations I mynde by Gods helpe to set forth when I shal come to the place. They are also demo<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>strated of Iordane that excellet learned authour in the first booke of his Arith<g ref="char:EOLhyphen"/>metike. In the meane tyme I thought it not amisse here to set forth the demon<g ref="char:EOLhyphen"/>strations of Barlaam, for that they geue great light to the seconde booke of Eu<g ref="char:EOLhyphen"/>clide, besides the inestimable pleasure, which they bring to the studious conside<g ref="char:EOLhyphen"/>rer. And now to declare the first Proposition by numbers. I haue put this exam<g ref="char:EOLhyphen"/>ple following.</p>
               <p>Take two numbers the one vndeuided as 74. the other deuided into what partes and how many you list, as 37. deuided into 20. 10. 5. and 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which altogether make the whole 37. Then if you multiply the number vndeuided, namely, 74, into all the partes of the number deuided as into 20. 10. 5. and 2. you shall produce 1480. 740. 370. 148. which added together make 2738: which self number is also produced if you multiplye the two numbers first geuen the one into the other. As you see in the exam<g ref="char:EOLhyphen"/>ple on the other side set.</p>
               <p>
                  <pb facs="tcp:7062:93"/>
                  <gap reason="math">
                     <desc>〈 math 〉</desc>
                  </gap>
               </p>
               <p>So by the aide of this Proposition is gotten a compendious way of multiplication by breaking of one of the numbers into his partes: which oftentimes serueth to great vse in working<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> chi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ly in the rule of proportions. The demonstration of which proposition followeth in Barlaam. But <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst are put of the author these principles following.</p>
               <div type="section">
                  <head>
                     <note place="margin">Barlaam.</note>¶ Principles.</head>
                  <p n="1">1. A number is s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>yd to multiply an other number: when the number multiplied is so oftentymes added to it selfe, as there be vnities in the number, which multiplieth: wherby is produced a certaine number which the number multiplied measureth by the vnities which are in the number which mul<g ref="char:EOLhyphen"/>tipli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th.</p>
                  <p n="2">2. And the number produced of that a multiplication is called a plaine or superficiall number.</p>
                  <p n="3">3. A square number is that which is produced of the multiplicatian of any number into it selfe.</p>
                  <p n="4">4. Euery lesse number compared to a greater is sayd to be a part of the greater, whether the lesse mea<g ref="char:EOLhyphen"/>sure the greater, or measure it not.</p>
                  <p n="5">5. Numbers, whome one and the selfe same number measureth equally, that is, by one and the selfe same number are equall the one to the othe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</p>
                  <p n="6">6. Numbers that are equemultipl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ces to one and the selfe same number, that is, which contayne one and the same number equally and alike, are equall the one to the other.</p>
               </div>
               <div type="section">
                  <head>The first Proposition.</head>
                  <p>
                     <hi>Two numbers beyng geuen, if the one of them be deuided into
<figure/>
any numbers how many soeuer:</hi>
                     <note place="margin">
                        <hi>Barlaam.</hi>
                     </note> 
                     <hi>the playne or superficiall number which is produced of the multiplication of the two numbers first geuen the one into the other, shall be equall to the superficiall nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>bers which are produced of the multiplication of the number not deuided into euery part of the number deuided.</hi>
                  </p>
                  <p>Suppose that there be two numbers <hi>AB</hi> and <hi>C.</hi> And deuide the number <hi>AB</hi> into certayne other numbers how many soeuer, as into <hi>AD, DE,</hi> and <hi>EB.</hi> Then I say that the superficiall number which is produced of the multiplication of the number <hi>C</hi> into the number <hi>AB</hi> is equall to the superficiall numbers which are produ<g ref="char:EOLhyphen"/>ced of the multiplication of the number <hi>C</hi> into the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber <hi>AD,</hi> and of <hi>C</hi> into <hi>DE,</hi> and of <hi>C</hi> into <hi>EB.</hi> For let <hi>F</hi> be the superficiall number produced of the multiplica<g ref="char:EOLhyphen"/>tion of the number <hi>C</hi> into the number <hi>AB,</hi> and let <hi>GH</hi> be the superficiall number produced of the multipli<g ref="char:EOLhyphen"/>cation of <hi>C</hi> into <hi>AD:</hi> And let <hi>HI</hi> be produced of the multiplication of <hi>C</hi> into <hi>DE:</hi> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d finally of the multi<g ref="char:EOLhyphen"/>plication of <hi>C</hi> into <hi>EB</hi> let there be produced the num<g ref="char:EOLhyphen"/>ber
<pb n="63" facs="tcp:7062:93"/>
                     <hi>IK.</hi> Now forasmuch as <hi>AB</hi> multiplying the number <hi>C</hi> produced the number <hi>F</hi>: therefore the number <hi>C</hi> measureth the number <hi>F</hi> by the vnities which are in the num<g ref="char:EOLhyphen"/>ber <hi>AB.</hi> And by the same reason may be proued that the number <hi>C</hi> doth also measure the number <hi>GH,</hi> by the vnities which are in the number <hi>AD,</hi> and that it doth mea<g ref="char:EOLhyphen"/>sure the number <hi>HI</hi> by the vnities which are in the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>DF</hi> and finally that it mea<g ref="char:EOLhyphen"/>sureth the number <hi>IK</hi> by the vnities which are in the number <hi>EB.</hi> Wherefore the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber <hi>C</hi> measureth the whole number <hi>GK</hi> by the vnities which are in the number <hi>AB.</hi> But it before measured the number <hi>F</hi> by the vnities which are in the number <hi>AB,</hi> wher<g ref="char:EOLunhyphen"/>fore either of these numbers <hi>F</hi> and <hi>GK</hi> is equemultiplex to the number <hi>C.</hi> But num<g ref="char:EOLhyphen"/>bers which are equemultiplices to one &amp; the selfe same numbers are equall the one to the other (by the 6. definition). Wherfore the number <hi>F</hi> is equall to the number <hi>GK.</hi> But the number <hi>F</hi> is the superficiall number produced of the multiplication of the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber <hi>C</hi> into the number <hi>AB</hi>: and the number <hi>GK</hi> is composed of the superficiall num<g ref="char:EOLhyphen"/>bers produced of the multiplication of the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>C</hi> not deuided into euery one of the numbers <hi>AD, DE,</hi> and <hi>EB.</hi> If therefore there be two numbers geuen and the one of them be deuided &amp;c. Which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Theoreme. The <hi>2.</hi> Proposition. If a right line be deuided by chaunce, the rectangles figures which are comprehended vnder the whole and euery one of the partes, are equall to the square whiche is made of the whole.</head>
               <p>SVppose that the right line AB be by chaunse de<g ref="char:EOLhyphen"/>nided
<figure/>
in the point C. Then I say that the rectan<g ref="char:EOLhyphen"/>gle figure comprehended vnder AB and BC to<g ref="char:EOLhyphen"/>gether with the rectangle comprehended vnder AB and AC is equall vnto the square made of AB.<note place="margin">Construction.</note> De<g ref="char:EOLhyphen"/>scribe (by the <hi>46.</hi> of the first) vpon AB a square ADEB: and (by the <hi>31</hi> of the first) by the point C draw a line paral<g ref="char:EOLhyphen"/>lel vnto either of these lines AD a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d BE, and let the same be C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                  </note> Now is the parallelogramme AE equall to the parallelogrammes AF and CE, by the first of this booke. But AE is the square made of AB. And AF is the rectangle parallelogramme comprehended vnder the lines BA and AC: for it is compre<g ref="char:EOLhyphen"/>hended vnder the lines DA and AC: but the line AD is equall vnto the line AB. And likewise the parrallelogramme CE is equall to that which is contayned vnder the lynes AB and BC: for the line BE is equal vnto the line AB. VVher<g ref="char:EOLhyphen"/>fore that which is contayned vnder BA and AC together with that which is contayned vnder the lines AB and BC, is equall to the square made of the line AB. If therefore a right line be deuided by chaunce, the recta<g ref="char:cmbAbbrStroke">̄</g>gle figures which are comprehended vnder the whole, and euery one of the partes, are equall to the square which is made of the whole: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration of Campane.</head>
                  <p>
                     <pb facs="tcp:7062:94"/>Suppose that the line <hi>AB</hi> be deuided into
<figure/>
the lines <hi>AC, CD,</hi> and <hi>DB.</hi> Then I say that the square of the whole line <hi>AB,</hi> which let be <hi>AEBF,</hi> is equal to the rectangle figures which are contayned vnder the whole and euery one of the partes: for take the line <hi>K,</hi> which let be e<g ref="char:EOLhyphen"/>qual to the line A<hi>B.</hi> Now then by the first pro<g ref="char:EOLhyphen"/>position the rectangle figure contained vnder the lines <hi>AB</hi> and <hi>K,</hi> is equall to the rectangle figures contayned vnder the line <hi>K</hi> and al the partes of the line <hi>AB.</hi> But that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines <hi>K</hi> and <hi>AB</hi> is equall to the square of the line <hi>AB,</hi> and the rectangle figures contayned vnder the line <hi>K</hi> and al the partes of <hi>AB,</hi> are equall to the rectangle fi<g ref="char:EOLhyphen"/>gures contayned vnder the line <hi>AB</hi> and all the partes of the line <hi>AB</hi>: for the lines <hi>AB</hi> and <hi>K</hi> are equall: wherefore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>An example of this Proposition in numbers.</head>
                  <p>Take a number, as 11. and deuide it into two partes, namely, 7. and 4: and multiply 11. into 7, and then into 4. and there shalbe produced 77. and 44: both which numbers added together make 121. which is equall to the square number produced of the mul<g ref="char:EOLhyphen"/>tiplication of the number 11. into himselfe, as you see in the example.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration whereof followeth in Barlaam.</p>
                  <div type="subsection">
                     <head>The second Proposition.</head>
                     <p>
                        <note place="margin">
                           <hi>Barlaam.</hi>
                        </note>
                        <hi>If a number geuen be deuided into two other numbers: the superficiall numbers, which are pro<g ref="char:EOLhyphen"/>duced of the multiplication of the whole into either part, added together, are equall to the square num<g ref="char:EOLhyphen"/>ber of the whole number geuen.</hi>
                     </p>
                     <p>Suppose that the number geuen be <hi>AB</hi>: and let it be deuided into two other num<g ref="char:EOLhyphen"/>bers <hi>AC</hi> and <hi>CB.</hi> Then I say that the two superficiall numbers, which are produced of the multiplication of <hi>AB</hi> into <hi>AC,</hi> and of <hi>AB</hi> into <hi>BC,</hi> those two superficiall num<g ref="char:EOLhyphen"/>bers (I say) beyng added together, shalbe equall to the square number produced of the multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of the number <hi>AB</hi> into it selfe. For let the number <hi>AB</hi> multiplying it selfe produce the number <hi>D.</hi> Let the number <hi>AC</hi> also multiplying the number <hi>AB</hi>
                        <pb n="64" facs="tcp:7062:94"/>
produce the number <hi>E</hi>F: agayne let the number <hi>CB</hi> multipli<g ref="char:EOLhyphen"/>ing
<figure/>
the selfe same number <hi>AB</hi> produce the number FG. Now forasmuch as the number <hi>AC</hi> multiplying the number <hi>AB</hi> produced the number EF: therefore the number AB measu<g ref="char:EOLhyphen"/>reth the number EF by the vnities which are in <hi>AC.</hi> Againe for<g ref="char:EOLhyphen"/>asmuch as the number <hi>CB</hi> multiplied the number <hi>AB</hi> and pro<g ref="char:EOLunhyphen"/>duced the number FG: therfore the number <hi>AB</hi> measureth the number FG by the vnities which are in the number <hi>CB.</hi> But the same number <hi>AB</hi> before measured the number EF by the vni<g ref="char:EOLhyphen"/>ties which are in the number <hi>AC.</hi> Wherefore the number <hi>AB</hi> measureth the whole number <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>G</hi> by the vnities whcih are in <hi>AB.</hi> Farther forasmuch as the number <hi>AB</hi> multiplying it selfe pro<g ref="char:EOLunhyphen"/>duced the number <hi>D</hi>: therefore the number <hi>AB</hi> measureth the number <hi>D</hi> by the vnities which are in himselfe. Wherfore it mea<g ref="char:EOLunhyphen"/>sureth either of these numbers, namely, the number <hi>D,</hi> and the number <hi>EG,</hi> by the vnities which are in himselfe. Wherfore how multiplex the number <hi>D</hi> is to the number <hi>AB,</hi> so multiplex is the number <hi>EG</hi> to the same number <hi>AB.</hi> But numbers which are equemultiplices to one and the selfe same number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> are equal the one to the other. Wherefore the number <hi>D</hi> is equall to the number <hi>EG.</hi> And the number <hi>D</hi> is the square number made of the number <hi>AB,</hi> and the number <hi>EG</hi> is composed of the two su<g ref="char:EOLhyphen"/>perficiall numbers produced of <hi>AB</hi> into <hi>BC,</hi> and of <hi>BA</hi> into <hi>AC.</hi> Wherefore the square number produced of the number <hi>AB</hi> is equall to the superficial numbers, produced of the number AB into the number <hi>BC,</hi> and of AB into AC, added together. If therefore a number be deuided into two other numbers &amp;c. which was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Theoreme. The <hi>3.</hi> Proposition. If a right line be deuided by chaunce: the rectangle figure co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>prehended vnder the whole and one of the partes, is equall to the rectangle figu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e comprehended vnder the partes, &amp; vnto the square which is made of the foresaid part.</head>
               <p>SVppose that the right line geuen AB be deuided by chaunce in the point C. Then I say that the rectangle figure comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder the lines AB and BC is equall vnto the rectangle figure comprehended vnder the lines AC and CB, and also vnto the square which is made of the line BC.<note place="margin">Construction.</note> Describe (by the <hi>46.</hi> of the first) vpon the line BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> square CDEB: and (by the second peticion) extend ED vnto F. And by the point A draw (by the <hi>31.</hi> of the first) a line parallel vnto either of these lines CD and BE, and let the same be AF. Now the parallelo<g ref="char:EOLhyphen"/>grame
<figure/>
AE is equall vnto the parallelo<g ref="char:EOLhyphen"/>grammes AD and CE.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                  </note> And AE is the rectangle figure comprehended vnder the lines AB and BC. For it is compre<g ref="char:EOLhyphen"/>hended vnder the lines AB and BE,
<pb facs="tcp:7062:95"/>
which line BE is equall vnto the line BC. And the parallelograme AD is e<g ref="char:EOLhyphen"/>quall to that which is contayned vnder the lines AC and CB: for the line DC is equall vnto the line CB And DB is the square which is made of the lyne CB. VVherfore the rectangle figure comprehended vnder the lynes AB and BC is equall to the rectangle figure comprehended vnder the lines AC and CB &amp; al<g ref="char:EOLhyphen"/>so vnto the square which is made of the line BC. If therfore a right line be de<g ref="char:EOLhyphen"/>uided by chaunce, the rectangle figure comprehended vnder the whole and one of the partes, is equall to the rectangle figure comprehended vnder the partes, and vnto the square which is made of the foresayd part: which was required to be proued.</p>
               <div type="section">
                  <head>An example of this Proposition in numbers.</head>
                  <p>Suppose a number, namely, 14. to be deuided into two partes 8. and 6. The whole number 14. multiplied into 8. one of his partes, produceth 112: the partes 8. &amp; 6. mul<g ref="char:EOLhyphen"/>tiplied the one into the other produce 48, which added to 64 (which is the square of 8. the former part of the number) amounteth also to 112: whiche is equall to the former summe. As you see in the example.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration hereof followeth in Barlaam.</p>
                  <div type="subsection">
                     <head>The third proposition.</head>
                     <p>
                        <note place="margin">
                           <hi>Barlaam.</hi>
                        </note>
                        <hi>If a number geuen be deuided into two numbers: the superficiall number which is produced of the multiplication of the whole into one of the partes, is equall to the superficiall number which is pro<g ref="char:EOLhyphen"/>duced of the partes the one into the other, and to the square number produced of the foresayd part.</hi>
                     </p>
                     <p>Suppose that the number geuen be <hi>AB,</hi> which let be deuided into two numbers <hi>AC</hi> and <hi>CB.</hi> Then <hi>I</hi> say that the superficiall number whiche is produced of the multipli<g ref="char:EOLhyphen"/>cation of the number <hi>AB</hi> into the number <hi>BC,</hi> is equall to the superficiall number which is produced of the multiplication of the number <hi>AC</hi> into the number <hi>CB,</hi> and to the square number produced of the number <hi>CB.</hi> For let the number <hi>AB</hi> multipli<g ref="char:EOLhyphen"/>eng the number <hi>CB</hi> produce the number <hi>D.</hi> And let the number <hi>AC</hi> multiplieng the number <hi>CB</hi> produce the number <hi>EF</hi>: and finally let the number <hi>CB</hi> multiplieng him<g ref="char:EOLhyphen"/>selfe produce the number <hi>FG.</hi> Now forasmuch as the number <hi>AB</hi> multiplieng the
<pb n="65" facs="tcp:7062:95"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <figure/>
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Theoreme. The <hi>4.</hi> Proposition. If a right line be deuided by chaunce, the square whiche is made of the whole line is equal to the squares which are made of the partes, &amp; vnto that rectangle figure which is compre<g ref="char:EOLhyphen"/>hended vnder the partes twise.</head>
               <p>SVppose that the right lyne AB be by chaunce deuided in the pointe C. Then I say that the square made of the line AB is equall vnto y<hi rend="sup">e</hi> squares which are made of the lines AC and CB, and vnto the rectangle figure contained vnder the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AC and CB <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wise.<note place="margin">Construction.</note> Describe
<figure/>
(by y<hi rend="sup">e</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. of the first) vpon the line AB a square ADEB: and draw a line from B to D, and (by the <hi>31.</hi> of the first) by the point C draw a line parallel vnto either of these lines AD and BE c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tting the diameter BD in the point G, and let the same be CF. And (by the point G (by the selfe same) draw a line parallel vnto eyther of these lines AB and DE, and let the same be HK.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>.</note> And forasmuch as the line CF is a parallel vnto
<pb facs="tcp:7062:96"/>
the line AD, and vpon them falleth a right line BD: therfore (by the <hi>29.</hi> of the first) the outward angle CGB is equall vnto the inward and opposite angle ADB. But the angle ADB is (by the <hi>5.</hi> of the first) equall vnto the angle ABD: for the side BA is equall vnto the side AD (by the definition of a square). VVherfore the angle CGB is equall vnto the angle GBC: wherfore (by the <hi>6.</hi> of the first) the side BC is equall vnto the side CG. But CB is equall vnto GK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and CG is equall vnto KB wherfore GK is equall vnto KB. VVherfore the figure CGKB consisteth of foure equall sides. I say also that it is a rectangle fi<g ref="char:EOLhyphen"/>gure. For forasmuch as CG is a parallel vnto BK &amp; vpon the<g ref="char:cmbAbbrStroke">̄</g> falleth a right line CB, therfore (by y<hi rend="sup">e</hi> 
                  <hi>29.</hi> of the <hi>1.</hi>) the angles KBC, and GCB are equal vnto two right angles. But the angle KBC is a right angle, wherfore y<hi rend="sup">e</hi> angle. BCG is also a right angle. VVherfore (by the <hi>34.</hi> of the first) the angles opposite vnto them, namely, CGK, and GKB are right angles. VVherfore CGKB is a rectan<g ref="char:EOLhyphen"/>gle figure. And it was before proued that the sides are equall. VVherfore it is a square, and it is described vpon the line B <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And by the same reason also HF is a square, and is described vpon the line HG, that is, v<g ref="char:EOLhyphen"/>pon
<figure/>
the line AC. VVherfore the squares HF and CK are made of the lines AC and CB. And forasmuch as the parallelograme AG is (by the <hi>43.</hi> of the first) e<g ref="char:EOLhyphen"/>quall vnto the parallelogramme GE. And AG is that which is contayned vnder AC and CB, for CG is equal vnto CB, wherfore GE is equall to that which is con<g ref="char:EOLhyphen"/>tained vnder AC and CB. VVherefore AG and GE are equall vnto that which is comprehended vnder AC and CB twise. And the squares HF and CK are made of the lines AC and CB. VVherfore these foure rectangle figures HF, CK, AG, and GE are equall vnto the squares whiche are made of the lines AC and CB, and to the rectangle figure which is comprehended vnder the lines AC and CB twise. But the rect<g ref="char:EOLhyphen"/>angle figures HF, CK, AG, and GE are the whole rectangle figure ADEB which is the square made of the line AB. VVherfore the square which is made of the line AB is equall to the squares which are made of the lines AC and CB, and vnto the rectangle figure which is comprehended vnder the lines AC and CB <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wise. If therfore a right line be deuided by chaunce, the square whiche is made of the whole line, is equall to the squares which are made of the partes, &amp; vnto the rectangle figure which is comprehended vnder the partes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wise: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration.</head>
                  <p>I say that the square of the line AB is equall vnto the squares whiche are made of the lines AC and CB, &amp; vnto the rectangle figure which is comprehe<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ded vnder the lines AC and CB twise. For the selfe same discription abiding, for<g ref="char:EOLhyphen"/>asmuch
<pb n="66" facs="tcp:7062:96"/>
as the line AB is equall vnto y<hi rend="sup">e</hi> line AD, y<hi rend="sup">e</hi> angle ABD is (by the <hi>5.</hi> of the first) equall vnto the angle ADB. And forasmuch as the three angles of euery triangle are equal to two right angles (by the <hi>32.</hi> of the first)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore y<hi rend="sup">e</hi> three angles of the triangle ABD, namely, the angles ADB, DBA, and BAD, are equall to two right angles. But the angle BAD is a right angle, wherefore the angles remayning ABD, and ADB, are equall vnto one right angle: and they are equall y<hi rend="sup">e</hi> one to the other, wherfore either of these
<figure/>
angles ABD, &amp; ADB, is the halfe of a right angle. And the angle BCG is a right angle, for it is equall vnto the opposite angle at the point A (by the <hi>29.</hi> of the first). VVherefore the angle remayning CGB is the halfe of a right angle. VVherefore the angle CGB is equal vnto the angle CBG: wherefore also the side BC is equall vnto the side CG. But BC is equall vnto GK, and CG is equal vnto BK. VVherefore the fi<g ref="char:EOLhyphen"/>gure CK consisteth of equall sides: and in it is a right angle CBK. VVhere<g ref="char:EOLhyphen"/>fore CK is a square, and is made of the line BC. And by the same reason HF is a square, and is equall vnto that square which is made of the line AC. VVhere<g ref="char:EOLhyphen"/>fore CK and HF are squares, and are equall to those squares which are made of the lines AC and CB. And forasmuch as AG is equall vnto EG: and AG is that which is contayned vnder AC and CB, for GC is equal vnto CB: where<g ref="char:EOLhyphen"/>fore EG also is equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>prehended vnder AC and CB: where<g ref="char:EOLhyphen"/>fore AG and EG are equall vnto that rectangle figure which is comprehended vnder AC, and CB twise. And CK, and HF are equal vnto the squares which are made of AC, and CB: wherefore CK, HF, AG, and GE are equal vn<g ref="char:EOLhyphen"/>to those squares which are made of AC, and CB, and vnto that rectangle figure which is comprehended vnder AC and CB twise. But CK, HF, AG, and GE are the whole square AE which is made of AB. VVherefore the square which is made of AB, is equall to the squares which are made of AC and CB, and vn<g ref="char:EOLhyphen"/>to the rectangle figure which is comprehended vnder AC and CB twise: which was required to be demonstrated.</p>
                  <p>Hereby it is manifest that the parallelogrames which consist about the diameter of a square must needes be squares.<note place="margin">A Corollary.</note>
                  </p>
                  <p>
                     <hi>This proposition is of infinite vse chiefely in surde numbers. By helpe of it is made in the<g ref="char:cmbAbbrStroke">̄</g> additio<g ref="char:cmbAbbrStroke">̄</g> &amp; substraction, also multiplicatio<g ref="char:cmbAbbrStroke">̄</g> in Binomials &amp; residu<g ref="char:EOLhyphen"/>als. And by helpe hereofalso is demoustrated that kinde of equation, which is, when there are three denominations in naturall order, or equally distant, and two of the greater denominations are equall to the thirde being lesse On this proposition is grounded the extraction of square roots. And many other things are also by it demonstrated.</hi>
                  </p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:97"/>
                  <head>An example of this Proposition in numbers.</head>
                  <p>Suppose a number namely, 17. to be deuided into two partes 9. and 8. The whole number 17. multiplied into himselfe, produceth 289. The square numbers of 9. and 8. are 81. and 64: the numbers produced of the multiplication of the partes the one in<g ref="char:EOLhyphen"/>to the other twise are 72. and 72: which two numbers added to the square numbers of 9. and 8. namely, to 81. and 64. make also 289. which is equall to the square number of the whole number 17. As you see in the example.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
                  <div type="subsection">
                     <head>The fourth Proposition.</head>
                     <p>
                        <note place="margin">
                           <hi>Barlaam.</hi>
                        </note>
                        <hi>If a number geuen be deuided into two numbers: the square number of the whole, is equall to the square numbers of the partes, and to the superficiall number which is produced of the multiplication of the partes the one into the other twise.</hi>
                     </p>
                     <p>Suppose that the number geuen be <hi>AB:</hi> which let be deuided into two numbers <hi>AC</hi> and <hi>CB.</hi> Then I say that the square number of the whole number <hi>AB,</hi> is equall to the squares of the partes, that is, to the squares of the numbers <hi>AC</hi> and <hi>CB,</hi> and to the superficiall number produced of the multiplication of the numbers <hi>AC</hi> and <hi>CB</hi> the one into the other twise. Let the square number produced of the multiplication of the whole number <hi>AB</hi> into himselfe be <hi>D.</hi> And let <hi>CA</hi> multiplied into himselfe produce the number <hi>EF</hi>: And <hi>CB</hi> multiplyed into it selfe let it produce <hi>GH:</hi> and fi<g ref="char:EOLhyphen"/>nally of the multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of the numbers <hi>AC</hi> and <hi>CB</hi> the one into the other twise let there be produced either of these superficiall numbers <hi>FG</hi> and <hi>HK.</hi> Now forasmuche as the number <hi>AC</hi> multiplying it self produced the number <hi>EF</hi>: therefore the num<g ref="char:EOLhyphen"/>ber <hi>AC</hi> measureth the number <hi>EF</hi> by the vnities which are in it selfe. And forasmuch as the number <hi>CB</hi> multiplyed the number <hi>CA</hi> and produced the number <hi>FG</hi>: there<g ref="char:EOLhyphen"/>fore
<pb n="67" facs="tcp:7062:97"/>
the number <hi>AC</hi> measureth the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>FG</hi> by the vnities
<figure/>
whiche are in the number <hi>CB.</hi> But it before also measured the number <hi>EF</hi> by the vnities which are in it selfe. Where<g ref="char:EOLhyphen"/>fore the number <hi>AB</hi> multiplying the number <hi>AC</hi> produ<g ref="char:EOLhyphen"/>ceth the number <hi>EG.</hi> And therefore the number <hi>EG</hi> is the superficiall number produced of the multiplication of the number <hi>BA</hi> into the number <hi>AC.</hi> And by the same rea<g ref="char:EOLhyphen"/>son may we proue that the number <hi>GK</hi> is the superficiall number produced of the multiplication of the number <hi>AB</hi> into the number <hi>BC.</hi> Farther the number <hi>D</hi> is the square of the number <hi>AB.</hi> But if a number be deuided into two num<g ref="char:EOLhyphen"/>bers, the square of the whole number is equall to the two superficiall numbers which are produced of the multipli<g ref="char:EOLhyphen"/>cation of the whole into either the partes (by the 2. Theo<g ref="char:EOLhyphen"/>reme.) Wherefore the square number <hi>D</hi> is equall to the su<g ref="char:EOLhyphen"/>perficiall number <hi>EK.</hi> But the number <hi>EK</hi> is composed of the squares of the numbers <hi>AC</hi> and <hi>CB,</hi> and of the superfi<g ref="char:EOLhyphen"/>cial number which is produced of the multiplication of the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>AC</hi> and <hi>CB</hi> the one into the other twise: &amp; the num<g ref="char:EOLhyphen"/>ber <hi>D</hi> is the square of the whole number <hi>AB.</hi> Wherfore the square number produced of the multiplication of the num<g ref="char:EOLhyphen"/>ber <hi>AB</hi> into himselfe, is equall to the square numbers of the partes, that is, to the square numbers of the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AC</hi> and <hi>CB,</hi> and to the superficiall number produced of the multiplication of the num<g ref="char:EOLhyphen"/>bers <hi>AC</hi> and <hi>CB,</hi> the one into the other twise. If therefore a number geuen be deui<g ref="char:EOLhyphen"/>ded into two numbers &amp;c. Which was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Theoreme. The <hi>5.</hi> Proposition. If a right line be deuided into two equall partes, &amp; into two vnequall partes: the rectangle figures comprehended vnder the vnequall part<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> of the whole, together with the square of that which is betwene the sectio<g ref="char:cmbAbbrStroke">̄</g>s, is equal to the square which is made of the halfe.</head>
               <p>SVppose that the right line AB be deuided into two equall partes in the point C, and into two vnequall partes in the point D. Then I say that the rectangle figure comprehended vnder AD and DB together with the square which is made of CD, is equall to the square which is made of CB.<note place="margin">Construction.</note> De<g ref="char:EOLhyphen"/>scribe (by the <hi>46.</hi> of the first)
<figure/>
vppon CB a square, and let the same be CEFB. And (by the first peticion) drawe a line from E to B. And by y<hi rend="sup">e</hi> point D drawe (by the <hi>3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> of the first) a line pa<g ref="char:EOLhyphen"/>rallel vnto ech of these lines CE and BF cutting the diameter BE in the point H, and let y<hi rend="sup">e</hi> same be DG. And agayne (by the selfe
<pb facs="tcp:7062:98"/>
same) by the point H drawe a line parallel vnto eche of these lines AB and EF, and let the same be KO: and let KO be equall vnto AB. And againe (by the selfe same) by the point A draw a line parallel vnto either of these lines CL and BO,<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                  </note> and let the same be AK. And forasmuch as (by the <hi>43.</hi> of the first) the supplement CH is equall to the suppleme<g ref="char:cmbAbbrStroke">̄</g>t HF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> put the figure DO co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon vnto them both. VVherefore the whole figure CO is equall to the whole fi<g ref="char:EOLhyphen"/>gure DF. But the figure CO is equall vnto the figure AL, for y<hi rend="sup">e</hi> line AC is equall vnto the line CB. VVherefore the figure AL also is equal vnto the figure DF. Put the figure CH common vnto them both. VVherfore the whole figure AH is equall vnto the figures DL and DF. But AH is equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned vnder the lines AD and DB, for DH is equall vnto DB. And the fi<g ref="char:EOLhyphen"/>gures FD &amp; DL are the Gno<g ref="char:EOLhyphen"/>mon
<figure/>
MNX. VVherfore y<hi rend="sup">e</hi> Gno<g ref="char:EOLhyphen"/>mon MNX is equall to that which is contayned vnder AD and DB. Put the figure LG co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon vnto them both, which is e<g ref="char:EOLhyphen"/>qual to the square which is made of CD. VVherefore the Gnomo<g ref="char:cmbAbbrStroke">̄</g> MNX and the figure LG are equall to the rectangle figure co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>prehended vnder AD and DB and vnto the square which is made of CD. But the Gnomon MNX, and the figure LG are the whole square CEFB, which is made of BC. VVherefore the rectangle figure comprehended vnder AD and DB, together with the square which is made of CD, is equall to the square which is made of CB. If therefore a right line be deuided into two equall parts, and into two vnequall partes, the rectangle figure comprehended vnder the vn<g ref="char:EOLhyphen"/>equall partes of the whole, together with the square of that which is betwene the sections, is equall to the square which is made of the halfe: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               <p>
                  <hi>This Proposition also is of greate vse in Algebra. By it is demonstrated that equation wherein the greatest and least karectes or numbers are equall to the middle.</hi>
               </p>
               <div type="section">
                  <head>An example of this proposition in numbers.</head>
                  <p>Take any number as 20: and deuide it into two equall partes 10. and 10. and then into two vnequall partes as 13. and 7. And take the differe<g ref="char:cmbAbbrStroke">̄</g>ce of the halfe to one of the vnequall partes which is 3. And multiply the vnequall partes, that is, 13 and 7. the one into the other, which make 91. take also the square of 3. which is 9. and adde it to the foresayde number 91: and so shall there be made 100. Then multiply the halfe of the whole number into himself, that is, take the square of 10. which is 100. which is equal to the number before produced of the multiplication of the vnequal parts the one in<g ref="char:EOLhyphen"/>to the other, &amp; of the difference into it selfe which is also 100. As you se in the example.
<pb n="68" facs="tcp:7062:98"/>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
                  <div type="subsection">
                     <head>The fifth proposition.</head>
                     <p>
                        <hi>If an euen number be deuided into two equall partes, and againe also into two vnequall partes: the superfic<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>all number which is produced of the multiplication of the vnequall partes the one into the other, together with the square of the number set betwene the parts, is equal to the square of halfe the number.</hi>
                     </p>
                     <p>Suppose that <hi>AB</hi> be an euen number: which let be
<figure/>
deuided into two equall numbers <hi>AC</hi> and <hi>CB,</hi> and into two vnequall numbers <hi>AD</hi> and <hi>DB.</hi> Then <hi>I</hi> say, that the square number which is produced of the multiplication of the halfe number <hi>CB</hi> into it selfe, is equall to the su<g ref="char:EOLhyphen"/>perficiall number produced of the multiplication of the vnequall numbers <hi>AD</hi> and <hi>DB</hi> the one into the other, and to the square number produced of the number <hi>CD</hi> which is set betwene the sayde vnequall partes. Let the square number produced of the multiplication of the halfe number <hi>CB</hi> into it selfe be <hi>E.</hi> And let the superfi<g ref="char:EOLhyphen"/>ciall number produced of the multiplication of the vne<g ref="char:EOLhyphen"/>qual nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AD</hi> and <hi>DB</hi> the one into the other, be the number <hi>FG</hi>: and let the square of the number <hi>DC</hi> which is set betwene the partes be <hi>GH.</hi> Now forasmuch as the number <hi>BC</hi> is deuided into the numbers <hi>BD</hi> and <hi>DC,</hi> therfore the square of the number <hi>BC,</hi> that is, the num<g ref="char:EOLhyphen"/>ber <hi>E,</hi> is equall to the squares of the numbers <hi>BD</hi> and <hi>DC,</hi> and to the superficiall number which is composed of the multiplication of the numbers <hi>BD</hi> and <hi>DC</hi> the one into the other twise, (by the 4. proposition of this boke) Let the square of the number <hi>BD</hi> be the number <hi>KL</hi>: &amp; let NX be the square of the number <hi>DC</hi>: and finally of
<pb facs="tcp:7062:99"/>
the multiplication of the numbers <hi>BD</hi> and <hi>DC</hi> the one into the other twi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e, let be pro<g ref="char:EOLunhyphen"/>duced either of these numbers <hi>LM</hi> and <hi>MN.</hi> Wherefore the whole number <hi>KX</hi> is equall to the number <hi>E</hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> And forasmuch as the number <hi>BD</hi> multiplying it selfe produ<g ref="char:EOLhyphen"/>ced the number <hi>KL,</hi> therefor it measureth it by the vnities which are in it selfe. More ouer forasmuch as the number <hi>CD</hi> multiplying the number <hi>BD</hi> produced the num<g ref="char:EOLhyphen"/>ber <hi>LM,</hi> therefore also <hi>DB</hi> measureth <hi>LM</hi> by the vnities which are in the number <hi>CD</hi>: but it before measured the number <hi>KL</hi> by the vnities which are in it selfe. Where<g ref="char:EOLhyphen"/>fore the number <hi>DB</hi> measureth the whole number <hi>KM</hi> by the vnities which are in <hi>CB.</hi> But the number <hi>CB</hi> is equall to the number <hi>CA.</hi> Wherefore the number <hi>DB</hi> mea<g ref="char:EOLhyphen"/>sureth the number <hi>KM</hi> by the vnities which are in <hi>CA.</hi> Agayne forasmuch as the nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber <hi>CD</hi> multipli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ng the number <hi>DB</hi> produced the number <hi>MN:</hi> therefore the num<g ref="char:EOLhyphen"/>ber <hi>DB</hi> measureth the number <hi>MN</hi> by the vnities which are in the number <hi>CD:</hi> but it before measured the number <hi>KM</hi> by the vnities which are in the number <hi>AC.</hi> Wher<g ref="char:EOLunhyphen"/>fore the number <hi>BD</hi> measureth the whole number <hi>KN</hi> by the vnities which are in the number <hi>AD.</hi> Wherefore the number <hi>FG</hi> 
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> equall to the number <hi>KN.</hi> For numbers which are equemultiplices to one and the selfe same number, are equall the one to the other. But the number <hi>GH</hi> is equall to the number NX: for either of them is suppo<g ref="char:EOLhyphen"/>sed to be the squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>re of the number <hi>CD.</hi> Wherefore the whole number KX is equall to the whole number <hi>FH.</hi> But the number KX is equall to the number <hi>E</hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> Wherefore also the number <hi>FH</hi> is equall to the number <hi>E.</hi> And the number <hi>FH</hi> is the superficial num<g ref="char:EOLhyphen"/>ber produced of the multiplication of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the o<g ref="char:EOLhyphen"/>ther together with the square of the number <hi>DC.</hi> And the number <hi>E</hi> is the square of the number <hi>CB.</hi> Wherfore the superficiall number produced of the multiplication of the vnequal partes <hi>AD</hi> and <hi>DB</hi> the one into the other, together with the square of the nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber <hi>DC</hi> which is set betwene those vnequall partes, is equall to the square of the num<g ref="char:EOLhyphen"/>ber <hi>CB,</hi> which is the halfe of the whole number <hi>AB.</hi> If therfore an euen number be de<g ref="char:EOLhyphen"/>uided into two equall partes, &amp;c. which was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Theoreme. The <hi>6.</hi> Proposition. If a right line be deuided into two equal partes, and if vnto it be added an other right line directly, the rectangle figure con<g ref="char:EOLhyphen"/>tayned vnder the whole line with that which is added, &amp; the line which is addedt, ogether with the square which is made of the halfe, is equall to the square which is made of the halfe line and of that which is added as of one line.</head>
               <p>SVppose that the righte
<figure/>
line AB be deuided in<g ref="char:EOLhyphen"/>to two equall partes in y<hi rend="sup">e</hi> point C: &amp; let there added vnto it an other right line DB directly, that is to say, which being ioyned vnto AB make both one right line AD. Then I say, that the rectangle figure compre<g ref="char:EOLunhyphen"/>hended vnder AD and DB, to<g ref="char:EOLhyphen"/>gether
<pb n="69" facs="tcp:7062:99"/>
with the square whiche is made of BC is equall to the square whiche, is made of DC. Describe (by the <hi>46.</hi> of the <hi>1.</hi>) vpon CD a square CEFD, and (by the first peticion) draw a line from D to E: and (by the <hi>31.</hi> of the first) by y<hi rend="sup">e</hi> point B draw a line parallel vnto either of these lines EC &amp; DF, cutting the diame<g ref="char:EOLhyphen"/>ter DE in the point H, and let the same be BG, &amp; (by the self same) by y<hi rend="sup">e</hi> point H draw to either of these lines AD and EF a parallel line KM: and moreouer by the pointe A drawe a line pa<g ref="char:EOLhyphen"/>rallel
<figure/>
to either of these lines CL and DM: and let the same be AK. And forasmuch as AC is equall vnto CB, therfore (by the <hi>36.</hi> of the first) the figure AL is equal vnto the figure CH. But (by the <hi>43.</hi> of the first) CH is equal vn<g ref="char:EOLhyphen"/>the figure HF, wherfore AL is equall vnto HF. Put the figure CM common to them both, wher<g ref="char:EOLunhyphen"/>fore the whole line AM is equal vnto the gnomon NXO. But AM is that which is contayned vnder AD and DB: for DM is equal vnto DB: wherfore the gnomon NXO, is equall vnto the rectangle figure contained vn<g ref="char:EOLhyphen"/>der AD and DB. Put the figure LG common to them both, which is equall to the square which is made of CB. VVherefore the rectangle figure which is con<g ref="char:EOLhyphen"/>tayned vnder AD and DB together with the square which is made of CB is e<g ref="char:EOLhyphen"/>quall to the gnomon NXO, and vnto LG. But the gnomon NXO and LG are the whole square CEFD which is made of CD. VVherfore the rectangle figure contayned vnder AD and DB together with the square which is made of CB is equall to the square which is made of CD. If therfore a right line be de<g ref="char:EOLhyphen"/>uided into two equall partes, and vnto it be added an other right line directly: the rectangle figure contayned vnder the whole line with that which is added, and the line which is added, together with the square which is made of the halfe, is equall to the square which is made of the halfe line and of that which is added as of one line: which was required to be demonstrated.</p>
               <p>
                  <hi>By this Proposition (besides many other vses) is in Algebra demonstrated that equation wherin the two lesse numbers be equall to the number of the grea<g ref="char:EOLhyphen"/>test denomination.</hi>
               </p>
               <div type="section">
                  <head>An example of this proposition in numbers.</head>
                  <p>Take any euen number as 18. and adde vnto it any other number as 3. which make in all 21. And multiply 21. into the number added, namely, into 3. which maketh 63. Take also the halfe of the whole euen number, that is, of 18. which is 9. And multiply 9. into it self which maketh 81. which adde vnto 63. (the number produced of the whole euen number, and the number added into the number added) and you shal make 144.
<pb facs="tcp:7062:100"/>
Then adde 9. the halfe of the whole euen number vnto 3. the number added which ma<g ref="char:EOLhyphen"/>keth 12. And multiply 12. into it selfe, that is, take the square of 12. which is 144. which is equall to the number composed of the multiplication of the whole number and the number added into the number added, and of the square of the number added, which is also 144. As you see in the example.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
               </div>
               <div type="section">
                  <head>The sixt Proposition.</head>
                  <p>
                     <hi>If an euen number be deuided into two equall numbers, and vnto it be added some other number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the superficiall number which is made of the multiplication of the number composed of the whole nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber and the number added, into the number added, together with the square of the halfe number, is equall to the square of the number composed of the halfe and the number added.</hi>
                  </p>
                  <p>Suppose that <hi>AB</hi> be an euen number, and let it be deuided into two equall num<g ref="char:EOLhyphen"/>bers <hi>AC</hi> and <hi>CB</hi>: and vnto it let there be added an other number <hi>BD.</hi> Then <hi>I</hi> say that the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iciall number produced of the multiplication of the number <hi>AD</hi> into the number <hi>DB</hi> is equall to the square of the number <hi>CD.</hi> For let the square number of the number <hi>CD</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be the number <hi>E,</hi> and let the superficial number produced of the mul<g ref="char:EOLhyphen"/>tiplication of the number <hi>AD</hi> into the number <hi>DB</hi> be the number <hi>FG:</hi> and finally let the square number of <hi>CB</hi> be the number <hi>GH.</hi> And forasmuch as the square of the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber <hi>CD</hi> is (by the 4. proposition) equall to the squares of the numbers <hi>DB</hi> and <hi>BC</hi> to<g ref="char:EOLhyphen"/>gether with the superficiall number which is produced of the multiplication of the numbers <hi>DB</hi> and <hi>BC</hi> the one into the other twise. Let the square of the number <hi>BD</hi> be the number KL: and let the superficiall numbers produced of the multiplication o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the numbers <hi>DB</hi> and <hi>BC</hi> the one into the other twise be either of these number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb n="70" facs="tcp:7062:100"/>
LM and MN<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and finally let the square of the number <hi>BC</hi> be the number NX. Wher<g ref="char:EOLhyphen"/>fore the whole number KX shall be equall to the square of the number <hi>CD.</hi> But the square of the number <hi>CD</hi> is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>mber <hi>E.</hi> Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore the number KX is equall to the number <hi>E.</hi> And forasmuch as the number <hi>BD</hi> multiplieng it s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>lfe produced the num<g ref="char:EOLhyphen"/>ber KL: therfore the number <hi>BD</hi> measureth the num<g ref="char:EOLhyphen"/>ber
<figure/>
KL, by the vnities which are in it selfe, but it also measureth the number LM by the vnities which are in the number <hi>CB.</hi> Wherfore the number <hi>DB</hi> measureth the whole number KM by the vnities which are in the number <hi>CD.</hi> The nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>DB</hi> also measureth the num<g ref="char:EOLhyphen"/>ber MN by the vnities which are in the number <hi>CB</hi>: &amp; the number <hi>CB</hi> is equall to the number <hi>CA</hi> by suppo<g ref="char:EOLhyphen"/>sition. Wherfore the number <hi>DB</hi> measureth the whole number KN by the vnities which are in the number <hi>AD.</hi> But the number <hi>DB</hi> doth also measure the number <hi>FG</hi> by the vnities which are in the number <hi>AD</hi>: for by supposition the number <hi>FG</hi> is the superficiall num<g ref="char:EOLhyphen"/>ber produced of the multipli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ation of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other. Wherfore the num<g ref="char:EOLhyphen"/>ber <hi>FG</hi> is equall to the number KN. But the number <hi>HG</hi> is equall to the number NX: for either of them is the square number of the number <hi>CB.</hi> Wherefore the whole number <hi>FH</hi> is equall to the number KX: and the number KX is proued to be equall to the number <hi>E.</hi> Wherfore the number <hi>FH</hi> shall also be equall to the number <hi>E.</hi> And the number <hi>FH</hi> is the superficiall num<g ref="char:EOLunhyphen"/>ber produced of the multipli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ation of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other, together wyth the square of the number <hi>CB</hi>: and the number <hi>E</hi> is the square of the number <hi>CD.</hi> Wher<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore the superficiall number produced of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> mul<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>plication of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other, together with the square of the number <hi>CB</hi>; is equall to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> square of the number <hi>CD.</hi> If therfore an euen number &amp;c.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Theoreme. The <hi>7.</hi> Proposition. If a right lyne be deuided by chaunce, the square whiche is made of the whole together with the square which is made of one of the partes, is equall to the rectangle figure which is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned vnder the whole and the said parte twise, and to the square which is made of the other part.</head>
               <p>SVppose that the right line AB be deuided by chaunce in the point C. Then I say that the square which is made of AB, together with the square which is made of BC, is equal vnto the rectangle figure which is contayned vnder the lines AB and BC twise, and vnto the square which is made of AC. Describe (by the <hi>46.</hi> of the first) vppon AB a square ADEB, and make complete the figure. And forasmuch as (by the <hi>43.</hi> of the first) the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure AG is equall vnto the figure GE. Put the figure CF common to the<g ref="char:cmbAbbrStroke">̄</g>
                  <pb facs="tcp:7062:101"/>
both: wherfore the whole figure AF is equall to the whole figure CE. VVher<g ref="char:EOLhyphen"/>fore the figures AF and CE are double to the figure AF. But the figures AF and CE are the gnomon KLM, and the square CF: wherfore the gnomon KLM, and the square CF is double to the figure AF. But the double to AF is that which is con<g ref="char:EOLhyphen"/>tayned
<figure/>
vnder AB and BC twise, for BF is e<g ref="char:EOLhyphen"/>quall vnto BC. VVherfore the gnomon KLM and the square CF is equall vnto the rectangle figure contayned vnder AB and BC twise. Put the figure DG common vnto them both, which is the square made of AC. VVherfore the gno<g ref="char:EOLhyphen"/>mon KLM and the squares BG and GD are equal vnto the rectangle figure which is contai<g ref="char:EOLhyphen"/>ned vnder AB &amp; BC twise, &amp; vnto the square which is made of AC. But the gnomon KLM, &amp; the squares BG, &amp; DG are y<hi rend="sup">e</hi> whole square BADE, &amp; y<hi rend="sup">e</hi> part or square CF, which squares are made of the lines AB &amp; of BC, therfore y<hi rend="sup">e</hi> squares which are made of AB &amp; BC are equall vnto the rectangle figure which is contayned vnder AB and BC twise, and also vnto the square of AC. If therfore a right line be deuided by chaunce: the square which is made of the whole together with the square which is made of one of the partes, is equall to the rectangle figure which is contayned vnder the whole and the sayd part twise, and to the square which is made of the other parte: whiche was required to be demonstrated.</p>
               <div type="section">
                  <head>Flussates addeth vnto this Proposition this Corollary.</head>
                  <p>
                     <hi>The squares of two vnequall lines do exceede the rectangle figures contayned vnder the said lines by the square of the excesse wherby the greater lyne excedeth the lesse.</hi>
                  </p>
                  <p>For if the line <hi>AB</hi> be the greater, and the line <hi>BC</hi> the lesse, it is manifest that the squares of <hi>AB</hi> and <hi>BC</hi> are equall to the rectangle figure contayned vnder the lynes <hi>AB</hi> and <hi>BC</hi> twise, and moreouer to the square of the line <hi>AC,</hi> wherby the line <hi>AB</hi> exce<g ref="char:EOLhyphen"/>deth the line <hi>BC.</hi>
                  </p>
                  <p>By this proposition most wonderfully was found out the extraction of roote squares in irrationall numbers, beside many other straunge thinges.</p>
               </div>
               <div type="section">
                  <head>An example of this proposition in numbers.</head>
                  <p>Take any number as 13. and deuide it into two partes as into 4. &amp; 9. Take the square of 13. which is 169. take also the square of 4. which is 16. and adde these two squares together which make 185. Then multiply the whole number 13. into 4. the foresayde part twise, and you shall produce 52. and 52<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> take also the square of the other part, that is, of 9. which is 81. And adde it to the productes of 13. into 4. twise, that is, vnto 52. and 52. and those three numbers added together shall make 185. whiche is equall to the number composed of the squares of the whole and of one of the partes, which is also 185. As you see in the example.</p>
                  <p>
                     <pb n="71" facs="tcp:7062:101"/>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
               </div>
               <div type="section">
                  <head>The seuenth proposition.</head>
                  <p>
                     <hi>If a number be deuided into two numbers: the square of the whole number together with the square of one of the partes, is equall to the superficiall number produced of the multiplication of the whole number into the foresaid part twise, together with the square of the other part.</hi>
                  </p>
                  <p>Suppose that the number <hi>AB</hi> be deuided into the numbers <hi>AC</hi> and <hi>CB.</hi> Then <hi>I</hi>
                     <figure/>
say that the square numbers of the numbers <hi>BA</hi> and <hi>AC</hi> are equall to the superficiall number produced of the multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of the number <hi>BA</hi> into the number <hi>AC</hi> twise, together with the square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the number <hi>BC.</hi> For forasmuch as (by the 4. of this booke) the square of the number <hi>AB</hi> is equall to the squares of the numbers <hi>BC</hi> and <hi>CA,</hi> and to the superficiall number produced of the multiplication of the numbers <hi>BC</hi> and <hi>CA</hi> the one into the other twise: adde the square of the number <hi>AC</hi> common to them both. Wherfore the square of the number <hi>AB</hi> together with the square of the num<g ref="char:EOLhyphen"/>ber <hi>AC</hi> is equall to two squares of the number <hi>AC</hi> and to one square of the number <hi>CB,</hi> and also to the superficiall number produced of the multiplication of the numbers <hi>BC</hi> and <hi>CA</hi> the one into the other twise. And forasmuch as the superficial number pro<g ref="char:EOLunhyphen"/>duced of the multiplication of the numbers <hi>BA</hi> and <hi>CA</hi> the one into the other once, is equall to the superficiall nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of the multiplication of <hi>BC</hi> into <hi>CA</hi> once, and to the square of the number <hi>CA</hi> (by the third of this booke)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therfore the number produced of the multiplication of <hi>BA</hi> into <hi>AC</hi> twise is equall to the number produ<g ref="char:EOLhyphen"/>ced of the multiplication of <hi>BC</hi> into <hi>CA</hi> twise, and also to two squares of the number <hi>CA.</hi> Adde the square number of <hi>BC</hi> common to them both. Wherfore two squares of the number <hi>AC</hi> and one square of the number <hi>CB</hi> together with the superficiall number pro<g ref="char:EOLhyphen"/>duced
<pb facs="tcp:7062:102"/>
of the multiplication of <hi>BC</hi> into <hi>CA</hi> twise are equall to the superficiall number produced of the multiplication of the number <hi>BA</hi> into the number <hi>AC</hi> twise toge<g ref="char:EOLhyphen"/>ther with the square of the number <hi>CB.</hi> Wherfore the square of the number <hi>AB</hi> toge<g ref="char:EOLhyphen"/>ther with the square of the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>AC</hi> is equal to the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icial nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of the multiplication of the number <hi>BA</hi> into the number <hi>AC</hi> twise, together with the square of the number <hi>CB.</hi> If therfore a number be deuided into two numbers &amp;c. which was required to be demonstrated.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Theoreme. The <hi>8.</hi> Proposition. If a right line be deuided by chau<g ref="char:cmbAbbrStroke">̄</g>ce, the rectangle figure com<g ref="char:EOLhyphen"/>prehended vnder the whole and one of the partes foure times, together with the square which is made of the other parte, is equall to the square which is made of the whole and the fore<g ref="char:EOLhyphen"/>said part as of one line.</head>
               <p>SVppose that there be a certayne right line AB, and let it be deuided by chaunce in the point C. Then I say that the rectangle figure comprehended vnder AB and BC foure tymes, together with the square which is made of AC is equall to the square made of AB and BC as of one line. Extend the line AB (by the second peticion). And (by the third of the first) vnto CB put an equall lyne BD. And (by the <hi>46.</hi> of the first) describe vppon AD a square AEFD. And describe a double figure. And forasmuch as CB is equall vnto BD, but CB is equall vnto GK (by the <hi>34.</hi> of the first) and likewise BD
<pb n="72" facs="tcp:7062:102"/>
is equall vnto KN, wherefore GK also is equall vnto KN: and by the same reason also PR, is equall vnto RO. And forasmuch as BC is equall vnto BD, and GK vnto KN, therfore (by the <hi>36.</hi> of the first) the figure CK is equall vnto the figure KD, and the
<figure/>
figure GR is equall vnto the figure RN. But (by y<hi rend="sup">e</hi> 
                  <hi>43.</hi> of y<hi rend="sup">e</hi> 
                  <hi>1.</hi>) the figure CK is equal vn<g ref="char:EOLunhyphen"/>to the figure RN: for they are the supplementes of the parallelograme CO. VVher<g ref="char:EOLhyphen"/>fore the figure KD also is e<g ref="char:EOLhyphen"/>quall vnto the figure NR. VVherefore these figures DK, CK, GR, RN, are equall the one to the other. VVhere<g ref="char:EOLhyphen"/>fore those foure are quadru<g ref="char:EOLhyphen"/>ple to the figure CK. Agayne forasmuch as CB is equall vn<g ref="char:EOLhyphen"/>to BD, but BD is equall vn<g ref="char:EOLhyphen"/>to BK, that is, vnto CG. And CB is equall vnto GK that is vnto GP: therfore CG is equall vnto GP. And forasmuch as CG is equall vnto GP, and PR is equall vnto RO, therefore the figure AG is equall vnto the figure MP, and the figure PL is equall vnto the figure RF. But the figure MP is (by the <hi>43.</hi> of the first) equall vnto the figure PL, for they are the supplementes of the parallelogramme ML: wherfore the figure also AG is equall vnto the figure RF. VVherfore these foure figures AG, MP, PL, and RF are equall the one to the other: wherfore those foure are quadruple to the figure AG. And it is proued, that these foure figures CK, KDGR, RN, are quadruple to the figure CK. VVherfore the eight figures which contayne the gnomon STV, are quadruple to the figure AK. And forasmuch as the figure AK, is that which is contayned vnder the lines AB and BD, for the line BK is equall vnto the line BD: therfore that whiche is contayned vn<g ref="char:EOLhyphen"/>der the lines AB and BD foure tymes is quadruple vnto the figure AK. And it is proued that the gnomon STV is quadruple to AK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> VVherfore that which is contayned vnder the lines AB and BD foure tymes is equall vnto the gnomo<g ref="char:cmbAbbrStroke">̄</g> STV. Put the figure XH which is equall to the square made of AC common vnto them both. VVherfore the rectangle figure comprehended vnder the lines AB and BD foure tymes together with the square which is made of the line AC, is equall to the gnomon STV, and vnto the figure XH. But the gnomon STV: and the figure XH are the whole square AEFD, which is made of AD: wherfore that which is contayned vnder the lines AB and BD foure times together with the square which is made of AC, is equall to the square which is
<pb facs="tcp:7062:103"/>
made of AD. But BD is equall vnto BC. VVherfore the rectangle figure con<g ref="char:EOLhyphen"/>tayned foure tymes vnder AB and BC together with the square which is made of AC, is equall vnto the square which is made of AD, that is, vnto that whiche is made of AB and BC as of ou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line. If therefore a right lyne be deuided by chaunce, the rectangle figure comprehended vnder the whole and one of the partes foure tymes, together with the square which is made on the other part, is equall to the square which is made of the whole and the foresaid part, as of one line, which was required to be demonstrated.</p>
               <div type="section">
                  <head>An example of this Proposition in numbers.</head>
                  <p>
                     <hi>Take any number as 17. and deuide it into two partes, as into 6. and 11. And multi<g ref="char:EOLhyphen"/>ply 17. into 6. namely one of the partes foure tymes, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> produce 102. 102. 102. and 102. Take also the square of 11. the other part, which is 121: and adde it vnto the foure numbers produced of the whole 17. into the part 6. foure tymes, &amp; you shall make 529. Then adde the whole number 17. to the foresaid part 6. which make 23. &amp; take the square of 23. which is 529. which is equall to the number composed of the whole into the sayd part foure tymes, aud of the square of the other part, which num<g ref="char:EOLhyphen"/>ber composed is also 529. As you see in the example.</hi>
                  </p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>
                     <pb n="73" facs="tcp:7062:103"/>The demonstration wherof followeth in Barlaam.</p>
               </div>
               <div type="section">
                  <head>The eight proposition.</head>
                  <p>
                     <hi>If a number be deuided into two numbers, the superficiall number produced of the multiplication of the whole into one of the partes foure tymes, together with the square of the other parte, is equall to the square of the number composed of the whole number and the foresayd part.</hi>
                  </p>
                  <p>Suppose that the number <hi>AB</hi> be deuided into two numbers <hi>AC</hi> and <hi>CB.</hi> Then <hi>I</hi> say that the superficiall number produced of the multiplication of the number <hi>AB</hi> in to the number <hi>CB</hi> foure tymes together with the square of the number <hi>AC,</hi> is equall to the square of the number composed of the numbers <hi>AB</hi> &amp; <hi>CB.</hi> For vnto the num<g ref="char:EOLhyphen"/>ber <hi>BC</hi> let the number <hi>BD</hi> be equall. Now forasmuch as the square of the
<figure/>
number <hi>AD</hi> is equal to the squares of the numbers <hi>AB</hi> and <hi>BD,</hi> &amp; to the superficiall number produced of the multiplication of the numbers <hi>AB</hi> &amp; <hi>BD</hi> the one into the other twise (by the 4. of this booke) And the number <hi>BD</hi> is equall to the number <hi>BC</hi>: therefore the square of the number <hi>AD</hi> is equall to the squares of the numbers <hi>AB</hi> and <hi>BC,</hi> and to the superficiall number produced of the multiplication of the numbers <hi>AB</hi> and <hi>BC</hi> the one into the other twise. But the squares of the numbers <hi>AB</hi> and <hi>BC</hi> are e<g ref="char:EOLhyphen"/>quall vnto the superficiall number produced of the multiplication of the numbers <hi>AB</hi> and <hi>BC</hi> the one into the other twise, and to the square of <hi>AC</hi> (by the former proposition) Wherfore the square of the number <hi>AD</hi> is equall to the superficial number produced of the multiplication of the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers <hi>AB</hi> and <hi>BC</hi> the one into the other foure tymes, and to the square of the number <hi>AC.</hi> But the square of the number <hi>AD</hi> is the square of the number composed of the numbers <hi>AB</hi> and <hi>BC</hi>: for the number <hi>BD</hi> is e<g ref="char:EOLhyphen"/>qual to the number <hi>BC.</hi> Wherfore the square of the number composed of the numbers <hi>AB</hi> and <hi>BC</hi> is equall to the superficiall number produced of the multiplication of the numbers <hi>AB</hi> and <hi>BC</hi> the one into the other foure tymes, &amp; to the square of the num<g ref="char:EOLhyphen"/>ber <hi>AC.</hi> If therfore a number be deuided into two numbers, &amp;c.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:104"/>
               <head>The <hi>9.</hi> Theoreme. The <hi>9.</hi> Proposition. If a right line be deuided into two equall partes, and into two vnequall partes, the squares which are made of the vne<g ref="char:EOLhyphen"/>quall partes of the whole, are double to the squares, which are made of the halfe lyne, and of that lyne which is betwene the sections.</head>
               <p>SVppose that a certayne right line AB be deuided into two equall partes in the pointe C, and into two vnequall partes in the pointe D. Then I say that the squares which are made of the lines AD and DB, are double to the squares whiche are made of the lynes AC and CD.<note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ction.</note> For (by the <hi>11.</hi> of the
<figure/>
first) erect from y<hi rend="sup">e</hi> point C to the right line AB a perpendiculer line CE. And let CE (by the <hi>3.</hi> of the first) be put equall vnto either of these lines AC &amp; CB: and (by the first peticio<g ref="char:cmbAbbrStroke">̄</g>) draw lines from A to E, and from E to B. And (by the <hi>31.</hi> of the first) by the point D draw vnto the line EC a parallel lyne, and let the same be DF: and (by the selfe same) by the point F draw vnto AB a line parallel, and let the same be FG. And (by the first peticion) draw a line from A to F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as AC is e<g ref="char:EOLhyphen"/>quall vnto CE, therfore (by the <hi>5.</hi> of the first) the angle EAC is equal vnto the angle CEA. And forasmuch as the angle at the point C is a right angle: therfore the angles remayning EAC, and AEC, are equall vnto one right angle, where<g ref="char:EOLhyphen"/>fore eche of these angles EAC and AEC is the halfe of a right angle. And by the same reason also eche of these angles EBC and CEB is the halfe of a rig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t angle. VVherfore the whole angle AEB is a right angle. And forasmuch as the angle GEF is the halfe of a right angle, but EGF is a right angle. For (by the <hi>29</hi> of the first) it is equall vnto the inward and opposite angle, that is, vnto ECB: wherfore the angle remayning EFG is the halfe of a right angle. VVhere<g ref="char:EOLhyphen"/>fore (by the <hi>6.</hi> common sentence) the angle GEF is equall vnto the angle EFG. VVherfore also (by the <hi>6.</hi> of the first) the side EG is equall vnto the side FG. A<g ref="char:EOLhyphen"/>gaine <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as the angle at the point B is the halfe of a right angle, but the angle FDB is a right angle, for it also (by the <hi>29.</hi> of the first) is equall vnto the
<pb n="74" facs="tcp:7062:104"/>
inwarde and opposite angle ECB. VVherefore the angle remayning BFD is the halfe of a right angle. VVherfore the angle at the point B is equall vnto the angle DFB. VVherfore (by the <hi>6.</hi> of the first) the side DF is equall vnto the side DB. And forasmuch as AC is equall vnto CE, therfore the square which is made of AC is equall vnto the square which is made of CE. VVherefore the squares which are made of CA and CE are double to the square which is made of AC. But (by the <hi>47.</hi> of the first) the square which is made of EA is equall to the squares which are made of AC and CE (For the angle ACE is a right an<g ref="char:EOLhyphen"/>gle) wherefore the square of AE is double to the square of AC. Agayne foras<g ref="char:EOLhyphen"/>much as EG, is equall vnto GF, the square therfore which is made of EG is e<g ref="char:EOLhyphen"/>qual to the square which is made of GF. VVherfore the squares which are made of GE and GF are double to the square which is made of GF. But (by the <hi>47.</hi> of the first) the square which is made of EF is equall to the squares which are made of EG and GF. VVherfore the square which is made of EF is double to the square which is made of GF. But GF is equall vnto CD. VVherefore the square which is made of
<figure/>
EF is double to the square which is made of CD. And the square whiche is made of AE is double to the square which is made of AC. VVherefore the squares which are made of AE and EF are double to y<hi rend="sup">e</hi> squares which are made of AC and CD. But (by the <hi>47.</hi> of the first) the square which is made of AF is equal to the squares which are made of AE and EF (For y<hi rend="sup">e</hi> angle AEF is a right angle). VVherfore the square which is made of AF is double to the squares which are made of AC &amp; CD. But (by the <hi>47.</hi> of the first) y<hi rend="sup">e</hi> squares which are made of AD and DF are e<g ref="char:EOLhyphen"/>quall to y<hi rend="sup">e</hi> square which is made of AF. For y<hi rend="sup">e</hi> angle ot y<hi rend="sup">e</hi> point D is a right angle. VVherfore the squares which are made of AD and DF are double to y<hi rend="sup">e</hi> squares which are made of AC and CD. But DF is equall vnto DB. VVherfore the squares which are made of AD and DB, are double to the squares which are made of AC and CD. If therfore a right line be deuided into two equall partes and into two vnequall partes, the squares which are made of the vnequall partes of the whole, are double to the squares which are made of the halfe lyne, and of that lyne which is betwene the sections: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An example of this proposition in numbers.</head>
                  <p>Take any euen number as 12. And deuide it first equally as into 6. and 6. &amp; then vn<g ref="char:EOLhyphen"/>equally as into 8. &amp; 4. And take the difference of the halfe to one of the vnequal partes
<pb facs="tcp:7062:105"/>
which is 2. And take the square numbers of the vnequall partes 8, and 4, which are 6<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> and 16: and adde them together, which make 80. Then take the squares of the halfe 6. and of the differe<g ref="char:cmbAbbrStroke">̄</g>ce 2: which are 36, and 4: which added together make 40. Vnto which number, the number composed of the squares of the vnequall partes, whiche i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 80, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> double. As you see in the example.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
               </div>
               <div type="section">
                  <head>The ninth Proposition.</head>
                  <p>
                     <hi>If a number be deuided into two equall numbers, and againe be deuided into two inequall partes: the square numbers of the vnequall numbers, are double to the square which is made of the multiplicati<g ref="char:EOLhyphen"/>on of the halfe number into it selfe, together with the square whiche is made of the number set be<g ref="char:EOLhyphen"/>twene them.</hi>
                  </p>
                  <p>For let the number <hi>AB</hi> being an euen number be deuided into two equall numbers <hi>AC</hi> &amp; <hi>CB</hi>: &amp; into two vnequall nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AD</hi> and <hi>DB.</hi> Then <hi>I</hi> say that the square num<g ref="char:EOLhyphen"/>bers of <hi>AD</hi> and <hi>DB,</hi> are double to the squares which are made of the multiplication of the numbers <hi>AC</hi> and <hi>CD</hi> into themselues. For forasmuch as the number <hi>AB</hi> is an euen number, and is deuided also into two equal numbers <hi>AC</hi> and <hi>CB,</hi> and afterward into two vnequal nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AD</hi> and <hi>DB</hi>: therefore the superficial nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of the multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AD</hi> &amp; <hi>DB,</hi> th'one into the other, together with the square of the number <hi>DC,</hi> is equal to the square of the number <hi>AC</hi> (by the fift proposition) Wherfore the superficiall number produced of the multiplication of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other twise, together with two squares of the number <hi>CD,</hi> is double to the square of the number <hi>AC.</hi> Forasmuch as also the number <hi>AB</hi> is deui<g ref="char:EOLhyphen"/>ded into two equal numbers <hi>AC</hi> and <hi>CB,</hi> therfore the square number of <hi>AB</hi> is qua<g ref="char:EOLhyphen"/>druple to the square number produced of the multiplication of the number <hi>AC</hi> into it selfe (by the 4. proposition). Moreouer forasmuch as the superficiall number produ<g ref="char:EOLhyphen"/>ced of the multiplication of the numbers <hi>AD</hi> &amp; <hi>DB</hi> the one into the other twise to<g ref="char:EOLhyphen"/>gether, with two squares of the number <hi>DC,</hi> is double to the square number of <hi>CA</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp;
<pb n="75" facs="tcp:7062:105"/>
forasmuch as there are two numbers, of whiche the one is quadruple to
<figure/>
one and the selfe same number, and the other is double to the same num<g ref="char:EOLunhyphen"/>ber: therefore that number whiche is quadruple shall be double to that number whiche is double. Wherefore the square of the number <hi>AB</hi> is double to the number produced of the multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other twise together with the two squares of the number <hi>DC.</hi> Wherfore the number which is produced of the mul<g ref="char:EOLhyphen"/>tiplication of the numbers <hi>AD</hi> and <hi>DB</hi> the one into the other twise, is lesse the<g ref="char:cmbAbbrStroke">̄</g> halfe of the square of the number <hi>AB</hi> by the two squares of the numbers <hi>DC.</hi> And forasmuch as the nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of the multiplica<g ref="char:EOLhyphen"/>tion of the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>AD</hi> &amp; <hi>DB</hi> the one into the other twise, together with the nu<g ref="char:cmbAbbrStroke">̄</g>ber co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of the numbers <hi>AD</hi> and <hi>DB</hi> is (by the 4. proposition) equall to the square of the number <hi>AB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therfore the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber composed of the squares of the numbers <hi>AD</hi> &amp; <hi>DB</hi> is greater then the halfe of the square nu<g ref="char:cmbAbbrStroke">̄</g>ber of <hi>AB,</hi> by the two squares of the number <hi>DC.</hi> And the square of the number <hi>AB</hi> is quadruple to the square of the number <hi>AC.</hi> Wherfore the number composed of the squares of the numbers <hi>AD</hi> and <hi>DB</hi> is greater then the double of the square of the number <hi>AC</hi> by two squares of the number <hi>DC.</hi> Wherfore the said num<g ref="char:EOLhyphen"/>ber is double to the squares of the numbers <hi>AC</hi> and <hi>CD.</hi> If therefore a number be deuided &amp;c. which was required to be demonstrated.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Theoreme. The <hi>10.</hi> Proposition. If a right line be deuided into two equal partes, &amp; vnto it be added an other right line directly: the square which is made of the whole &amp; that which is added as of one line, together with
<pb facs="tcp:7062:106"/>
the square whiche is made of the lyne whiche is added, these two squares (I say) are double to these squares, namely, to the square which is made of the halfe line, &amp; to the square which is made of the other halfe lyne and that whiche is added, as of one lyne.</head>
               <p>SVppose that a certayne right line AB be deuided into two equall partes in the point C. And vnto it let there be added an other right line direct<g ref="char:EOLhyphen"/>ly, namely, BD. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the squares which are made of the lines AD and DB are double to the squares which are made of the lines AC and CD.<note place="margin">Construction.</note> Rayse vp (by the <hi>11.</hi> of the first) from the point C vnto the right line ACD a perpendiculer lyne, and let the same be CE. And let CE (by the <hi>3.</hi> of the first) be made equall vnto either of these lines AC and CB. And (by the first petiti<g ref="char:EOLhyphen"/>on) draw right lines from E to A, and from E to B. And (by the <hi>31.</hi> of the first) by the point E, draw a line parallel vnto CD, and let the same be EF. And (by y<hi rend="sup">e</hi> self same) by the point D draw a line parallel vnto CE and let the same be DF. And forasmuch as vpon these parallel lines CE &amp; DF lighteth a certain right line EF, therfore (by the <hi>29.</hi> of the first) the angles CEF and EFD are equal vnto two right angles. VVherfore the angles FEB, and EFD are lesse then two right angles. But lines produced from angles lesse then two right angles (by the fifth peticion) at the length meete together. VVherfore the lines EB and FD beyng produced on that side that the line BD is, will at the length meete to<g ref="char:EOLhyphen"/>gether. Produce them and let them meete together in the point G. And (by the first peticion) draw a line from A to G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AC is equall vnto the line CE, the angle
<figure/>
also AEC is (by the <hi>5.</hi> of the first) equall vnto the angle EAC. And the angle at y<hi rend="sup">e</hi> point C is a right angle. VVherfore eche of these angles EAC &amp; and AEC is the halfe of a right angle. And by the same reason eche of these angles CEB, and EBC is the halfe of a right angle: VVherefore the angle AEB is a right angle. And forasmuch as the angle EBC is the halfe of a right angle, therfore (by the <hi>15.</hi> of the first) the angle DBC is the half of a right angle. But y<hi rend="sup">e</hi> angle BD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is a right angle (for it is equal vnto the angle DCE, for they are alternate angles) VVherfore the angle remaining DGB is the halfe of a right angle. VVherfore (by the <hi>6</hi> common sentence of the first) the an<g ref="char:EOLhyphen"/>gle DGB is equall to the angle DBG. VVherfore (by the <hi>6.</hi> of the first) the side
<pb n="76" facs="tcp:7062:106"/>
BD is equall vnto the side GD. Agayne forasmuch as the angle EGF is the halfe of a right angle: and the angle at the pointe F is a right angle: for (by the <hi>34.</hi> of the first) it is equall vnto the opposite angle ECD. VVherefore the angle remayning FEG is the halfe of a right angle. VVherfore the angle EGF is e<g ref="char:EOLhyphen"/>quall to the angle FEG. VVherfore (by the <hi>6.</hi> of the first the side, FE is equall vnto the side FG. And forasmuch as EC is equall vnto CA, the square also which is made of EC is equall to the square which is made of CA. VVherefore the squares which are made of CE and CA are double to the square which is made of AC. But the square which is made of EA is (by the <hi>47.</hi> of the first) e<g ref="char:EOLhyphen"/>quall vnto the squares which are made of EC and CA. VVherefore the square which is made of EA is double to the square which is made of AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Againe for<g ref="char:EOLhyphen"/>asmuch as GF is equall vnto EF, the square also which is made of GF is equall to the square which is made of FE. VVherfore the squares which are made of GF and EF are double to the square which is made of EF. But (by the <hi>47.</hi> of the first) the square which is made of EG is equall to the squares which are made of GF and EF. VVherefore
<figure/>
the square which is made of EG is double to the square which is made of EF. But EF is equall vnto CD, wher<g ref="char:EOLhyphen"/>fore y<hi rend="sup">e</hi> square which is made of EG is double to the square which is made of CD. And it is proued the square which is made of EA is double to the square which is made of AC. VVherfore the squares which are made of AE and EG are double to the squares which are made of AC and CD. But (by the <hi>47.</hi> of ths first) the square which is made of AG is equall to the squares which are made of AE and EG. VVherefore the square which is made of AG is double to the squares which are made of AC and CD. But vnto the square whiche is made of AG are equall the squares which are made of AD and DG. VVherfore the squares which are made of AD and DG are double to the squares which are made of AC and DC. But DG is equall vnto DB. VVherfore the squares which are made of AD and DB are double to the squares which are made of AC and DC. If therfore a right line be deuided into two equall partes, and vnto it be added an other lyne directly, the square which is made of the whole and that which is added, as of one line, together with the square which is made of the line which is added, these two squares (I say) are double to these squares, namely, to the square which is made of the halfe lyne, and to the square which is made of the other halfe lyne and that which is added, as of one lyne: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other demonstration after Pelitarius.</head>
                  <p>
                     <pb facs="tcp:7062:107"/>Suppose that the lyne AB be deuided into two equall partes in the poynte C. And vnto it let there be added an other right lyne directly, namely, BD. Then I say that the square of AD together with the square of BD is double to the squares of AC and CD.</p>
                  <p>Vpon the whole line AD describe a square ADEF. And vpon the halfe lyne AC de<g ref="char:EOLhyphen"/>scribe the square ACGH. And produce the sides GH and CH till they cut the sides EF &amp; DF, wherby shalbe described the figure HLKF, which shalbe the square of the line CD: as (by the Corollary of the 4. of this boke, &amp; by the 34. Proposition of the 1.) it is manifest if we draw the diameter CD. For the lyne KF is equall to the line CD. And making also the lines HM and HN equall to either of these lynes AC and CB, drawe the lynes MO and NP cutting the one the other right angled wise in the point Q. Ei<g ref="char:EOLhyphen"/>ther of which lynes let cut the sides of the square AD:
<figure/>
EF in the pointes O and P. Now it nedeth not to proue that the figure HQ is the square of the lyne AC, seyng that it is the square of the line CB: as the figure QF is the square of the line BD: neither also needeth it to proue that the parallelograme HP is equall to either of the supplementes EH and HD: nor that the supple<g ref="char:EOLhyphen"/>mentes NO and QL are equall. For all this is manifest eue<g ref="char:cmbAbbrStroke">̄</g> by the forme of the figure, for that all the angles a<g ref="char:EOLhyphen"/>bout the diameter are half right angles, &amp; the sides are equall. Wherfore if we diligently marke of what partes the square HF which is the square of CD, is compo<g ref="char:EOLhyphen"/>sed, we may thus reaso<g ref="char:cmbAbbrStroke">̄</g>. Forasmuch as the whole square ED is composed of the two squares AH and HF and of the two supplementes EH and HD, we must proue that these supplementes with the square QF (which is the square of the line BD) are equall to the two squares AH and HF. For then shall we proue that these two squares AH &amp; HF taken twise are equall to the whole square DE together with the square of QF, which thing we tooke first in hand to proue. And thus do I proue it.</p>
                  <p>The Supplement EH is equall to the parallelograme HP. And the square AH to<g ref="char:EOLhyphen"/>gether with the lesser suppleme<g ref="char:cmbAbbrStroke">̄</g>t, NO, is equall to the other suppleme<g ref="char:cmbAbbrStroke">̄</g>t HD (by the first common sentence so oftentymes repeted as is neede) wherfore the two supplementes EH and HD are equall to the square AH and to the Gnomon KHLPQO. If therfore vnto either of them be added the square QF: the two supplementes EH and HD to<g ref="char:EOLhyphen"/>gether with the square of QF shal be equal to the square AH, &amp; to the Gnomon KHLPQO and to the square QF. But these three figures do make the two squares AH and HF. Wherfore the two supplementes EH and HD together with the square QF are e<g ref="char:EOLhyphen"/>quall to the two squares AH and HF, which was the second thing to be proued. Wher<g ref="char:EOLhyphen"/>fore the two squares AH and HF beyng taken twise are equall to the whole square DE together with the square of QF. Wherfore the square DE together with the square QF is double to the squares AH and HF: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ An example of this Proposition in numbers.</head>
                  <p>Take any euen number as 18: and take the halfe of it which is 9. and vnto 18. the whole, adde any other number as 3. which maketh 21. Take the square number of 21. (the whole number and the number added) which maketh 441. Take also the square of 3 (the number added) which is 9. which two squares added together make 450. Then adde the halfe number 9. to the number added 3. which maketh 12. And take the square of 9. the halfe number and of 12. the halfe number and the number added which squares are 81. and 144. and which two squares also added together make 225: vnto which summe the foresayd number 450. is double. As you see in the example.</p>
                  <p>
                     <pb n="77" facs="tcp:7062:107"/>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
                  <p>The demonstration wherof followeth in Barlaam.</p>
               </div>
               <div type="section">
                  <head>The tenth Proposition.</head>
                  <p>
                     <hi>If an euen nomber be deuided into two equall nombers, and vnto it be added any other nomber: the square nomber of the whole nomber composed of the no<g ref="char:cmbAbbrStroke">̄</g>ber and of that which is added, and the square nomber of the no<g ref="char:cmbAbbrStroke">̄</g>ber added: these two square no<g ref="char:cmbAbbrStroke">̄</g>bers (I say) added together, are double to these square nombers, namely, to the square of the halfe nomber, and to the square of the nomber composed of the halfe nomber and of the nomber added.</hi>
                  </p>
                  <p>Suppose that the nomber <hi>AB</hi> being an euen nomber be deuided into two equall nombers <hi>AC</hi> and <hi>CB</hi>: and vnto it let be added an other nomber <hi>BD.</hi> Then I say, that the square nombers of the nombers <hi>AD</hi> and <hi>DB</hi> are double to the square nombers of <hi>AC</hi> and <hi>CD.</hi> For forasmuch as the nomber <hi>AD</hi> is deuided into the nombers <hi>AB</hi> and <hi>BD</hi>: therefore the square nombers of the nombers <hi>AD</hi> and <hi>DB</hi> are equall to the su<g ref="char:EOLhyphen"/>perficiall nomber produced of the multiplication of the nombers <hi>AD</hi> and <hi>DB</hi> the on into the other twise, together with the square of the nomber <hi>AB</hi> (by the 7 propositio<g ref="char:cmbAbbrStroke">̄</g>) But the square of the nomber <hi>AB</hi> is equal to fower squares of either of the nombers <hi>AC</hi> or <hi>CB</hi> (for <hi>AC</hi> is equall to the nomber <hi>CB</hi>): wherfore also the squares of the nom<g ref="char:EOLhyphen"/>bers <hi>AD</hi> and <hi>DB</hi> are equall to the superficiall nomber produced of the multiplication of the nombers <hi>AD</hi> and <hi>DB</hi> the one into the other twise, and to fower squares of the nomber <hi>BC</hi> or <hi>CA.</hi> And forasmuch as the superficiall nomber produced of the multi<g ref="char:EOLhyphen"/>plication of the nombers <hi>AD</hi> and <hi>DB</hi> the one into the other, together with the square of the nomber <hi>CB,</hi> is equal to square of the nomber <hi>CD</hi> (by the 6 propositio<g ref="char:cmbAbbrStroke">̄</g>): therfore the nomber produced of the multiplication of the nomber<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>AD</hi> and <hi>DB</hi> the one into the other twise together with two squares of the nomber <hi>CB,</hi> is equall to two squares of the nomber <hi>CD.</hi> Wherefore the squares of the nombers <hi>AD</hi> and <hi>DB</hi> are equall to
<pb facs="tcp:7062:108"/>
two squares of the nomber <hi>CD,</hi> and to two squares of the nomber <hi>AC.</hi> Where<g ref="char:EOLhyphen"/>fore they are double to the squares of the numbers <hi>AC</hi> and <hi>CD.</hi> And the square of the nomber <hi>AD</hi> is the square of the whole and of the nomber added. And the square of <hi>DB</hi> is the square of the nomber added: the square also of the nomber <hi>CD</hi> is the square of the nomber composed of the halfe and of the nomber added: If there<g ref="char:EOLhyphen"/>fore an euen nomber be deuided. &amp;c. Which was required to be proued.</p>
                  <p>
                     <gap reason="math">
                        <desc>〈 math 〉</desc>
                     </gap>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Probleme. The <hi>11.</hi> Proposition. To deuide a right line geuen in such sort, that the rectangle figure comprehended vnder the whole, and one of the partes, shall be equall vnto the square made of the other part.</head>
               <p>
                  <hi>SVppose that the right line geuen be AB. Now it is required to deuide the line AB in such sort, that the rectangle figure contayned vnder the whole and one of the partes, shall be equall vnto the square which is made of the other part. Describe (by the</hi> 46. <hi>of the first) vpon AB a square ABCD.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>And (by the</hi> 10. <hi>of the
<figure/>
first) deuide the line AC into two equall partes in the point E, and draw a line from B to E. And (by the second petition) extend CA vnto the point F. And (by the</hi> 3. <hi>of the first) put the line EF equall vnto y<hi rend="sup">e</hi> line BE. And (by the</hi> 46. <hi>of the first) vpon the line AF describe a square FGAH. And (by the</hi> 2. <hi>petition) extend GH vnto the point K. Then I say that the line AB is deuided in the point H in such sort, that the rectangle figure which is comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder AB and BH is equall to the square which is made of AH.</hi>
                  <note place="margin">
                     <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                     </hi>
                  </note> 
                  <hi>For forasmuch as the right line AC is deuided into two equall partes in the poynt E, and vnto it is added an other right line
<pb n="78" facs="tcp:7062:108"/>
AF. Therefore (by the</hi> 6. <hi>of the second) the rectangle figure contayned vnder CF and FA together with the square which is made of AE is equall to y<hi rend="sup">e</hi> square which is made of EF. But EF is equall vnto EB. VVherefore the rectangle figure contayned vnder CF, and FA together with the square which is made of EA is equall to the square which is made of EB. But (by</hi> 47. <hi>of the first) vnto the square which is made of EB are equall the squares which are made of BA and AE. For the angle at the poynt A is a right angle. VVherefore that which is contayned vnder CF and FA, together with the square which is made of AE, is equall to the squares which are made of BA and AE. Take away the square which is made of AE which is common, to them both: VVherfore the rectangle figure remayning contayned vnder CF and FA is equall vnto the square which is made of AB. And that which is contained vnder the lines CF and FA is the figure FK. For the line FA is equall vnto the line FG. And the square which is made of AB is the figure AD. VVherefore the fi<g ref="char:EOLhyphen"/>gure FK is equall vnto the figure AD. Take away the figure AK which is common, to them both. VVherefore the residue, namely, the figure FH is equall vnto the residue, namely, vnto the figure HD. But the figure HD is that, which is contayned vnder the lines AB and BH, for AB is equall vn<g ref="char:EOLhyphen"/>to BD. And the figure FH is the square which is made of AH. VVherfore the rectangle figure comprehended vnder the lines AB and BH is equall to the square which is made of the line HA. VVherefore the right line geuen AB is deuided in the point H, in such sort that the rectangle figure contayned vnder AB and BH is equall to the square which is made of AH: which was required to be done.</hi>
               </p>
               <p>Thys proposition hath many singular vses. Vpon it dependeth the demonstration of that worthy Probleme the 10. Proposition of the 4. booke:<note place="margin">
                     <hi>Many and singuler vses of this propo<g ref="char:EOLhyphen"/>sition.</hi>
                  </note> which teacheth to de<g ref="char:EOLhyphen"/>scribe an Isosceles triangle, in which eyther of the angles at the base shall be double to the angle at the toppe. Many and diuers vses of a line so deuided shall you finde in the 13. booke of <hi>Euclide.</hi>
               </p>
               <p>Thys is to be noted that thys Proposition can not as the former Propositions of thys second booke be reduced vnto numbers.<note place="margin">
                     <hi>This proposi<g ref="char:EOLhyphen"/>tion can not be reduced vn<g ref="char:EOLunhyphen"/>to numbers.</hi>
                  </note> For the line EB hath vnto the line AE no proportion that can be named, and therefore it can not be expressed by numbers. For forasmuch as the square of EB is equall to the two squares of AB and AE (by the 47. of the first) and AE is the halfe of AB, therefore the line BE is irrationall. For euen as two equall square numbers ioyned together can not make a square number: so also two square numbers, of which the one is the square of the halfe roote of the other, can not make a square number. As by an example. Take the square of 8. which is 64. which doubled, that is, 128. ma<g ref="char:EOLhyphen"/>keth not a square number. So take the halfe of 8. which is 4. And the squares of 8. and 4. which are 64. and 16. added together likewyse make not a square num<g ref="char:EOLhyphen"/>ber. For they make 80. who hath no roote square. Which thyng must of necessi<g ref="char:EOLhyphen"/>tie be if thys Probleme should haue place in numbers.</p>
               <p>But in Irrational numbers it is true, and may by thys example be declared.</p>
               <p>
                  <pb facs="tcp:7062:109"/>Let 8. be so deuided, that that which is produced of the whole into one of his partes shall be equall to the square number produced of the other part. Multiply 8. into him selfe and there shall be produced 64. that is, the square <hi>ABCD.</hi> Deuide 8. into two equall partes, that is, into 4, and 4. as the line <hi>AE</hi> or <hi>EC.</hi> And multiply 4. into hym selfe, and there is produced 16, which adde vnto 64, and there shall be produced 80: whose roote is √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80: which is the line <hi>EB</hi> or the line <hi>EF</hi> by the 47. of the first. And forasmuch as the line <hi>EF</hi> is √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80. &amp; the lyne <hi>EA</hi> is 4. therfore the lyne <hi>AF</hi> is √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80—4. And so much shall the line AH be. And the line <hi>BH</hi> shall be 8—√<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80—4, that is, 12—√<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80. Now the<g ref="char:cmbAbbrStroke">̄</g> 12—√<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80 multiplied into 8 shal be as much as √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 80—4. multiplied into it selfe. For of either of them is produced 96—√ 5120.</p>
            </div>
            <div type="part">
               <head>The <hi>11.</hi> Theoreme. The <hi>12.</hi> Proposition. In obtuseangle triangles, the square which is made of the side subtending the obtuse angle, is greater then the squares which are made of the sides which comprehend the obtuse angle, by the rectangle figure, which is comprehended twise vnder one of those sides which are about the obtuse angle, vpon which being produced falleth a perpendicular line, and that which is outwardly taken betwene the perpendicular line and the obtuse angle.</head>
               <p>SVppose that ABC be an obtuseangle triangle hauing the angle BAC obtuse, and from the point B (by the <hi>12.</hi> of the first) draw a perpendicular line vnto CA produced and let the same be BD. Then I say that the square which is made of the side BC, is greater then the squares which are made of the sides BA and AC, by the rectan<g ref="char:EOLhyphen"/>gle figure comprehended vnder the lines CA and AD twise.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as the right line CD is
<figure/>
by chaunce deuided in the poynt A, therefore (by the <hi>4.</hi> of the second) the square which is made of CD is equall to the squares which are made of CA and AD, and vnto the rectangle figure contayned vnder CA and AD twise. Put the square which is made of DB com<g ref="char:EOLhyphen"/>mon vnto them both. VVherefore the squares which are made of CD and DB are equall to the squares which are made of the lines CA, AD, and DB, and vnto the rectangle figure contayned vnder the lines CA and AD twise. But (by the <hi>47.</hi> of the first) the square which is made of CB is equall to the squares which are made of the lines
<pb n="77" facs="tcp:7062:109"/>
CD and DB. For the angle at the point D is a right angle. And vnto the squares which are made of AD and DB (by the selfe same) is equall the square which is m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de of AB. VVherfore the square which is made of CB, is e<g ref="char:EOLhyphen"/>quall to the squares which are made of CA and AB and vnto the rectangle fi<g ref="char:EOLhyphen"/>gure contayned vnder the lines CA and AD twise. VVherfore y<hi rend="sup">e</hi> square which is made of CB, is greater then the squares which are made of CA and AB by the rectangle figure contayned vnder the lines CA and AD twise. In obtuse<g ref="char:EOLhyphen"/>angle triangles therefore, the square which is made of the side subtending the ob<g ref="char:EOLhyphen"/>tuse angle, is greater then the squares which are made of the sides vvhich com<g ref="char:EOLhyphen"/>prehend the obtuse angle, by the rectangle figure vvhich is comprehended twise vnder one of those sides which are about the obtuse angle, vpon which being pro<g ref="char:EOLhyphen"/>duced falleth a perpendiculer lyne, and that which is outwardly taken betwene the perpendiculer lyne and the obtuse angle: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
               <p>
                  <hi>Of what force thys Proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and the Proposition following, touching the measuring of the obtuseangle triangle and the acuteangle triangle, with the ayde of the 47. Proposition of the first booke touching the rightangle triangle, he shall well perceaue, which shall at any time neede the arte of triangles in which by thre thinges knowen is euer searched out three other thinges vnknowen, by helpe of the table of arkes and cordes.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Theoreme. The <hi>13.</hi> Proposition. In acuteangle triangles, the square which is made of the side that subtendeth the acute angle, is lesse then the squares which are made of the sides which comprehend the acute an<g ref="char:EOLhyphen"/>gle, by the rectangle figure which is co<g ref="char:cmbAbbrStroke">̄</g>prehended twise vnder one of those sides which are about the acuteangle, vpo<g ref="char:cmbAbbrStroke">̄</g> which falleth a perpendiculer lyne, and that which is inwardly ta<g ref="char:EOLhyphen"/>ken betwene the perpendiculer lyne and the acute angle.</head>
               <p>SVppose that ABC be an acuteangle triangle ha<g ref="char:EOLhyphen"/>uyng
<figure/>
the angle at y<hi rend="sup">e</hi> point B acute, &amp; (by the <hi>12.</hi> of the first) from the point A draw vnto the lyne BC a perpendiculer lyne AD. Then I say that the square which is made of the lyne AC is lesse then the squares which are made of the lyne CB and BA, by the rectangle figure conteyned vnder the lines CB and BD twise.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                  </note> For forasmuch as the right lyne BC is by chaunce deuided in the point D, therfore (by the <hi>7.</hi> of the second) the squares
<pb facs="tcp:7062:110"/>
which are made of the lines CB and BD are equall to the rectangle figure con<g ref="char:EOLhyphen"/>tained vnder the lines CB and DB twise and vnto the square whiche is made of line CD. Put the square which is made of the line DA common vnto them both. VVherfore the squares which are made of the lines CB, BD, and DA, are equall vnto the rectangle figure contayned vnder
<figure/>
the lines CB and BD twise, and vnto the squares which are made of AD and DC. But to the squares whiche are made of the lines BD and DA is equal y<hi rend="sup">e</hi> square which is made of the line AB: for th'angle at y<hi rend="sup">e</hi> point D is a right angle. And vnto the squares whiche are made of the lines AD and DC is equall the square whiche is made of the line AC (by the <hi>47.</hi> of y<hi rend="sup">e</hi> first): wherfore the squares which are made of the lines CB and BA are equal to the square which is made of the line AC, and to that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines CB and BD twise. VVherfore the square which is made of the line AC beyng taken alone, is lesse then the squares which are made of the lines CB and BA by the rectangle figure, which is con<g ref="char:EOLhyphen"/>tained vnder the lines CB and BD twise. In rectangle triangles therfore the square which is made of the side that subtendeth the acute angle, is lesse then the squares which are made of the sides which comprehend the acute angle, by the rectangle figure which is comprehended twise vnder one of those sides which are about the acute angle, vpon which falleth a perpendicular line, and that which is inwardly taken betwene the perpendicular line and the acute angle: which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary added by Orontius.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note>Hereby is easily gathered, that such a perpendicular line in rectangle triangles falleth of necessitie vpon the side of the triangle, that is, neyther within the trian<g ref="char:EOLhyphen"/>gle, nor without. But in obtuseangle triangles it falleth without, and in acuteangle triangles within. For the perpendicular line in obtuseangle triangles, and acute<g ref="char:EOLhyphen"/>angle triangles can not exactly agree with the side of the triangle: for then an ob<g ref="char:EOLhyphen"/>tuse &amp; an acuteangle should be equal to a right angle, contrary to the eleuenth and twelfth definitions of the first booke. Likewise in obtuseangle triangles it can not fall within, nor in acuteangle triangles without: for then the outward angle of a triangle should be lesse then the inward and opposite angle, which is contrary to the 16. of the first.</p>
                  <p>
                     <note place="margin">
                        <hi>This Proposi<g ref="char:EOLhyphen"/>tion true in all kindes of triangles.</hi>
                     </note>And this is to be noted, that although properly an acuteangle triangle, by the definition therof geue<g ref="char:cmbAbbrStroke">̄</g> in the first booke, be that triangle, whose angles be all acute: yet forasmuch as there is no triangle, but that it hath an acute angle, this propositi<g ref="char:EOLhyphen"/>on is to be vnderstanded, &amp; is true generally in all kindes of triangles whatsoeuer, and may be declared by them, as you may easily proue.</p>
               </div>
            </div>
            <div type="part">
               <pb n="80" facs="tcp:7062:110"/>
               <head>The <hi>2.</hi> Probleme. The <hi>14.</hi> Proposition. Vnto a rectiline figure geuen, to make a square equall.</head>
               <p>SVppose that the rectiline figure geuen be A. It is required to make a square equall vnto the rectiline figure A.<note place="margin">Construction.</note> Make (by the <hi>45.</hi> of y<hi rend="sup">e</hi> first) vnto the rectiline figure A an equall rectangle parallelogramme BCDE. Now if y<hi rend="sup">e</hi> line BE be equall vnto the line ED, then is y<hi rend="sup">e</hi> thyng done whiche was required: for vnto the
<figure/>
rectiline figure A is made an equal square BD. But if not, one of these lines BE &amp; is ED the greater. Let BE be the grea<g ref="char:EOLhyphen"/>ter, and let it be produced vnto y<hi rend="sup">e</hi> point F. And (by the <hi>3.</hi> of the first) put vnto ED an equall line EF. And (by the <hi>10.</hi> of the first) deuide the line BF into two equall partes in the point G. And making the centre the point G, and the space GB or GF describe a semicircle BHF. And (by the <hi>2.</hi> peticion) extend the line DE vnto y<hi rend="sup">e</hi> point H.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                  </note> And (by the <hi>1.</hi> peticion) draw a line from G to H. And forasmuch as the right line FB is deuided into two equall partes in the point G, and into two vnequal partes in the point E, therfore (by the <hi>5.</hi> of the second) the rectangle figure comprehended vnder the lines BE and EF together with the square which is made of the line EG, is equall to the square which is made of the line GF. But the line GF is equall vnto the line GH. VVherfore the rectangle figure comprehended vnder the lines BE and EF together with the square which is made of the line GE is equall to square which is made of the line GH. But vnto the square which is made of the line GH are equall the squares whiche are made of the lines HE and GE (by the <hi>47.</hi> of the first.) VVherfore y<hi rend="sup">t</hi> which is contained vnder y<hi rend="sup">e</hi> lines BE and EF together with y<hi rend="sup">e</hi> square which is made of GE is equall to y<hi rend="sup">e</hi> squares which are made of HE and GE. Take away the square of the line EG common to them both. VVherfore the rectangle figure contained vnder the lines BE &amp; EF is equall to the square which is made of the line EH. But that whiche is contained vnder the lines BE and EF is the parallelogramme BD, for the line EF is equall vnto the line ED. VVherfore the parallelogramme BD is equall to y<hi rend="sup">e</hi> square whiche is made of the line HE. But the parallelograme BD is equall vnto the rectiline figure A. VVherfore y<hi rend="sup">e</hi> rectiline figure A is equall to the square which is made of y<hi rend="sup">e</hi> line HE. VVherfore vnto the rectiline figure geuen A, is made an equall square described of the line EH: which was required to be done.</p>
            </div>
            <trailer>¶<hi>The ende of the second Booke of Euclides Elementes.</hi>
            </trailer>
         </div>
         <div n="3" type="book">
            <pb facs="tcp:7062:111"/>
            <head>¶The third booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>
               <note place="margin">
                  <hi>The argument of this booke.</hi>
               </note>THis third booke of Euclide entreateth of the most perfect figure, which is a circle. Where<g ref="char:EOLhyphen"/>fore it is much more to be estemed then the two bookes goyng before, in which he did set forth the most simple proprieties of rightlined figures. For sciences take their dignities of the worthynes of the matter that they entreat of. But of al figures the circle is of most absolute perfection, whose proprieties and passions are here set forth, and most certainely demo<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>strated. Here also is entreated of right lines subten<g ref="char:EOLhyphen"/>ded to arkes in circles: also of angles set both at the circumference and at the centre of a circle, and of the varietie and differences of them. Wherfore the readyng of this booke, is very profitable to the attayning to the knowledge of chordes and arkes. It teacheth moreouer which are circles con<g ref="char:EOLhyphen"/>tinge<g ref="char:cmbAbbrStroke">̄</g>t, and which are cutting the one the other: and also that the angle of contin<g ref="char:EOLhyphen"/>gence is the least of all acute rightlined angles: and that the diameter in a circle is the longest line that can be drawen in a circle. Farther in it may we learne how, three pointes beyng geuen how soeuer (so that they be not set in a right line), may be drawen a circle passing by them all three. Agayne, how in a solide body, as in a Sphere, Cube, or such lyke, may be found the two opposite pointes. Whiche is a thyng very necessary and commodious: chiefly for those that shall make instru<g ref="char:EOLhyphen"/>mentes seruyng to Astronomy, and other artes.</p>
            <div type="part">
               <head>Definitions. <note place="margin">
                     <hi>The first defi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note>Equall circles are such, whose diameters are equall, or whose lynes drawen from the centres are equall.</head>
               <p>The circles A and B are equal, if theyr diameters, namely, EF and CD be equall: or if their semidiameters, whiche are lynes drawen from the center to the circumference<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> namely AF and BD be equall.</p>
               <p>
                  <note place="margin">
                     <hi>Why circles take their equality of their diame<g ref="char:EOLhyphen"/>ters or semi<g ref="char:EOLhyphen"/>diameters.</hi>
                  </note>The reason why circles
<figure/>
take theyr equalitie, of the e<g ref="char:EOLhyphen"/>qualitie of their diameters or semidiameters is, for that a circle is described by one re<g ref="char:EOLhyphen"/>uolution or turnyng about of the semidiameter, hauing one of his endes fixed. As if you imagine the lyne AE to haue his one point namely A fastened, and the other end namely E to moue round till
<pb n="81" facs="tcp:7062:111"/>
it come to the place where it bega<g ref="char:cmbAbbrStroke">̄</g> to moue, it shal fully describ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the whole circle. Wherefore if the semidiameters bee equall, the circles of necessitye must also be equall: and also the diameters.</p>
               <p>By thys also is knowen the definition of vnequall circles.<note place="margin">
                     <hi>Definition of vnequall cir<g ref="char:EOLhyphen"/>cles.</hi>
                  </note>
               </p>
               <p>
                  <hi>Circles whose diameters or semidiameters are vnequall, are also vnequall. And that circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which hath the greater diameter or semidiameter, is the greater circle: and that circle which hath the lesse diameter or semidiameter, is the lesse circle.</hi>
               </p>
               <p>As the circle LM is greater
<figure/>
then the circle IK, for that the diameter LM is greater then the diameter IK: or for that the semi<g ref="char:EOLhyphen"/>diameter GL is greater then the semidiameter HI.</p>
            </div>
            <div type="part">
               <head>A right line is sayd to touch a circle,<note place="margin">
                     <hi>Second defi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> which touching the cir<g ref="char:EOLhyphen"/>cle and being produced cutteth it not.</head>
               <p>As the right lyne EF drawen from the point E, and passyng by a point of the circle, namely, by the point G to the point F on<g ref="char:EOLhyphen"/>ly
<figure/>
toucheth the circle GH, and cutteth it not, nor entreth within it. For a right line entryng within a circle, cutteth and deui<g ref="char:EOLhyphen"/>deth the circle. As the right lyne KL de<g ref="char:EOLhyphen"/>uideth and cutteth the circle KLM, and entreth within it: and therfore toucheth it in two places. But a right lyne tou<g ref="char:EOLhyphen"/>chyng a circle, which is commonly called a co<g ref="char:cmbAbbrStroke">̄</g>tingent lyne,<note place="margin">
                     <hi>A contigent line.</hi>
                  </note> toucheth the circle one<g ref="char:EOLhyphen"/>ly in one point.</p>
            </div>
            <div type="part">
               <head>Circles are sayd to touch the one the other,<note place="margin">
                     <hi>Third defini<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion.</hi>
                  </note> which touching the one the other, cut not the one the other.</head>
               <p>As the two circles AB and BC touch the
<figure/>
one the other. For theyr circumferences touch together in the poynt B. But neither of them cutteth or deuideth, the other. Neither doth any part of the one enter within the o<g ref="char:EOLhyphen"/>ther.<note place="margin">
                     <hi>The touch of circles is <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> in one po<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> onely.</hi>
                  </note> And such a touch of circles is euer in one poynt onely: which poynt onely is common to them both. As the poynt B is in the confe<g ref="char:EOLhyphen"/>rence of the circle AB, and also <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the <gap reason="illegible" resp="#PDCC" extent="5 letters">
                     <desc>•••••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>ference of the circle BC.</p>
               <p>
                  <pb facs="tcp:7062:112"/>
                  <note place="margin">
                     <hi>Circles may touch toge<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>her two ma<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> of wayes.</hi>
                  </note>Circles may touch together two maner of wayes, either outwardly the one wholy without the other: or els the one being contayned within the other.</p>
               <p>As the circles <hi>DE</hi> and <hi>DF:</hi> of which the one <hi>DE</hi> contay<g ref="char:EOLhyphen"/>neth
<figure/>
the other, namely <hi>DF:</hi> and touch the one the other in the poynt <hi>D:</hi> and that onely poynt is common to them both: neither doth the one enter into the other. If any part of the one enter into any part of the other, then the one cutteth and deuideth the other, and toucheth the one the other not in one poynt onely as in the other before, but in two point<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, and haue also a superficies common to them both. As the cir<g ref="char:EOLhyphen"/>cles <hi>GHK</hi> and <hi>HL</hi>K cut the one the other in two poyntes <hi>H</hi> and <hi>K</hi>: and the one entreth into the other: Al<g ref="char:EOLhyphen"/>so the superficies <hi>HK</hi> is common to them both: For it is a part of the circle <hi>GHK,</hi> and also it is a part of the circle <hi>HLK.</hi>
               </p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Fourth defi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note>Right lines in a circle are sayd to be equally distant from the cen<g ref="char:EOLhyphen"/>tre, when perpendicular lines drawen from the centre vnto those lines are equall. And that line is sayd to be more di<g ref="char:EOLhyphen"/>stant, vpon whom falleth the greater perpendicular line.</head>
               <p>As in the circle <hi>ABCD</hi> whose centre is <hi>E,</hi> the two lynes
<figure/>
                  <hi>AB</hi> and <hi>CD</hi> haue equall distance from the centre <hi>E:</hi> bycause that the lyne <hi>EF</hi> drawen from the centre <hi>E</hi> perpendicularly vpon the lyne <hi>AB,</hi> and the lyne <hi>EG</hi> drawen likewise perpendi<g ref="char:EOLhyphen"/>larly from the centre <hi>E</hi> vpon the lyne <hi>CD</hi> are equall the one to the other. But in the circle <hi>HKLM</hi> whose centre is <hi>N</hi> the lyne <hi>HK</hi> hath greater distance from the centre <hi>N</hi> then hath the lyne <hi>LM</hi>: for that the lyne <hi>ON</hi> drawen from the centre <hi>N</hi> perpendicularly vppon the lyne <hi>HK</hi> is greater then the lyne <hi>NP</hi> which is drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre <hi>N</hi> perpendicularly vpon the lyne <hi>LM.</hi>
               </p>
               <p>So likewise in the other figure the lynes <hi>AB</hi> and <hi>DC</hi> in the circle <hi>ABCD</hi> are equidist<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nt from the centre <hi>G</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> bycause the lynes <hi>OG</hi> and <hi>GP</hi> perpendicularly drawen from the centre <hi>G</hi> vppon the sayd lynes <hi>AB</hi> and <hi>DC</hi> are equall. And the lyne <hi>AB</hi> hath greater distance from the centre <hi>G</hi> then hath the the lyne <hi>EF,</hi> bycause the lyne <hi>OG</hi> perpendicul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rly dr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wen from the centre <hi>G</hi> to the lyne <hi>AB</hi> is gre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter then the lyne <hi>HG</hi> whiche is perpendicularly drawen from the c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tre <hi>G</hi> to the lyne <hi>EF.</hi>
               </p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Fift defini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>A section or segment of a circle, is a figure co<g ref="char:cmbAbbrStroke">̄</g>prehended vnder a right line and a portion of the circumference of a circle.</head>
               <p>
                  <pb n="82" facs="tcp:7062:112"/>As the figure <hi>ABC</hi> is a section of a circle
<figure/>
bycause it is comprehended vnder the right lyne <hi>AC</hi> and the circumference of a circle <hi>ABC.</hi> Likewise the figure <hi>DEF</hi> is a section of a circle, for that it is comprehended vnder the right lyne <hi>DF,</hi> and the circu<g ref="char:cmbAbbrStroke">̄</g>ference <hi>DEF.</hi> And the figure <hi>ABC</hi> for that it co<g ref="char:cmbAbbrStroke">̄</g>taineth within it the centre of the circle is called the greater section of a circle: and the figure <hi>DEF</hi> is the lesse section of a circle, bycause it is wholy without the centre of the circle as it was noted in the 16. Definition of the first booke.</p>
            </div>
            <div type="part">
               <head>An angle of a section or segment, is that angle which is con<g ref="char:EOLhyphen"/>tayned vnder a right line and the circu<g ref="char:cmbAbbrStroke">̄</g>ference of the circle.<note place="margin">
                     <hi>Sixt defini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>
               </head>
               <p>As the angle ABC in the section ABC is an angle of a sec<g ref="char:EOLhyphen"/>tion,
<figure/>
bycause it is contained of the circumference BAC and the right lyne BC. Likewise the angle CBD is an angle of the section BDC bycause it is contayned vnder the circumference BDC, and the right lyne BC. And these angles are commonly called mixte angles,<note place="margin">
                     <hi>Mixt angles.</hi>
                  </note> bycause they are contayned vnder a right lyne and a crooked. And these portions of circumferences are commonly called arkes,<note place="margin">
                     <hi>Arkes.</hi>
                  </note> and the right lynes are called chordes,<note place="margin">
                     <hi>Chordes.</hi>
                  </note> or right lynes subtended. And the greater section hath euer the greater angle, and the lesse section the lesse angle.</p>
            </div>
            <div type="part">
               <head>An angle is sayd to be in a section,<note place="margin">
                     <hi>Seuenth de<g ref="char:EOLhyphen"/>finition.</hi>
                  </note> whe<g ref="char:cmbAbbrStroke">̄</g> in the circumference is taken any poynt, and from that poynt are drawen right lines to the endes of the right line which is the base of the segment, the angle which is contayned vnder the right lines drawen from the poynt, is (I say) sayd to be an angle in a section.</head>
               <p>As the angle ABC is an angle is the section ABC, bycause
<figure/>
from the poynt B beyng a poynt in the circumference ABC are drawen two right lynes BC and BA to the endes of the lyne AC which is the base of the section ABC. Likewise the angle ADC is an angle in the section ADC, bycause from the poynt D beyng in the circu<g ref="char:cmbAbbrStroke">̄</g>ference ADC are drawen two right lynes, namely, DC &amp; DA to the endes of the right line AC which is also the base to the sayd section ADC. So you see, it is not all one to say,<note place="margin">
                     <hi>Difference of an angle of a Section, and of an angle in a Section.</hi>
                  </note> an an<g ref="char:EOLhyphen"/>gle of a section, and an angle in a section. An angle of a section co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>sisteth of the touch of a right lyne and a crooked. And an angle in a section is placed on the circumference, and is contayned of two right lynes. Also the greater section hath in it the lesse angle, and the lesse section hath in it the greater angle.</p>
            </div>
            <div type="part">
               <head>But when the right lines which comprehend the angle do re<g ref="char:EOLhyphen"/>ceaue any circumference of a circle,<note place="margin">
                     <hi>Eight defi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> then that angle is sayd to be correspondent, and to pertaine to that circumference.</head>
               <p>
                  <pb facs="tcp:7062:113"/>As the right lynes BA and BC which containe the angle AB
<figure/>
C, and receaue the circumference ADC therfore the angle ABC is sayd to subtend and to pertaine to the circu<g ref="char:cmbAbbrStroke">̄</g>ference ADC. And if the right lynes whiche cause the angle, concurre in the centre of a circle: then the angle is sayd to be in the centre of a circle. As the angle EFD is sayd to be in the centre of a circle, for that it is comprehended of two right lynes FE and FD: whiche concurre and touch in the centre F. And this angle likewise subtendeth the circumference EGD: whiche circumference also, is the measure of the greatnes of the angle EFD.</p>
            </div>
            <div type="part">
               <head>A Sector of a circle is (an angle being set at the centre of a circle) a figure contayned vnder the right lines which make that angle,<note place="margin">
                     <hi>Ninth defi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> and the part of the circumference re<g ref="char:EOLhyphen"/>ceaued of them.</head>
               <p>As the figure ABC is a sector of a circle, for that it hath an angle
<figure/>
at the centre, namely the angle BAC, &amp; is co<g ref="char:cmbAbbrStroke">̄</g>tained of the two right lynes AB and AC (whiche contayne that angle and the circumfe<g ref="char:EOLhyphen"/>rence receaued by them.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Tenth defini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>Like segmentes or sections of a circle are those, which haue equall angles, or in whom are equall angles.</head>
               <p>
                  <note place="margin">
                     <hi>Two defini<g ref="char:EOLhyphen"/>tions.</hi>
                  </note>Here are set two definitions of like sections of
<figure/>
a circle. The one pertaineth to the angles whiche are set in the centre of the circle and receaue the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce of the sayd sections:<note place="margin">
                     <hi>First.</hi>
                  </note> the other per<g ref="char:EOLhyphen"/>taineth to the angle in the section, whiche as be<g ref="char:EOLhyphen"/>fore was sayd is euer in the circumference. As if the angle BAC, beyng in the centre A and re<g ref="char:EOLhyphen"/>ceaued of the circumference BLC be equall to the angle FEG beyng also in the centre E and receaued of the circumference FKG, then are the two sections BCL and FGK lyke by the first definition. By the same definition also are the other two sections like, name<g ref="char:EOLhyphen"/>ly BCD, and FGH, for that the angle BAC is equall to the angle FEG.</p>
               <p>
                  <note place="margin">
                     <hi>Second.</hi>
                  </note>Also by the second definition if B
<figure/>
AC beyng an angle placed in the cir<g ref="char:EOLhyphen"/>cumference of the section BCA be e-angle EDF beyng an angle in the se<g ref="char:EOLhyphen"/>ction EFD placed in the circumfe<g ref="char:EOLhyphen"/>rence, there are the two sections BCA, and EFD lyke the one to the o<g ref="char:EOLhyphen"/>ther. Likewise also if the angle BGC beyng in the section BCG be equall to the angle EHF beyng in the sectio<g ref="char:cmbAbbrStroke">̄</g> EHF the two sections BCG and EFH are lyke. And so is it of angles beyng equall in any poynt of the circumference.</p>
               <p>
                  <pb n="83" facs="tcp:7062:113"/>Euclide defineth not equall Sections:<note place="margin">
                     <hi>Why Euclide defineth not equall Secti<g ref="char:EOLhyphen"/>ons.</hi>
                  </note> for they may infinite wayes be described. For there may vppon vnequall right lynes be set equall Sections (but yet in vne<g ref="char:EOLhyphen"/>quall circles) For from any circle beyng the greater, may be cut of a portion equall to a portion of an other circle beyng the lesse. But when the Sections are equall, and are set vpon equall right lynes, theyr circumferences also shalbe equall. And right lynes beyng deuided into two equall partes, perpendicular lynes drawen from the poyntes of the diuision to the cir<g ref="char:EOLhyphen"/>cumfere<g ref="char:cmbAbbrStroke">̄</g>ces
<figure/>
shalbe equall. As if the two secti<g ref="char:EOLhyphen"/>ons <hi>ABC</hi> and <hi>DEF,</hi> beyng set vppon equall ryght lynes <hi>AC</hi> &amp; <hi>DF,</hi> be equall: then if ech of the two lynes <hi>AC</hi> &amp; <hi>DF</hi> be deuided into two e<g ref="char:EOLhyphen"/>quall partes in the poyntes <hi>G</hi> and <hi>H,</hi> &amp; from the sayd poyntes be drawen to the circumferences two perpendicular lynes <hi>BG</hi> and <hi>EH,</hi> the sayd perpendicular lynes shalbe equall.</p>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Probleme. The <hi>1.</hi> Proposition. To finde out the centre of a circle geuen.</head>
               <p>SVppose that there be a circle geuen ABC. It is requi<g ref="char:EOLhyphen"/>red to finde out the centre of the circle ABC. Draw in it a right line at all aduentures, and let the same be AB.<note place="margin">Constuction.</note> And (by the <hi>10.</hi> of the first) deuide the line AB into two equall partes in the poynt D. And (by the <hi>11.</hi> of the same) fro<g ref="char:cmbAbbrStroke">̄</g> the poynt D raise vp vnto AB a perpendicular line DC, &amp; (by the second petition) extend DC vnto y<hi rend="sup">e</hi> point E. And (by the <hi>10.</hi> of the first) deuide
<figure/>
the line CE into two equall partes in the poynt F. Then I say that the point F is the centre of the circle ABC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> For if it be not, let some other point, namely G, be the centre. And (by the first peti<g ref="char:EOLhyphen"/>tion) draw these right lines GA, GD, and GB. And for asmuch as AD is equall vnto DB, and DG is common vnto the<g ref="char:cmbAbbrStroke">̄</g> both, therefore these two lines AD and DG are equall to these two lines GD and DB, the one to the other, and (by the <hi>15.</hi> definition of the first) the base GA is equall to the base GB. For they are both drawen from the cen<g ref="char:EOLhyphen"/>tre G to the circumference: therefore (by the <hi>8.</hi> of the first) the angle ADG is equall to the angle BDG. But when a right line standing vpon a right line maketh the angles on eche side equall the one to the other, eyther of those angles (by the <hi>10.</hi> definition of the first) is a right angle. VVherefore the angle BDG
<pb facs="tcp:7062:114"/>
is a right angle: but y<hi rend="sup">e</hi> angle FDB is also a right angle by construction. VVher<g ref="char:EOLhyphen"/>fore (by the <hi>4.</hi> petition) the angle FDB is equall to the angle BDG, the grea<g ref="char:EOLhyphen"/>ter to the lesse, which is impossible. VVherefore the poynt G is not the centre of the circle ABC. In like wise may we proue that no other poynt besides F is the centre of the circle ABC. VVherefore the poynt F is the centre of the circle ABC: which was required to be done.</p>
               <div type="section">
                  <head>Correlary.</head>
                  <p>
                     <note place="margin">Correlary.</note> Hereby it is manifest, that if in a circle a right line do deuide a right line into two equall partes, and make right angles on eche side: in that right line which deuideth the other line into two e<g ref="char:EOLhyphen"/>quall partes is the centre of the circle.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Theoreme. The <hi>2.</hi> Proposition. If in the circu<g ref="char:cmbAbbrStroke">̄</g>ference of a circle be take<g ref="char:cmbAbbrStroke">̄</g> two poyntes at all ad<g ref="char:EOLhyphen"/>uentures: a right line drawen from the one poynt to the other shall fall within the circle.</head>
               <p>SVppose that there be a circle ABC. And in the circumference ther<g ref="char:EOLhyphen"/>of, let there be take<g ref="char:cmbAbbrStroke">̄</g> at all aduentures these two poyntes A &amp; B. Then I say that a right line drawen from A to B shall fall within the circle ABC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if it do not, let it fall without the circle, as the line AEB doth, which if it be possible imagine to be a right line. And (by the Proposition going before) take the centre of the circle, and let the same be D. And (by the first petition) draw lines from D to A, and from D to B. And extend DF to E. And for asmuch as (by the <hi>15.</hi> definition of y<hi rend="sup">e</hi> first)
<figure/>
DA is equall vnto DB. Therefore the an<g ref="char:EOLhyphen"/>gle DAE is equall to the angle DBE. And for asmuch as one of the sides of the triangle DAE, namely the side AEB is produced, therefore (by the <hi>16.</hi> of the first) the angle DEB, is greater then the angle DAE. But the angle DAE is equall vnto the angle DBE. VVherfore the angle DEB is grea<g ref="char:EOLhyphen"/>ter then the angle DBE. But (by the <hi>18.</hi> of the first) vnto the greater angle is subtended the greater side. VVherefore the side DB is greater then the side DE. But (by the <hi>15.</hi> definition of the first) the line DB is
<pb n="84" facs="tcp:7062:114"/>
equall vnto the line DF. VVherfore the line DF is greater then the line DE, namely, the lesse greater then the greater: which is impossible. VVherfore a right line drawen from A to B falleth not without the circle. In like sort also may we proue that it falleth not in the circumference. VVherefore it falleth within the circle. If therefore in the circumference of a circle be taken two poyntes at all ad<g ref="char:EOLhyphen"/>uentures: a right line drawen from the one poynt to the other shall fall within the circle: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Theoreme. The <hi>3.</hi> Proposition. If in a circle a right line passing by the centre do deuide an o<g ref="char:EOLhyphen"/>ther right line not passing by the ce<g ref="char:cmbAbbrStroke">̄</g>tre into two equall partes: it shall deuide it by right angles. And if it deuide the line by right angles, it shall also deuide the same line into two equall partes.</head>
               <p>SVppose that there be a circle ABC,<note place="margin">The first para of this Propo<g ref="char:EOLhyphen"/>sition.</note> and let there be in it drawen a right line passing by the centre, and let the same be CD, deuiding an other right line AB not passing by y<hi rend="sup">e</hi> centre into two equall partes in the poynt F. Then I say that the angles at the poynt of the deuision are right angles.<note place="margin">Construction.</note> Take (by y<hi rend="sup">e</hi> first of the third)
<figure/>
the centre of the circle ABC, and let the same be E. And (by the first petition) drawe lines from E to A &amp; from E to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for asmuch as the line AF is equall vnto the line FB, and the line FE is common to them both, therfore these two lines EF and FA are equall vnto these two lines EF &amp; FB. And the base EA is equall vnto the base EB (by the <hi>15.</hi> defini<g ref="char:EOLhyphen"/>tion of the first). VVherefore (by the <hi>8.</hi> of the first) the angle AFE is equall to the angle BFE. But when a right line standing vpon a right line doth make the angles on eche side equall the one to the other, eyther of those angles is (by the <hi>10.</hi> defini<g ref="char:EOLhyphen"/>tion of the first) a right angle. VVherfore either of these angles AFE, &amp; BFE is a right angle. VVherefore the line CD passing by the centre, and deuiding the line AB not passing by the centre into two equall partes, maketh at the point of the deuision right angles.</p>
               <p>But now suppose that the line CD do deuide the line AB in such sort that it maketh right angles.<note place="margin">The second part conuerst of the first.</note> Then I say that it deuideth it into two equall partes, that is, y<hi rend="sup">t</hi> the line AF is equall vnto the line FB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For the same order of construction remayning, for asmuch as the line EA is equall vnto the line EB (by the <hi>15.</hi> de<g ref="char:EOLhyphen"/>finition
<pb facs="tcp:7062:115"/>
of the first). Therefore the angle EAF is equall vnto the angle EBF (by the <hi>5.</hi> of the first). And the right angle AFE is (by the <hi>4.</hi> petition) equall to the right angle BFE. VVherefore there are two triangles EAF, &amp; EBF hauing two angles equall to two angles, &amp; one side equall to one side, namely the side EF which is common to them both, and subtendeth one of the equall angles, wherefore (by the <hi>26.</hi> of the first) the sides remayning of the one, are equall vn<g ref="char:EOLhyphen"/>to the sides remayning of the other. VVherefore the line AF is equall vnto the line FB. If therefore in a circle a right line passing by the centre do deuide an other right line not passing by the centre into two equall partes, it shall deuide it by right angles. And if it deuide the line by right angles it shall also deuide the same line into two equall partes: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Theoreme. The <hi>4.</hi> Proposition. If in a circle two right lines not passing by the centre, deuide the one the other: they shall not deuide eche one the other into two equall partes.</head>
               <p>
                  <hi>SVppose that there be a circle ABCD, and let there be in it drawen two right lines not passing by the centre and deuiding the one the other, and let the same be AC and BD, which let deuide the one the other in the poynt E. Then I say that they deuide not eche
<figure/>
the one the other into two equall partes.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</hi>
                  </note> 
                  <hi>For if it be possible let them deuide eche the one the other into two equall partes, so that let AE be equall vnto EC, &amp; BE vnto ED. And take the centre of the circle ABCD, which let be F. And (by the first petition) draw a line from F to E. Now for asmuch as a certaine right line FE passing by the centre deuideth an other line AC not passing by the centre into two equall partes, it maketh where the deuisi<g ref="char:EOLhyphen"/>on is right angles (by the</hi> 3. <hi>of the third). VVherfore the angle FEA is a right angle. Againe for asmuch as the right line FE, passing by the centre, deuideth the right line BD not passing by the centre into two equall partes, therefore (by the same) it maketh where y<hi rend="sup">e</hi> deuision is right angles. VVherfore the angle FEB is a right angle. And it is proued that the angle FEA is a right angle. VVher<g ref="char:EOLhyphen"/>fore (by the</hi> 4. <hi>petition) the angle FEA is equall vnto the angle FEB, namely the lesse angle vnto the greater: which is impossible. VVherefore the right lines AC and BD deuide not eche one the other into two equall partes. If therfore in a circle two right lines not passing by the centre, deuide the one the other, they
<pb n="85" facs="tcp:7062:115"/>
shall not deuide eche one the other into two equall partes: which was required to be demonstrated.</hi>
               </p>
               <p>In this Proposition are two cases.<note place="margin">
                     <hi>Two cases in this Propo<g ref="char:EOLhyphen"/>sition.</hi>
                  </note> For the lines cutting the one the other, do ey<g ref="char:EOLhyphen"/>ther, neyther of them passe by the centre, or the one of them doth passe by the cen<g ref="char:EOLhyphen"/>tre, &amp; the other not. The first is declared by the author. The second is thus proued.</p>
               <p>Suppose that in the circle <hi>ABCD</hi> the line
<figure/>
                  <hi>BD</hi> passing by the centre doe cut the line <hi>AC</hi> not passing by the centre.<note place="margin">
                     <hi>Construction for the second case.</hi>
                  </note> Then I say that the lines <hi>AC</hi> and <hi>BD</hi> do not deuide the one the other in<g ref="char:EOLhyphen"/>to two equall partes. For by the former Proposi<g ref="char:EOLhyphen"/>tion the line <hi>BD</hi> passing by the centre and deui<g ref="char:EOLhyphen"/>ding the line <hi>AC</hi> into two equall partes,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> it shall also deuide it perpendicularly. And for asmuch as the line <hi>AC</hi> deuideth the line <hi>BD</hi> into two equall partes &amp; right angled wise: therfore by the Correl<g ref="char:EOLhyphen"/>lary of the first of thys booke, the line <hi>AC</hi> passeth by the centre of the circle: which is co<g ref="char:cmbAbbrStroke">̄</g>trary to the supposition. Wherfore the lines <hi>AC</hi> and <hi>BD</hi> do not deuide the one the other into two equall partes: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Theoreme. The <hi>5.</hi> Proposition. If two circles cut the one the other, they haue not one and the same centre.</head>
               <p>SVppose that these two circles
<figure/>
ABC,<note place="margin">Construction.</note> and CBG do cut the one the other in the poyntes C and B. Then I say that they haue not one &amp; the same centre.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if it be possi<g ref="char:EOLhyphen"/>ble let E be centre to them both. And (by the first petition) draw a line from E to C. And draw an other right line EFG at all aduentures. And for asmuch as the poynt E is the centre of the circle ABC, therefore (by the <hi>15.</hi> definition of the first) the line EC is equall vnto the line EF. Agayne for asmuch as the poynt E is the centre of the circle CBG, there<g ref="char:EOLhyphen"/>fore (by the same definition) the line EC is equall vnto the line EG. And it is proued that the line EC is equall vnto the line EF: wherefore the line EF also is equall vnto the line EG, namely the lesse vnto the greater: which is impossi<g ref="char:EOLhyphen"/>ble. VVherfore the poynt E is not the centre of both the circles ABC, &amp; CBG. I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> like sort also may we proue that no other poynt is the centre of both the sayd
<pb facs="tcp:7062:116"/>
circles. If therefore two circles cut the one the other, they haue not one and the same centre: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Theoreme. The <hi>6.</hi> Proposition. If two circles touch the one the other, they haue not one and the same centre.</head>
               <p>SVppose that these two circles ABC, &amp; CDE do touch the one the other in the poynt C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>on leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> Then I say that they haue not one and the same centre. For if it be possible let the point F be centre vnto them both. And (by the first petition) draw a line from F to C: and
<figure/>
drawe the line FEB at all aduentures. And for asmuch as the poynt F is the cen<g ref="char:EOLhyphen"/>tre of the circle ABC, therfore (by the <hi>15.</hi> definition of the first) the line FC is equall vnto the line FB. Agayne forasmuch as the poynt F is the centre of y<hi rend="sup">e</hi> circle CDE, therefore (by the same definition) the line FC is equall vnto the line FE. And it is proued, that the line FC is equall vnto the line FB, wherefore the line FE also is e<g ref="char:EOLhyphen"/>quall vnto the line FB, namely the lesse vnto y<hi rend="sup">e</hi> greater: which is impossible. VVher<g ref="char:EOLhyphen"/>fore the poynt F is not the centre of both the circles ABC and CDE. In like sort also may we proue that no other poynt is the centre of both the sayd circles. If therefore two circles touch the one the other: they haue not one and the same centre: which was required to be demonstrated.</p>
               <p>
                  <note place="margin">Two case<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in thys Pro<g ref="char:EOLhyphen"/>position.</note>
                  <hi>In thys Proposition are two cases: for the circles
<figure/>
touchyng the one the other, may touch eyther within or without. If they touch the one the other within, then is it by the former demonstration manifest, that they haue not both one and the selfe same centre. It is also manifest if they touch the one the other without: for that euery cen<g ref="char:EOLhyphen"/>tre is in the middest of hys circle.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Theoreme. The <hi>7.</hi> Proposition. If in the diameter of a circle be taken any poynt, which is not
<pb n="86" facs="tcp:7062:116"/>
the centre of the circle, and from that poynt be drawen vnto the circumference certaine right lines: the greatest of those lines shall be that line wherein is the centre, and the lest shall be the residue of the same line. And of all the other lines, that which is nigher to the line which passeth by the centre is greater then that which is more distant. And from that point can fall within the circle on ech side of the least line onely two equall right lines.</head>
               <p>SVppose that there be a circle ABCD: and let the diameter thereof be AD. And take in it any poynt besides the centre of the circle, and let the same be F.<note place="margin">Construction.</note> And let the centre of the circle (by the <hi>1.</hi> of y<hi rend="sup">e</hi> third) be the poynt E. And from the poynt F let there be drawen vnto the circumference ABCD these right lines FD, FC, and FG. Then I say that the line FA is the greatest: and the line
<figure/>
FD is the lest. And of the other lines, the line FB is greater then the line FC, and the line FC is greater then the line FG.<note place="margin">The first part of this Propo<g ref="char:EOLhyphen"/>sition.</note> Drawe (by the first petition) these right lines BE, CE, and GE. And for asmuch as (by the <hi>20.</hi> of the first) in eue<g ref="char:EOLhyphen"/>ry triangle two sides are greater then the third,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore y<hi rend="sup">e</hi> lines EB and EF are greater then the residue, namely then the line FB. But the line AE is e<g ref="char:EOLhyphen"/>quall vnto the line BE (by the <hi>15.</hi> defi<g ref="char:EOLhyphen"/>nition of the first). VVherefore the lines BE and EF are equall vnto the line AF. VVherefore the line AF is greater then then the line BF. Agayne for asmuch as the line BE is equall vnto CE (by the <hi>15.</hi> definition of the first) and the line FE is common vnto them both, therefore these two lines BE and EF are equall vnto these two CE and EF. But the angle BEF is greater then the angle CEF. VVherefore (by the <hi>24.</hi> of the first) the base BF is greater then the base CF: and by the same reason the line CF is greater then the line FG. Agayne for asmuch as the lines GF and FE are greater then the line EG (by the <hi>20.</hi> of the first).<note place="margin">Second part.</note> But (by the <hi>15.</hi> definition of the first) the line EG is equall vnto the line ED: VVherefore the lines GF and FE are greater then the line ED, take away EF, which is co<g ref="char:cmbAbbrStroke">̄</g>mon to the<g ref="char:cmbAbbrStroke">̄</g> both, wherfore y<hi rend="sup">e</hi> residue GF is grea<g ref="char:EOLhyphen"/>ter then the residue FD. VVherefore the line FA is the greatest, and the line FD is the lest, and the line FB is greater then the line FC, and the line FC
<pb facs="tcp:7062:117"/>
is greater then the line FG.<note place="margin">Third part.</note> Now also I say that from the poynt F there can be drawen onely two equall right lines into the circle ABCD on eche side of the least line, namely FD. For (by the <hi>23.</hi> of the first) vpon the right line geuen EF and to the poynt in it, namely E, make vnto the angle GEF an equall an<g ref="char:EOLhyphen"/>gle FEH: and (by the first petition) draw a line from F to H. Now for asmuch as (by the <hi>15.</hi> definition of the first) the line EG is equall vnto the line EH, and the line EF is common vnto them both, therefore these two lines GE and EF are equall vnto these two lines HE and EF, and (by construction) the angle GEF is equall vnto the angle HEF. VVherefore (by the <hi>4.</hi> of y<hi rend="sup">e</hi> first) the base FG is equall vnto the base FH.<note place="margin">This demon<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rated by an argument lea<g ref="char:EOLhyphen"/>ding to an im<g ref="char:EOLhyphen"/>possibilie.</note> I say moreouer that from the poynt F can be drawen into the circle no other right line equall vnto the line FG. For if it possible let the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> FK be equall vnto the line FG. And for asmuch as FK is equall vnto FG. But the line FH is equall vnto the line FG, therefore the line FK is equall vnto the line FH. VVherfore the line which is nigher to the line which passeth by the centre is equall to that which is farther of, which we haue before proued to be impossible.</p>
               <p>
                  <note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the latter part of the Proposition leading also to an impossi<g ref="char:EOLhyphen"/>bilitie.</note>Or els it may thus be demonstrated.
<figure/>
Draw (by the first petition) a line from E to K: and for asmuch as (by y<hi rend="sup">e</hi> 
                  <hi>15.</hi> de<g ref="char:EOLhyphen"/>finitio<g ref="char:cmbAbbrStroke">̄</g> of y<hi rend="sup">e</hi> first) y<hi rend="sup">e</hi> line GE is equall vnto y<hi rend="sup">e</hi> line EK, and the line FE is common to them both, and the base GF is equall vnto the base FK, therefore (by the <hi>8.</hi> of the first) the angle GEF is equall to the angle KEF. But the angle GEF is equall to the angle HEF. VVhere<g ref="char:EOLhyphen"/>fore (by the first common sentence) the angle HEF is equall to the angle KEF the lesse vnto the greater: which is impossible. VVherefore from the poynt F there can be drawen into the circle no other right line equall vnto the line GF. VVherefore but one onely. If therefore in the diameter of a circle be taken any poynt, which is not the centre of the cir<g ref="char:EOLhyphen"/>cle, and from that poynt be drawen vnto the circumference certaine right lines: the greatest of those right lines shall be that wherein is the centre: and the least shall be the residue. And of all the other lines, that which is nigher to the line which passeth by the centre is greater then that which is more distant. And from that poynt can fall within the circle on ech side of the least line onely two equall right lines: which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note>Hereby it is manifest, that two right lines being drawen fro<g ref="char:cmbAbbrStroke">̄</g> any one poynt of the diameter, the one of one side, and the other of the other side, if with the diame<g ref="char:EOLhyphen"/>ter they make equall angles, the sayd two right lines are equall. As in thys place are the two lines FG and FH.</p>
               </div>
            </div>
            <div type="part">
               <pb n="87" facs="tcp:7062:117"/>
               <head>The <hi>7.</hi> Theoreme. The <hi>8.</hi> Proposition. If without a circle be taken any poynt, and from that poynt be drawen into the circle vnto the circumference certayne right lines, of which let one be drawen by the centre and let the rest be drawen at all aduentures: the greatest of those lines which fall in the concauitie or hollownes of the circumference of the circle, is that which passeth by the centre: and of all the other lines that line which is nigher to the line which passeth by the centre is greater then that which is more distant. But of those right lines which end in the conuexe part of the circumfe<g ref="char:EOLhyphen"/>rence, that is the least which is drawen from the poynt to the diameter: and of the other lines that which is nigher to the least is alwaies lesse then that which is more distant. And from that poynt can be drawen vnto the circumference on ech side of the least onely two equall right lines.</head>
               <p>SVppose y<hi rend="sup">t</hi> the circle geuen be ABC,
<figure/>
&amp; without y<hi rend="sup">e</hi> circle ABC, take the point D: and fro<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> same point draw certain right lines into y<hi rend="sup">e</hi> circle vnto the cir<g ref="char:EOLhyphen"/>cumference, &amp; let the<g ref="char:cmbAbbrStroke">̄</g> be DA, DE, DF, &amp; DC: &amp; let y<hi rend="sup">e</hi> line DA passe by y<hi rend="sup">e</hi> centre. Then I say, of y<hi rend="sup">e</hi> right lines which fall in the concauitie of y<hi rend="sup">e</hi> circumference AEFC, y<hi rend="sup">t</hi> is, within y<hi rend="sup">e</hi> circle, y<hi rend="sup">e</hi> greatest is y<hi rend="sup">t</hi> which passeth by y<hi rend="sup">e</hi> centre, that is, DA. And of those lines which fall vpon y<hi rend="sup">e</hi> conuex part of y<hi rend="sup">e</hi> circumfe<g ref="char:EOLhyphen"/>rence, y<hi rend="sup">e</hi> lest is y<hi rend="sup">t</hi> which is drawen fro<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> point D vnto y<hi rend="sup">e</hi> end of y<hi rend="sup">e</hi> diameter AG. And of the right lines falling w<hi rend="sup">t</hi>in the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce, the line DE is greater then y<hi rend="sup">e</hi> line DF, &amp; the line DF is greater then y<hi rend="sup">e</hi> line DC. And of the right lines which end in y<hi rend="sup">e</hi> conuex part of the circumference, y<hi rend="sup">t</hi> is, without y<hi rend="sup">e</hi> circle, that (which is nigher vnto DG y<hi rend="sup">e</hi> lest, is alwayes lesse then y<hi rend="sup">t</hi> which is more dista<g ref="char:cmbAbbrStroke">̄</g>t, that is, the line DK is lesse then the line DL, and the line DL is lesse then the line DH. Take (by the first of the third) the centre of the circle ABC,<note place="margin">Construction.</note> and let the
<pb facs="tcp:7062:118"/>
same be M:<note place="margin">The first part of this Propo<g ref="char:EOLhyphen"/>sition.</note> and (by the first petition) drawe these right lines ME, MF, MC, MH, ML, and MK. And for asmuch as (by the <hi>15.</hi> definition of the first) the line AM is equall vnto the line EM, put the line MD common to them both. VVherefore the line AD is equall vnto the lines EM and MD. But the lines EM and MD are (by the <hi>20.</hi> of the first) greater then the line ED. VVherefore the line AD also is greater then the line ED. Agayne for asmuch as (by the <hi>15.</hi> definition of the first) the line ME is equall vnto the line MF, put the line MD common to them both: VVherefore the lines EM and MD are equall to the lines FM and MD, and the angle EMD is greater then the angle FMD: VVherefore (by the <hi>19.</hi> of the first) the base ED is greater then the base FD. In like sort also may we proue that the line FD is greater then the line CD. VVherefore the line DA is the greatest, and the line DE is grea<g ref="char:EOLhyphen"/>ter then the line DF, and the line DF is greater then the line DC.</p>
               <p>
                  <note place="margin">Second part.</note>And for asmuch as (by the <hi>20.</hi> of the
<figure/>
first) <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e lines MK and KD are greater then the line MD. But (by the <hi>15.</hi> defini<g ref="char:EOLhyphen"/>tion of the first) the line MG is equall vn<g ref="char:EOLhyphen"/>to the line MK. VVherefore the residue KD is greater then y<hi rend="sup">e</hi> residue GD. VVher<g ref="char:EOLhyphen"/>fore the line GD is lesse then the line KD. And for asmuch as from the endes of one of the sides of the triangle MLD, namely, MD are drawen two right lines MK and KD meeting within the triangle, therfore (by the <hi>21.</hi> of the first) the lines MK and KD are lesse then the lines ML &amp; LD, of which the line MK is equall vnto the line ML. VVherefore the residue DK is lesse then the residue DL. In like sort also may we proue that the line DL is lesse then the line DH. VVherefore the line DG is the lest, and the line DK is lesse then the line DL, and the line DL is lesse then the line DH.</p>
               <p>
                  <note place="margin">Third part.</note>Now also I say that from the poynt D can be drawen vnto the circumference on eche side of DG the least onely two equall right lines. Vpon the right line MD, and vnto the poynt in it M make (by the <hi>23.</hi> of the first) vnto the an<g ref="char:EOLhyphen"/>gle KMD an equall angle DMB. And (by the first petition) drawe a line from D to B. And for asmuch as (by the <hi>15.</hi> definition of the first) the line MB is equall vnto the line MK put the line MD common to the both, wher<g ref="char:EOLhyphen"/>fore these two lines MK and MD are equall to these two lines BM and MD the one to the other, and the angle KMD is (by the <hi>23.</hi> of the first) equall to the angle BMD: VVhere<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore (by the <hi>4.</hi> of the first) the base DK is equall
<pb n="88" facs="tcp:7062:118"/>
to the base DB.</p>
               <p>Now I say that from the poynt D on
<figure/>
that side that the line DB is,<note place="margin">This is demo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>strated by an argument lea<g ref="char:EOLhyphen"/>ding to an ab<g ref="char:EOLhyphen"/>surdity.</note> can not be drawen vnto the circumference any other line besides DB equall vnto the right line DK. For if it be possible let there be drawen an other line besides DB, and let the same be DN. And for asmuch as the line DK is equall vnto the line DN. But vnto the line DK is equall the line DB. Therfore (by the first common sentence) the line DB is equall vnto the line DN. VVherefore that which is nigher vnto DG the least, is equall to y<hi rend="sup">e</hi> which is more distant<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> VVhich we haue before proued to be impossible.</p>
               <p>Or it may thus be demonstrated.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the latter part, leading also to an im<g ref="char:EOLhyphen"/>possibility.</note> Draw (by the first petition) a line from M to N. And for asmuch as (by the <hi>15.</hi> definition of the first) the line KM is equall vnto the line MN, and the line MD is common to them both. And the base KD is e<g ref="char:EOLhyphen"/>quall to the base DN (by supposition) therefore (by the <hi>8.</hi> of the first) the an<g ref="char:EOLhyphen"/>gle KMD is equall to the angle DMN. But the angle KMD is equall to the angle BMD. Wherfore the angle BMD is equall to the angle NMD, the lesse vnto the greater: which is impossible. Wherefore from the poynt D can not be drawen vnto the circumference ABC on eche side of DG the lest, more then two equall right lines. If therefore without a circle be taken any poynt and from that poynt be drawen into the circle vnto the circumference certaine right lines, of which let one be drawen by the centre, and let the rest be drawen at all adventures: the greatest of those right lines which fall in y<hi rend="sup">e</hi> concauitie or hollow<g ref="char:EOLhyphen"/>nes of the circumference of the circle is that which passeth by the centre. And of all the other lines, that line which is nigher to the line which passeth by the cen<g ref="char:EOLhyphen"/>tre, is greater then that which is more distant. But of those right lines which end in the conuexe part of the circumference, that line is the lest which is drawen from the poynt to the dimetient: and of the other lines that which is nigher to the least is alwayes lesse then that which is more distant. And from that poynt can be drawen vnto the circumference on ech side of the lest only two equall right lines: which was required to be proued.</p>
               <p>
                  <hi>Thys Proposition is called commonly in old bookes amongest the barbarous,</hi> Ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Panonis,<note place="margin">This Proposi<g ref="char:EOLhyphen"/>on is commo<g ref="char:cmbAbbrStroke">̄</g>ly called Ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Panonis.</note> 
                  <hi>that is, the Peacockes taile.</hi>
               </p>
               <div type="section">
                  <head>¶ A Corollary.</head>
                  <p>Hereby it is manifest,<note place="margin">
                        <hi>A Corollary.</hi>
                     </note> that the right lines, which being drawen from the poynt
<pb facs="tcp:7062:119"/>
geuen without the circle, and fall within the circle, are equally distant from the least, or from the greatest (which is drawen by the centre) are equall the one to the other: but contrarywyse if they be vnequally distant, whether they light vpon the concaue or conuexe circumference of the circle, they are vnequall.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Theoreme. The <hi>9.</hi> Proposition. If within a circle be taken a poynt, and from that poynt be drawen vnto the circumference moe then two equall right lines, the poynt taken is the centre of the circle.</head>
               <p>SVppose that the circle be ABC, and within it let there be taken the poynt D. And from D let there be drawen vnto the circumference ABC moe then two equall right lines,<note place="margin">Construction.</note> that is, DA, DB, and DC. Then I say that the poynt D is the centre of the circle ABC. Draw (by the first petition) these right lines
<figure/>
AB and BC:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and (by the <hi>10.</hi> of the first) deuide the<g ref="char:cmbAbbrStroke">̄</g> into two equall partes in the poyntes E and F: namely, the line AB in the poynt E, and the line BC in the poynt F. And draw y<hi rend="sup">e</hi> lines ED and FD, and (by the second pe<g ref="char:EOLhyphen"/>tition) extend the lines ED and FD on eche side to the poyntes K, G, and H, L. And for asmuch as the line AE is equall vnto the line EB, and the line ED is common to them both, there<g ref="char:EOLhyphen"/>fore these two sides AE and ED are equall vnto these two sides BE, and ED: and (by supposition) the base DA is equall to the base DB. Wherfore (by the <hi>8.</hi> of the first) the angle AED is equall to the angle BED. Wherfore eyther of these angles AED and BED is a right angle. Wherefore the line GK deuideth y<hi rend="sup">e</hi> line AB into two equall partes and maketh right angles. And for asmuch as, if in a circle a right line deuide an other right line into two equall partes in such sort that it maketh also right angles, in y<hi rend="sup">e</hi> line that deuideth is the centre of the circle (by the Correllary of the first of the third). Therfore (by the same Correllary) in the line GK is the centre of the circle ABC. And (by the same reason) may we proue that in y<hi rend="sup">e</hi> line HL is the centre of the circle ABC, and the right lines GK, and HL haue no other poynt common to them both besides the poynt D. Wherefore the poynt D is the centre of the circle ABC. If therefore within a circle be taken a poynt, and from that point be drawen vnto the circumference more then two equall right lines, the poynt taken is the centre of the circle: which was required to be proued.</p>
               <div type="section">
                  <pb n="89" facs="tcp:7062:119"/>
                  <head>¶ An other demonstration.</head>
                  <p>Let there be taken within the circle ABC the poynt D.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the same leading also to an impossi<g ref="char:EOLhyphen"/>bilitie.</note> And from the poynt D let there be drawen vnto the circumference more then two equall right lines, namely, DA, DB, and DC. Then I say that the poynt D is the centre of the circle. For if not, then if it be possible
<figure/>
let the point E be the centre: and draw a line from D to E, and extend DE to the poyntes F and G. Wherefore the line FG is the diameter of the circle ABC. And for asmuch as in FG the diameter of the circle ABC is taken a poynt, namely D, which is not the centre of that circle, therefore (by the <hi>7.</hi> of the third) the line DG is y<hi rend="sup">e</hi> grea<g ref="char:EOLhyphen"/>test, and the line DC is greater then the line DB, and the line DB is grea<g ref="char:EOLhyphen"/>te<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then the line DA. But the lines DC, DB, DA, are also equall (by supposi<g ref="char:EOLhyphen"/>tion): which is impossible. Wherefore the poynt E is not the centre of the circle ABC. And in like sort may we proue that no other poynt besides D. Wherefore the poynt D is the centre of the circle ABC: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Theoreme. The <hi>10.</hi> Proposition. A circle cutteth not a circle in moe pointes then two.</head>
               <p>FOr if it be possible let the circle ABC cut the circle DEF in mo pointes then two,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> that is, in B, G, H, &amp; F. And drawe lines fro<g ref="char:cmbAbbrStroke">̄</g> B to G, and from B to H. And (by y<hi rend="sup">e</hi> 
                  <hi>10.</hi> of the first) deuide either of the lines BG &amp; BH into two equall partes, in y<hi rend="sup">e</hi> pointes
<figure/>
K and L. And (by the <hi>11.</hi> of the first) from the poynt K raise vp vnto y<hi rend="sup">e</hi> line BH a perpendicular line KC, and likewise from the poynt L raise vp vnto y<hi rend="sup">e</hi> line BG a perpendicular line LM, and extend the line CK to the poynt A, and LNM to the poyntes X and E. And for asmuch as in the circle ABC, the right line AC deuideth the right line BH into two equall partes and maketh right angles, therfore (by the <hi>3.</hi> of the third)
<pb facs="tcp:7062:120"/>
in the line AC is the centre of the circle ABC. Agayne, for asmuch as in the selfe same circle ABC the right line NX, that is, the line ME deuideth the right line BG into two equall partes and maketh right angles, therefore (by the third of the third) in the line NX is the centre of the circle ABC. And it is proued that it is also in the line
<figure/>
AC. And these two right lines AC and NX meete together in no other poynt besides O. Where<g ref="char:EOLhyphen"/>fore the poynt O is the centre of the circle ABC. And in like sort may we proue that the poynt O is the centre of the circle DEF. Wherefore the two circles ABC and DEF deuiding the one the other haue one and the same cen<g ref="char:EOLhyphen"/>tre: which (by the <hi>5.</hi> of the third) is impossible. A circle therfore cutteth not a circle in moe poyntes then two: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration to proue the same.</head>
                  <p>Suppose that the circle ABC do cut the circle DGF in mo poyntes then two,<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the same leading also to an impossi<g ref="char:EOLhyphen"/>bilitie.</note> that is, in B, G, F, and H. And (by the first of the third) take the centre of the circle ABC and let the same be the poynt K. And draw these right lines KB, KG, and KF. Now for asmuch as
<figure/>
within the circle DEF is taken a cer<g ref="char:EOLhyphen"/>taine poynt K, and from that poynt are drawen vnto the circumference moe then two equall right lines, namely, KB, KG, and KF: therefore (by the <hi>9.</hi> of the third) K is the centre of the circle DEF. And the poynt K is the centre of the circle ABC. Wherefore two cir<g ref="char:EOLhyphen"/>cles cutting the one the other haue one and the same centre: which (by the <hi>5.</hi> of the third) is impossible. A circle therfore cutteth not a circle in moe pointes then two: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Theoreme. The <hi>11.</hi> Proposition. If two circles touch the one the other inwardly, their centres
<pb n="90" facs="tcp:7062:120"/>
being geuen: a right line ioyning together their centres and produced, will fall vpon the touch of the circles.</head>
               <p>SVppose that these two circles ABC, and ADE do touch the one the other in the poynt A.<note place="margin">Construction.</note> And (by the first of the third) take the centre of the circle ABC, and let the same be F: and likewise y<hi rend="sup">e</hi> centre of the circle ADE, and let the same be G. Then I say that a right line drawen from F to G and being produced, will fall vpon the poynt A. For if not,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> then if it be possible let it fall as the line FGDH doth. And draw these right lines AF, &amp; AG. Now for asmuch as the lines AG and
<figure/>
GF are (by the <hi>20.</hi> of the first) grea<g ref="char:EOLhyphen"/>ter then the line FA, that is, then the line FH, take away the line GF which is common to them both. Wherefore the residue AG is greater then the residue GH. But the line DG is equall vnto the line GA (by the <hi>15.</hi> definition of the first). Wherefore the line GD is greater then y<hi rend="sup">e</hi> line GH: the lesse then the greater: which is impossible. Wher<g ref="char:EOLhyphen"/>fore a right line drawen from the poynt F to the poynt G and produced, falleth not besides the poynt A, which is y<hi rend="sup">e</hi> point of the touch. Wherefore it fallet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vpon the touch. If therefore two circles touch the one y<hi rend="sup">e</hi> other inwardly, their centres being geuen, a right line ioyning together their centres and produced, will fall vpon the touch of the circles: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
               <div type="section">
                  <head>An other demonstration to proue the same.</head>
                  <p>
                     <hi>But now let it fall as GFC falleth, and extend y<hi rend="sup">e</hi> line GFC to the poynt H:</hi>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the same leading also to an impossi<g ref="char:EOLhyphen"/>bilitie.</hi>
                     </note> 
                     <hi>and drawe these right lines AG and AF. And for asmuch as the lines AG and GF are (by the</hi> 20. <hi>of the first) greater then the line AF. But the line AF is equall vnto the line CF, that is, vnto the line FH. Take away the line FG common to them both. Wherfore the residue AG is greater then y<hi rend="sup">e</hi> residue GH, that is, the line GD is greater then the line GH: the lesse greater then y<hi rend="sup">e</hi> grea<g ref="char:EOLhyphen"/>ter: which is impossible.</hi>
                  </p>
                  <p>Which thing may also be proued by the 7. Proposition of this booke. For for asmuch as the line HC is the diameter of the circle ABC,<note place="margin">
                        <hi>The same <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>gaine demon<g ref="char:EOLhyphen"/>strated by an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rgument lea<g ref="char:EOLhyphen"/>ding to an ab<g ref="char:EOLhyphen"/>surdititie.</hi>
                     </note> &amp; in it is taken a poynt which is not the centre, namely, the poynt G, therefore the line GA is greater then the line GH by the sayd 7. Proposition. But the line GD is equall to the line GA (by the definition of a circle). Wherefore the line GD is greater then the line GH, namely, the part gr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ater then the whole: which is impossible.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:121"/>
               <head>The <hi>11.</hi> Theoreme. The <hi>12.</hi> Proposition. If two circles touch the one the other outwardly, a right line drawen by their centres shall passe by the touch.</head>
               <p>SVppose that these two circles ABC and ADE do touch the one the o<g ref="char:EOLhyphen"/>ther outwardly in the poynt A. And (by the third of the third) take the centre of the circle ABC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> leading to an impossibi<g ref="char:EOLhyphen"/>li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ie.</note> and let the same be the poynt F: and likewise the centre of the circle ADE, and let the same be the poynt G. Then I say that a right line drawen from the poynt F to the poynt G shall passe by the poynt of the touch, namely, by the poynt A. For if not, then if it be possible, let it passe as the right line FCDG doth. And
<figure/>
draw these right lines AF &amp; AG. And for asmuch as the poynt F is the centre of y<hi rend="sup">e</hi> circle ABC, ther<g ref="char:EOLhyphen"/>fore the line FA is equall vnto the line FC. Againe for asmuch as the poynt G is the centre of the circle ADE, therefore the line GA is equall to the line GD. And And it is proued that the line FA is equall to the line FC. Wherefore the lines FA and AG are equall vnto the lines FC and GD: Wherefore the whole line FG is greater then the lines FA and AG. But it is also lesse (by the <hi>20.</hi> of the first): which is impossible. Wherfore a right line drawen from the poynt F to the poynt G shall passe by the poynt of the touch, namely, by the poynt A. If there<g ref="char:EOLhyphen"/>fore two circles touch the one the other outwardly, a right line drawen by their centres, shall passe by the touch: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶ An other demonstration after Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Pelita<g ref="char:EOLhyphen"/>rius leading also to an ab<g ref="char:EOLhyphen"/>surditie.</hi>
                     </note>Suppose that the two circles <hi>ABC</hi> and <hi>DEF</hi> do touch the one the other out<g ref="char:EOLhyphen"/>wardly in the poynt <hi>A</hi>: And let <hi>G</hi> be the centre of the circle <hi>ABC</hi>: From which poynt produce by the touch of the circles the line <hi>GA</hi> to the poynt <hi>F</hi> of the circum<g ref="char:EOLhyphen"/>ference <hi>DEF.</hi> Which for asmuch as it passeth not by the centre of the circle <hi>DEF</hi> (as the aduersary affirmeth) draw
<figure/>
from the same centre G an other right line <hi>GK,</hi> which if it be possi<g ref="char:EOLhyphen"/>ble let passe by the centre of the cir<g ref="char:EOLhyphen"/>cle <hi>DEF,</hi> namely, by the poynt <hi>H</hi>: cutting the circumference <hi>ABC</hi> in the poynt <hi>B,</hi> &amp; the circu<g ref="char:cmbAbbrStroke">̄</g>ference <hi>DEF</hi> in the poynt <hi>D,</hi> &amp; let the op<g ref="char:EOLhyphen"/>posite poynt therof be in the point <hi>K.</hi> And for asmuch as fro<g ref="char:cmbAbbrStroke">̄</g> the poynt <hi>G</hi> taken without the circle <hi>DEF</hi> is drawen the line <hi>GK</hi> passing by the centre <hi>H,</hi> and fro<g ref="char:cmbAbbrStroke">̄</g> the same poynt is<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> drawen also an
<pb n="91" facs="tcp:7062:121"/>
other line not passing by the contre, namely, the line <hi>GF.</hi> Therefore (by the 8. of thys booke) the outward part <hi>GD.</hi> of the line <hi>GK</hi> shall be lesse then the outward part <hi>GA</hi> of the line <hi>GF</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> But the line <hi>GA</hi> is equall to the line <hi>GB.</hi> Wherfore the line <hi>GD</hi> is lesse then the line <hi>GB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> namely, the whole lesse then the part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which is absurde.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Theoreme. The <hi>13.</hi> Proposition. A circle can not touch an other circle in moe poyntes then one, whether they touch within or without.</head>
               <p>FOr if it be possible, let the circle ABCD touch y<hi rend="sup">e</hi> circle EBFD first inwardly in moe poyntes then one, that is, in D and B.<note place="margin">Of circles which touch the one the other inward<g ref="char:EOLhyphen"/>ly.</note> Take (by the first of the third) the centre of the circle ABCD, and let the same be y<hi rend="sup">e</hi> point G: and likewise y<hi rend="sup">e</hi> centre of the circle EBFD, and let the same be y<hi rend="sup">e</hi> poynt H. Wherefore (by the <hi>11.</hi> of the same) a right line drawen from the poynt G to the poynt H and produced, will fall vp<g ref="char:EOLhyphen"/>on the poyntes B and D: let it so fall as the line BGHD doth. And for asmuch as the poynt G is the centre of the circle ABCD, therefore (by the <hi>15.</hi> definiti<g ref="char:EOLhyphen"/>on of the first) the line BG is equall to
<figure/>
the line DG. Wherfore the line BG is greater then then the line HD: Wher<g ref="char:EOLhyphen"/>fore the line BH is much greater then the line HD. Agayne for asmuch as the poynt H is the centre of the circle EBFD, therefore (by the same defini<g ref="char:EOLhyphen"/>tion) the line BH is equall to the line HD: and it is proued that it is m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ch greater then it: which is impossible. A circle therefore can not touch a circle in<g ref="char:EOLhyphen"/>wardly in moe poyntes then one.</p>
               <p>Now I say that neither outwardly also a circle toucheth a circle in moe poyntes then one.<note place="margin">Of circles which touch the one the other out<g ref="char:EOLhyphen"/>wardly.</note> For if it be possible, let the circle ACK touch y<hi rend="sup">e</hi> circle ABCD outwardly in moe poyntes the<g ref="char:cmbAbbrStroke">̄</g> one, that is, in A and C: And (by the first petition) draw a line from the poynt A to the poynt C. Now for asmuch as in y<hi rend="sup">e</hi> circumference of either of the circles ABCD, and ACK, are taken two poyntes at all aduentures, namely, A and C, therefore (by the second of the third) a right line ioyning together those poyntes shall fall within both the circles. But it falleth within the circle ABCD, &amp; without the circle ACK: which is absurde. Wherefore a circle shall not touch a circle out<g ref="char:EOLhyphen"/>wardly in moe pointes then one, and it is proued y<hi rend="sup">t</hi> neither also inwardly. Where<g ref="char:EOLhyphen"/>fore a circle can not touch an other circle in moe poyntes then one, whether they
<pb facs="tcp:7062:122"/>
touch within or without: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶ An other demonstration after Pelitarius and Flussates.</head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Pelita<g ref="char:EOLhyphen"/>rius &amp; Flus<g ref="char:EOLhyphen"/>sates, of circles which tooch the one the other out<g ref="char:EOLhyphen"/>wardly.</hi>
                     </note>Suppose that there be two circles ABG and ADG, which if it be possible, let touch the one the other outwardly in moe poyntes then one, namely, in A and G. Let the centre of the circle ABG be the poynt I, and let the centre of the circle ADG be the poynt K. And draw a right line from the poynt I to the poynt K, which (by the 12. of thys booke) shall passe both by the
<figure/>
poynt A and by the poynt G: which is not possible: for then two right lines should include a superficies, contrary to the last common sentence. It may also be thus demonstrated. Draw a line from the centre I to the centre K, which shall passe by one of the touches, as for example by the poynt A. And draw these right lines GK and GI, and so shall be made a tri<g ref="char:EOLhyphen"/>angle, whose two sides GK and GI shall not be greater then the side IK: which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> contrary to the 20. of the first.</p>
                  <p>But now if it be possible, let the foresayd circle ADG touch the circle ABC inward<g ref="char:EOLhyphen"/>ly in moe poyntes then one,<note place="margin">
                        <hi>Of circles which tooch the one the other in<g ref="char:EOLhyphen"/>wardly.</hi>
                     </note> namely, in the pointes A and G: and let the centre of the circle ABC be the poynt I, as before: and let
<figure/>
the centre of the circle ADG be the poynt K, as also before. And extend a line from the poynt I to the poynt K, which shall fall vpon the touch (by the 11. of thys booke). Draw also these lines KG, and IG. And for asmuch as the line KG is equall to the line KA (by the 15. definition of the first) adde the line KI common to them both. Wherefore the whole line AI is equall to the two lines KG and KI: but vnto the line AI is equall the line IG (by the definition of a circle). Wher<g ref="char:EOLhyphen"/>fore in the triangle IKG the side IG is not lesse then the two sides IK and KG: which is con<g ref="char:EOLhyphen"/>trary to the 20. of the first.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Theoreme. The <hi>14.</hi> Proposition. In a circle, equall right lines, are equally distant from the ce<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tre. And lines equally distant from the centre, are equall the one to the other.</head>
               <p>
                  <note place="margin">The first part of this Theo<g ref="char:EOLhyphen"/>reme.</note>SVppose that there be a circle
<figure/>
ABCD, and let there be in it drawen these equall right lines AB and CD. Then I say that they are equally distant from the centre.<note place="margin">Construction.</note> Take (by the first of the third) the centre of the cir<g ref="char:EOLhyphen"/>cle ABCD, and let the same be y<hi rend="sup">e</hi> poynt E. And (by the <hi>12.</hi> of the first) from the point E draw vnto the lines AB &amp; CD
<pb n="92" facs="tcp:7062:122"/>
perpendicular lines EF and EG. And (by the first petition) draw these right lines AE and CE.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now for asmuch as a certaine right line EF drawen by the centre cutteth a certaine other right line AB not drawen by the centre, in such sort that it maketh right angles, therefore (by the third of the third) it deuideth it into two equall partes. Wherefore the line AF is equall to the line FB. Wher<g ref="char:EOLhyphen"/>fore the line AB is double to the line AF: and by the same reason also the line CD is double to the line CG. But the line AB is equall to the line CD. Wher<g ref="char:EOLhyphen"/>fore the line AF is also equall to the line CG. And for asmuch as (by the <hi>15.</hi> de<g ref="char:EOLhyphen"/>finition of the first) the line AE is equall to the line EC, therefore the square of the line EC is equall to the square of the line AE. But vnto the square of the line AE, are equall (by the <hi>47.</hi> of the first) the squares of the lines AF &amp; FE: for the angle at the poynt F is a right angle. And (by y<hi rend="sup">e</hi> selfe same) to the square of the line EC are equall the squares of the lines EG and GC: for the angle at the poynt G is a right angle. Wherefore the squares of the lines AF and FE are equall to the squares of the lines CG and GE: of which the square of the line AF is equall to the square of the line
<figure/>
CG: for the line AF is equall to the line CG. Wherefore (by the third common sentence) the square remayning, namely, the square of the line FE, is equall to the square remayning, namely, to the square of the line EG. Wherefore the line EF is equall to the line EG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> But right lines are sayd to be equally distant from y<hi rend="sup">e</hi> cen<g ref="char:EOLhyphen"/>tre, when perpendicular lines drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre to those lines, are equall (by the <hi>4.</hi> definition of the third). Wherfore the lines AB and CD are equally distant from the centre.</p>
               <p>But now suppose that the right lines AB and CD be equally distant from the centre,<note place="margin">The second part which is the conuerse of the first.</note> that is, let the perpendicular line EF be equall to the perpendicular line EG. Then I say that the line AB is equall to the line CD. For the same order of construction remayning, we may in like sort proue that the line AB is double to the line AF, and that the line CD is double to the line CG. And for asmuch as the line AE is equall to the line CE, for they are drawen from y<hi rend="sup">e</hi> cen<g ref="char:EOLhyphen"/>tre to the circumference, therfore the square of the line AE is equall to y<hi rend="sup">e</hi> square of the line CE. But (by the <hi>47.</hi> of the first) to the square of the line AE are equall the squares of the lines EF and FA. And (by the selfe same) to y<hi rend="sup">e</hi> square of the line CE are equall the squares of the lines EG and GC. Wherfore the squares of the lines EF and FA are equall to the squares of the lines EG and GC. Of which the square of the line EG is equall to the square of the line EF, for the line EF is equall to the line EG. Wherefore (by the third common sen<g ref="char:EOLhyphen"/>tence) the square remayning, namely, the square of the line AF, is equall to the square of the line CG. Wherefore the line AC is equall vnto the line CG. But
<pb facs="tcp:7062:123"/>
the line AB is double to the line AF, and the line CD is double to the line CG. Wherefore the line AB is equall to the line CD. Wherefore in a circle equall right lines are equally distant from the centre. And lines equally distant from the centre, are equall the one to y<hi rend="sup">e</hi> other: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other demonstration for the first part after Campane.</head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the first part after Campane.</hi>
                     </note>Suppose that there be a circle <hi>ABDC,</hi> whose centre let be the poynt <hi>E.</hi> And draw in it two equall lines <hi>AB</hi> and <hi>CD.</hi> Then I say that they are equally distant from the centre. Draw from the centre vnto the lines <hi>AB</hi>
                     <figure/>
and <hi>CD,</hi> these perpendicular lines <hi>EF</hi> and <hi>EG.</hi> And (by the 2. part of the 3. of this booke) the line <hi>AB</hi> shall be equally deuided in the poynt <hi>F.</hi> and the line <hi>CD</hi> shall be equally deuided in the poynt <hi>G.</hi> And draw these right lines <hi>EA, EB, EC,</hi> and <hi>ED.</hi> And for asmuch as in the triangle <hi>AEB</hi> the two sides <hi>AB</hi> and <hi>AE</hi> are equall to the two sides <hi>CD</hi> and <hi>CE</hi> of the triangle <hi>CED,</hi> &amp; the base <hi>EB</hi> is equall to the base <hi>ED.</hi> therefore (by the 8. of the first) the angle at the point <hi>A</hi> shall be equall to the angle at the point <hi>C.</hi> And for asmuch as in the triangle <hi>AEF</hi> the two sides <hi>AE</hi> and <hi>AF</hi> are equall to the two sides <hi>CE</hi> and <hi>CG</hi> of the triangle <hi>CEG,</hi> and the angle <hi>EAF</hi> is equall to the angle <hi>CEG,</hi> therefore (by the 4. of the first) the base <hi>EF</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall to the base <hi>EG</hi>: which for asmuch as they are perpendicular lines, therefore the lines <hi>AB</hi> &amp; <hi>CD</hi> are equally distant fro<g ref="char:cmbAbbrStroke">̄</g> the centre, by the 4. definition of this booke.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Theoreme. The <hi>15.</hi> Proposition. In a circle, the greatest line is the diameter, and of all other lines that line which is nigher to the centre is alwayes greater then that line which is more distant.</head>
               <p>SVppose that there be a circle
<figure/>
ABCD, and let the diameter thereof be the line AD, and let the centre thereof be the poynt E. And vnto the diameter. AD let the line BC be nigher then the line FG. Then I say that the line AD is the greatest, and the line BC is greater then y<hi rend="sup">e</hi> line FG. Draw (by the <hi>12.</hi> of the first) from the centre E to the lines BC and FG per<g ref="char:EOLhyphen"/>pendicular lines EH and EK.<note place="margin">Construction.</note> And for asmuch as the line BC is nigher vn<g ref="char:EOLhyphen"/>to the centre then the line FG, therfore
<pb n="93" facs="tcp:7062:123"/>
(by the <hi>4.</hi> definition of the third) the line EK is greater then the line EH. And (by the third of the first) put vnto the line EH an equall line EL. And (by the <hi>11.</hi> of the first) from the point L raise vp vnto the line EK a perpen<g ref="char:EOLhyphen"/>dicular line LM: and extend the line LM to the poynt N. And (by the first petition) draw these right lines, EM, EN, EF, and EG. And for asmuch as the line EH is equall to the line EL, therefore (by the <hi>14.</hi> of the third,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and by the <hi>4.</hi> definition of the same) the line BC is equall to the line MN. Againe for asmuch as the line AE is equall to
<figure/>
the line EM, and the line ED to the line EN, therefore the line AD is e<g ref="char:EOLhyphen"/>quall to the lines ME and EN. But the lines ME and EN are (by the <hi>20.</hi> of the first) greater then the line MN. Wherefore the line AD is greater then the line MN. And for asmuch as these two lines ME and EN are equall to these two lines FE and EG (by the <hi>15.</hi> definition of the first) for they are drawen from the centre to the circumfe<g ref="char:EOLhyphen"/>rence, and the angle MEN is greater then the angle FEG, therefore (by the <hi>24.</hi> of the first) the base MN is greater then the base FG. But it is proued that the line MN is equall to the line BC: Wherefore the line BC also is grea<g ref="char:EOLhyphen"/>ter then the line FG. Wherefore the diameter AD is the greatest, and the line BC is greater then the line FG. Wherefore in a circle, the greatest line is the diameter, and of all the other lines, that line which is nigher to y<hi rend="sup">e</hi> centre is alwaies greater then that line which is more distant: which was required to be proued.</p>
               <div type="section">
                  <head>¶An other demonstration after Campane.</head>
                  <p>In the circle <hi>ABCD,</hi> whose centre let be the poynt E, draw these lines, <hi>AB, AC, AD, FG,</hi> and <hi>HK,</hi> of which let the line <hi>AD</hi> be the diameter of the circle.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Cam<g ref="char:EOLhyphen"/>pane.</hi>
                     </note> Then I say that the line <hi>AD</hi> is the greatest of all the lines.
<figure/>
And the other lines eche of the one is so much greater then ech of the other, how much nigher it is vnto the centre. Ioyne together the endes of all these lines with the centre, by drawing these right lines <hi>EB, EC, EG, EK, EH,</hi> and <hi>EF.</hi> And (by the 20. of the first) the two sides <hi>EF</hi> and <hi>EG</hi> of the triangle <hi>EFG,</hi> shall be greater then the third side <hi>FG.</hi> And for asmuch as the sayd sides <hi>EF</hi> &amp; <hi>EG</hi> are equall to the line <hi>AD</hi> (by the definition of a circle) therefore the line <hi>AD</hi> is greater then the line <hi>FG.</hi> And by the same reason it is greater then euery one of the rest of the lines, if they be put to be bases of tri<g ref="char:EOLhyphen"/>angles: for that euery two sides drawen fro<g ref="char:cmbAbbrStroke">̄</g> the
<pb facs="tcp:7062:124"/>
centre are equall to the line <hi>AD.</hi> Which is
<figure/>
the first part of the Proposition. Agayne, for as<g ref="char:EOLhyphen"/>much as the two sides <hi>EF</hi> and <hi>EG</hi> of the tri<g ref="char:EOLhyphen"/>angle <hi>EFG,</hi> are equall to the two sides <hi>EH</hi> and <hi>EK</hi> of the triangle <hi>EHK,</hi> and the angle <hi>FEG</hi> is greater then the angle <hi>HEK,</hi> therfore (by the 24. of the first) the base <hi>FG</hi> is greater then the base <hi>HK.</hi> And by the same reason may it be proued, that the line <hi>AC</hi> is greater then the line <hi>AB.</hi> And so is manifest the whole Pro<g ref="char:EOLhyphen"/>position.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>15.</hi> Theoreme. The <hi>16.</hi> Proposition. If from the end of the diameter of a circle be drawen a right line making right angles: it shall fall without the circle: and betwene that right line and the circumference can not be drawen an other right line: and the angle of the semicircle is greater then any acute angle made of right lines, but the o<g ref="char:EOLhyphen"/>ther angle is lesse then any acute angle made of right lines.</head>
               <p>SVppose that there be a circle ABC: whose centre let be the point D, and let the diameter therof be AB. Then I say y<hi rend="sup">t</hi> a right line drawen from the poynt A, making with the diameter AB right angles, shall fall without the circle.<note place="margin">The first part of this Theo<g ref="char:EOLhyphen"/>reme.</note> For if it do not, then if it be possible, let it fall within the circle as the line AC doth,
<figure/>
and draw a line from the point D to the point C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurdi<g ref="char:EOLhyphen"/>tie.</note> And for asmuch as (by the <hi>15.</hi> definition of the first) the line DA is equall to the line DC, for they are drawen from the centre to the circum<g ref="char:EOLhyphen"/>ference, therefore the angle DAC is equall to the angle ACD. But the an<g ref="char:EOLhyphen"/>gle DAC is (by supposition) a right angle: Wherfore also the angle ACD is a right angle. Wherefore the angles DAC and ACD, are equall to two right angles: which (by the <hi>17.</hi> of the first) is impossible. Wherefore a right line drawen from the poynt A, making with the diameter AB right angles, shall not fall within y<hi rend="sup">e</hi> circle. In like sort also may we proue, that it falleth not in
<pb n="94" facs="tcp:7062:124"/>
the circumference. Wherefore it falleth without, as the line AE doth.</p>
               <p>I say also,<note place="margin">Second part.</note> that betwene the right line AE, and the circumference ACB, can not be drawen an other right line. For if it be possible, let the line AF so be drawen. And (by the <hi>12.</hi> of the first) from the poynt D draw vnto the line FA a perpendicular line DG: And for asmuch as AGD is a right angle, but DAG is lesse then a right angle, therefore (by the <hi>19.</hi> of the first) the side AD is greater then the side DG. But the line DA is equall to the line DH, for they are drawen from the centre to the circumference. Wherefore the line DH is greater then the line DG: namely, the lesse greater then the greater: which is impossible. Wherefore betwene the right line AE and the circumference ACB, can not be drawen an other right line.</p>
               <p>I say moreouer,<note place="margin">Third part.</note> that the angle of
<figure/>
the semicircle contayned vnder y<hi rend="sup">e</hi> right line AB and the circu<g ref="char:cmbAbbrStroke">̄</g>ference CHA, is greater then any acute angle made of right lines. And the angle remayning co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder y<hi rend="sup">e</hi> circumference CHA and the right line AE, is lesse then any acute angle made of right lines. For if there be any angle made of right lines greater then that angle which is con<g ref="char:EOLhyphen"/>tayned vnder the right line BA and the circumference CHA, or lesse then that which is contayned vnder the cir<g ref="char:EOLhyphen"/>cumference CHA and the right line AE, then betwene the circumference CHA and the right line AE, there shall fall a right line, which maketh the angle contayned vnder the right lines, grea<g ref="char:EOLhyphen"/>ter then that angle which is contayned vnder the right line BA and the cir<g ref="char:EOLhyphen"/>cumference CHA, and lesse then the angle which is contayned vnder the cir<g ref="char:EOLhyphen"/>cumference CHA and the right line AE. But there can fall no such line, as it hath before bene proued. Wherfore no acute angle contained vnder right lines, is greater then the angle contayned vnder the right line BA and the circumfe<g ref="char:EOLhyphen"/>rence CHA, nor also lesse then the angle contayned vnder the circumference CHA and the line AE.</p>
               <div type="section">
                  <head>Correlary.</head>
                  <p>Hereby it is manifest that a perpendicular line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the end of the diameter of a circle toucheth the circle: and that a right line toucheth a circle in one poynt onely. For it was proued (by the <hi>2.</hi> of the third) that a right line drawen from two pointes taken in
<pb facs="tcp:7062:125"/>
the circumference of a circle, shall fall within the circle. Which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Probleme. The <hi>17.</hi> Proposition. From a poynt geuen, to draw a right line which shall touch a circle geuen.</head>
               <p>
                  <note place="margin">Construction.</note>SVppose that the poynt geuen be A, and let the circle geuen be BCD. It is required from the poynt A to draw a right line which shall touch the circle BCD. Take (by the first of the third) the centre of the circle, and let the same be E. And (by the first petition) draw the right line ADE. And making the centre E, and the space AE, describe (by y<hi rend="sup">e</hi> third petition) a circle AFG. And from the poynt D raise vp (by the <hi>11.</hi> of the first) vnto the line EA a perpendicular line DF. And (by the first petition) drawe these lines EBF and AB. Then I
<figure/>
say, that from the point A is drawen to the circle BCD a touch line AB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For for asmuch as the point E is the centre of the circle BCD, and also of the circle AFG, therfore the line EA is equall to the line EF, and the line ED to y<hi rend="sup">e</hi> line EB, for they are drawen from the centre to the circumference. Wherefore to these two lines AE and EB, are equall these two lines EF &amp; ED, and the angle at the poynt E is common to them both: Wherefore (by the <hi>4.</hi> of the first) the base DF is equall to the base AB, and y<hi rend="sup">e</hi> triangle DEF is equall to the triangle EBA, and the rest of the angles remayning to the rest of the angles remayning. Wherefore the angle EDF is equall to y<hi rend="sup">e</hi> angle EBA. But the angle EDF is a right angle: Wherfore also the angle EBA is a right angle, and the line EB is drawen from the centre. But a perpendicular line drawen from the end of the diameter of a circle, toucheth the circle (by y<hi rend="sup">e</hi> Corella<g ref="char:EOLhyphen"/>ry of the <hi>16.</hi> of the third). Wherefore the line AB toucheth the circle BCD. Wherfore from the point geuen, namely, A, is drawen vnto y<hi rend="sup">e</hi> circle geue<g ref="char:cmbAbbrStroke">̄</g> BCD, a touch line AB: which was required to be done.</p>
               <div type="section">
                  <head>¶ An addition of Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>An addition of Pelitarius.</hi>
                     </note>Vnto a right lyne which cutteth a circle, to drawe a parallel line which shall touch the circle.</p>
                  <p>
                     <pb n="95" facs="tcp:7062:125"/>S<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ppose that the right lyne AB do out the circle ABC in the poyntes A and B. It is required to drawe vnto the line AB a parallel lyne
<figure/>
which shall touche the circle. Let the centre of the circle be the point D. And deuide the lyne AB into two equall partes in the point E. And by the point E and by the centre D, draw the diameter CDEF. And from the point F (which is the ende of the dia<g ref="char:EOLhyphen"/>meter) rayse vp (by the 11. of the first) vnto the dia<g ref="char:EOLhyphen"/>meter CF a perpendicular line GFH. Then I say that the lyne GFH (which by the correllary of the 16. of this booke toucheth the circle) is a parallel vnto the line AB.<note place="margin">
                        <hi>This Pro<g ref="char:EOLhyphen"/>bleme commo<g ref="char:EOLhyphen"/>dious for the inscribing and circumscri<g ref="char:EOLhyphen"/>bing of figures in or abou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> circles.</hi>
                     </note> For forasmuch as the right line CF fallyng vpon either of these lines AB &amp; GH ma<g ref="char:EOLhyphen"/>keth all the angles at the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> right angles (by the 3. of this boke) and the two angles at the point Fare supposed to be right angles: therfore (by the 29. of the first) the lines AB and GH are parallels: which was required to be done. And this Probleme is very commodious for the inscribing or circumscribing of figures in or about circles.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>16.</hi> Theoreme. The <hi>18.</hi> Proposition. If a right lyne touch a circle, and from the centre to the touch be drawen a right line, that right line so drawen shalbe a per<g ref="char:EOLhyphen"/>pendicular lyne to the touche lyne.</head>
               <p>SVppose that the right line DE do touch the circle ABC in the point C. And take the centre of the circle ABC, and let the same be F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossi<g ref="char:EOLhyphen"/>bilitie.</note> And (by the first petition) from the poynt F to the poynt C drawe a right line FC. Then I say, that CF is a perpendicular line to DE. For if not, draw (by the <hi>12.</hi> of the first) from the
<figure/>
poynt F to the line DE a perpendicular line FG. And for asmuch as the angle FGC is a right angle, therefore the angle GCF is an acute angle: Wherefore the angle FGC is greater then y<hi rend="sup">e</hi> angle FCG, but vnto the greater angle is subtended the greater side (by the <hi>19.</hi> of the first). Wherefore the line FC is greater then the line FG. But the line FC is equall to the line FB, for they are drawen from the centre to the circumference: Wherfore the line FB also is greater then the line FG, namely, the lesse then the greater: which is impossible. Wherefore the line FG is not a perpendicular line vnto the line DE. And in like sort may we proue, that no other line is a perpendicular line vnto y<hi rend="sup">e</hi> line DE besides the line FC: Wher<g ref="char:EOLhyphen"/>fore the line FC is a perpendicular line to DE. If therefore a right line touch
<pb facs="tcp:7062:126"/>
a circle, &amp; from y<hi rend="sup">•</hi> centre to y<hi rend="sup">e</hi> touch be drawen a right line, y<hi rend="sup">t</hi> right line so drawen shall be a perpendicular line to y<hi rend="sup">e</hi> touch line: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other demonstration after Orontius.</head>
                  <p>Suppose that the circle geuen be ABC, which let the right lyne DE touch in the point C.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>onstration after Oron<g ref="char:EOLhyphen"/>tius.</hi>
                     </note> And let the centre of the circle be the point F. And draw a right line from F to C. Then
<figure/>
I say that the line FC is perpendicular vnto the line DE. For if the line FC be not a perpe<g ref="char:cmbAbbrStroke">̄</g>diculer vnto the line DE, then, by the conuerse of the x. definition of the first boke, the angles DCF &amp; FCE shal be vnequall: &amp; therfore the one is grea<g ref="char:EOLhyphen"/>ter then a right angle, and the other is lesse then a right angle. (For the angles DCF and FCE are by the 13. of the first equall to two right an<g ref="char:EOLhyphen"/>gles) Let the angle FCE, if it be possible, be grea<g ref="char:EOLhyphen"/>ter then a right angle, that is, let it be an obtuse angle. Wherfore the angle DCF <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hal be an acute angle. And forasmuch as by suppositio<g ref="char:cmbAbbrStroke">̄</g> the right line DE touche<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h the circle ABC, therefore it cutteth not the circle. Wherefore the circumfe<g ref="char:EOLhyphen"/>rence BC falleth betwene the right lines DC &amp; CF: &amp; therfore the acute and rectiline angle DCF shall be greater then the angle of the semicircle BCF which is contayned vnder the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce BC &amp; the right line CF. And so shall there be geue<g ref="char:cmbAbbrStroke">̄</g> a rectiline &amp; acute angle greater then the angle of a semicircle: which is contrary to the 16. propo<g ref="char:EOLhyphen"/>sition of this booke. Wherfore the angle DCF is not lesse then a right angle. In like sort also may we proue that it is not greater then a right angle. Wherfore it is a right angle, and therfore also the angle FCE is a right angle. Wherefore the right line FC is a per<g ref="char:EOLhyphen"/>pendicular vnto the right line DE by the 10. definition of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>17.</hi> Theoreme. The <hi>19.</hi> Proposition. If a right lyne doo touche a circle, and from the point of the touch be raysed vp vnto the touch lyne a perpendicular lyne, in that lyne so raysed vp is the centre of the circle.</head>
               <p>SVppose that the right line DE do
<figure/>
touch the circle ABC in the point C. And from C raise vp (by y<hi rend="sup">e</hi> 
                  <hi>11.</hi> of the first) vnto the line DE a perpendicu<g ref="char:EOLhyphen"/>lar line CA. Then I say, that in the line CA is the centre of the circle. For if not, then if it be possible, let y<hi rend="sup">e</hi> centre be with<g ref="char:EOLhyphen"/>out the line CA, as in y<hi rend="sup">e</hi> poynt F. And (by the first petition) draw a right line from C to F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> And for asmuch as a certaine right line DE toucheth the circle ABC, and from the centre to the touch is drawen a right line CF, therefore (by the <hi>18.</hi> of the third) FC is a perpendicular line to
<pb n="96" facs="tcp:7062:126"/>
DE. Wherefore the angle FCE is a right angle. But the angle ACE is also a right angle: Wherefore the angle FCE is equall to the angle ACE, namely, the lesse vnto the greater: which is impossible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore the poynt F is not the centre of the circle ABC. And in like sort may we proue, y<hi rend="sup">t</hi> it is no other where but in the line AC. If therefore a right line do touch a circle, and from the point of the touch be raised vp vnto the touch line a perpendicular line, in that line so raised vp is the centre of the circle: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>18.</hi> Theoreme. The <hi>20.</hi> Proposition. In a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles haue to their base one and the same circumference.</head>
               <p>SVppose that there be a circle ABC, and at the centre thereof,<note place="margin">Two cases in thys Pro<g ref="char:EOLhyphen"/>position the one when the angle set at the circumfe<g ref="char:EOLhyphen"/>rence inclu<g ref="char:EOLhyphen"/>deth the cen<g ref="char:EOLhyphen"/>ter.</note> namely, the poynt E, let y<hi rend="sup">e</hi> angle BEC be set, &amp; at the circumference let there be set the angle BAC, and let them both haue one and the same base, namely, the circumference BC. Then I say, that the angle BEC is double to the angle BAC. Draw y<hi rend="sup">e</hi> right
<figure/>
line AE, and (by the second petition) ex<g ref="char:EOLhyphen"/>tend it to the poynt F. Now for asmuch as the line AE is equall to the line EB,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> for they are drawen from the centre vnto the circumference, the angle EAB is e<g ref="char:EOLhyphen"/>quall to the angle EBA (by the <hi>5.</hi> of the first). Wherefore the angles EAB and EBA are double to the angle EAB. But (by the <hi>32.</hi> of the same) the angle BEF is equall to the angles EAB and EBA: Wherefore the angle BEF is double to the angle EAB. And by the same reason the angle FEC is double to the angle EAC. Wherefore the whole angle BEC is double to the whole angle BAC.</p>
               <p>Againe, suppose that there be set an other angle at the circumference,<note place="margin">The other whe<g ref="char:cmbAbbrStroke">̄</g> the same angle set at the circumfe<g ref="char:EOLhyphen"/>rence inclu<g ref="char:EOLhyphen"/>deth not the center.</note> and let the same be BDC. And (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst petition) draw a line from D to E. And (by the second petition) extend the line DE vnto the poynt G. And in like sort may we proue, that the angle GEC is double to the angle EDC. Of which the an<g ref="char:EOLhyphen"/>gle GEB is double to the angle EDB. Wherfore the angle remayning BEC is double to the angle remayning BDC. Wherfore in a circle an angle set at the centre, is double to an angle set at the circumference, so that both the angles haue to their base one and the same circumference: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:127"/>
               <head>The <hi>19.</hi> Theoreme. The <hi>21.</hi> Proposition. In a circle the angles which consist in one and the selfe same section or segment, are equall the one to the other.</head>
               <p>
                  <hi>SVppose y<hi rend="sup">t</hi> there be a circle ABCD, &amp; in the segment therof BAED, let there consist these angles BAD and BED. Then I say, that the angles BAD and BED are equall the one to the other. Take (by the first of the third) the centre of the circle
<figure/>
ABCD,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>and let the same be the point F. And (by the first petition) draw these lines BF and FD.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>Now for asmuch as the angle BFD is set at the centre, and the angle BAD at the circumference, and they haue both one and y<hi rend="sup">e</hi> same base, name<g ref="char:EOLhyphen"/>ly, the circumference BCD, therefore the angle BFD is (by the Proposition going before) double to the angle BAD: and by the same reason the angle BFD is al<g ref="char:EOLhyphen"/>so double to the angle BED. Wherefore (by the</hi> 7. <hi>common sentence) the angle BAD is equall to the angle BED. Wherefore in a circle the angles which consiste in one and the selfe same segment, are equall the one to the other: which was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <p>
                        <hi>Three cases in this Proposi<g ref="char:EOLhyphen"/>tion.</hi>
                     </p>
                     <p>
                        <hi>The first case.</hi>
                     </p>
                     <p>
                        <hi>The second case.</hi>
                     </p>
                  </note>In this proposition are three cases. For the angles consisting in one and the self same segment, the segment may either be greater the<g ref="char:cmbAbbrStroke">̄</g> a semicircle, or lesse then a semicircle, or els iust a semicircle. For the first case the demonstration before put serueth.</p>
               <p>But now suppose that the angles <hi>BAD</hi> and <hi>BED</hi> do consist in the sectio<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>BAD,</hi> which let be lesse then a semicircle. Euen in this case al<g ref="char:EOLhyphen"/>so
<figure/>
I say that the angles <hi>BAD</hi> and <hi>BED</hi> are e<g ref="char:EOLhyphen"/>quall. For draw a right line from <hi>A</hi> to <hi>E.</hi> And let the lines <hi>AD</hi> and <hi>BE</hi> cutte the one the other in the poynt <hi>G,</hi> wherefore the segment <hi>ACE</hi> is greater then a semicircle. And therfore by the first part of this proposition the angles whiche are in it, namely, the angles <hi>ABE</hi> and <hi>EDA</hi> are equall the one to the other. And forasmuch as in the triangle <hi>ABG</hi> the inward and oppo<g ref="char:EOLhyphen"/>site angles <hi>ABG</hi> and <hi>GAB</hi> are equall to the outwarde angle <hi>BGD,</hi> and by the same reason the two angles <hi>EDG</hi> and <hi>GED</hi> of the triangle <hi>DEG</hi> are equall to the selfe same outward angle <hi>BGD.</hi> Wherfore the two angles <hi>ABG</hi> and <hi>GAB</hi> are equall to the two angles <hi>EDG</hi> and <hi>GED</hi>
                  <pb n="97" facs="tcp:7062:127"/>
by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> first commo<g ref="char:cmbAbbrStroke">̄</g> senten<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. From which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>f there
<figure/>
be taken equall angles, namely, <hi>ABG,</hi> and <hi>EDG,</hi> the angl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> remainyng <hi>BAG</hi> shall be equall to the angle remayning <hi>DEG,</hi> that is, the angle <hi>BAD</hi> to the angle <hi>DEB</hi> (by the third common sen<g ref="char:EOLhyphen"/>tence) which was required to be proued.</p>
               <p>The selfe same construction and demonstrati<g ref="char:EOLhyphen"/>on will also serue,<note place="margin">
                     <hi>The third case.</hi>
                  </note> if the angles were set in a semi<g ref="char:EOLhyphen"/>circle as it is pla<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne to see, in the figure here set.</p>
            </div>
            <div type="part">
               <head>The <hi>20.</hi> Theoreme. The <hi>22.</hi> Proposition. If within a circle be described a figure of fower sides, the an<g ref="char:EOLhyphen"/>gles therof which are opposite the one to the other, are equall to two right angles.</head>
               <p>SVppose that there be a circle ABCD, and let there be described in it a figure of fower sides, namely, ABCD. Then I say, that the angles thereof which are opposite the one to the other, are equall to two right angles. Draw (by the first petition) these right lines AC and BD.<note place="margin">Construction.</note> Now for asmuch as (by the <hi>32.</hi> of the first) the three angles of euery triangle are equall to two right angles: therfore y<hi rend="sup">•</hi> three angles of the triangle ABC, namely,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> CAB, ABC, and BCA, are equall to two right angles. But (by the <hi>21.</hi> of the third) the angle CAB is equall to the
<figure/>
angle BDC, for they consist in one and the self same segme<g ref="char:cmbAbbrStroke">̄</g>t, namely, BADC. And (by the same Proposition) the an<g ref="char:EOLhyphen"/>gle ACB is equall to the angle ADB, for they consist in one and the same seg<g ref="char:EOLhyphen"/>ment ADCB. Wherefore the whol<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> angle ADC is equall to y<hi rend="sup">e</hi> angles BAC and ACB: put the angle ABC com<g ref="char:EOLhyphen"/>mon to them both. Wherefore the angles ABC, BAC, and ACB, are equall to the angles ABC and ADC. But the angles ABC, BAC, and ACB, are equall to two right angles. Wherefore the angles ABC and ADC are equall to two right angles. And in like sort also may we proue, that the angles BAD and DCB are equall to two right
<pb facs="tcp:7062:128"/>
angles. If therefore within a circle be described a figure of fower sides, the an<g ref="char:EOLhyphen"/>gles thereof which are opposite the one to y<hi rend="sup">e</hi> other, are equall to two right angles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>21.</hi> Theoreme. The <hi>23.</hi> Proposition. Vpon one and the selfe same right line can not be described two like and vnequall segmentes of circles, falling both on one and the selfe same side of the line.</head>
               <p>
                  <note place="margin">
                     <hi>Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</hi>
                  </note>
                  <hi>FOr if it be possible, let there be described vpon the right line AB two like &amp; vnequall sections of circles, namely, ACB &amp; ADB, falling both on one and the selfe same side of the line AB. And (by the first petition) drawe the right line ACD, and (by the third petition) drawe right lines from C to B, and from D to B. And for asmuch as the segment ACB is like to the segment ADB: and like
<figure/>
segme<g ref="char:cmbAbbrStroke">̄</g>tes of circles are they which haue equall angles (by the</hi> 10. <hi>definition of the third). Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore the angle ACB is equall to the angle ADB, namely, the outward angle of y<hi rend="sup">e</hi> triangle CDB to the inward angle: which (by the</hi> 16. <hi>of the first) is impossible. Wherfore vp<g ref="char:EOLhyphen"/>on one and the self same right line can not be described two like &amp; vnequall seg<g ref="char:EOLhyphen"/>mentes of circles, falling both on one &amp; the selfe same side of the line: which was required to be demonstrated.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>An ad<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ition of Campane d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>mo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>strated by Pelitari<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s.</hi>
                  </note>Here Campane addeth that vpon one and the selfe same right lyne cannot be described two like and vnequall sections neither on one and the selfe same side of the lyne, nor on the opposite side. That they can not be described on one and the selfe same side, hath bene before demonstrated, and that neither also on the oppo<g ref="char:EOLhyphen"/>site side, Pelitarius thus demonstrateth.</p>
               <p>Let the section ABC be set vppon the lyne AC, and vpon the other side let be set the section ADC vppon the selfe same lyne AC,
<figure/>
and let the section ADC be lyke vnto the secti<g ref="char:EOLhyphen"/>on ABC. Then I say that the sections ABC and ADC being thus set are not vnequal. For if it be pos<g ref="char:EOLhyphen"/>sible let the section ADC be the greater. And de<g ref="char:EOLhyphen"/>uide the line AC into two equal partes in the point E. And draw the right lyne BED deuiding the lyne AC right angled wise. And draw these right lynes AB, CB, AD and CD.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLunhyphen"/>tion leading to an impossi<g ref="char:EOLhyphen"/>bilitie.</hi>
                  </note> And forasmuch as the section ADC is greater then the section ABC, the perpen<g ref="char:EOLunhyphen"/>dicular lyne also ED shall be greater then the per<g ref="char:EOLhyphen"/>pendicular lyne EB: as is before declared in the ende of the definitions of this third booke. Wher<g ref="char:EOLhyphen"/>fore
<pb n="98" facs="tcp:7062:128"/>
from the lyne ED <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut of a ly<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e equall to the lyne EB: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hich <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> be EF. And draw these right lynes AF and CF. Now then (by the 4. of the first) the triangle AEB shall be equall to the triangle AEF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the angle EBA shall be equall to the angle EFA. And by the same reason the angle EBC shall be equall to the angle EFC. Wherefore the whole angle ABC is equall to the whole angle AFC. But by the 21. of the first, the an<g ref="char:EOLhyphen"/>gle AFC is greater then the angle ADC. Wherfore also the angle ABC is greater then the angle ADC. Wherefore by the definition the sections ABC and ADC are not lyke, which is contrary to the supposition. Wherefore they are not lyke and vnequall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>22.</hi> Theorme. The <hi>24.</hi> Proposition. Like segmentes of circles described vppon equall right lines, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that vpon these equall right lines AB and CD be described these like segmentes of circles, namely, AEB and CFD. Then I say, that the segment AEB is equall to the segment CFD. For putting the segment AEB vpon the segment CFD, and the poynt A vpon y<hi rend="sup">e</hi> poynt C, and the right line AB vpon the right line CD, the poynt B also shall fall vpon the poynt D, for y<hi rend="sup">e</hi> line AB is equall to the line CD. And the right line AB ex<g ref="char:EOLhyphen"/>actly agreing with
<figure/>
the right line CD, y<hi rend="sup">•</hi> segment also AEB shall exactly agree with the segment CFD.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</hi>
                  </note> 
                  <hi>For if the right line AB do exactly agree with the right line CD, and the segment AEB do not exactly agree with the segment CFD, but dif<g ref="char:EOLhyphen"/>fereth as the segment CGD doth: Now (by the</hi> 13. <hi>of the third) a circle cutteth not a circle in more pointes then two, but the circle CGD cutteth y<hi rend="sup">e</hi> circle CFD in more pointes the<g ref="char:cmbAbbrStroke">̄</g> two, that is, in the points C, G, and D: which is (by the same) impossible. Wherefore the right line AB exactly agreing with the right line CD, the segment AEB shall not but exactly agree with the segment CFD: Wherefore it exactly agreeth with it, and is equall vnto it. Wherefore like seg<g ref="char:EOLhyphen"/>mentes of circles described vpon equall right lines, are equall the one to the other: which was required to be proued.</hi>
               </p>
               <p>This Proposition may also be demonstrated by the former proposition For if the se<g ref="char:EOLhyphen"/>ctions <hi>AEB</hi> and <hi>CFD</hi> being like and set vpon equall right lines <hi>AB</hi> and <hi>CD,</hi>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>monstration.</hi>
                  </note> should be vnequall, then the one beyng put vpon the other, the great<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r shall exceede the lesse: but the line <hi>AB</hi> is one line with the line <hi>CD</hi>: so that therby shal follow the contrary of the former Proposition.</p>
               <p>
                  <pb facs="tcp:7062:129"/>Suppose that there be two right lines <hi>AB</hi> &amp; <hi>CD</hi> which let be equall: and vpon the<g ref="char:cmbAbbrStroke">̄</g> let there be set these like sections <hi>ABK,</hi> and <hi>CDE.</hi>
                  <note place="margin">
                     <hi>An other de<g ref="char:EOLhyphen"/>mo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>stratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>ter Pelitarius.</hi>
                  </note> Then I say that the said sections are equall. For if not, then let <hi>CED</hi> be the greater section. And deuide the two lines <hi>AB</hi> and <hi>CD</hi> into two equall partes, the line <hi>AB</hi> in the pointe <hi>F,</hi> and the line <hi>CD</hi> in the
<figure/>
point <hi>G.</hi> And erect two perpendicular lines <hi>FK</hi> and <hi>GE.</hi> And draw these right lines <hi>AK</hi> &amp; <hi>KB: EC, ED.</hi> And forasmuch as the section <hi>CED</hi> is the greater, therefore the perpendicular line <hi>GE</hi> is greater then the perpendicular <hi>FK</hi>: From the lyne <hi>GE</hi> cut of a line equall to the line <hi>FK,</hi> which let be <hi>GH</hi>: and draw these right lines <hi>CH</hi> and <hi>HD.</hi> And forasmuch as in the triangle <hi>AKF</hi> the two sides <hi>AF</hi> and <hi>FK</hi> are equall to the two sides <hi>CG</hi> &amp; <hi>GH</hi> of the triangle <hi>CH<g ref="char:EOLhyphen"/>G,</hi> and the angles at the pointes <hi>F</hi> and <hi>G</hi> are equal (for that they are right angles) ther<g ref="char:EOLhyphen"/>fore (by the 4. of the first) the base <hi>A</hi> K is equall to the base <hi>CH,</hi> and the angle <hi>AKF</hi> to the angle <hi>CHG.</hi> And by the same reason the angle <hi>BKF</hi> is equall to the angle <hi>DHG.</hi> Wherfore the whole angle <hi>AKB</hi> is equall to the whole angle <hi>CHD.</hi> But the angle <hi>CHD</hi> is greater then the angle <hi>CED</hi> by the 21. of the first. Wherfore also the angle <hi>AKD</hi> is greater then the angle <hi>CED.</hi> Wherfore the sections are not lyke, which is contrary to the supposition.</p>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Probleme. The <hi>25.</hi> Proposition. A segment of a circle beyng geuen to describe the whole cir<g ref="char:EOLhyphen"/>cle of the same segment.</head>
               <p>SVppose that y<hi rend="sup">e</hi> segment geuen be
<figure/>
ABC. It is required to describe the whole circle of the same seg<g ref="char:EOLhyphen"/>ment ABC.<note place="margin">Construction.</note> Deuide (by the <hi>10.</hi> of the first) the line AC into two equall partes in the poynt D. And (by the <hi>11.</hi> of the same) from the poynt D raise vp vnto the line AC a perpendicular line BD. And (by the first petition) draw a right line from A to B. Now then the angle ABD being compared to y<hi rend="sup">e</hi> angle BAD, is either greater then it, or equall vnto it, or lesse then it.</p>
               <p>
                  <note place="margin">
                     <p>Three cases in this Proposi<g ref="char:EOLhyphen"/>tion.</p>
                     <p>The first case.</p>
                  </note>First let it be greater. And (by the <hi>23.</hi> of the same) vpon the right line BA, and vnto the poynt in it A, make vnto the angle ABD an equall angle BAE. And (by the second petition) extend the line BD vnto the poynt E. And (by the first petition) draw a line from E to C. Now for asmuch as the angle ABE is e<g ref="char:EOLhyphen"/>quall to the angle BAE,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> therefore (by the <hi>6.</hi> of the first) the right line EB is equall to the right line AE. And for asmuch as the line AD is equall to y<hi rend="sup">e</hi> line DC, and the line DE is common to them both: therefore these two lines AD
<pb n="99" facs="tcp:7062:129"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd DE, are equall to these t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o lines CD and DE the one to the other. And the angle ADE <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> (by the <hi>4.</hi> petition) equall to the angle CDE, for either of them is a right angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (by the <hi>4.</hi> of the first) the base AE is equall to the base CE. But it is pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ed, that the line AE is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall to y<hi rend="sup">e</hi> line BE. Wher<g ref="char:EOLhyphen"/>fore the line BE also is equall to the line CE. Wherefore these three lines AE, EB, and EC, are equall the one to the other. Wherefore making the centre E, and the space either AE, or EB, or EC, describe (by the third petition) a cir<g ref="char:EOLhyphen"/>cle, and it shall passe by the poyntes A, B, C. Wherefore there is described the whole circle of the segment geuen, And it is manifest, that the segment ABC is les<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e then a semicircle, for the centre E falleth witho<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t it.</p>
               <p>The like demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>stration also will
<figure/>
serue if the angle ABD be equall to the angle BAD.<note place="margin">The second case.</note> For the line AD being equall to either of these lines, BD, and DC, there are three lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> DA, DB, and DC, equall the one to the other. So that the point D shall be the centre of the circle being complete, and ABC shall be a semicircle.</p>
               <p>But if the angle ABD be lesse then the angle
<figure/>
BAD,<note place="margin">The third case.</note> then (by the <hi>23.</hi> of the first) vpon the right line BA, and vnto the point in it A, make vnto the angle ABD an equall angle within y<hi rend="sup">•</hi> segment ABC. And so the ce<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tre of the circle shall fall in y<hi rend="sup">e</hi> line DB, and it shall be the point E: and the segment ABC shall be greater then a semicircle, Wherefore a segment being geuen, there is described the whole circle of the same segment: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hich was required to be done.</p>
               <div type="section">
                  <head>¶ A Corollary. Hereby it is manifest, that in a semicircle the angle BAD is equall to the angle DBA: but in a section lesse then a semicir<g ref="char:EOLhyphen"/>cle, it is lesse: in a section greater then a semicircle, it is greater.</head>
                  <p>There is also an other generall w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y to finde out the
<figure/>
foresaid centre,<note place="margin">
                        <hi>An other more ready way.</hi>
                     </note> which will serue indifferently for any section whatsoeuer: And that is thus. Take in the cir<g ref="char:EOLhyphen"/>cumference geuen or section <hi>ABC,</hi> three pointes at all auentures which let be <hi>A, B, C.</hi> A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d draw these lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s <hi>AB</hi> and <hi>BC</hi> (by the first peticion) And (by the 10. of the first) deuide into two equall partes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ith<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r of the sayd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lines. the line <hi>AB</hi> in the point <hi>D,</hi> &amp; <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> line <hi>BC</hi> in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> point <hi>E.</hi> And (by the 11. of the first) from th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> point<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s
<pb facs="tcp:7062:130"/>
                     <hi>D</hi> and <hi>E</hi> rayse vp vnto the lines <hi>AB</hi> and <hi>BC</hi> perpendicular lynes <hi>DF</hi> and <hi>EF.</hi> Now forasmuch as either of these angles <hi>BDF,</hi>
                     <note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> and <hi>BEF</hi> is a right angle, a right line produ<g ref="char:EOLhyphen"/>ced from the point <hi>D</hi> to the point <hi>E,</hi> shall deuide either of the said angles: and foras<g ref="char:EOLhyphen"/>much as it falleth vppon the right lines <hi>DF</hi> and <hi>EF,</hi> it shall make the inward angles on one and the selfe same side, namely, the angles <hi>DEF</hi>
                     <figure/>
and <hi>EDF</hi> lesse then two right angles. Wherefore (by the fift peticion) the lines <hi>DF</hi> and <hi>EF</hi> being produced shall concurre. Let them concurre in the point <hi>F.</hi> And forasmuch as a certaine right line <hi>DF</hi> deuideth a cer<g ref="char:EOLhyphen"/>taine right lyne <hi>AB</hi> into two equall partes and per<g ref="char:EOLhyphen"/>pendicularly, therfore (by the corollary of the first of this booke) in the line <hi>DF</hi> is the centre of the circle, &amp; by the same reason the centre of the selfe same circle shalbe in the right line EF. Wherfore the centre of the circle wherof <hi>ABC</hi> is a section, is in the point <hi>F,</hi> which is commo<g ref="char:cmbAbbrStroke">̄</g> to either of the lines <hi>DF</hi> and <hi>EF.</hi> Wherfore a section of a circle being geue<g ref="char:cmbAbbrStroke">̄</g>, namely, the section <hi>ABC,</hi> there is described the circle of the same section: which was required to be done.</p>
                  <p>And by this last generall way,<note place="margin">
                        <hi>An addition.</hi>
                     </note> if there be geuen three pointes, set howsoeuer, so that they be not all three in one right line, a man may describe a circle which shall passe by all the said three pointes. For as in the example before put, if you suppose onely the 3. pointes <hi>A, B, C,</hi> to be geuen and not the circumference <hi>ABC</hi> to be drawen, yet follow<g ref="char:EOLhyphen"/>ing the selfe <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ame order you did before, that is, draw a right line from <hi>A</hi> to <hi>B</hi> and an other from <hi>B</hi> to <hi>C</hi> and deuide the said right lines into two equall parts, in the points <hi>D</hi> and <hi>E,</hi> and erect the perpendicular lines <hi>DF</hi> and <hi>EF</hi> cutting the one the other in the point <hi>F,</hi> and draw a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ight line from <hi>F</hi> to <hi>B</hi>: and making the centre the point <hi>F,</hi> and the space <hi>FB</hi> describe a circle, and it shall passe by the pointes <hi>A</hi> &amp; <hi>C</hi>: which may be pro<g ref="char:EOLhyphen"/>ued by drawing right lines from <hi>A</hi> to <hi>F,</hi> and from <hi>F</hi> to <hi>C.</hi> For forasmuch as the two si<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es <hi>AD</hi> and <hi>DF</hi> of the triangle <hi>ADF</hi> are equall to the two sides <hi>BD</hi> and <hi>DF</hi> of the triangle <hi>BDF</hi> (for by supposition the line <hi>AD</hi> is equall to the line <hi>DB,</hi> and the lyne <hi>DF</hi> is common to them bot<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>) and the angle <hi>ADF</hi> is equall to the angle <hi>BDF</hi> (for they are both right angles) therfore (by the 4. of the first) the base <hi>AF</hi> is equall to the base <hi>BF.</hi> And by the same reason the line <hi>FC</hi> is equall to the line <hi>FB.</hi> Wherefore these three lines <hi>FA, FB</hi> and <hi>FC</hi> are equall the one to the other. Wherefore makyng the centre the point <hi>F</hi> and the space <hi>FB,</hi> it shall also passe by the pointes <hi>A</hi> and <hi>C.</hi> Which was required to be done. This proposition is very necessary for many things as you shal afterward see.</p>
                  <p>
                     <hi>Campane</hi> putteth an other way, how to describe the
<figure/>
whole circle of a sectio<g ref="char:cmbAbbrStroke">̄</g> geuen.<note place="margin">
                        <hi>An other con<g ref="char:EOLhyphen"/>struction and demonstration of this Propo<g ref="char:EOLhyphen"/>sition after, Campane</hi>
                     </note> Suppose that the section be <hi>AB.</hi> It is required to describe the whole circle of the same section. Draw in the section two lines at all aduen<g ref="char:EOLhyphen"/>tures <hi>AC</hi> and <hi>BD</hi>: which deuide into two equall parts <hi>AC</hi> in the point <hi>E,</hi> and <hi>BD</hi> in the pointe <hi>F.</hi> Then from the two pointes of the deuisions draw within the secti<g ref="char:EOLhyphen"/>on two perpendicular lines <hi>EG</hi> and <hi>FH</hi> which let cutte the one the other in the point <hi>K.</hi> And the centre of the circle shall be in either of the said perpendicular lines by the corollary of the first of this booke. Wherfore the point <hi>K</hi> is the centre of the cir<g ref="char:EOLhyphen"/>cle: which was required to be done.</p>
                  <p>But if the lines <hi>EG</hi> &amp; <hi>FH</hi> do not cut the one the o<g ref="char:EOLhyphen"/>ther,
<figure/>
but make one right line as doth <hi>GH</hi> in the seco<g ref="char:cmbAbbrStroke">̄</g>d figure: which happeneth when the two lines <hi>AC</hi> and <hi>BD</hi> are equidistant. Then the line <hi>GH,</hi> being applyed to either part of the circumference geuen, shall passe by the centre of the circle, by the selfe same Corollary. For the lines <hi>EG</hi> and <hi>FH</hi> cannot be equidistant. For then one and the self same circumference should haue two centres. Wherfore the line <hi>HG</hi> being deuided in<g ref="char:EOLhyphen"/>to
<pb n="100" facs="tcp:7062:130"/>
two equall partes in the point <hi>K,</hi> the said point <hi>K</hi> shall be the centre of the section.</p>
                  <p>Pelitarius here addeth a briefe way how to finde out the centre of a circle, which is commonly vsed of Artificers.</p>
                  <p>Suppose that the circumference be ABCD, whose centre it is required to finde out.<note place="margin">
                        <hi>A ready way to finde out the center of a circle commo<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>ly vsed a<g ref="char:EOLhyphen"/>mongst a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ti<g ref="char:EOLhyphen"/>ficers.</hi>
                     </note> Take a point in the circumference geuen which let be A, vppon which describe a circle with what openyng of the compasse you will, which let be EFG. Then take an other point in the circumference geuen which let be B, vpon which describe an other circle with the same opening of the compasse that the cir<g ref="char:EOLhyphen"/>cle
<figure/>
EFG was described, and let the same be EHG, which let cut the circle EFG in the two pointes E and G. (I haue not here drawen the whole circles, but onely those partes of them which cut the one the other for auoyding of confusion) And drawe from those centres these right lines AE, BE, AG, and BG, which foure lines shall be equall, by reason they are semidiameters o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall circles. And draw a right line from A to B, and so shall there be made two Isosceles triangles AEB, and AGB vnto whom the line AB is a common base. Now then deuide the line AB into two equal partes in the point K which must nedes fall betwene the two circumference EFG and EHG, otherwise the part should be greater then his whole. Drawe a line from E to K and pro<g ref="char:EOLhyphen"/>duce it to the point G. Now you see that there are two Isosceles triangles deuided into foure equall triangles EAK, EBK, GAK and GBK. For the two sides AE and AK of the triangle AEK are equall to the two sides BE and BK of the triangle BEK, and the base EK is common to them both. Wherefore the two an<g ref="char:EOLhyphen"/>gles at the point K of the two triangles AEK and BEK are by the 8. of the first equall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and therfore are right angles. And by the same reason the other angles at the poynte K are right angles. Wherfore EG is one right lyne by the 14. of the first. Which foras<g ref="char:EOLhyphen"/>much as it deuideth the line AB perpendicularly, therefore it passeth by the center by by the corollary of the first of this booke. And so if you take two other poyntes, name<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> C and D in the circumference geuen, and vpon the<g ref="char:cmbAbbrStroke">̄</g>
                     <figure/>
describe two circles cuttyng the one the other in the pointes L and M, and by the said poyntes pro<g ref="char:EOLhyphen"/>duce a right line, it shall cutte the lyne EG beyng produced in the pointe N, which shall be the cen<g ref="char:EOLhyphen"/>tre of the circle by the same Corollary of the first of this booke, if you imagine the light line CD to be drawen and to be deuided perpendicularly by the lyne LM, which it must needes be as we haue before proued. And here note that to do this me<g ref="char:EOLhyphen"/>chanically not regardyng demonstration, you neede onely to marke the poyntes where the cir<g ref="char:EOLhyphen"/>cles cut the one the other, namely, the poyntes E, G, and L, M, and by those poyntes to produce the lines EG and LM till they cut the one the other, and where they cut the one the other, there is the centre of the circle, as you see here in the seconde figure.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:131"/>
               <head>The <hi>23.</hi> Theoreme. The <hi>26.</hi> Proposition. Equall angles in equall circles consist in equall circu<g ref="char:cmbAbbrStroke">̄</g>ferences, whether the angles be drawen from the centres, or from the circumferences.</head>
               <p>SVppose that these circles ABC and DEF, be equall. And from their centres, namely, the pointes G and H, let there be drawen these equall angles BGC and EHF: and likewise from their circumferences these equall angles BAC and EDF. Then I say, that the circumference BKC is equall to the circumference ELF. Draw (by the first petition) right lines from B to C,<note place="margin">Construction.</note> and from E to F. And for asmuch as the circles ABC and DEF are equall, the right lines also drawen from their centres to their circumferences,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> are (by the first definition of the third) equall the one to the other. Wherefore these two lines BG and
<figure/>
GC, are equall to these two lines EH and HF. And the angle at the poynt G is equall to the angle at the point H: Wherfore (by the <hi>4.</hi> of the first) the base BC is equall to y<hi rend="sup">e</hi> base EF. And for asmuch as the angle at the poynt A is equall to the angle at the point D, therefore the segment BAC is like to the segment EDF. And they are described vpon equall right lines BC and EF. But like segmentes of circles described vpon equall right lines, are (by the <hi>24.</hi> of the third) equall the one to the other. Wherefore the segment BAC is equall to the segment EDF. And the whole circle ABC is equall to y<hi rend="sup">e</hi> whole circle DEF. Wherefore (by the third common sentence) the circumference re<g ref="char:EOLhyphen"/>mayning BKC is equall to the circumference remayning ELF. Wherefore equall angles in equall circles consist in equall circumferences, whether the angles be drawen from the centres or from the circumferences: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>24.</hi> Theoreme. The <hi>27.</hi> Proposition. In equall circles the angles which consist in equall circumfe<g ref="char:EOLhyphen"/>rences, are equall the one to the other, whether the angles be drawen from the centres, or from the circumferences.</head>
               <p>
                  <pb n="101" facs="tcp:7062:131"/>SVppose y<hi rend="sup">t</hi> these circles ABC, and DEF, be equall. And vpon these equall circumferences of the same circles, namely, vpon BC and EF, let there consist these angles BGC and EHF drawen from the cen<g ref="char:EOLhyphen"/>tres, and also these angles BAC and EDF drawen from the cir<g ref="char:EOLhyphen"/>cumferences. Then I say, that the angle BGC is equall to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equall to the an<g ref="char:EOLhyphen"/>gle EHF, then it is manifest, that the angle BAC is equall to y<hi rend="sup">e</hi> angle EDF (by the 20. of the third). But if the angle BGC be not equall to y<hi rend="sup">e</hi> angle EHF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> then is the one of them greater then the other. Let the angle BGC be greater And (by the
<figure/>
                  <hi>23.</hi> of the first) vpon the right line BG, and vnto the point geuen in it G, make vnto the angle EHF an equall an<g ref="char:EOLhyphen"/>gle BGK. But (by the <hi>26.</hi> of y<hi rend="sup">e</hi> third) equall angles in equall circles consist vpo<g ref="char:cmbAbbrStroke">̄</g> equall circumferences, whether they be drawen from the centres or from the circumferences. Wherefore the circumference BK is equall to the circumference EF. But the circumference EF is equall to the circumference BC: Wherefore the circumference BK also is equall to the cir<g ref="char:EOLhyphen"/>cumference BC, the lesse to the greater: which is impossible. Wherfore the an<g ref="char:EOLhyphen"/>gle BGC is not vnequall to the angle EHF: Wherefore it is equall. And (by the <hi>20.</hi> of the third) the angle at the point A is the halfe of the angle BGC: and (by the same) the angle at the point D is the halfe of the angle EHF. Where<g ref="char:EOLhyphen"/>fore the angle at the point A is equall to the angle at the point D. Wherefore in equall circles, the angles which consist in equall circumferences, are equall the one to the other, whether the angles be drawen from the centres or from the circumfe<g ref="char:EOLhyphen"/>rences: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>25.</hi> Theoreme. The <hi>28.</hi> Proposition. In equall circles, equall right lines do cut away equall cir<g ref="char:EOLhyphen"/>cumferences, the greater equall to the greater, and the lesse e<g ref="char:EOLhyphen"/>quall to the lesse.</head>
               <p>
                  <pb facs="tcp:7062:132"/>SVppose that these circles ABC, and DEF, be equall. And in them let there be drawen these equall right lines BC and EF, which let cut away these circumferences BAC and DEF being the greater, &amp; also these circumferences BGC and EHF being the lesse. Then I say, that the greater circumference BAC is equall to the greater circumference EDF: and the lesse circumference BGC is equall to the lesse circumference EHF.<note place="margin">Construction.</note> Take (by the first of the third) the centres of the circles, and let the same be the pointes K and L. And draw these right lines, KB, KC, LE, and LF. And for asmuch as the circles are equall,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore (by the first definition of the third) the lines which are drawen fro<g ref="char:cmbAbbrStroke">̄</g> the cen<g ref="char:EOLhyphen"/>tres
<figure/>
are equall. Where<g ref="char:EOLhyphen"/>fore these two lines BK and KC, are equall to these two lines EL and LF. And (by suppositi<g ref="char:EOLhyphen"/>on) the base BC is equall to the base EF. Where<g ref="char:EOLhyphen"/>fore (by the <hi>8.</hi> of y<hi rend="sup">e</hi> first) the angle BKC is equall to the angle ELF. But (by the <hi>26.</hi> of the third) equall angles drawen from the centres, consist vpon equall circumferences. Wher<g ref="char:EOLhyphen"/>fore the circumference BGC is equall to the circumference EHF: and y<hi rend="sup">e</hi> whole circle ABC is equall to the whole circle DEF. Wherefore the circumference remayning BAC, is (by the third common sentence) equall to the circumference remayning EDF. Wherefore in circles, equall right lines do cut away equall circumferences, the greater equall to the greater, and the lesse equall to the lesse: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>26.</hi> Theoreme. The <hi>29.</hi> Proposition. In equall circles, vnder equall circumferences are subtended equall right lines.</head>
               <p>
                  <note place="margin">The conuerse of the former Proposition.</note> SVppose that these circles ABC and DEF, be equall. And in them let there be taken these equall circumferences, BGC and EHF: and drawe these right lines BC and EF. Then I say, that the right line BC is equall to the right line EF. Take (by the first of y<hi rend="sup">e</hi> third) the centres of the circles, and let them be the pointes K and L,<note place="margin">Construction.</note> and draw these right lines KB, KC, LE, LF. And for asmuch as the circumference BGC is equall to the circumference EHF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> the angle BKC is equall to y<hi rend="sup">e</hi> angle ELF (by the <hi>27.</hi> of the third). And for asmuch as the circles ABC and DEF are equall the one to the other, therefore (by the first definition of the third) the lines
<pb n="102" facs="tcp:7062:132"/>
whiche are
<figure/>
drawe<g ref="char:cmbAbbrStroke">̄</g> from the centres are equall. Wherefore these y. lines BK and KC, are e<g ref="char:EOLhyphen"/>quall to these y. lines LE and LF, &amp; they comprehend equall angles. Wherefore (by the <hi>4.</hi> of the first) the base BC is e<g ref="char:EOLhyphen"/>quall to the base EF. Wherefore in equall circles vnder equall circumferences, are subtended equall right lines: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Probleme. The <hi>30.</hi> Proposition. To deuide a circumference geuen into two equall partes.</head>
               <p>SVppose that the circumference geuen be ADB. It is required to de<g ref="char:EOLhyphen"/>uide the circumference ADB into two equall partes.<note place="margin">Construction.</note> Draw a right line from A to B. And (by the <hi>10.</hi> of the first) deuide the line AB into two equall partes in the point C. And (by the <hi>11.</hi> of the first) from the point C rayse vp vnto AB a perpendicular line CD. And draw these right lines AD and DB. And forasmuch as the line AC is equall to the line CB,<note place="margin">Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> &amp; the line CD is common to them both, there<g ref="char:EOLhyphen"/>fore
<figure/>
these two lines AC and CD are equall to these two lines BC and CD. And (by the <hi>4.</hi> peti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ion) the angle ACD is equall to the angle BCD, for either of them is a right right angle. Wherfore (by the <hi>4.</hi> of the first) the base AD is equall to the base DB. But equall right lines do cut away equall circum<g ref="char:EOLhyphen"/>ferences, the greater equall to the greater, &amp; the lesse equall to the lesse (by the <hi>28.</hi> of the third) And either of these circum<g ref="char:EOLhyphen"/>ferences AD and DB is lesse then a semicircle. Wherfore the circumference AD is equall to the circumference DB. Wherfore the circumference geuen is deui<g ref="char:EOLhyphen"/>ded into two equall partes: Which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>27.</hi> Theoreme. The <hi>31.</hi> Proposition. In a circle an angle made in the semicircle is a right angle:
<pb facs="tcp:7062:133"/>
but an angle made in the segment greater then the semicircle is lesse then a right angle, and an angle made in the segment lesse then the semicircle, is greater then a right angle. And moreouer the angle of the greater segment is greater then a right angle: and the angle of the lesse segment is lesse then a right angle.</head>
               <p>SVppose that the circle be ABCD, and let the dimetient of the circle be the right line BC, and the ce<g ref="char:cmbAbbrStroke">̄</g>tre therof the point E. And take in the se<g ref="char:EOLhyphen"/>micircle a point at all auentures, and let the same be D. And draw these right lines BA, AC, AD, and DC.<note place="margin">The first <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>art of this Theo<g ref="char:EOLhyphen"/>reme.</note> Then I say that the angle in the semicircle BAC,
<figure/>
namely, the angle BAC is a right angle. And the angle ABC which is in the segment ABC being greater then the semicircle, is lesse then a right angle. And the angle ADC which is in the segment ADC being lesse then the semicircle is greater the<g ref="char:cmbAbbrStroke">̄</g> a right an<g ref="char:EOLhyphen"/>gle. Draw a line from the point A to the point E, and extend the line BA vnto the point F. And forasmuch as the line BE is equall to the line EA, (for they are drawen from the centre to the circumference) therfore the an<g ref="char:EOLhyphen"/>gle EAB is equall to the angle EBA (by the <hi>5.</hi> of the first). Againe forasmuch as the line AE is equall to the line EC, the angle ACE is (by the same) equall to the angle CAE. Wherfore the whole angle BAC is equall to these two angles ABC and ACB. But the angle FAC which is an outward angle of the triangle ABC is (by the <hi>32.</hi> of the first) equall to the two angles ABC &amp; ACB. Wherfore the angle BAC is equall to the angle F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>AC. Wherfore either of them is a right angle. Wherfore the angle BAC which is in the semicircle BAC is a right angle.</p>
               <p>
                  <note place="margin">Second part.</note>And forasmuch as (by the <hi>17.</hi> of the first) the two angles of the triangle ABC, namely, ABC and BAC are lesse then two right angles, and the angle BAC is a right angle. Therfore the angle ABC is lesse then a right angle, and it is in the segment ABC which is greater then the semicircle.</p>
               <p>
                  <note place="margin">Thir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> part.</note>And forasmuch as in the circle there is a figure of foure sides, namely, ABCD. But if within a circle be described a figure of foure sides, the angles therof which are opposite the one to the other are equall to two right angles (by the <hi>22.</hi> of the third) Wherfore (by the same) the angles ABC and ADC are equall
<pb n="103" facs="tcp:7062:133"/>
to two right angles. But the angle ABC is lesse then a right angle. Wherfore the angle remayning ADC is greater then a right angle, and it is in a segment which is lesse then the semicircle.</p>
               <p>Now also I say that the angle of the greater segment, namely,<note place="margin">The fourth part.</note> the angle which is comprehended vnder the circumference ABC and the right line AC is greater then a right angle, and the angle of the lesse segment comprehended vnder the circumference ADC, and the right line AC is lesse the<g ref="char:cmbAbbrStroke">̄</g> a right angle, which
<figure/>
may thus be proued. Forasmuch as the angle comprehended vnder the right lines BA and AC is a right angle, therfore the angle com<g ref="char:EOLhyphen"/>prehended vnder the circumference ABC and the right line AC is greater then a right angle: for the whole is euer greater then his part (by the <hi>9.</hi> common sentence.</p>
               <p>Againe forasmuch as the angle compre<g ref="char:EOLhyphen"/>hended vnder the right lines AC and AF is a right angle,<note place="margin">The fift and last part.</note> therfore the angle com<g ref="char:EOLhyphen"/>prehended vnder the right line CA and the circumference ADC is lesse then a right an<g ref="char:EOLhyphen"/>gle. Wherfore in a circle an angle made in the semicircle is a right angle, but an angle made in the segment greater then the semicircle is lesse then a right angle, and an an<g ref="char:EOLhyphen"/>gle made in the segment lesse then the semicircle, is greater then a right angle. And moreouer the angle of the greater segment is greater then a right angle: &amp; the angle of the lesse segment is lesse then a right angle: which was required to be demonstrated.</p>
               <p>An other demonstration to proue that the angle BAC is a right angle.<note place="margin">An other De<g ref="char:EOLhyphen"/>monstration to proue that the ang<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e in a semicircle is a right angle.</note> For<g ref="char:EOLhyphen"/>asmuch as the angle AEC is double to the angle BAE (by the <hi>32.</hi> of the first) for it is equall to the two inward angles which are opposite. But the inwarde an<g ref="char:EOLhyphen"/>gles are (by the <hi>5.</hi> of the first) equall the one to the other, and the angle AEB is double to the angle EAC. Wherfore the angles AEB and AEC are double to the angle BAC. But the angles AEB and AEC are equall to two right an<g ref="char:EOLhyphen"/>gles: Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rfore the angle BAC is a right angle. Which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
               <div type="section">
                  <head>Correlary.</head>
                  <p>Hereby it is manifest,<note place="margin">A Corollary.</note> that if in a triangle one angle be equall to the two other angles remayning the same angle is a right
<pb facs="tcp:7062:134"/>
angle: for that the side angle to that one angle (namely, the angle which is made of the side produced without the trian<g ref="char:EOLhyphen"/>gle) is equall to the same angles, but when the side angles are equall the one to the other, they are also right angles.</p>
                  <div type="subsection">
                     <head>¶ An addition of Pelitarius.</head>
                     <p>If in a circle be inscribed a rectangle triangle, the side opposite vnto the right angle shall be the diameter of the circle.</p>
                     <p>
                        <note place="margin">
                           <hi>An addition of P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>litarius.</hi>
                        </note>Suppose that in the circle ABC be inscribed a
<figure/>
rectangle triangle ABC, whose angle at the point B let be a right angle. Then I say, that the side AC is the diameter of the circle. For if not, then shall the centre be without the line AC, as in the point E.<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ing to an absurdi<g ref="char:EOLhyphen"/>t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>.</hi>
                        </note> And draw a line from the poynt A to the point E, &amp; produce it to the circumference to the point D: and let AED be the diameter: and draw a line from the point B to the point D. Now (by this 31. Propositio<g ref="char:cmbAbbrStroke">̄</g>) the angle ABD shall be a right angle, and therefore shall be equall to the right angle ABC, namely, the part to the whole: which is ab<g ref="char:EOLhyphen"/>surde. Euen so may we proue, that the centre is in no other where but in the line AC. Wherfore AC is the diameter of the circle: which was required to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ An addition of Campane.</head>
                     <p>
                        <note place="margin">
                           <hi>An addition of Campane.</hi>
                        </note>By thys 31. Proposition, and by the 16. Proposition of thys booke, it is mani<g ref="char:EOLhyphen"/>fest, that although in mixt angles, which are contayned vnder a right line and the circumference of a circle, there may be geuen an angle lesse &amp; greater then a right angle, yet can there neuer be geue<g ref="char:cmbAbbrStroke">̄</g> an angle equall to a right angle. For euery secti<g ref="char:EOLhyphen"/>on of a circle is eyther a semicircle, or greater then a semicircle, or lesse, but the an<g ref="char:EOLhyphen"/>gle of a semicircle is by the 16. of thys booke, lesse then a right angle, and so also is the angle of a lesse section by thys 31. Proposition: Likewise the angle of a greater section, is greater then a right angle, as it hath in thys Proposition bene proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>28.</hi> Theoreme. The <hi>32.</hi> Proposition. If a right line touch a circle, and from the touch be drawen a right line cutting the circle: the angles which that line and the touch line make, are equall to the angles which consist in the alternate segmentes of the circle.</head>
               <p>SVppose that the right line EF do touch the circle ABCD in the point B: and from the point B let there be drawen into the circle ABCD a right line cutting the circle, and let the same be BD. Then I say, that the angles which the line BD together with the
<pb n="104" facs="tcp:7062:134"/>
touch line EF do make, are equall to the angles which are in the alternate seg<g ref="char:EOLhyphen"/>mentes of the circle, that is, the angle FBD is equall to the angle which consi<g ref="char:EOLhyphen"/>steth in the segment BAD, and the angle EBD is equall to the angle which consisteth in the segment BCD.<note place="margin">Construction.</note> Raise vp (by the <hi>11.</hi> of the first) from y<hi rend="sup">e</hi> point B vnto the right line EF a perpendicular line BA. And in the circumference BD take a point at all aduentures, and let the same be C. And draw these right lines AD, DC, and CB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for asmuch as a certaine right line EF tou<g ref="char:EOLhyphen"/>cheth the circle ABC in the point B, and from the point B where the touch is, is raysed vp vnto the touch line a perpen<g ref="char:EOLhyphen"/>dicular
<figure/>
BA. Therfore (by the <hi>19.</hi> of the third) in the line BA is the centre of the circle ABCD. Wherfore y<hi rend="sup">e</hi> angle ADB being in the semicircle, is (by the <hi>31.</hi> of the third) a right angle. Wherefore the an<g ref="char:EOLhyphen"/>gles remayning BAD and ABD, are equall to one right angle. But the angle ABF is a right angle. Wherefore the an<g ref="char:EOLhyphen"/>gle ABF is equall to the angles BAD and ABD. Take away y<hi rend="sup">e</hi> angle ABD which is common to them both. Wherefore the angle remayning DBF, is equall to the angle remayning BAD, which is in the alternate segment of the circle. And for asmuch as in the circle is a figure of fower sides, namely, ABCD, therfore (by the <hi>22.</hi> of the third) the angles which are opposite the one to the other, are equall to two right angles. Wherfore the an<g ref="char:EOLhyphen"/>gles BAD and BCD, are equall to two right angles. But the angles DBF and DBE, are also equall to two right angles. Wherefore the angles DBF and DBE, are equall to the angles BAD and BCD. Of which we haue proued that the angle BAD is equall to the angle DBF. Wherefore the an<g ref="char:EOLhyphen"/>gle remayning DBE, is equall to the angle remayning DCB, which is in the alternate segment of the circle, namely, in the segment DCB. If therfore a right line touch a circle, and from the touch be drawen a right line cutting the circle: the angles which that line and the touch line make, are equall to y<hi rend="sup">e</hi> angles which consist in y<hi rend="sup">e</hi> alternate segmentes of the circle: which was required to be proued.</p>
               <p>
                  <hi>In thys Proposition may be two cases.</hi>
                  <note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> 
                  <hi>For the line drawen from the touch and cutting the circle, may eyther passe by the centre or not. If it passe by the centre, then is it manifest (by the 18. of thys booke) that it falleth perpendicularly vpon the touch line, and deuideth the circle into two equall partes, so that all the angles in eche semicircle, are by the former Proposition, right angles, and therfore equall to the alternate angles made by the sayd perpendicular line and the touch line. If it passe not by the centre, then followe the construction and demonstration be<g ref="char:EOLhyphen"/>fore put.</hi>
               </p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:135"/>
               <head>The <hi>5.</hi> Probleme. The <hi>33.</hi> Proposition. Vppon a right lyne geuen to describe a segment of a circle, which shall contayne an angle equall to a rectiline angle geue<g ref="char:cmbAbbrStroke">̄</g>.</head>
               <p>SVppose that the right line geuen be AB, and let the rectiline angle ge<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>n be C. It is required vpon the right line geue<g ref="char:cmbAbbrStroke">̄</g> AB to describe a seg<g ref="char:EOLhyphen"/>ment of a circle which shall contayne an angle equall to the angle C. Now the angle C is either an acute angle, or a right angle, or an ob<g ref="char:EOLhyphen"/>tuse angle.</p>
               <p>
                  <note place="margin">
                     <p>Three cases in this Proposi<g ref="char:EOLhyphen"/>tion.</p>
                     <p>The first case.</p>
                  </note>First, let it be an acute angle as
<figure/>
in the first description. And (by the <hi>23</hi> of the first) vpon the right line AB and to the point in it A describe an angle equal to the angle C,<note place="margin">Construction.</note> and let the same be DAB. Wherfore the angle DAB is an acute angle. From the point A raise vp (by the <hi>11.</hi> of y<hi rend="sup">e</hi> first) vnto the line AD a perpendiculer line AF. And (by the <hi>10.</hi> of the first) deuide the line AB into two equall partes in the point F. And (by the <hi>11.</hi> of the same) from the point F raise vp vnto the line AB a perpendicular lyne FG, and draw a line from G to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> And forasmuch as the line AF is equall to the line FB and the line FG is common to them both, therfore these two lines AF and FG are equall to these two lines FB and FG: and the angle AFG is (by the <hi>4.</hi> peticion) equall to the angle GFB. Wherfore (by the <hi>4.</hi> of the same) the base AG is equall to the base GB. Wherfore making the centre G and the space GA describe (by the <hi>3.</hi> peticion) a circle and it shall passe by the point B: de<g ref="char:EOLhyphen"/>scribe such a circle &amp; let the same be ABE: And draw a line from E to B. Now forasmuch as from the ende of the diameter AE, namely, from the point A is <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> a right line AD making together with the right line AE a right an<g ref="char:EOLhyphen"/>gle, therfore (by the correllary of the <hi>16.</hi> of the third) the line AD toucheth the circle ABE. And forasmuch as a certaine right line AD toucheth the circle ABE, &amp; from the point A where the touch is, is drawen into y<hi rend="sup">e</hi> circle a certaine right line AB: therfore (by the <hi>32.</hi> of the third) the angle DAB is equall to the angle AEB which is in the alternate segment of the circle. But the angle DAB is equall to the angle C, wherfore the angle C is equall to the angle AEB. Wher<g ref="char:EOLhyphen"/>fore vpon the right line geuen AB is described a segment of a circle which con<g ref="char:EOLhyphen"/>tayneth the angle AEB, which is equall to the angle geuen, namely, to C.</p>
               <p>
                  <note place="margin">The second case.</note>But now suppose that the angle C be a right angle. It is againe required v<g ref="char:EOLhyphen"/>pon
<pb n="105" facs="tcp:7062:135"/>
the right line AB to describe a seg<g ref="char:EOLhyphen"/>ment
<figure/>
of a circle, which shall contayne an angle equal to the right angle C.<note place="margin">Construction.</note> Describe againe vpon the right line AB and to the point in it A an angle BAD equal to the rectiline angle geuen C (by the <hi>23.</hi> of the first) as it is set forth in the second de<g ref="char:EOLhyphen"/>scription. And (by the <hi>10.</hi> of the first) de<g ref="char:EOLhyphen"/>uide the line AB into two equall partes in the point F. And making the centre the point F and the space FA or FB describe (by the <hi>3.</hi> peticion) y<hi rend="sup">e</hi> circle AEB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wher<g ref="char:EOLhyphen"/>fore the right line AD toucheth the cir<g ref="char:EOLhyphen"/>cle AEB: for that the angle BAD is a right angle. Wherfore y<hi rend="sup">e</hi> angle BAD is equall to the angle which is in the segment AEB, for the angle which is in a semicircle is a right angle (by the <hi>31.</hi> of the third) But the angle BAD is equal to the angle C. Wherfore there is againe described vpon the line AB a segment of a circle, namely, AEB, which containeth an angle equall to the angle geuen namely, to C.</p>
               <p>But now suppose that the angle C be an obtuse angle.<note place="margin">The third case.</note> Vpon the right line AB and to the point in it A describe (by the <hi>23.</hi> of the first) an angle BAD equall to the angle C: as it is in the third description.<note place="margin">Construction.</note> And from the point A rayse vp vnto the line AD a perpendiculer line AE
<figure/>
(by the <hi>11.</hi> of the first) And agayne by the <hi>10.</hi> of the first) deuide the line AB into two equall partes in the point F. And from the point F. ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se vp vnto the line AB a perpe<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>dicular line FG (by the <hi>11.</hi> of the same) &amp; drawe a line from G to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And now foras<g ref="char:EOLhyphen"/>much as the line AF is equal to the line FB, and the line FG is common to them both, therfore these two lines AF and FG are e<g ref="char:EOLhyphen"/>quall to these two lines BF and FG: and the angle AFG is (by the <hi>4.</hi> peticion) equall to the angle BFG: wherfore (by the <hi>4.</hi> of the same) the base AG is equall to the base GB. Wherfore making the centre G, and the space GA describe (by the <hi>3.</hi> peticion) a circle and it shall passe by the point B: let it be described as the circle AEB is. And forasmuch as from the ende of the diameter AE is drawen a perpendiculer line AD, therefore (by the correllary of the <hi>16.</hi> of the third) the line AD toucheth the circle AEB &amp; from the point of the touche, namely; A, is extended the line AB. Wherfore (by the <hi>32.</hi> of the third) the angle BAD is equall to the angle AHB which is in the alternate segment of the circle. But the angle BAD is equall to the angle C.
<pb facs="tcp:7062:136"/>
Wherefore the angle which is in the segment AHB is equall to the angle C. Wherfore vpon the right line geuen AB, is described a segment of a circle AHB, which contayneth an angle equall to the angle geuen, namely, C: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Probleme. The <hi>34.</hi> Proposition. From a circle geuen to cut away a section which shal containe an angle equall to a rectiline angle geuen.</head>
               <p>SVppose that the circle geuen be AC and let the rectiline angle geuen be D. It is required fro<g ref="char:cmbAbbrStroke">̄</g> the circle ABC to cut away a segment which shall contayne an angle equall to the angle D.<note place="margin">Cons<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>uction.</note> Draw (by the <hi>17.</hi> of the third) a line touching the circle, and let the same be EF: and let it touche in the point B. And (by the <hi>23.</hi> of
<figure/>
the first) vpon the right line EF and to the point in it B describe the angle FBC equall to the angle D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as a certayne right line EF toucheth the cir<g ref="char:EOLhyphen"/>cle ABC in the point B: and frora y<hi rend="sup">e</hi> point of the touche, namely, B, is drawn into the circle a certaine right line BC, there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore (by the <hi>32.</hi> of the third) the angle FBC is equall to the angle BAC which is in the alternate segment. But the angle FBC is equall to the angle D. Wherfore the angle BAC which consisteth in the segment BAC is equall to the angle D. Where<g ref="char:EOLhyphen"/>fore from the circle geuen ABC is cut away a segment BAC, which containeth an angle equall to the rectiline angle geuen: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>29.</hi> Theoreme. The <hi>35.</hi> Proposition. If in a circle two right lines do cut the one the other, the rect<g ref="char:EOLhyphen"/>angle parallelograme comprehended vnder the segmentes or parts of the one line is equall to the rectangle parallelograme comprehended vnder the segment or partes of the other line.</head>
               <p>
                  <hi>LEt the circle be ABCD, and in it let these two right lines AC and BD c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the one the other in the point E. Then I say that the rectangle parallelogramme contayned vnder the partes AE and EC is equall to
<pb n="106" facs="tcp:7062:136"/>
the rectangle parallelogramme contained vnder the
<figure/>
partes DE and EB.</hi>
                  <note place="margin">
                     <p>
                        <hi>Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</hi>
                     </p>
                     <p>
                        <hi>First case.</hi>
                     </p>
                  </note> 
                  <hi>For if the line AC and BD be drawen by the centre, then is it manifest, that for as much as the lines AE and EC are equall to the lines DE and EB by the definition of a circle,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> the rectangle parallelograme also contayned vnder the lines AE and EC is equall to y<hi rend="sup">e</hi> rectangle paralle<g ref="char:EOLhyphen"/>lograme contained vnder the lines DE and EB.</hi>
               </p>
               <p>
                  <hi>But now suppose that the lines AC and DB be not extended by the centre,</hi>
                  <note place="margin">
                     <hi>The second c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>se.</hi>
                  </note> 
                  <hi>and take (by the</hi> 1. <hi>of the third) the centre of the cir<g ref="char:EOLhyphen"/>cle ABCD, and let the same be the point F,<note place="margin">
                        <hi>Construction.</hi>
                     </note> and from the point F draw to the right lines AC and DB perpendicular lines FG and FH (by the</hi> 12. <hi>of the first) and draw these right lines FB, FC, and FE.
<figure/>
And forasmuch as a certaine right line FG drawen by the centre,</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>cutteth a certaine right line AC not drawen by the centre in such sorte that it maketh right angles, it therfore deuideth the line AC into two equall partes (by the</hi> 3. <hi>of the third). Wherfore the line AG is equall to the line GC. and foras<g ref="char:EOLhyphen"/>much as the right line AC is deuided into two e<g ref="char:EOLhyphen"/>quall partes in the point G, and into two vnequall partes in the point E: therfore (by the</hi> 5. <hi>of the second) the rectangle parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines. AE and EC together with the square of the line EG is equall to the square of the line GC. Put the square of the line GF common to them both, wherfore that which is contained vnder the lines AE &amp; EC together with the squares of the lines EG and GF is equall to the squares of the lines GF &amp; GC. But vnto y<hi rend="sup">e</hi> squares of y<hi rend="sup">e</hi> lines EG &amp; GF is equall y<hi rend="sup">e</hi> square of y<hi rend="sup">e</hi> line FE (by the</hi> 47. <hi>of the first): and to the squares of the lines GC and GF is equall the square of the line FC (by the same) Wherfore that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AE and EC, together with the square of the line FE is e<g ref="char:EOLhyphen"/>quall to the square of the line FC. But the line FC is equall to the line FB. For they are drawen from the centre to the circumference. Wherfore that which is contained vnder the lines AE and E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> together with the square of the lyne FE is equal to the square of the line FB. And by the same demonstration that which is contained vnder the lines DE and EB together with the square of the line FE is equall to the square of the line FB. Wherfore that which is contained vn<g ref="char:EOLhyphen"/>der the lines AE and EC together with the square of the line EF is equall to that which is contayned vnder the lines DE and EB together with the square of the line EF. Take away the square of the line FE which is common to them both. Wherfore the rectangle parallelogramme remayning which is contayned vnder the lines AE and EC is equall to the rectangle parallelogramme remay<g ref="char:EOLhyphen"/>ning, which is contayned vnder the lines DE and EB. If therefore in a circle two right lines do cut the one the other: the rectangle parallelogramme compre<g ref="char:EOLhyphen"/>hended
<pb facs="tcp:7062:137"/>
vnder the segmentes or parts of the one line is equall to the rectangle pa<g ref="char:EOLhyphen"/>rallelograme comprehended vnder the segmentes or parts of the other line: which was required to be demonstrated.</hi>
               </p>
               <p>In thys Proposition are three cases:<note place="margin">
                     <hi>Three cases in this <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>roposi<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For eyther both the lines passe by the cen<g ref="char:EOLhyphen"/>tre, or <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eyther of them passeth by the centre: or the one passeth by the centre and the other not. The two first cases are before demonstrated.</p>
               <p>But now let one of the lines onely, namely, the line <hi>AC</hi> passe by the centre, which let be the poynt <hi>F,</hi>
                  <note place="margin">
                     <hi>The third case.</hi>
                  </note> and let it cut the other line, namely, <hi>BD,</hi> in the poynt <hi>E.</hi> Now then the line <hi>AC</hi> deuideth the line <hi>BD</hi> eyther into two equall partes, or into two vn<g ref="char:EOLhyphen"/>equall partes. Fyrst let it deuide it into two equall partes: Wherefore also it deuideth it <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ight angled wyse by the 3. of thys booke. Drawe a right line from <hi>B</hi> to <hi>F.</hi> Where<g ref="char:EOLhyphen"/>fore <hi>BEF</hi> is a right angled triangle. And for asmuch as the right line <hi>AC</hi> is deuided into two equall partes in the poynt <hi>F,</hi> &amp; into two vnequall partes in the poynt <hi>E.</hi> Ther<g ref="char:EOLhyphen"/>fore the rectangle figure contayned vnder the
<figure/>
lines <hi>AE</hi> and <hi>EC</hi> together with the square of the line <hi>EF,</hi> is equall to the square of the line <hi>FC</hi> (by the 5. of the second). But vnto the square of the line <hi>FC</hi> is equall the square of the line <hi>BF</hi> (for that the lines <hi>FB</hi> and <hi>FC</hi> are equall). Ther<g ref="char:EOLhyphen"/>fore that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines <hi>AE</hi> and <hi>EC</hi> together with the square of the line <hi>EF,</hi> is <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>uall to the square of the line <hi>BF.</hi> But vnto the square of the line <hi>BF,</hi> are equall the squares of the lines <hi>BE</hi> and <hi>EF</hi> (by the 47. of the first). Wherefore that which is contayned vnder the lines <hi>AE</hi> and <hi>EC</hi> together with the square of the line <hi>EF,</hi> is equall to the squares of the lines <hi>BE</hi> and <hi>EF.</hi> Take away the square of the line <hi>EF</hi> which is common to them both: Wherefore that which remayneth, namely, that which is contayned vnder the lines <hi>AE</hi> and <hi>EC,</hi> is equall to the residue, namely, to the square of the line <hi>BE.</hi> But the square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line <hi>BE</hi> is that which is contained vn<g ref="char:EOLhyphen"/>der the lines <hi>BE</hi> and <hi>ED</hi>: for (by supposition) the line <hi>BE</hi> is equall to the line <hi>ED.</hi> Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore that which is contayned vnder the lines <hi>AE</hi> &amp; <hi>EC,</hi> is equall to that which is contayned vnder the lines <hi>BE</hi> and <hi>ED</hi>: which was required to be proued.</p>
               <p>But now let the line <hi>AC</hi> passing by the centre,
<figure/>
deuide, the line <hi>BD</hi> not passing by the centre, vn<g ref="char:EOLhyphen"/>equally in the poynt <hi>E.</hi> And fro<g ref="char:cmbAbbrStroke">̄</g> the poynt <hi>E</hi> raise vp vnto the line <hi>AC</hi> a perpendicular line <hi>EH,</hi> which produce on the other side to the poynt <hi>G.</hi> Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore (by the 3. of this booke) the line <hi>EH</hi> is equall to the line <hi>EG.</hi> Wherfore as we haue be<g ref="char:EOLhyphen"/>fore proued, that which is contayned vnder the lines <hi>AE</hi> and <hi>EC,</hi> is equall to that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines <hi>GE</hi> &amp; <hi>EH:</hi> but that which is contayned vnder the lines <hi>BE</hi> and <hi>ED,</hi> is also equall to that which is contayned vnder the lines <hi>GE</hi> and <hi>EH,</hi> by the second case of thys Propositi<g ref="char:EOLhyphen"/>on: Wherfore that which is contayned vnder the lines <hi>AE</hi> and <hi>EC,</hi> is equall to that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines <hi>BE</hi> and <hi>ED</hi>: which was agayne required to be proued.</p>
               <p>Amongest all the Propositions in this third booke, doubtles thys is one of the chiefest. For it setteth forth vnto vs the wonderfull nature of a circle. So that by
<pb n="107" facs="tcp:7062:137" rendition="simple:additions"/>
it may be done many goodly conclusions in Geometry, as shall afterward be de<g ref="char:EOLhyphen"/>clared when occasion shall serue.</p>
            </div>
            <div type="part">
               <head>The <hi>30.</hi> Theoreme. The <hi>36.</hi> Proposition. If without a circle be taken a certaine point, and from that point be drawen to the circle two right lines, so that the one of them do cut the circle, and the other do touch the circle: the rectangle parallelogramme which is comprehended vnder the whole right line which cutteth the circle, and that portion of the same line that lieth betwene the point and the vtter cir<g ref="char:EOLhyphen"/>cu<g ref="char:cmbAbbrStroke">̄</g>ference of the circle, is equall to the square made of the line that toucheth the circle.</head>
               <p>SVppose that the circle be ABC: and without the same circle take a<g ref="char:EOLhyphen"/>ny point at all aduentures, and let the same be D.<note place="margin">Construction.</note> And from the point D let there be drawen to the circle two right lines DCA and DB, and let the right line DCA cut the circle ACB in the point C, and let the right line BD touch the same. Then I say, that the rectangle parallelo<g ref="char:EOLhyphen"/>gramme contayned vnder the lines AD and DC,<note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> is equall to the square of the line BD. Now the line DCA is either drawen by the centre, or not.</p>
               <p>First let it be drawen by the centre.<note place="margin">The first case.</note> And (by
<figure/>
the first of the third) let the poynt F be y<hi rend="sup">e</hi> centre of the circle ABC, and drawe a line from F to B. Wherefore the angle FBD is a right angle.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for asmuch as y<hi rend="sup">e</hi> right line AC is deuided into two equall partes in the poynt F, and vnto it is added directly a right line CD, therfore (by the <hi>6.</hi> of the second) that which is contayned vnder the lines AD and DC together with the square of y<hi rend="sup">e</hi> line CF, is equall to the square of the line FD. But the line FC is equall to the line FB, for they are drawen from the centre to y<hi rend="sup">e</hi> circumference: Wher<g ref="char:EOLhyphen"/>fore that which is contayned vnder the lines AD and DC together with the square of the line FB, is equall to the square of the line FD. But y<hi rend="sup">e</hi> square of the line FD, is (by the <hi>47.</hi> of the first) equall to the squares of the lines FB and BD (for the angle FBD is a right angle). Wherefore that which is contayned vnder the lines AD and DC together with the square of the line FB, is equall to the
<pb facs="tcp:7062:138" rendition="simple:additions"/>
squares of the lines FB and BD. Take away the square of the line FB which is common to them both. Wherefore that which remayneth, namely, that which is contayned vnder the lines AD and DC, is equall to the square made of the line DB which toucheth the circle.</p>
               <p>But now suppose that the right line DCA be
<figure/>
not drawen by the centre of the circle ABC.<note place="margin">The second case.</note> And (by the first of the third) let the point E be y<hi rend="sup">e</hi> cen<g ref="char:EOLhyphen"/>tre of the circle ABC.<note place="margin">Construction.</note> And from y<hi rend="sup">e</hi> poynt E, draw (by the 12. of the first) vnto the line AC a per<g ref="char:EOLhyphen"/>pendicular line EF, and draw these right lines EB, EC, and ED.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now the angle EFD is a right angle. And <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or asmuch as a certaine right line EF drawen by the centre, cutteth a certayne other right line AC not drawen by the centre, in such sort that it maketh right angles, it deuideth it (by y<hi rend="sup">e</hi> third of the third) into two equall partes. Wherefore the line AF is equall to the line FC. And for asmuch as the right line AC is deuided into two equall partes in the poynt F, &amp; vnto it is added directly an other right line making both one right line, therefore (by the <hi>6.</hi> of the second) that which is contayned vnder the lines DA and DC together with the square of the line FC, is equall to the square of the line FD: put the square of the line FE common to them both. Wherefore that which is contayned vnder the lines DA and DC together with the squares of the lines CF and FE, is equall to the squares of the lines FD and FE. But to the squares of the lines FD and FE, is equall the square of the line DE (by the <hi>47.</hi> of the first) for the angle EFD is a right angle. And to the squares of the lines CF and FE, is equall the square of the line CE (by the same). Wherfore that which is contayned vn<g ref="char:EOLhyphen"/>der the lines AD and DC together with the square of the line EC, is equall to the square of the line ED. But the line EC is equall to the line EB: for they are drawen from the centre to the circumference. Wherefore that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines AD and DC together with the square of the line EB, is equall to the square of the line ED. But to the square of the line ED, are e<g ref="char:EOLhyphen"/>quall the squares of the lines EB and BD (by the <hi>47.</hi> of the first) for the an<g ref="char:EOLhyphen"/>gle EBD is a right angle: Wherefore that which is contayned vnder the lines AD and DC together with the square of the line EB, is equall to the squares of the lines EB and BD. Take away the square of the line EB which is com<g ref="char:EOLhyphen"/>mon to them both: Wherefore the residue, namely, that which is contayned vn<g ref="char:EOLhyphen"/>der the lines AD and DC, is equall to the square of the line DB. If therfore without a circle be taken a certaine point, and from that poynt be drawen to the circle two right lines, so that the one of them do cut the circle, and the other do
<pb n="108" facs="tcp:7062:138"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ouch the circle: the rectangle parallelogramme which is comprehended vnder the whole right line which cutteth the circle and that portion of the same line that lieth betwene the poynt and the vtter circumference of the circle, is equall to the square made of the line that toucheth the circle: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶Two Corollaries out of Campane.</head>
                  <p>
                     <hi>If from <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> and the selfe same poynt taken without a circle be drawen into the circle lines how many soeuer:</hi>
                     <note place="margin">
                        <hi>First Corolla<g ref="char:EOLhyphen"/>ry.</hi>
                     </note> 
                     <hi>the rectangle Parallelogrammes contayned vnder euery one of them and hys outward par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, are equall the one to the other.</hi>
                  </p>
                  <p>And thys is hereby manifest, for that euery one of those rectangle Parallelo<g ref="char:EOLhyphen"/>grammes are equall to the square of the line which is drawen from that poynt and toucheth the circle by thys 36. Proposition. Hereunto he addeth.</p>
                  <p>
                     <hi>If two lines drawen from one and the selfe same point do touch a circle,</hi>
                     <note place="margin">
                        <hi>Second Co<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note> 
                     <hi>they are equall the one to the other.</hi>
                  </p>
                  <p>Which although it neede no demonstration, for that the square of eyther of them is equall to that which is contayned vnder the line drawen from the same poynt and hys outward part: yet he thus proueth it.</p>
                  <p>Suppose that there be a circle <hi>BCD,</hi> whose
<figure/>
centre let be <hi>E,</hi> and without it take the point <hi>A.</hi> And from the poynt <hi>A</hi> drawe two lines <hi>AB</hi> and <hi>AD,</hi> which let touch the circle in the poyntes <hi>B</hi> and <hi>D.</hi> Then I say, that they are equall. Draw these right lines <hi>EB, ED,</hi> and <hi>AE.</hi> And by the 18. of thys booke, eyther of the angles at the poyntes <hi>B</hi> and <hi>D</hi> is a right angle. Wherefore (by the 47. of the first) the square of the line <hi>AE,</hi> is equall to the two squares of the lines <hi>AB</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd <hi>EB:</hi> and by the same reason, to the two squares of the lines <hi>AD</hi> and <hi>ED.</hi> Wherefore the two squares of the lines <hi>AB</hi> and <hi>EB,</hi> are equall to the two squares of the lines <hi>AD</hi> and <hi>ED.</hi> And for asmuch as the squares of the lines <hi>EB</hi> and <hi>ED</hi> are equall, therefore the two other squares of the lines <hi>AB</hi> and <hi>AD</hi> are also equall. Wher<g ref="char:EOLhyphen"/>fore the line <hi>AB</hi> is equall to the line <hi>AD</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be proued.</p>
                  <p>The same may be proued an other way: Draw a line from <hi>B</hi> to <hi>D.</hi> And (by the 5. of the first) the angle <hi>EBD</hi> is equall to the angle <hi>EDB.</hi> And for asmuch as the two an<g ref="char:EOLhyphen"/>gles <hi>ABE</hi> and <hi>ADE</hi> are equall, namely, for that they are right angles: if you take from them the equall angles <hi>EBD</hi> &amp; <hi>EDB,</hi> the two other angles remayning, namely, the angles <hi>ABD</hi> and <hi>ADB</hi> shall be equall. Wherefore (by the 6. of the first) the line <hi>AB</hi> is equall to the line <hi>AD.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>¶ Hereunto also Pelitarius addeth this Corollary.</head>
                  <p>
                     <hi>From a poynt geuen without <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> circle, can be drawen vnto a circle onely two touch lines.</hi>
                     <note place="margin">
                        <hi>Third Corol<g ref="char:EOLhyphen"/>lary.</hi>
                     </note>
                  </p>
                  <p>The former description remayning, I say that from the poynt A can be drawen vnto
<pb facs="tcp:7062:139"/>
the circle <hi>BCD</hi> no more touch lines, but the two lines <hi>AB</hi> and <hi>AD.</hi> For if it be pos<g ref="char:EOLhyphen"/>sible, let <hi>AF</hi> also be in the former figure a touch line, touching the circle in the poynt <hi>F.</hi> And prawe a line from <hi>E</hi> to <hi>F.</hi> And the angle at the point <hi>F</hi> shall be a right angle, by the 18. of this booke: Wherefore it is equall to the angle <hi>EBA,</hi> which is contrary to the 20. of the first.</p>
                  <p>This may also be thus proued. For asmuch as all the lines drawen from one and the selfe same poynt &amp; touching a circle are equall, as we haue before proued, but the lines <hi>AB</hi> and <hi>AF</hi> can not be equall, by the 8. Proposition of this booke, therefore the line <hi>AF</hi> can not touch the circle <hi>BCD.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>31.</hi> Theoreme. The <hi>37.</hi> Proposition. If without a circle be taken a certaine point, and from that point be drawen to the circle two right lines, of which, the one doth cut the circle and the other falleth vpon the circle, and that in such sort, that the rectangle parallelogramme which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the whole right line which cutteth the circle, and that portion of the same line that lieth betwene the point and the vtter circumfere<g ref="char:cmbAbbrStroke">̄</g>ce of the circle, is equall to the square made of the line that falleth vpon the circle: then that line that so falleth vpon the circle shall touch the circle.</head>
               <p>LEt the circle be ABC: and with<g ref="char:EOLhyphen"/>out
<figure/>
the same circle take a point, and let the same be D,<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>uerse of the former.</note> &amp; from the point D let there be drawen to the circle ABC two right lines DCA and DB: and let DCA cut the circle, and DB fall vpon the circle. And that in such sort, that that which is contayned vnder the lines AD and DC, be equall to the square of the line DB. Then I say, that y<hi rend="sup">e</hi> line DB toucheth the circle ABC. Drawe (by the <hi>17.</hi> of the third) from the poynt D a right line touching the circle ABC, and let the same be DE.<note place="margin">Construction.</note> And (by the first of the same) let the point F be the centre of the circle ABC: and draw these right lines FE, FB, and FD. Wherfore the angle FED is a right angle.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for asmuch as the right line DE toucheth the circle ABC, and the right line DCA cutteth the same, therfore (by the Proposition going before) that which is contayned vnder the lines AD and DC, is equall to the square of the line DE. But that which is contayned
<pb n="109" facs="tcp:7062:139"/>
vnder the lines AD and DC, is supposed to be equall to the square of the line DB. Wherefore the square of the line DE is equall to the square of the line DB. Wherefore also the line DE is equall to the line DB. And the line FE is equall to the line FB, for they are drawen from the centre to y<hi rend="sup">e</hi> circumference. Now therefore these two lines DE and EF are equall to these two lines DB and BF, and FD is a common base to them both. Wherefore (by the <hi>8.</hi> of the first) the angle DEF is equall to the angle DBF. But the angle DEF is a right angle. Wherefore also the angle DBF is a right angle. And y<hi rend="sup">e</hi> line FB being produced, shall be the diameter of the circle. But if from the end of the di<g ref="char:EOLhyphen"/>ameter of a circle be drawen a right line making right angles, the right line so drawen toucheth the circle (by the Correllary of the <hi>16.</hi> of the third). Wherfore the right line DB toucheth the circle ABC. And the like demonstration will serue if the centre be in the line AC. If therefore without a circle be taken a cer<g ref="char:EOLhyphen"/>taine point, and from that poynt be drawen to the circle two right lines, of which the one doth cut the circle, and the other falleth vpon the circle, and that in such sort, that the rectangle parallelogramme which is contayned vnder the whole right line which cutteth the circle, and that portion of the same line that lieth be<g ref="char:EOLhyphen"/>twene the poynt and the vtter circumference of the circle, is equall to the square made of the line that falleth vpon the circle: then the line that so falleth vpon the circle shall touch the circle: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other demonstration after Pelitarius.</head>
                  <p>Suppose that there be a circle <hi>BCD,</hi> whose
<figure/>
centre let be <hi>E:</hi>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Pelita<g ref="char:EOLhyphen"/>rius.</hi>
                     </note> and take a point without it, name<g ref="char:EOLhyphen"/>ly, <hi>A</hi>: And fro<g ref="char:cmbAbbrStroke">̄</g> the poynt <hi>A</hi> drawe two right lines <hi>ABD,</hi> and <hi>AC</hi>: of which let <hi>ABD</hi> cut the circle in the poynt <hi>B,</hi> &amp; let the other fall vpon it. And let that which is contained vnder the lines <hi>AD</hi> and <hi>AB,</hi> be equall to the square of the line <hi>AC.</hi> Then I say, that the line <hi>AC</hi> toucheth the circle. For first if the line <hi>ABD</hi> do passe by the centre, draw the right line <hi>CE.</hi> And (by the 6. of the second) that which is contayned vnder the lines <hi>AD</hi> and <hi>AB</hi> together with the square of the line <hi>EB,</hi> that is, with the square of the line <hi>EC</hi> (for the lines <hi>EB</hi> and <hi>EC</hi> are equall) is equall to the square of the line <hi>AE.</hi> But that which is contained vnder the lines <hi>AD</hi> and <hi>AB,</hi> is supposed to be equall to the square of the line <hi>AC</hi>: Wherefore the square of the line <hi>AC</hi> together with the square of the line <hi>CE,</hi> is equall to the square of the line <hi>AE.</hi> Wherefore (by the last of the first) the angle at the point <hi>C</hi> is a right angle. Wherfore (by the 18. of this boke) the line <hi>AC</hi> toucheth the circle.</p>
                  <p>
                     <pb facs="tcp:7062:140"/>But if the line <hi>ABD</hi> doo not passe by the centre, drawe from the point <hi>A</hi> the line <hi>AD,</hi> in which let be the centre <hi>E.</hi> And forasmuch as that which is contained vnder this whole line and his outward part, is equall to that which is contained vnder the lines <hi>AD</hi> and <hi>AB</hi> by the first Corollary before put, therefore the same is equall to the square of the line <hi>AC,</hi> wherefore the angle <hi>ECA</hi> is a right angle as hath before bene proued in the first part of this Proposition. And therfore the line <hi>AC</hi> toucheth the circle: Which was required to be proued.</p>
               </div>
            </div>
            <trailer>The ende of the third booke of Euclides Elementes.</trailer>
         </div>
         <div n="4" type="book">
            <pb n="110" facs="tcp:7062:140"/>
            <head>¶The fourth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>THIS FOVRTH BOOKE intreateth of the inscrip<g ref="char:EOLhyphen"/>tion &amp; circumscription of rectiline figures:<note place="margin">
                  <hi>The argument of this booke.</hi>
               </note> how one right lined figure may be inscribed within an other right lined figure, and how a right lined figure may be circumscribed about an other right lined figure, in such as may be inscribed and circumscribed within or about the other. For all right lined figures cannot so be inscribed or circumscribed within or about the other. Also it teacheth how a triangle, a square, and certayne other rectiline figures being regular may be inscribed within a circle. Also how they may be cir<g ref="char:EOLhyphen"/>cumscribed about a circle. Likewise how a circle may be inscribed within them. And how it may be circumscribed about them. And be<g ref="char:EOLhyphen"/>cause the maner of entreatie in this booke is diuers from the entreaty of the for<g ref="char:EOLhyphen"/>mer bookes, he vseth in this other wordes and termes then he vsed in them. The definitions of which in order here after follow.</p>
            <div type="part">
               <head>Definitions.</head>
               <div type="section">
                  <head>A rectiline figure is sayd to be inscribed in a rectiline figure,<note place="margin">
                        <hi>First defini<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> when euery one of the angles of the inscribed figure toucheth euery one of the sides of the figure wherin it is inscribed.</head>
                  <p>As the triangle <hi>ABC</hi> is inscribed in the triangle <hi>DEF,</hi> because that euery angle of the triangle inscribed, namely, the triangle <hi>ABC</hi> toucheth euery side of the triangle within which it is described, namely, of the triangle <hi>DEF.</hi> As the angle <hi>CAB</hi> toucheth
<figure/>
the side <hi>ED</hi> the angle <hi>ABC</hi> toucheth the side <hi>DF,</hi> and the angle <hi>ACB</hi> toucheth the side <hi>EF.</hi> So likewise the square <hi>ABCD</hi> is said to be inscribed within the square <hi>EFGH.</hi> for euery angle of it toucheth some one side of the other. So also the Pentagon or fiue angled figure <hi>ABCDE</hi> is inscribed within the Pentagon or fiue angled figure <hi>FGHIK</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> As you see in the figure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:141"/>
                  <head>Likewise a rectiline figure is said to be circumscribed about a rectiline figure,<note place="margin">
                        <hi>Second de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> when euery one of the sides of the figure cir<g ref="char:EOLhyphen"/>cumscribed, toucheth euery one of the angles of the figure a<g ref="char:EOLhyphen"/>bout which it is circumscribed.</head>
                  <p>As in the former descriptions the triangle <hi>DEF</hi> is said to be circumscribed about the triangle <hi>ABC,</hi> for that euery side of the figure circumscribed, namely, of the trian<g ref="char:EOLhyphen"/>gle <hi>DEF</hi> toucheth euery angle of the figure wherabout it is circumscribed. As the side <hi>DF</hi> of the triangle <hi>DEF</hi> circumscribed, toucheth the angle <hi>ABC</hi> of the triangle <hi>ABC</hi> about which it is circumscribed: and the side <hi>EF</hi> toucheth the angle <hi>BCA,</hi> and the side <hi>CD</hi> toucheth the angle <hi>CAB.</hi> Likewise vnderstand you of the square <hi>EFGH</hi> which is circumscribed about the square <hi>ABCD</hi>: for euery side of the one toucheth some one side of the other. Eue<g ref="char:cmbAbbrStroke">̄</g> so by the same reason the Pentagon <hi>FGHIK</hi> is circumscribed a<g ref="char:EOLhyphen"/>bout the Pentagon <hi>ABCDE,</hi> as you see in the figure on the other side. And thus may you of other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ectiline figures consider.</p>
                  <p>
                     <note place="margin">
                        <hi>The inscripti<g ref="char:EOLhyphen"/>tion and cir<g ref="char:EOLhyphen"/>cumscription of rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>res pertai<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>eth only to regular fi<g ref="char:EOLhyphen"/>gures.</hi>
                     </note>By these two definitions it is manifest, that the inscription and circumscription of rectiline figures here spoken of, pertayne to such rectiline figures onely, which haue equall sides and equall angles, which are commonly called regular. It is also to be noted that rectiline figures only of one kinde or forme can be inscribed or circumscribed the one within or about the other. As a triangle within or about a triangle: A square within or about a square: and so a Pentagon within or about a Pentago<g ref="char:cmbAbbrStroke">̄</g>, &amp; likewise of others of one forme. But a triangle can not be inscribed or circumscribed within or aboute a square: nor a square within or about a Penta<g ref="char:EOLhyphen"/>gon. And so of others of diuers kyndes. For euery playne rectiline figure hath so many angles as it hath sides. Wherfore the figure inscrided must haue so many an<g ref="char:EOLhyphen"/>gles as the figure in which it is inscribed hath sides: and the angles of the one (as is sayd) must touche the sides of the other. And contrariwise in circumscription of figures, the sides of the figure circumscribed must touch the angles of the figure a<g ref="char:EOLhyphen"/>bout which it is circumscribed.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The third de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>A rectiline figure is sayd to be inscribed in a circle, when eue<g ref="char:EOLhyphen"/>ry one of the angles of the inscribed figure toucheth the cir<g ref="char:EOLhyphen"/>cumference of the circle.</head>
                  <p>A circle by reason of his vniforme and regular distance which it hath from the centre to the circumference may easily touche all the angles of any regular recti<g ref="char:EOLhyphen"/>line figure within it: and also all the sides of any figure without it. And therfore a<g ref="char:EOLhyphen"/>ny regular rectiline figure may be inscribed within it, and also be circumscribed a<g ref="char:EOLhyphen"/>bout it. And agayne a circle may be both inscribed within any regular rectiline fi<g ref="char:EOLhyphen"/>gure, and also be circumscribed about it.</p>
                  <p>As the triangle <hi>ABC</hi> is inscribed in the circle <hi>ABC</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for that euery angle toucheth some one pointe of the circumference of the circle. As the angle <hi>CAB</hi> of the triangle <hi>ABC</hi> toucheth the point <hi>A</hi> of the circumference of the circle. And the angle <hi>ABC</hi>
                     <pb n="111" facs="tcp:7062:141"/>
of the triangle toucheth the pointe <hi>B</hi> of the circumfe<g ref="char:EOLhyphen"/>rence
<figure/>
of the circle. And also the angle <hi>ACB</hi> of the tri<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ngle <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the pointe <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the circumference of the circle. In like manner the square <hi>ADEF</hi> is inscri<g ref="char:EOLhyphen"/>bed in the same circle <hi>ABC</hi>: for that euery angle of the square inscribed, toucheth some one poynte of the circle in which it is inscribed. And so imagine you of rectilined figures.</p>
               </div>
               <div type="section">
                  <head>A circle is sayd to be circumscribed about a rectiline figure,<note place="margin">
                        <hi>The fourth definition.</hi>
                     </note> whe<g ref="char:cmbAbbrStroke">̄</g> the cir<g ref="char:EOLhyphen"/>cumference of the circle toucheth euery one of the angles of the figure about which it is circumscribed.</head>
                  <p>As in the former example of the third definition. The circle <hi>ADEF</hi> is circumscri<g ref="char:EOLhyphen"/>bed about the triangle <hi>ABC,</hi> because the circumference of the circle which is circum<g ref="char:EOLhyphen"/>scribed toucheth euery angle of the triangle about which it is circumscribed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> namely, the angles <hi>CAB, ABC,</hi> and <hi>BCA.</hi> Likewise the same circle <hi>ADEF</hi> is circumscribed about the square <hi>ADEF</hi> by the same definition as you may see.</p>
               </div>
               <div type="section">
                  <head>A circle is sayd to be inscribed in a rectiline figure,<note place="margin">
                        <hi>The fift de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> when the circumference of the circle toucheth euery one of the sides of the figure within which it is inscribed.</head>
                  <p>As the circle <hi>ABCD</hi> is inscribed within the triangle
<figure/>
                     <hi>EFG,</hi> because the circumference of the circle toucheth euery side of the triangle in which it is inscribed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> namely the side <hi>EF</hi> in the point <hi>B,</hi> and the side <hi>GF</hi> in the pointe <hi>C,</hi> and the side <hi>GE</hi> in the point <hi>D.</hi> Likewise by the same reason the same circle is inscribed within the square <hi>HIKL.</hi> And so may you iudge of other rectiline figures.</p>
               </div>
               <div type="section">
                  <head>A rectilined figure is said to be circum<g ref="char:EOLhyphen"/>scribed about a circle,<note place="margin">
                        <hi>The sixt de<g ref="char:EOLhyphen"/>uition.</hi>
                     </note> when euery one of the sides of the fi<g ref="char:EOLhyphen"/>gure circumscribed toucheth the circumference of the circle.</head>
                  <p>As in the former figure of the fift definition, the triangle <hi>EFG</hi> is circumscribed a<g ref="char:EOLhyphen"/>bout the circle <hi>ABCD,</hi> for that euery side of the same triangle beyng circumscribed toucheth the circumference of the circle, about which it is circumscribed. As the side <hi>EG</hi> of the triangle <hi>EFG</hi> toucheth the circumference of the circle in the point <hi>D</hi>: and the side <hi>EF</hi> toucheth it in the point <hi>B</hi>: and the side <hi>GF</hi> in the point <hi>C.</hi> Likewise also the square <hi>HIKL</hi> is circumscribed about the circle <hi>ABCD,</hi> for euery one of his sides toucheth the circumference of the circle, namely, in the pointes <hi>A, B, C, D.</hi> And thus consider of all other regular right lined figures (for of them onely are vnderstanded these definitions) to be circumscribed about a circle, or to be inscribed within a cir<g ref="char:EOLhyphen"/>cle: or of a circle to be circumscribed or inscribed about or within any of them.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:142"/>
                  <head>
                     <note place="margin">
                        <hi>Seuenth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>A right lyne is sayd to be coapted or applied in a circle, when the extremes or endes therof, fall vppon the circumference of the circle.</head>
                  <p>As the line <hi>BC</hi> is sayd to be coapted or to be appli<g ref="char:EOLhyphen"/>ed
<figure/>
to the circle <hi>ABC,</hi> for that both his extremes fall vpon the circumference of the circle in the pointes <hi>B</hi> and <hi>C.</hi> Likewise the line <hi>DE.</hi> This definition is very necessary, and is properly to be taken of any lyne ge<g ref="char:EOLhyphen"/>uen to be coapted and applied into a circle, so <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hat it exceede not the diameter of the circle geuen.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Probleme. The <hi>1.</hi> Proposition. In a circle geuen, to apply a right line equall vnto a right line geuen, which excedeth not the diameter of a circle.</head>
               <p>SVppose that the circle geuen be ABC, and let the right line geuen, exceding not the diameter of the same circle, be D. Now it is required in the circle geuen ABC, to ap<g ref="char:EOLhyphen"/>ply a right line equall vnto the right line D.<note place="margin">Construction.</note> Draw the diameter of the circle ABC, and let the same be BC. Now if the line BC be equall vnto the line D, then is that done which was required. For in the circle geuen ABC is applyed a right line
<figure/>
BC equall vnto y<hi rend="sup">e</hi> right line D.<note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> But if not,<note place="margin">First case<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> then is the line BC greater then y<hi rend="sup">e</hi> line D.<note place="margin">Second case.</note> And (by the third of the first) put vnto the line D an equall line CE. And making the centre C, and the space CE, describe (by the third petition) a circle EGF, cut<g ref="char:EOLhyphen"/>ting the circle ABC in the point F, &amp; draw a line from C to F. And for asmuch as the point C is y<hi rend="sup">e</hi> centre of the circle EGF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore (by the <hi>13.</hi> definition of the first) the line CF is equall vnto the line CE. But the line CE is equall vnto the line D. Wherefore (by the first common sentence) the
<pb n="111" facs="tcp:7062:142" rendition="simple:additions"/>
line CF also is equall vnto the line D. Wherefore in the circle geuen ABC, is applyed a right line CA equall vnto the right line geuen D: which was requi<g ref="char:EOLhyphen"/>red to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Probleme. The <hi>2.</hi> Proposition. In a circle geuen, to describe a triangle equiangle vnto a tri<g ref="char:EOLhyphen"/>angle geuen.</head>
               <p>SVppose that the circle geuen be ABC: and let the triangle geuen be DEF. Now it is required in the circle geuen ABC to describe a tri<g ref="char:EOLhyphen"/>angle equiangle vnto the triangle geuen DEF.<note place="margin">Construction.</note> Draw (by the <hi>17.</hi> of the third) a right line touching the circle ABC, and let the same be GAH, and let it touch in the point A. And (by the <hi>23.</hi> of the first) vnto y<hi rend="sup">e</hi> right line AH, and vnto the point in it A, describe an angle HAC equall vnto the angle DEF. And (by the selfe same) vnto the right line AG, and vnto the point in it A, make an an<g ref="char:EOLhyphen"/>gle
<figure/>
GAB equall vnto the angle DFE. And draw a right line from B to C. And for asmuch as a cer<g ref="char:EOLhyphen"/>taine right line GAH toucheth y<hi rend="sup">e</hi> circle ABC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and from the point where it toucheth, namely, A, is drawen into the circle a right line AC; therefore (by the <hi>31.</hi> of the third) the angle HAC is equall vnto the angle ABC which is in the alternate seg<g ref="char:EOLhyphen"/>ment of the circle. But the angle HAC is equall to the angle DEF. Wherfore the angle ABC is equall to the angle DEF. And by the same reason, the an<g ref="char:EOLhyphen"/>gle ACB is equall to the angle DFE. Wherefore the angle remayning, BAC, is equall vnto the angle remayning, EDF. Wherefore the triangle ABC is e<g ref="char:EOLhyphen"/>quiangle vnto the triangle DEF. And it is described in the circle geuen ABC. Wherefore in a circle geuen, is described a triangle equiangle vnto a triangle ge<g ref="char:EOLhyphen"/>uen: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Probleme. The <hi>3.</hi> Proposition.
About a circle geuen, to describe a triangle equiangle vnto a triangle geuen.</head>
               <p>
                  <pb facs="tcp:7062:143"/>SVppose y<hi rend="sup">t</hi> the circle geuen be ABC, and let the triangle geuen be DEF. It is required about the circle ABC to describe a triangle equiangle vn<g ref="char:EOLhyphen"/>to the triangle DEF.<note place="margin">Construction.</note> Extend the line EF on ech side to the poyntes G and H. And (by the first of the third) take the centre of the circle ABC, and let the same be the point K. And then draw a right line KB. And (by the <hi>23.</hi> of the first) vnto the right line KB, and vnto the point in it K, make an angle BKA equall vnto the angle DEG, and likewise make the angle BKC equall vnto the angle DFH. And (by the <hi>17.</hi> of the third) draw right lines touching the circle A, B, C, in the pointes A, B, C. And let y<hi rend="sup">e</hi> same be LAM, MBN, and NCL.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for asmuch as the right lines LM, MN, &amp; NL, do touch the circle ABC in the pointes A, B, C, and from the centre K vnto the pointes A, B, C, are drawen right lines KA, KB, and KC, therefore the angles which are at the pointes A, B, C,
<figure/>
are right angles (by y<hi rend="sup">e</hi> 
                  <hi>18.</hi> of the third). And for as<g ref="char:EOLhyphen"/>much as the fower angles of the fower sided figure AMBK, are equall vnto fower right angles: whose angles KAM, &amp; KBM, are two right angles: ther<g ref="char:EOLhyphen"/>fore the angles remayning AKB, and AMB, are equall to two right angles. And the angles DEG, &amp; DEF, are (by the <hi>13.</hi> of the first) equall to two right angles. Wherefore the angles AKB, and AMB, are equall vnto the angles DEG, and DEF: of which two angles the angle AKB is equall vnto the angle DEG: Wherefore the angle remayning, AMB, is equall vnto the an<g ref="char:EOLhyphen"/>gle remayning, DEF. In like sort may it be proued, that the angle LNM, is equall to the angle DFE. Wherfore the angle remayning MLN, is equall vn<g ref="char:EOLhyphen"/>to the angle remayning EDF. Wherefore the triangle LMN, is equiangle vnto the triangle DEF: and it is described about the circle ABC. Wherefore about a circle geuen is described a triangle equiangle vnto a triangle geuen: which was required to be done.</p>
               <div type="section">
                  <head>¶ An other way after Pelitarius.</head>
                  <p>
                     <note place="margin">
                        <hi>An other way after Peli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>a<g ref="char:EOLhyphen"/>rius.</hi>
                     </note>In the circle ABC inscribe a triangle GHK equiangle to the triangle EDF (by the former Proposition): so that let the angle at the poynt G be equall to the angle D, and let the angle at the point H be equall to the angle E: and let also the angle at the poynt K be equall to the angle F. Then drawe the line LM parallel to the line GH, which let touch the circle in the poynt A (which may be done by the Proposition ad<g ref="char:EOLhyphen"/>ded of the sayd Pelitarius after the 17. Proposition).<note place="margin">
                        <hi>Construction.</hi>
                     </note> Draw likewyse the line MN pa<g ref="char:EOLhyphen"/>rallel vnto the line HK and touching the circle in the poynt B: And also draw the line
<pb n="112" facs="tcp:7062:143"/>
LN parallel vnto the line
<figure/>
GK and touching the cir<g ref="char:EOLhyphen"/>cle in the poynt C. And these three lines shall vn<g ref="char:EOLhyphen"/>doubtedly concurre, as in the poyntes L, M, and N,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> which may easily be pro<g ref="char:EOLhyphen"/>ued, if you produce on ei<g ref="char:EOLhyphen"/>ther side the lines GH, GK, and HK, vntill they cut the lines LM, LN, and MN, in the poyntes O, P, Q, R, S, T. Now I say, that the triangle LMN circu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>scribed about the circle ABC, is equiangle to the triangle DEF. For it is manifest, that it is equian<g ref="char:EOLhyphen"/>gle vnto the tria<g ref="char:cmbAbbrStroke">̄</g>gle GHK, by the proprietie of parallel lines. For the angle MTQ is equall to the angle at the poynt G of the triangle GHK (by the 29. of the first) and therefore also the angle at the poynt L, is equall to the selfe same angle at the poynt G (for the angle at the point L, is by the same 29. Proposition, equall to the angle MTQ). And by the same rea<g ref="char:EOLhyphen"/>son the angle at the poynt M, is equall to the angle at the poynt H of the selfe same tri<g ref="char:EOLhyphen"/>angle: and the angle at the poynt N, to the angle at the poynt K. Wherfore the whole triangle LMN, is equiangle to the whole triangle GHK: Wherfore also it is equiangle to the triangle DEF: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Probleme. The <hi>4.</hi> Proposition. In a triangle geuen, to describe a circle.</head>
               <p>SVppose that the triangle geuen be ABC. It is required to describe a cir<g ref="char:EOLhyphen"/>cle in the triangle ABC.<note place="margin">Construction.</note> Deuide (by the <hi>9.</hi> of the first) the angles ABC, and ACB into two equall par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es by two right lines BD and CD. And let these right lines meete together in the point D. And (by the <hi>12.</hi> of the first) from the point D draw vnto the right lines AB, BC and CA perpendi<g ref="char:EOLhyphen"/>cular lines, namely, DE, DF, and DG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the angle ABD is equall to tho angle CBD, and the right angle
<figure/>
BED is equall vnto the right angle BFD. Now then there are two triangles EBD and FBD hauing two angles equall to two angles, and one side equall to one side, namely, BD which is common to them both, and subtendeth one of the equall angles. Wherfore (by the <hi>26.</hi> of the first) the rest of the sides are equall vnto the rest of the sides. Wherfore the line DE is equall vnto the line DF: and by the same rea<g ref="char:EOLhyphen"/>son also the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne DG is equall vnto y<hi rend="sup">e</hi> line DF.
<pb facs="tcp:7062:144"/>
Wherfore these three right lines DE, DF, &amp;
<figure/>
DG, are equall the one to the other (by the first common sentence). Wherefore making the cen<g ref="char:EOLhyphen"/>tre the point D, and the space DE, or DF, or DG, describe a circle and it will passe through the pointes E, F, G, and will touch y<hi rend="sup">e</hi> right lines AB, BC, and CA. For the angles made at the pointes E, F, G, are right angles. For if y<hi rend="sup">e</hi> circle cut those right lines, then fro<g ref="char:cmbAbbrStroke">̄</g> the end of the di<g ref="char:EOLhyphen"/>ameter of the circle shall be drawen a right line making two right angles,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impos<g ref="char:EOLhyphen"/>sibilitie.</note> &amp; falling within the circle: which is impossible, as it was manifest (by the <hi>16.</hi> of the third). Where<g ref="char:EOLhyphen"/>fore the circle described, D being the centre therof, and the space therof being ei<g ref="char:EOLhyphen"/>ther DE, or DF, or DG, cutteth not these right lines AB, BC, &amp; CA. Wher<g ref="char:EOLhyphen"/>fore (by the Corollary of the same) it toucheth them, and the circle is described in the triangle ABC. Wherfore in the triangle geuen ABC, is described a circle EFG: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Probleme. The <hi>5.</hi> Proposition. About a triangle geuen, to describe a circle.</head>
               <p>SVppose that the triangle geuen be ABC. It is required about the trian<g ref="char:EOLhyphen"/>gle ABC to describe a circle. Deuide (by the <hi>10.</hi> of the first) the right l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nes AB and AC into two equall partes in the pointes D and E. And from the pointes D and E (by the <hi>11.</hi> of the first) rayse vp vnto the lines AB &amp; AC two perpendicular lines DF and EF.<note place="margin">Three cases in this Proposi<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n.</note> Now these perpendicular lines meete together either within the triangle ABC, or in the right line BC, or els witho<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the right line BC.</p>
               <p>
                  <note place="margin">The <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> case.</note>First let them meete together within
<figure/>
the triangle in the point F. And (by y<hi rend="sup">e</hi> first peticion) drawe right lines from F to B, from F to C, and from F to A. And foras<g ref="char:EOLhyphen"/>much as the lyne AD is equall vnto the line DB, and the line DF is common vn<g ref="char:EOLhyphen"/>to them both, and maketh the angles on ech side of him right angles, therfore (by the <hi>4.</hi> of the first) the base AF is equall vnto the base FB. In like sorte may we proue that the line CF is equall vnto the lyne AF. Wherfore the line FB is equall vnto the line CF. Wherfore these three right lines FA, FB, and FC are equall the one <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>
                  <pb n="113" facs="tcp:7062:144"/>
the other. Wherfore making the centre the point F, and the space FA, or FB, or FC describe a circle and it shall passe by the poyntes ABC. And so there is a cir<g ref="char:EOLhyphen"/>cle described about the triangle ABC, as ye see in the first description.</p>
               <p>But now suppose that the right lines
<figure/>
DF and EF do meete together vpon the right line BC in the point F, as it is in the second description,<note place="margin">The second case.</note> and draw a right lyne from A to F, and in like sorte may we proue that the poynte F is the centre of the circle described aboute the triangle ABC.</p>
               <p>But now suppose that the right lines DF and EF do meete together without the triangle ABC in the point F.<note place="margin">The third case.</note> Againe as it is in the third description draw right lines from F to A, from F to B, and from F to C. And forasmuch as the line AD is equall vnto the line DB, and the line DF is com<g ref="char:EOLhyphen"/>mon vnto them both, and maketh a right an<g ref="char:EOLhyphen"/>gle on eche side of him, wherfore (by the <hi>4.</hi> of the first) the base AF is equall vnto the base BF. And in like sort may we proue that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line CF is equal vnto the line AF. Wherfore agayne making F the centre, and the space FA, or FB, or FC, describe a circle and it shal passe by the pointes A, B, C, and so is there a circle described about the triangle ABC, as ye see it i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the third description. Wherfore about a triangle geuen is described a circle: which was required to be done.</p>
               <div type="section">
                  <head>Correlary.</head>
                  <p>Hereby it is manifest, that when the centre of the circle fal<g ref="char:EOLhyphen"/>leth within the triangle, the angle BAC being in a greater segment of a circle is lesse the<g ref="char:cmbAbbrStroke">̄</g> a right angle. But when it fal<g ref="char:EOLhyphen"/>leth vpon the right line BC the angle BAC being in a semi<g ref="char:EOLhyphen"/>circle is a right angle. But when the centre falleth without the right line BC, the angle BAC being in a lesse segment of a circle, is greater then a right angle. Wherfore also when the angle geuen is lesse then a right angle, the right lines DF
<pb facs="tcp:7062:145"/>
and EF will meete together within the sayd triangle. But when it is a right angle they will meete together vpon the line BC. But whe<g ref="char:cmbAbbrStroke">̄</g> it is greater then a right angle, they will meete together without the right line BC.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Probleme. The <hi>6.</hi> Proposition. In a circle geuen, to describe a square.</head>
               <p>SVppose that the circle geuen be ABCD. It is required in the circle ABCD to describe a square. Draw in the circle ABCD two diameters making right angles,<note place="margin">Construction.</note> and let the same be AC and BD, and drawe right lines from A to B, from B to C, from C to D,
<figure/>
and from D to A. And forasmuch as the line BE is equall vnto the line ED (by the <hi>15.</hi> definition of the first) for the point E is the centre.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And the line EA is common to them both, making on eche side a right angle: ther<g ref="char:EOLhyphen"/>fore (by the <hi>4.</hi> of the first) the base AB is e<g ref="char:EOLhyphen"/>quall vnto the base AD. And by the same reason also either of these lines BC and CD is equall to either of these lines AB and AD: wherefore ABCD is a figure of foure equal sides. I say also that it is a rectangle figure. For forasmuch as the right line BD is the diameter of the circle ABCD, therfore the angle BAD beyng in the se<g ref="char:EOLhyphen"/>micircle is a right angle (by the <hi>31.</hi> of the third) And by the same reason euery one of these angles ABC, BCD and CDA is a right angle. Wherfore the foure sided figure ABCD is a rectangle figure, and it is proued that it consisteth of equall sides. Wherfore (by the <hi>30.</hi> definition of the first) it is a square, and it is described in the circle ABCD: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Probleme. The <hi>7.</hi> Proposition. About a circle geuen, to describe a square.</head>
               <p>SVppose that the circle geuen be ABCD. It is required about the cir<g ref="char:EOLhyphen"/>cle ABCD to describe a square. Draw in the circle ABCD two diameters making right angles,<note place="margin">Construction.</note> where they cut the one the other, and let the same be AC and BD. And by the pointes A, B, C, D, draw
<pb n="114" facs="tcp:7062:145"/>
(by the <hi>17.</hi> of the third) right lines touching
<figure/>
the circle ABCD, and let the same be FG, GH, HK, and KF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now for asmuch as the right line FG toucheth the circle ABCD in the point A, and from the centre E to the point A where the touch is, is drawen a right line EA, therfore (by the <hi>18.</hi> of the third) the angles at the point A are right angles, and by the same reason the angles which are at the pointes B, C, D, are also right angles. And forasmuch as the angle AEB is a right an<g ref="char:EOLhyphen"/>gle, &amp; the angle EBG is also a right angle, therfore (by the <hi>28.</hi> of the first) the line GH is a parallel vnto the line AC: and by the same reason the lyne AC is a parallel vnto the lyne FK. In like sorte also may we proue that either of these lines GF and HK is a parallel vnto the lyne BED. Wherfore these figures GK, GC, AK, FB, and BK are parallelogrames<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore (by the <hi>34.</hi> of the first) the line GF is equall vnto the line HK, and the line GH is equall vnto the line FK. And forasmuch as the line AC is equall vnto the lyne BD, but the line AC is equall vnto either of these lines GH and FK: an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line BD is equall to either of these lines GF and HK. Wherfore either of these lines GH and FK is equall to either of these lines GF and HK. Wherfore the figure FGHK consisteth of foure equall sides. I say also that it is a rectangle figure. For foras<g ref="char:EOLhyphen"/>much as GBEA is a parallelogramme, and the angle AE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is a right angle: therfore (by the <hi>34.</hi> of the first) the angle AGB is a right angle. In like sorte may we proue that the angles at the poyntes H, K, and F are right angles. Wherfore FGHK is a rectangle foure sided figure, and it is proued that it con<g ref="char:EOLhyphen"/>sisteth of equall sides: wherfore it is a square, and it is described about the circle ABCD. Wherfore about a circle geuen is described a square: which was requi<g ref="char:EOLhyphen"/>red to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Probleme. The <hi>8.</hi> Proposition. In a square geuen, to describe a circle.</head>
               <p>SVppose that the square geuen be ABCD. It is required in the square ABCD to describe a circle.<note place="margin">Construction.</note> Deuide (by the <hi>10.</hi> of the first) either of these lines AB and AD into two equall partes in the pointes E and F. And by the point E (by the <hi>31.</hi> of the first) draw a line EH parallel vnto either of these lines AB and DC:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and (by the same) by the point F draw a line FK parallel vnto either of these lines AD and BC. Wherfore euery one of these fi<g ref="char:EOLhyphen"/>gures
<pb facs="tcp:7062:146"/>
AK, KB, AH, HD, AG, GC, BG, and GD is a parallelograme, and the sides which are opposite the one to the other, are (by the <hi>34.</hi> of the first) equall the one to the other. And forasmuch as the line AD is equall vnto the line AB, and the halfe of the line AD is the line AE, and the halfe of the line AB, is the line
<figure/>
AF, therefore the line AE is equall vnto the line AF: wherefore (by the same) the sides which are opposite are equall. Wherefore the line FG is equal vnto the line EG. In like sort may we proue that either of these lines GH, and GK is equall to either of these lines FG and GE. Wherfore (by the first common sen<g ref="char:EOLhyphen"/>tence) these foure lines GE, GF, GH, and GK are equall the one to the other. Wherfore making the centre G, and the space either GE, or GF, GH, or GK, describe a circle and it will passe by the pointes E, F, H, K, and will touche the right lines AB, BC, CD, and DA. For the angles at the pointes E, F, H, K, are right angles. For if the circle do cut the right lines AB, BC, CD, and DA, then the line which is drawen by the ende of the diameter of the circle making right angles should fall within the circle, which is impossible (by the <hi>16.</hi> of the third) Wherfore the cen<g ref="char:EOLhyphen"/>tre being the poynt G and the space beyng GE, or GF, or GH, or GK if a cir<g ref="char:EOLhyphen"/>cle be described, it shall not cut the rig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t lines AB, BC, CD, and DA. Where<g ref="char:EOLhyphen"/>fore it shall touch them. And it is described in the square ABCD: wherefore in a square geuen is described a circle: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Probleme. The <hi>9.</hi> Proposition. About a square geuen, to describe a circle.</head>
               <p>SVppose that the square geuen be AB
<figure/>
CD. It is required about the square ABCD to describe a circle. Drawe right lines from A to C, and from D to B, &amp; let them cut the one the other in the poynt E.<note place="margin">Construction.</note> And forasmuch as the lyne DA is equall vn<g ref="char:EOLhyphen"/>to the lyne AB,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and the line AC is common vnto them both, therfore these two lines DA and AC are equall vnto these two lynes BA and AC, the one to the other. And the base DC is equall vnto the base BC. Wherefore (by the <hi>8.</hi> of the first) the angle DAC is equall vnto the angle BAC. Where<g ref="char:EOLhyphen"/>fore the angle DAB is deuided into two equall partes by the line AC. And in
<pb n="115" facs="tcp:7062:146"/>
like sor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> may we proue that euery one of these angles ABC, BCD, and CDA is deuided into two equall partes by the right lines AC and DB. And foras<g ref="char:EOLhyphen"/>much as the angle DAB is equall vnto the angle ABC, and of the angle DAB the angle EAB is the halfe; and of the angle ABC the angle EBA is the halfe: Therfore the angle EAB is equall vnto the angle EBA: wherfore (by the <hi>6.</hi> of the first) the side EA is equall vnto the side EB. In like sorte may we proue that either of these right lines EA and EB is equall vnto either of these lines EC and ED. Wherfore these foure lines EA, EB, EC, and ED are e<g ref="char:EOLhyphen"/>quall the one to the other. Wherfore making the centre E, and the space any of these lines EA, EB, EC, or ED. Describe a circle and it will passe by the pointes A, B, C, D, and shall be described about the square ABCD, as it is euident in the figure ABCD. Wherfore about a square geue<g ref="char:cmbAbbrStroke">̄</g> is described a cir<g ref="char:EOLhyphen"/>cle: which was required to be done.</p>
               <div type="section">
                  <head>¶ A Proposition added by Pelitarius.</head>
                  <p>
                     <hi>A square circumscribed about a circle, is double to the square inscribed in the same circle.</hi>
                  </p>
                  <p>Suppose that the square <hi>ABCD</hi> be cir<g ref="char:EOLhyphen"/>cumscribed about the circle <hi>EFGH,</hi>
                     <note place="margin">
                        <hi>A Propos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tion added by Pe<g ref="char:EOLhyphen"/>litarius.</hi>
                     </note> whose centre let be <hi>K.</hi> And let the poyntes of the
<figure/>
touches be <hi>E, F, G, H.</hi> And drawing these two diameters <hi>EG,</hi> and <hi>FH,</hi> and these right lines <hi>EF, FG, GH,</hi> and <hi>HB,</hi> there shall be inscri<g ref="char:EOLhyphen"/>bed in the circle a square <hi>EFGH</hi> (by the sixt of this booke). Then I say, that the square <hi>ABCD,</hi> is double to the square <hi>EFGH.</hi> For forasmuch as the side <hi>AB</hi> of the greater square, is (by the 34. of the first) equall to <hi>FH,</hi> which is the diameter of the lesse square: but the square of <hi>FH</hi> is double to the square whose diameter it is, namely, to the square <hi>EFGH</hi> (by the 47. of the first). Wherefore also the square of <hi>AB</hi> which is <hi>ABCD,</hi> is double to the square <hi>EFGH</hi>: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
                  <p>Thys may also be demonstrated by the equalite of the triangles and squares contayned in the great squares.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Probleme. The <hi>10.</hi> Proposition. To make a triangle of two equall sides called Isosceles, which shall haue eyther of the angles at the base double to the o<g ref="char:EOLhyphen"/>ther angle.</head>
               <p>
                  <pb facs="tcp:7062:147" rendition="simple:additions"/>
                  <hi>TAke a right line at all aduentures which let be AB, &amp; (by y<hi rend="sup">e</hi>
                  </hi> 11. <hi>of the second) let it be so deuided in y<hi rend="sup">e</hi> pointe C,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>y<hi rend="sup">t</hi> the rectangle figure comprehended vnder the lines AB and BC be equall vnto y<hi rend="sup">e</hi> square which is made of the line AC. And making the centre the point A, &amp; the space AB, describe
<figure/>
(by the</hi> 3. <hi>peticion) a circle BDE, and (by the</hi> 1. <hi>of the fourth) into the circle BDE apply a right line BD equall to the right lyne AC which is not greater then the diameter of the circle BDE. And draw lines from A to D and from D to C. And (by the</hi> 5. <hi>of y<hi rend="sup">e</hi> fourth) about the triangle ACD describe a circle ACDF.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>And forasmuch as the rect<g ref="char:EOLhyphen"/>angle figure contained vn<g ref="char:EOLhyphen"/>der the lines AB and BC is equall to the square which is made of the line AC: (For that is by supposition) But the line AC is equall vnto the line BD. Wherfore that which is contayned vnder the lines AB and BC is equall to the square which is made of the line BD. And forasmuch as without the circle ACDF is taken a poynt B, and from B vnto the circle ACDF are drawen two right lines BCA, and BD, in such sort that the one of them cutteth the circle, and the other endeth at the circumfe<g ref="char:EOLhyphen"/>rence, and that which is contained vnder the lines AB and BC is equall to the square which is made of the line BD, therfore (by the</hi> 17. <hi>of the third) the line BD toucheth the circle ACDF. And forasmuch as the line BD toucheth in the point D, and from D where the touche is, is drawen a right line DC, there<g ref="char:EOLhyphen"/>fore (by the</hi> 32. <hi>of the same) the angle BDC is equall vnto the angle DAC, which is in y<hi rend="sup">e</hi> alternate segment of y<hi rend="sup">e</hi> circle. And forasmuch as y<hi rend="sup">e</hi> angle BDC is e<g ref="char:EOLhyphen"/>qual vnto y<hi rend="sup">e</hi> angle DAC, put the angle CDA common vnto the<g ref="char:cmbAbbrStroke">̄</g> both. Wherfore y<hi rend="sup">e</hi> whole angle BDA is equal to these two angles CDA, &amp; DAC. But vnto y<hi rend="sup">e</hi> angles CDA, &amp; DAC is equall the outward angle BCD (by the</hi> 32. <hi>of the</hi> 1.<hi>) Wherfore y<hi rend="sup">e</hi> angle BDA is equal vnto y<hi rend="sup">e</hi> angle BCD. But y<hi rend="sup">e</hi> angle BDA is (by y<hi rend="sup">e</hi>
                  </hi> 5. <hi>of the first) equall vnto the angle CBD, for (by the</hi> 15. <hi>definition of y<hi rend="sup">•</hi> first) the side AD is equall vnto the side AB: wherfore (by the</hi> 1. <hi>common sentence) the angle DBA is equall vnto the angle BCD. Wherefore these three angles BDA, DBA, and BCD are equall the one to the other. And forasmuch as the angle DBC is equall vnto the angle BCD, the side therfore BD is equall vnto the side DC. But the line BD is by supposition equall vnto the lyne CA.</hi>
                  <pb n="117" facs="tcp:7062:147"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 page">
                     <desc>〈1 page〉</desc>
                  </gap>
                  <pb facs="tcp:7062:148"/>
two right angles. And either of the Angles at the base<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift partes of two right angles, or foure fift partes of one right angle. Which shall manifestly appeare, if we deuide two right angles into fiue partes. For then in thys kinde of triangle, the angle at the toppe shall be one fift part, and eyther of the two angles at the base shall be two fift partes.</p>
               <p>Thys also is to be noted, that the line <hi>AC</hi> is the side of an equilater Pentagon to be inscribed in the circle <hi>ACD.</hi> For by the latter construction it is manifest, that the three arkes <hi>AC, CD,</hi> and <hi>DE,</hi> of the lesse circle, are equall. And forasmuch as by the same it is manifest that the two lines <hi>AD</hi> and <hi>AE</hi> are equall, the arke also <hi>AE</hi> shall be equall to the arke <hi>AD</hi> (by the 20. of the third). Wherefore their halfes also are e<g ref="char:EOLhyphen"/>quall. If therefore the arke <hi>AE</hi> be (by the 30. of the third) deuided into two equall partes, the whole circu<g ref="char:cmbAbbrStroke">̄</g>ference <hi>ACDEA</hi> shall be deuided into fiue equall arkes. And forasmuch as the lines subtending the sayd equall arkes are (by the 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the same) e<g ref="char:EOLhyphen"/>quall, therefore euery one of the sayd sides shall be the side of an equilater Pentagon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued. And the same line <hi>AC</hi> shall be the side of an equila<g ref="char:EOLhyphen"/>ter ten angled figure to be inscribed in the circle <hi>BDE</hi>: the demonstration wherof I omitte, for that it is demonstrated by Propositions following.</p>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>A Proposition added by Pe<g ref="char:EOLhyphen"/>tarilius.</hi>
                     </note>¶ A Proposition added by Pelitarius.</head>
                  <p>
                     <hi>Vpon a right line geuen being finite, to describe an equilater and equiangle Pentagon figure.</hi>
                  </p>
                  <p>Suppose that the right line geuen be <hi>AB,</hi> vpon which it is required to describe an equilater and equiangle Pentagon. Vpon the line <hi>AB</hi> describe (by the 23. and 32. of the first) an Isosceles triangle <hi>ABC</hi> equiangle to the Isosceles triangle described by the former Proposition: namely, let the angles <hi>CAB</hi> and <hi>CBA,</hi> at the base <hi>AB,</hi> be equall to the two angles <hi>ABD</hi> and <hi>ADB</hi> in the former construction: so that eyther of them shall be two fift partes of two right angles, and the angle at the toppe, namely, the angle <hi>C,</hi> shall be one fift part. Then deuide the angle <hi>C</hi> into two equall parte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by drawing the right line <hi>CD.</hi> And vpon the line <hi>AC,</hi> and vnto the poynt <hi>A,</hi> describe the angle <hi>CAD</hi> equall to the angle <hi>ACD,</hi> by drawing the line <hi>AD,</hi> which line <hi>AD</hi> let concurre with the line <hi>CD,</hi> in the poynt <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and that within the triangle <hi>ABC,</hi> for the line <hi>CD</hi> being produced, shall fall vpon the base <hi>AB,</hi>
                     <figure/>
and the line <hi>AD</hi> vpon the side <hi>BC.</hi> And draw a line from the poynt <hi>D</hi> to the poynt <hi>B.</hi> And for asmuch as in the triangle <hi>ACD</hi> the two angles <hi>A</hi> and <hi>C</hi> are equall, ther<g ref="char:EOLhyphen"/>fore (by the 6. of the first) the two sides <hi>AD</hi> and <hi>CD</hi> are equall. Againe forasmuch as the two sides <hi>CB</hi> and <hi>CD</hi> of the triangle <hi>CBD,</hi> are equall to the two sides <hi>CA</hi> and <hi>CD</hi> of the triangle <hi>ACD,</hi> and the angle <hi>C</hi> of the one, is equall to the angle <hi>C</hi> of the other (by construction) therefore (by the 4. of the first) the base <hi>DB</hi> is equall to the base <hi>DA,</hi> and so is equall to the line <hi>DC.</hi> Wherefore (by the 9. of the third) the poynt <hi>D</hi> shall be the centre of the circle described about the triangle <hi>ABC.</hi> Describe the sayd circle and let it be <hi>ABECF.</hi> Now then the an<g ref="char:EOLhyphen"/>gle <hi>ADB</hi> is double to the angle <hi>ACD</hi> (by the 20. of the third). Wherefore the angle <hi>ADB</hi> maketh two fift partes of two right angles, that is, one fift part of foure right angles. And for asmuch as the space about the centre <hi>D</hi> is equall to foure right angles, then if the sayd space be deuided into fiue angles equall to the angle <hi>ADB,</hi> namely, into fine fift partes, by drawing the right lines <hi>DE</hi> &amp; <hi>DF,</hi> which with the lines <hi>DA, DB,</hi> and <hi>DC,</hi> will cause the sayd space to be deuided into fiue equall partes, and if also there be drawen these right lines <hi>AF, FC, CE,</hi> and <hi>EB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> there shall be described a rectiline Pentagon figure <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ECF,</hi> which shall be equila<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er, by the rule of a circle and of a circumference, and helpe of the 4. Proposition of the
<pb n="118" facs="tcp:7062:148"/>
first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and shall also be equiangle (by the 4. and 5. of the same). For the fiue angles <hi>A, B, E, C, F,</hi> are deuided ech into ten equall partes: which was required to be done.</p>
                  <p>If we consider well thys demonstration of Pelitarius, it will not be hard for vs, vpon a right line geuen to describe the rest of the figures whose inscriptions here<g ref="char:EOLhyphen"/>after followe.<note place="margin">
                        <hi>Note.</hi>
                     </note>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>11.</hi> Probleme. The <hi>11.</hi> Proposition. In a circle geuen to describe a pentagon figure aequilater and equiangle.</head>
               <p>SVppose that the circle geuen be ABCDE.<note place="margin">Construction.</note> It is required in the circle ABCDE to inscribe a figure of fiue angles of equall sides and of e<g ref="char:EOLhyphen"/>quall angles. Take (by the proposition going before) an Isosceles trian<g ref="char:EOLhyphen"/>gle FGH ha<g ref="char:EOLhyphen"/>uing
<figure/>
eyther of the angles at y<hi rend="sup">e</hi> base GH dou<g ref="char:EOLhyphen"/>ble to the other angle, namely, vnto the angle F. And (by the <hi>2.</hi> of y<hi rend="sup">e</hi> fourth) in the circle ABCDE in<g ref="char:EOLhyphen"/>scribe a trian<g ref="char:EOLhyphen"/>gle ACD e<g ref="char:EOLhyphen"/>quiangle vnto the triangle FGH. So that let the angle CAD be equall to the angle F, &amp; the angle ACD vnto the angle G, and likewise the angle CDA to the angle H. Wherefore ey<g ref="char:EOLhyphen"/>ther of these angles ACD, and CDA is double to the angle CAD. Deuide (by the <hi>9.</hi> of the first) either of these angles ACD, and CDA into two equall partes by the right lines CE and DB. and draw right lines from A to B, from B to C, from C to D, from D to E, and from E to A.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as either of these angles ACD, and CDA is double to the angle CAD: and they are deuided into two equall partes by the right lines CE and DB, therfore the fiue angles DAC, ACE, ECD, CDB, and BDA are equall the one to the o<g ref="char:EOLhyphen"/>ther. But equall angles (by the <hi>26.</hi> of the third) subtend equall circumferences. Wherfore the fiue circumferences AB, BC, CD, DE, and EA are equall the one to y<hi rend="sup">e</hi> other. And (by y<hi rend="sup">e</hi> 
                  <hi>29,</hi> of y<hi rend="sup">e</hi> same) vnto equall circumferences are subte<g ref="char:cmbAbbrStroke">̄</g>ded equal right lines: wherfore y<hi rend="sup">e</hi> fiue right lines AB, BC, CD, DE, &amp; EA are equal y<hi rend="sup">e</hi> one to the other. Wherfore y<hi rend="sup">e</hi> figure ABCDE hauing fiue angles is equilater.</p>
               <p>Now also I say that it is equiangle. For forasmuch as the circumference AB
<pb facs="tcp:7062:149"/>
is equall to the circumference DE, put the circumference BCD common vn<g ref="char:EOLhyphen"/>to them both. Wherfore the whole circumference ABCD is equall to the whole circumference EDCB: and vpon the circumference ABCD consisteth the angle AED,
<figure/>
and vppon the circumference EDCB, con<g ref="char:EOLhyphen"/>sisteth the an<g ref="char:EOLhyphen"/>gle BAE. Wherefore the angle BAE is equall to y<hi rend="sup">e</hi> an<g ref="char:EOLhyphen"/>gle AED (by the <hi>27.</hi> of the third) and by the same rea<g ref="char:EOLhyphen"/>son euery one of these angles ABC, and BCD, and CDE, is equall to euery one of these angles BAE and AED. Wherfore the fiue angled figure ABCDE, is equiangle, and it is pro<g ref="char:EOLhyphen"/>ued, that is also equilater. Wherfore in a circle geuen is described a figure of fiue angles equilater and equiangle: which was required to be done.</p>
               <div type="section">
                  <head>¶ An other way to do the same after Pelitarius.</head>
                  <p>Suppose that the Isosceles triangle described by the former Proposition, be <hi>DE<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>
                     </hi> so that let eyther of the angles <hi>E</hi> and <hi>F</hi> be double to the angle <hi>D</hi>:<note place="margin">
                        <hi>An other way to do the same after Pelita<g ref="char:EOLhyphen"/>rius.</hi>
                     </note> And let the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>uen be <hi>ABC:</hi> the ce<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tre
<figure/>
wherof let be <hi>K.</hi> And vpon the centre <hi>K</hi> describe the angle <hi>BKC</hi> equall to one of the angles <hi>E</hi> or <hi>F</hi> of the triangle <hi>DEF.</hi> And draw a right line from <hi>B</hi> to <hi>C.</hi> Then I say, that the line <hi>BC</hi> is the side of the Pen<g ref="char:EOLhyphen"/>tagon figure to be in<g ref="char:EOLhyphen"/>scribed in the circle <hi>ABC.</hi> Deuide the an<g ref="char:EOLhyphen"/>gle <hi>BKC</hi> into two e<g ref="char:EOLhyphen"/>quall partes by draw<g ref="char:EOLhyphen"/>ing the diameter <hi>A<g ref="char:EOLhyphen"/>KL.</hi> And draw these right lines <hi>BA</hi> &amp; <hi>AC.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> it is manifest (by the 20. of the third) that the angle
<pb n="119" facs="tcp:7062:149"/>
                     <hi>BKC</hi> is double to the whole angle <hi>BAC.</hi> Wherefore the whole angle <hi>BAC</hi> is equall to the angle <hi>D,</hi> vnto which angle, the angle <hi>BKC</hi> also is double. And forasmuch as in the triangle <hi>ABK</hi> the two angles <hi>A</hi> and <hi>B</hi> are (by the 5. of the first) equall (for the lines <hi>KA</hi> and <hi>KB</hi> are drawen from the centre) therefore (by the 32. of the same) the outward angle <hi>BKL</hi> is double to eyther of the inward angles <hi>KAB</hi> and <hi>KBA.</hi> And by the same reason, the angle <hi>CKL</hi> is double to eyther of the angles <hi>KAC</hi> and <hi>KCA.</hi> Wherfore forasmuch as the two angles at the poynt <hi>K</hi> are equall, the two an<g ref="char:EOLhyphen"/>gles <hi>A</hi> and <hi>B</hi> of the triangle <hi>ABK,</hi> are equall to the two angles <hi>A</hi> and <hi>C</hi> of the trian<g ref="char:EOLhyphen"/>gle <hi>ACK,</hi> the one to the other: and therefore (by the 26. of the first) the two bases <hi>AB</hi> and <hi>AC</hi> are equall. Wherefore <hi>ABC</hi> is an Isosceles triangle. And forasmuch as the whole angle <hi>BAC</hi> is equall to the angle <hi>D,</hi> the two angles remayning <hi>ABC</hi> and <hi>ACB,</hi> shall (by the 32. of the first) be equall to the two angles remayning <hi>E</hi> &amp; <hi>F.</hi> Wherefore the triangle <hi>ABC</hi> is equiangle to the triangle <hi>DEF.</hi> And now you may procede in the demonstration as you did in the former, imagining first the lines <hi>BG</hi> and <hi>CH</hi> to be drawen.</p>
                  <p>Here it is a pleasant thing to beholde the varietie of triangles: for in the trian<g ref="char:EOLhyphen"/>gle ABC either of the angles at the point A is one fift part of a right angle. Wher<g ref="char:EOLhyphen"/>by is produced the side of a ten angled figure to be inscribed in the selfsame circle: Which is manifest if we imagine the lines BL and LC to be drawen. For the arke BC is deuided into two equall partes in the poynt L (by the 26. of the third). So then by the inscription of an equilater triangle, is knowen how to inscribe an Hexagon figure, namely, by deuiding ech of the arkes subtended vnder the sides of the triangle into two equall partes. And so alwayes by the simple number of the side, is knowen the double thereof: as by a square is knowen an eight angled figure: and by an eight angled figure a sixtene angled figure. And so continually in the rest.</p>
                  <p>
                     <hi>Pelitarius</hi> teacheth yet an other way how to inscribe a Pentagon.<note place="margin">
                        <hi>An other way also after Pelitarius.</hi>
                     </note> Take the same cir<g ref="char:EOLhyphen"/>cle that was before, namely, <hi>ABC,</hi> and the same triangle also <hi>DEF</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And (by the 17. of the third) draw the line <hi>MAN</hi> touching the circle in the poynt <hi>A.</hi> And vpon the line <hi>AM</hi> and to the poynt <hi>A,</hi> describe (by the 23. of the first) the angle <hi>MAB</hi> equall to one of these two angles <hi>E</hi> or <hi>F</hi> (eyther of which, as it is manifest, is lesse then a right angle) by drawing the right line <hi>AB</hi>: which let out the circumference in the poynt <hi>B.</hi> Agayne, vpon the line <hi>AN</hi> and to the poynt in it <hi>A,</hi> describe the angle <hi>NAC</hi> equall to the angle <hi>MAB,</hi> by drawing the right line <hi>AC</hi>: which let cut the circumference in the poynt <hi>C.</hi> And draw a right line from <hi>B</hi> to <hi>C.</hi> Then I say, that <hi>BC</hi> is the side of a Pentagon figure to be inscribed in the circle <hi>ABC.</hi> Which is manifest, if we deuide the arke <hi>AB</hi> into two equall partes in the poynt <hi>H,</hi> and draw these right lines <hi>AH</hi> and <hi>BH,</hi> and if also we deuide the arke <hi>AC</hi> into two equall partes in the poynt <hi>G,</hi> and draw these right lines <hi>AG</hi> and <hi>CG.</hi> For taking the quadrangle figure <hi>ABCG,</hi> it is manifest (by the 32. of the third) that the angle <hi>ABC</hi> is equall to the al<g ref="char:EOLhyphen"/>ternate angle <hi>NAC,</hi> and therfore is equall to the angle <hi>E.</hi> Likewise taking the quadran<g ref="char:EOLhyphen"/>gle figure <hi>ACBH,</hi> the angle <hi>ACB</hi> shall be equall to the alternate angle <hi>MAB,</hi> and therefore is equall to the angle <hi>F.</hi> Wherefore (by the 32. of the first) as before, the tri<g ref="char:EOLhyphen"/>angle <hi>ABC</hi> is equiangle to the triangle <hi>DEF:</hi> And now may you procede in the de<g ref="char:EOLhyphen"/>monstration as you did in the former.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Probleme. The <hi>12.</hi> Proposition. About a circle geuen, to describe an equilater and aquian<g ref="char:EOLhyphen"/>gle pentagon.</head>
               <p>
                  <pb facs="tcp:7062:150"/>SVppose that the circle geuen be ABCDE. It is required about the cir<g ref="char:EOLhyphen"/>cle ABCDE to describe a figure of fiue angles consisting of equal sides and of equall angles.<note place="margin">Construction.</note> Take the pointes of the angles of a fiue angled figure described (by the <hi>11.</hi> of the fourth) so that by the proposition goyng be<g ref="char:EOLhyphen"/>fore, let y<hi rend="sup">e</hi> circumfere<g ref="char:cmbAbbrStroke">̄</g>ces AB, BC, CD, DE, and EA be equall the one to the o<g ref="char:EOLhyphen"/>ther. And by the pointes A, B, C, D, E, draw (by the <hi>17.</hi> of the third) right lines touching the circle, and
<figure/>
let the same be GH, HK, KL, LM, and MG. And (by the <hi>1.</hi> of the third) take the centre of the circle &amp; let the same be F. And drawe right lines from F to B, from F to K, from F to C, from F to L, and from F to D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the right lyne KL, toucheth the circle ABCDE in the pointe C, and from the centre F vnto the point C where the touche is, is drawen a right line FC: therefore (by the <hi>18.</hi> of the third) FC is a perpendicular line vn<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o KL. Wherfore either of the angles which are at the point C is a right angle, and by the same reason the angles which are at the pointes D and B are right angles. And forasmuch as the angle FCK is a right angle, therfore the square which is made of FK is (by the <hi>47.</hi> of the first) equall to the squares which are made of FC and CK. And by the same reason also the square which is made of FK is equall to the squares which are made of FB and BK. Wherefore the squares which are made of FC and CK are equall to the squares which are made of FB and BK, of which the square which is made of FC is equall to the square which is made of FB. Wherfore the square which is made of CK is e<g ref="char:EOLhyphen"/>quall to the square which is made of BK. Wherfore the line BK is equall vnto the line CK. And forasmuch as FB is equall vnto FC, and FK is common to them both, therfore these two BF and FK are equall to these two CF &amp; FK. And the base BK is equall vnto the base CK. Wherfore (by the <hi>8.</hi> of the first) the angle BFK is equall vnto the angle KFC: and the angle BKF to the angle FKC. Wherefore the angle BFC is double to the angle KFC. And the angle BKC is double to the angle FKC. And by the same reason the angle CFD is double to the angle CFL, and the angle DLC is double to the angle FLC. And forasmuch as the circumference BC is equall vnto the circumference
<pb n="120" facs="tcp:7062:150"/>
CD, therfore (by the <hi>27.</hi> of the third) the angle BFC is equall to the angle CFD. And the angle BFC is double to the angle KFC, and the angle DFC is double to the angle LFC, wherfore the angle KFC is equall vnto the angle LFC. Now then there are two triangles FKC, and FLC, hauing two angles e<g ref="char:EOLhyphen"/>quall to two angles, and one side equall to one side, namely, FC, which is common to them both. Wherfore (by the <hi>26.</hi> of the third, the other sides remayning are e<g ref="char:EOLhyphen"/>quall vnto the sides remayning, and the angle remayning vnto the angle remay<g ref="char:EOLhyphen"/>ning. Wherfore the right line KC is equall to the right line CL, and the angle FKC to the angle FLC. And forasmuch as KC is equall to CL: therefore KL is double to KC, and by the same reason also may it be proued that HK is double to BK. And forasmuch as it is proued that BK is equall vnto KC, and <hi>K</hi>L is double to <hi>K</hi>C, and H<hi>K</hi> double to B<hi>K,</hi> therfore H<hi>K</hi> is equall vnto <hi>K</hi>L. In like sort may we proue y<hi rend="sup">t</hi> euery one of these lines HG, GM, &amp; ML is equall vnto either of these lines H<hi>K</hi> and <hi>K</hi>L. Wherefore the fiue angled figure GH<hi>K</hi>LM is of equall sides. I say also that it is of equall angles. For forasmuch as the angle F<hi>K</hi>C is equall vnto the angle FLC, and it is proued that the angle H<hi>K</hi>L is double to the angle F<hi>K</hi>C, and the angle <hi>K</hi>LM is double to the angle FLC, therefore the angle H<hi>K</hi>L is equall to the angle <hi>K</hi>LM. In like sort may it be proued that euery one of these angles <hi>K</hi>HG, HGM, and GML, is equall to either of these angles H<hi>K</hi>L, and <hi>K</hi>LM. Where<g ref="char:EOLhyphen"/>fore the fiue angles GH<hi>K,</hi> H<hi>K</hi>L, <hi>K</hi>LM, LMG, and MGH are equall the one to the other. Wherfore the fiue angled figure GH<hi>K</hi>LM is equiangle, and it is also proued that it is equilater, and it is described about the circle ABCDE: which was required to be done.</p>
               <div type="section">
                  <head>¶An other way to do the same after Pelitarius, by parallel lines.</head>
                  <p>Suppose that the circle ge<g ref="char:EOLhyphen"/>uen
<figure/>
be ABC,<note place="margin">
                        <hi>An other way to do the sam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> after Pelita<g ref="char:EOLhyphen"/>rius.</hi>
                     </note> whose ce<g ref="char:cmbAbbrStroke">̄</g>tre let be the poynt F: and in it (by the former Proposition) in<g ref="char:EOLhyphen"/>scribe an equilater and equi<g ref="char:EOLhyphen"/>angle Pentagon ABCDE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by whose fiue angles drawe from the centre beyond the circu<g ref="char:cmbAbbrStroke">̄</g>ference, fiue lynes, FG, FH, FK, FL, and FM. And it is m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nifest, that the fiue an<g ref="char:EOLhyphen"/>gles at the ce<g ref="char:cmbAbbrStroke">̄</g>tre F, are equall, when as the fiue sides of the triangles within are equall, and also their bases. It is ma<g ref="char:EOLhyphen"/>nifest also, that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> fiue an<g ref="char:EOLhyphen"/>gles of the Pentagon which are at the circumference, are deuided into <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>n equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n<g ref="char:EOLhyphen"/>gles (by the 4. of the fir<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>): Now then betwene the two lines FG and FH, draw the
<pb facs="tcp:7062:151"/>
line GH parallel to the side AB, and touching the circle ABC (which is done by a Pro<g ref="char:EOLhyphen"/>position added by <hi>Pelitarius</hi> after the 17. of the third). And so likewyse draw these lines HK, KL, and LM, parallel to ech of these sides BC, CD, and DE, and touching the circle. And for asmuch as the lines FG and FH fall vpon the two parallel lines AB and GH,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> the two angles FGH, &amp; FHG, are equall to the two angles FAB and FBA, the one to the other (by the 29. of the first). Wherefore (by the sixt of the same) the two lines FG and FH are equall. And by the same reason, the two angles FHK &amp; FKH, are equall to the two angles FGH and FHG the one to the other: and the line FK is equall to the line FH, and therefore is equall to the line FG. And forasmuch as the angles at the poynt F are equall, therefore (by the 4. of the first) the base HK is equall to the base GH. In like sort may we proue, that the three lines FK, FL, and FM, are equall to the two lines FG and FH. And also that the two bases KL and LM, are e<g ref="char:EOLhyphen"/>quall to the two bases GH and HK: and that the angles which they make with the lines FK, FL, and FM, are equall the one to the other. Now then draw the fift line MG: which shall be equall to
<figure/>
the foure former lines (by the 4. of the first) for that as we haue proued, the two lines FG &amp; FM, are equall, &amp; the angle GFM is equall to euery one of the angles at the poynt F. Thys line also MG toucheth the circle. For vnto the point where the line LM toucheth the circle whi<g ref="char:EOLhyphen"/>che let be N, drawe the lyne FN. And it is manifest (by the 18. of the third) that either of the angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> at the poynt N, is a right angle. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore for asmuch as the angle L of the triangle FLN, is equall to the angle M of the trian<g ref="char:EOLhyphen"/>gle FMN, &amp; the angle N of the one, is equall to the angl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> N of the other: and the lyne FN is co<g ref="char:cmbAbbrStroke">̄</g>mon to the<g ref="char:cmbAbbrStroke">̄</g> both, the line NL shall (by the 26. of the first) be equall to the line NM. And so is the line ML deuided equally in the poynt N. And forasmuch as the three sides of the triangle FGP are equall to the three sides of the triangle FMP, the angle P of the one shal be equall to the angle P of the other (by the 8. of the first). Wher<g ref="char:EOLhyphen"/>fore either of the<g ref="char:cmbAbbrStroke">̄</g> is a right angle (by the 13. of the same). And forasmuch as the two an<g ref="char:EOLhyphen"/>gles FMP and FPM of the triangle FMP, are equall to the two angles FMN &amp; FNM of the triangle FMN, and the side FM is common to them both, therefore the line FP is equall to the line FN. But the line FN is drawen from the centre to the circumfe<g ref="char:EOLhyphen"/>rence. Wherefore also the line FP is drawen from the centre to the circumference. And forasmuch as the line MG is perpendicular to the line FP, therefore (by the Corolla<g ref="char:EOLhyphen"/>ry of the 16. of the third) it toucheth the circle. Wherfore the Pentagon GHKLM circumscribed about the circle is equilater: it is also equiangle, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> it is easie to proue by the equalitie of the halfes: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Probleme. The <hi>13.</hi> Proposition. An equilater and equiangle pentagon figure beyng geuen, to describe in it a circle.</head>
               <p>
                  <pb n="121" facs="tcp:7062:151"/>SVppose y<hi rend="sup">t</hi> y<hi rend="sup">e</hi> equilater &amp; equiangle Pe<g ref="char:cmbAbbrStroke">̄</g>tago<g ref="char:cmbAbbrStroke">̄</g> figure geue<g ref="char:cmbAbbrStroke">̄</g> be ABCDE. It is required in the said fiue angled figure ABCDE to describe a circle. Deuide (by the the <hi>9.</hi> of the first) either of these angles BCD, and CDE into two equall partes, by these right lines CF and FD,<note place="margin">Demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>tion.</note> and from the point F where th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se right lines CF and DF meete. Draw these right lines FB, FA, &amp; F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And forasmuch as BC is equal vnto CD <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd CF is common to them both. Wherefore these two
<figure/>
lines BC and CF are equal to these two lines DC and CF: and the angle BCF is equall to the angle DCF. Wher<g ref="char:EOLunhyphen"/>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e (by the <hi>4.</hi> of the first) the base BF is e<g ref="char:EOLhyphen"/>quall to the base DF, and the triangle BCF is equall to the triangle DCF, and the an<g ref="char:EOLhyphen"/>gles remayning are e<g ref="char:EOLhyphen"/>quall vnto the angles remaynyng the one to the other, vnder which are subtended e<g ref="char:EOLhyphen"/>quall sides. Wherefore the angle CBF is equall to the angle CDF. And forasmuch as the angle CDE is double to the angle CDF. But the angle CDE is equall to the angle ABC, and the angle CDF is equall to the angle CBF. Wherefore the angle CBA is double to the angle CBF. Wherefore the angle ABF is equall to the angle FBC. Wherefore the angle ABC is deui<g ref="char:EOLhyphen"/>ded into two equall partes by the right line BF. In lyke sorte also may it be proued<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that either of these angles BAE, and AED are deuided into two e<g ref="char:EOLhyphen"/>quall partes by either of these lines FA and EF. Drawe (by the <hi>12.</hi> of the first) from the pointe E to the right lines AB, BC, CD, and EA perpendi<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ular <milestone type="tcpmilestone" unit="unspecified" n="F"/> line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> FG, FH, FK, FL, and FM. And forasmuch as the angle HCF is eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the angl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> KCF, and the right angle FHC is equall to the right angle FKC: now then there are two triangles FHC, and FKC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hauyng two angles equall to two angles the one to the other, and one side equall to one side: for FC is commo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, vnto them both, and is subtended vnder one of the equal angles. Wherfore (by the <hi>2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> of the first) the sides remay<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ng are eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all vnto the sides remayning. Wherfore the perpendiculer FH is equall vnto the perpendicu<g ref="char:EOLhyphen"/>ler FK. And in like sort also may it be prou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d that euery one of these lines FL, FM, and FG is equall to euery one of th<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> lines FH and FK. Wh<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>fore these fiue right lines FG, FH, FK, FL, and FM are equall the one to the other.
<pb facs="tcp:7062:152"/>
Wherfore making the centre F and the space FG, or FH, or FK, or FL, or FM. Describe a circle, and it will passe by the poyntes G, H, K, L, M. And shall touche the right lines AB, BC, CD, DE, and EA (by the correllary of the <hi>16.</hi> of the third) for the angles which are at the pointes G, H, K, L, M, are right angles.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absur<g ref="char:EOLhyphen"/>ditie.</note> For if it do not touche them but cut, then from the ende of the diameter of the circle, shall be drawen a right line making two right angles, and falling with in the circle: whiche is
<figure/>
(by the <hi>16.</hi> of y<hi rend="sup">e</hi> third) proued to be impossible Wherefore makyng F the centre, and y<hi rend="sup">e</hi> space one of these lynes, FG, FH, FK, FL, FM, describe a circle and it shall not cut the right lines AB, BC, CD, DE, and EA. Wherefore (by the corollary of the <hi>16.</hi> of the third) it shall touche them as it is ma<g ref="char:EOLhyphen"/>nifest in the circle GHKLM. Wherefore in the equilater and equi<g ref="char:EOLhyphen"/>angle pentagon figure geuen, is described a circle: which was required to be done<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </p>
               <p>
                  <hi>By this proposition and the former, it is manifest that perpendicular lines drawn from the middle poyntes of the sides of an equilater and equiangle Pentagon fi<g ref="char:EOLhyphen"/>gure,</hi>
                  <note place="margin">A Corollary.</note> 
                  <hi>and produced, shall passe by the centre of the circle, in which the sayd Penta<g ref="char:EOLhyphen"/>gon figure is inscribed, and shall also deuide the opposite angles equally, as here, AK is one right line, and deuideth the side CD and the angle A equally. And so of the rest which is thus proued. The space about the centre F is equall to foure right angles, which are deuided into ten equall angles by ten right lines metyng together in the point F. Wherfore the fiue angles AFM, MFE, EFL, LFD, and DFK are equall to two right angles. Wherfore (by the 14. of the first) the lynes AF and FK make one right lyne. The lyke proofe also will serue touching the re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t of the lynes. And this is alwayes true in all equilater figures which consist of vne<g ref="char:EOLhyphen"/>uen sides.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Probleme. The <hi>14.</hi> Proposition. About a pentagon or figure of fiue angles geuen beyng equi<g ref="char:EOLhyphen"/>later and equiangle, to describe a circle.</head>
               <p>
                  <pb n="122" facs="tcp:7062:152"/>SVppose that y<hi rend="sup">e</hi> Pentagon or figure of fiue angles geuen, being of equall sides and of equall angles, be ABCDE. It is required about the sayd Pentagon ABCDE, to describe a circle. Deuide (by the <hi>9.</hi> of the first) eyther of these angles BCD,<note place="margin">Construction.</note> &amp; CDE into two equall partes by either of these right lines CF, and DF. And from the poynt F where those right lines meete, draw vnto the pointes B, A, E, these right lines FB, FA, FE.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And in like sort (by the Pro<g ref="char:EOLhyphen"/>position
<figure/>
going before) may it be proued, that euery one of these angles CBA, BAE, and AED is deuided into two equall partes, by these right lines FB, FA, &amp; FE. And for asmuch as the angle BCD is equall to the angle CDE, &amp; the halfe of the an<g ref="char:EOLhyphen"/>gle BCD is the angle FCD, and likewise the halfe of the angle CDE is y<hi rend="sup">e</hi> angle CDF. Wherfore the angle FCD is equall to the angle FDC. Wherefore y<hi rend="sup">e</hi> side FC is equall to the side FD. In like sort al<g ref="char:EOLhyphen"/>so may it be proued, that euery one of these lines FB, FA, and FE is equall to e<g ref="char:EOLhyphen"/>uery one of these lines FC and FD. Wherefore these fiue right lines FA, FB, FC, FD, and FE are equall the one to the other. Wherefore making the centre F; and the space FA, or FB, or FC, or FD, or FE. Describe a circle and it will passe by the pointes A, B, C, D, E, and shall be described <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>bout the fiue angled figure ABCDE which is equiangle and equilater. Let this circle be descri<g ref="char:EOLhyphen"/>bed and let the same be ABCDE. Wherefore about the Pentagon geuen be<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>g both equiangle and equilater, is described a circle: which was required to be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>one.</p>
            </div>
            <div type="part">
               <head>The <hi>15.</hi> Probleme. The <hi>15.</hi> Proposition. In a circle geuen to describe an hexagon or figure of sixe an<g ref="char:EOLhyphen"/>gles equilater and equiangle.</head>
               <p>SVppose that the circle geuen be ABCDEF. It is required in the cir<g ref="char:EOLhyphen"/>cle geuen ABCDE to describe a figure of sixe angles of equall sides and of equall angles. D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>aw the diameter of the circle ABCDEF, and
<pb facs="tcp:7062:153"/>
let the same be AD. And (by the first of the third) take the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re of the circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let the same be G. And making the centre D, and the space DG, describe (by the third petition) a circle CGEH: and drawing right lines from E to G, and from G to C, extend them to the pointes B and F of the circumference of the circle geuen. And draw these
<figure/>
right lines AB, BC, CD, DE, EF, and FA. Then I say, that ABCDEF is an Hexagon figure of e<g ref="char:EOLhyphen"/>quall sides and of equall an<g ref="char:EOLhyphen"/>gles.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as the point G is the centre of the circle ABCDEF, there<g ref="char:EOLhyphen"/>fore (by the <hi>15.</hi> definition of the first) the line GE is e<g ref="char:EOLhyphen"/>quall vnto the line GD. A<g ref="char:EOLhyphen"/>gaine forasmuch as the point D is the centre of the circle CGEH (therfore (by the selfe same) the line DE is e<g ref="char:EOLhyphen"/>quall vnto the line DG: And it is proued that the line GE is equall vnto the lyne G<g ref="char:EOLhyphen"/>D. Wherfore the line GE is equall vnto the line ED (by the first common sentence) wherfore the triangle EGD is equilater, (and bi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> three angles, namely, EGD, GDE, DEG, are equall the one to the other. And forasmuch as (by the <hi>5.</hi> of the first) in triangles of two equal sides c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> called Isosceles, the angles at the base are equall the one to the other, and the three angles of a triangle are (by the <hi>30.</hi> of the first) equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nto two right an<g ref="char:EOLhyphen"/>gles, therfore the angle EGD is the third part of two right angles. And in lyke sor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e may it be proued, that the angle DGC is the third parte of two right an<g ref="char:EOLhyphen"/>gles. And forasmuch as the right line CG standing vpon the right line EB doth (by the <hi>13.</hi> of the fi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>t) make the two side angles <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>GC, and CG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equall to two right angles, therfore the angle remayning CGB is the third part of two right angles. Wherfore the angles EGD, DGC, and CGB are equall the one to the other. Wherfore their hed angles, that is, BGA, AGF, and FGE are (by the <hi>15.</hi> of the first) equall to these angles EGD, DGC, and CGB. Wherfore these sixe angles EGD, DGC, CGB, BGA, AGF, and FGE are equall the one to the other. But equall angles consist vpon equall circumferences (by <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the third. Therfore these sixe circumferences AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> DE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> EF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>A are equall the one to the other. But vnder equall circumferences are <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>bt<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>ded
<pb n="123" facs="tcp:7062:153"/>
equall right lines (by the <hi>29.</hi> of the same.) Wherefore these sixe right lynes AB, BC, CD, DE, EF, and FA are equall the one to the other. Wherfore the Hexagon ABCDEF is equilater. I say also that it is equiangle. For foras<g ref="char:EOLhyphen"/>much as the circumference AF is equall vnto the circumference ED adde the circumference ABCD common to them both. Wherefore the whole circumfe<g ref="char:EOLhyphen"/>rence FABCD is equall to the whole circumference EDCBA. And vppon the circumference FABCD consisteth the angle FED: and vppon the cir<g ref="char:EOLhyphen"/>cumference EDCBA consisteth the angle AFE. Wherefore the angle AFE is equall to the angle DEF. In like sort also may it be proued that the rest of the angles of the Hexagon ABCDEF, that is euery one of these angles FAB, ABC, BCD, and CDE is equall to euery one of these angles AFE, and FED. Wherfore the Hexagon figure ABCDEF is equiangle, and it is pro<g ref="char:EOLhyphen"/>ued that it is also equilater, and it is described in the circle ABCDEF. Wher<g ref="char:EOLhyphen"/>fore in the circle geuen ABCDEF is described a figure of sixe angles of equal sides and of equall angles: Which was required to be done<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </p>
               <div type="section">
                  <head>¶ An other way to do the same after Orontius.</head>
                  <p>Suppose that the circle geuen be <hi>ABCDEF</hi> in which first let there be described an equilater and equiangle triangle <hi>ACE</hi> (by the second of thys booke).<note place="margin">
                        <hi>An <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ther way to do the same after Oro<g ref="char:cmbAbbrStroke">̄</g>tius.</hi>
                     </note> Wherefore the arkes <hi>ABC, CDE, EFA</hi> are (by the 28. of the third) equall the one to the other. Deuide euery one of those three arkes into two equall partes (by the 30. of the same) in the pointes <hi>B, D,</hi> &amp; <hi>F.</hi> And draw these right
<figure/>
lines <hi>AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> BC, CD, DE, EF,</hi> and <hi>FA.</hi> Now then by the 2. definition of this booke there shall be described in the circle geue an Hex<g ref="char:EOLhyphen"/>agon figure <hi>ABCDEF,</hi> which must nedes be equilater: for that euery one of the arkes which subtend the sides thereof are equall the one to the other. I say also that it is e<g ref="char:EOLhyphen"/>quiangle. For euery angle of the Hexagon figure is set vpon equall arkes, namely, vpon foure such partes of the circu<g ref="char:cmbAbbrStroke">̄</g>ference wher<g ref="char:EOLhyphen"/>of the whole circumference co<g ref="char:cmbAbbrStroke">̄</g>tayneth sixe. Wherfore the angles of the Hexagon figure are equall the one to the other (by the 27. of the third). Wherefore in the circle geuen <hi>ABCDEF</hi> is inscribed an equilater and equiangle Hexagon figure: which was re<g ref="char:EOLhyphen"/>quired to be done.</p>
               </div>
               <div type="section">
                  <head>¶ An other way to do the same after Pelitarius.</head>
                  <p>Suppose that the circle in which is to be inscribed <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>uil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ter &amp; <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>iangle Hexa<g ref="char:EOLhyphen"/>gon figure be <hi>ABCDE,</hi>
                     <note place="margin">
                        <hi>An other way after Pelitarius.</hi>
                     </note> whose c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ntre let be <hi>F.</hi> And from the centre draw the semidia<g ref="char:EOLhyphen"/>meter <hi>FA.</hi> And from the poynt <hi>A</hi> apply (by the first o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hys booke) the line <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> equall to the semidiameter. Which I say is the side of an equilater and eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iangle Hexagon fi<g ref="char:EOLhyphen"/>gure to be inscribed in the circle <hi>ABCDE.</hi> Draw <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ght <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ine from E to <hi>B</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And for asmuch as the line <hi>AB</hi> is equall to the line <hi>FA</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp; it is also equall to the line <hi>FB,</hi> ther<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:154"/>
the triangle <hi>AFB</hi> is equilater, and by the 5. of the first equiangle. Now then vp<g ref="char:EOLhyphen"/>on the centre <hi>F</hi> describe the angle <hi>BFC</hi> equall to the angle <hi>AF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,</hi> or to the angle <hi>FBA</hi> (which is all one) by drawing the right line <hi>FC.</hi> And draw a line from <hi>B</hi> to <hi>C.</hi> And for asmuch as the angle <hi>AFB</hi> is the third part of two
<figure/>
right angles by the 5. and 32. of the first, the angle <hi>BFC</hi> also shall be the third part of two right angles. Where<g ref="char:EOLhyphen"/>fore either of the two angles remaining <hi>FBC</hi> and <hi>FCB,</hi> for asmuch as they are equall by the 5. of the first, shall be two third partes of two right angles (by the 32. of the same). Or (by the 4. of the first) forasmuch as the an<g ref="char:EOLhyphen"/>gle <hi>BFC</hi> is equall to the angle <hi>FBA,</hi> and the two sides <hi>FB</hi> and <hi>FC</hi> are equall to the two sides <hi>AB</hi> and <hi>BF,</hi> the base <hi>BC</hi> shall be equall to the base <hi>BF,</hi> and therefore is equall to the line <hi>FC.</hi> Wherefore the triangle <hi>FBC</hi> is equilater and equiangle. Lastly make the angle <hi>CFD</hi> equall to eyther of the angles at the poynt <hi>F,</hi> by drawing the line <hi>FD.</hi> And draw a line from <hi>C</hi> to <hi>D.</hi> Now then by the former reason the triangle <hi>FCD</hi> shall be equilater and equiangle. And for as<g ref="char:EOLhyphen"/>much as the three angles at the point <hi>F</hi> are equall to two right angles (for ech of them is the third part of two right angles) therefore (by the 14. of the first) <hi>AD</hi> is one right line: and for that cause is the diameter of the circle. Wherefore if the other semicircle <hi>AFD</hi> be deuided into so many equall partes as the semicircle <hi>ABCD</hi> is deuided into, it shall comprehend so many equall lines subtended vnto it. Wherefore the line <hi>AB</hi> is the side of an equilater Hexagon figure to be inscribed in the circle: which Hex<g ref="char:EOLhyphen"/>agon figure also shall be equiangle: For the halfe of the whole angle <hi>B</hi> is equall to the halfe of the whole angle <hi>C</hi>: which was required to be done.</p>
                  <p>Now then if we draw from the centre <hi>F</hi> a perpendicular line vnto <hi>AD,</hi> which let be <hi>FE,</hi> and draw also these right lines <hi>BE</hi> and <hi>CE</hi>: there shall be described a triangle <hi>BEC</hi> whose angle <hi>E</hi> which is at the toppe shall be the 6. part of two right angles by the 20. of the third. For the angle <hi>BFC</hi> is double vnto it. And either of the two angles at the base, namely, the angles <hi>EBC,</hi> and <hi>ECB</hi> is dupla sesquialter to the angle <hi>E</hi>: that is, eyther of them contayneth the angle <hi>E</hi> twise and halfe the angle <hi>E.</hi> And by this rea<g ref="char:EOLhyphen"/>son was found out the side of an Hexagon figure.</p>
                  <div type="subsection">
                     <head>Correlary.</head>
                     <p>Hereby it is mani<g ref="char:EOLhyphen"/>fest,
<figure/>
that the side of an Hexagon figure described in a circle is equall to a right line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the ce<g ref="char:cmbAbbrStroke">̄</g>tre of the said cir<g ref="char:EOLhyphen"/>cle vnto the circum<g ref="char:EOLhyphen"/>ference. And if by the pointes <hi>A, B, C, D, E, F,</hi> be
<pb n="124" facs="tcp:7062:154"/>
drawen right lines touching the circle, then shall there be described about the circle an Hexagon figure equilater and equiangle, which may be demonstrated by that which hath bene spoken of the describing of a Pentagon about a circle. And moreouer by those thinges which haue ben spoke<g ref="char:cmbAbbrStroke">̄</g> of Pen<g ref="char:EOLhyphen"/>tagons we may in a Hexagon geuen either describe or circum<g ref="char:EOLhyphen"/>scribe a circle: which was required to be done.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The <hi>16.</hi> Probleme. The <hi>16.</hi> Proposition. In a circle geuen to describe a quindecagon or figure of fiftene angles, equilater and equiangle.</head>
               <p>SVppose that the circle geuen be ABCD. It is required in the circle ABCD to describe a figure of fiftene angles consisting of equall sides and of equall angles. Describe in the circle ABCD the sides of an e<g ref="char:EOLhyphen"/>quilater triangle, and let the same be AC, and in y<hi rend="sup">e</hi> arke AC describe the side of an equilater pentagon and let the same be AB.<note place="margin">Construction.</note> Now then of such e<g ref="char:EOLhyphen"/>quall partes wherof the whole circle ABCD containeth fiftene, of such partes I say, the circumference ABC being the third parte of the circle shall contayne fiue. And the
<figure/>
circumference AB being the fift part of a circle shall contain thre, wherefore the residue BC shall containe two. Deuide (by the 30. of the first) the arke BC into two equall partes in the point E.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wher<g ref="char:EOLunhyphen"/>fore either of these circum<g ref="char:EOLhyphen"/>ferences BE &amp; EC is the fiftene part of the circle ABCD. If therfore there be drawn right
<pb facs="tcp:7062:155"/>
lines from B to E, and from E to C, and then beginning at the point B or at the point C there be applied into the circle ABCD right lines equall vnto EB or EC, and so continuing till ye come to the point C if you began at B, or to y<hi rend="sup">e</hi> point B if you began at C, and there shall be described in the circle ABCD a figure of fiftene angles equilater and equiangle: which was required to be done. And in like sort as in a pentagon, if by the pointes where the circle is deuided, be drawen right lines touching the circle in the said pointes, there shall be described about y<hi rend="sup">e</hi> circle a figure of fiftene angles equilater &amp; equiangle. And in like sort by y<hi rend="sup">e</hi> selfe same obseruations that were in Pentagons, we may in a figure of fiftene angles geuen being equilater and equiangle either inscribe, or circumscribe a circle.</p>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>An addition of Flussates.</hi>
                     </note>¶ An addition of Flussates to finde out infinite figures of many angles.</head>
                  <p>
                     <q>If into a circle from one poynt be applyed the sides of two<note n="*" place="margin">
                           <hi>A Poligonon figure is a fi<g ref="char:EOLhyphen"/>gure consisting of many sides.</hi>
                        </note> Poligonon figures: the ex<g ref="char:EOLhyphen"/>cesse of the greater arke aboue the lesse, shall comprehend an arke contayning so many sides of the Poligonon figure to be inscribed by how many vnities the denomination of the Poligonon figure of the lesse side excedeth the denomination of the Poligonon figure of the greater side: and the number of the sides of the Poligonon figure to be inscribed is produced of the multiplication of the denominations of the foresayd Poligonon figures the one into the other.</q>
                  </p>
                  <p>As for example. Suppose that into the circle <hi>ABE</hi> be applyed the side of an equi<g ref="char:EOLhyphen"/>later and equiangle Hexagon figure (by the 15. of thys booke) which let be <hi>AB</hi>: and likewise the side of a Pentagon (by the 11. of this booke) which let be <hi>AC:</hi> and the side of a square (by the 6. of thys booke) which let be <hi>AD:</hi> and the side of an equilater tri<g ref="char:EOLhyphen"/>angle (by the 2. of this booke) which let be <hi>AE.</hi> Then I say, that the excesse of the arke <hi>AD</hi> aboue the arke <hi>AB,</hi> which excesse is the arke <hi>BD,</hi> contayneth so many sides of the Poligonon figure to be inscribed, of how many vnities the denominator of the Hex<g ref="char:EOLhyphen"/>agon <hi>AB,</hi> which is sixe, excedeth the denominator of the square <hi>AD,</hi> which is foure. And forasmuch as that excesse it two vni<g ref="char:EOLhyphen"/>ties,
<figure/>
therfore in <hi>BD</hi> there shall be two sides. And the denominator of the Poligonon fi<g ref="char:EOLhyphen"/>gure which is to be inscribed shall be pro<g ref="char:EOLhyphen"/>duced of the multiplication of the deno<g ref="char:EOLhyphen"/>minators of the foresayd Poligonon fi<g ref="char:EOLhyphen"/>gures, namely, of the multiplication of 6. into 4. which maketh 24. which number is the denominator of the Poligonon fi<g ref="char:EOLhyphen"/>gure, whose two sides shall subtend the arke <hi>BD.</hi> For of such equall partes wherof the whole circumference co<g ref="char:cmbAbbrStroke">̄</g>tayneth 24, of such partes <hi>I say,</hi> the circumference <hi>AB</hi> con<g ref="char:EOLhyphen"/>tayneth 4, and the circumference <hi>AD</hi> contayneth 6. Wherefore if from <hi>AD</hi> which subtendeth 6. partes be taken away 4. which <hi>AB</hi> subtendeth, there shall re<g ref="char:EOLhyphen"/>mayne vnto <hi>BD</hi> two of such partes of which the whole contayneth 24. Wherfore of an Hexagon and a square is made a Poligonon figure of 24. sides. Likewyse of the Hexagon <hi>AB</hi> and of the Pentagon <hi>AC</hi> shall be made a Poligonon figure of 30.
<pb n="125" facs="tcp:7062:155"/>
sides, one of whose sides shall subtend the arke <hi>BC.</hi> For the denomination of <hi>AB</hi> which is 6. excedeth the denomination of <hi>AC</hi> which is 5. onely by vnitie. So also forasmuch as the denomination of <hi>AB</hi> which is 6. excedeth the de<g ref="char:EOLhyphen"/>nomination of <hi>AE</hi> which is 3. by 3. therefore the arke <hi>BE</hi> shall contayne 3. sides of a Poligonon figure of .18. sides. And obseruing thys selfe same methode and order, a man may finde out infinite sides of a Poligonon figure.</p>
               </div>
            </div>
            <trailer>The end of the fourth booke of Euclides Elementes.</trailer>
         </div>
         <div n="5" type="book">
            <pb facs="tcp:7062:156"/>
            <head>¶The fifth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>THIS FIFTH BOOKE of <hi>Euclide</hi> is of very great commoditie and vse in all Geometry,<note place="margin">
                  <hi>The argument of this fift booke.</hi>
               </note> and much dili<g ref="char:EOLhyphen"/>gence ought to be bestowed therin. It ought of all o<g ref="char:EOLhyphen"/>ther to be throughly and most perfectly and readily knowne. For nothyng in the bookes followyng can be vnderstand without it: the knowledge of them all depende of it. And not onely they and other wri<g ref="char:EOLhyphen"/>tinges of Geometry, but all other Sciences also and artes: as <hi>Musike, Astronomy, Perspectiue, Arithmetique,</hi> the arte of accomptes and reckoning, with other such like. This booke therefore is as it were a chiefe trea<g ref="char:EOLhyphen"/>sure, and a peculiar iuell much to be accompted of. It entreateth of proportion and Analogie, or proportionalitie, which pertayneth not onely vnto lines, figures, and bodies in Geometry: but also vnto soundes &amp; voyces, of which Musike entreateth, as witnesseth <hi>Boetius</hi> and others which write of Musike. Also the whole arte of Astronomy teacheth to measure proportions of tymes and mouinges. <hi>Archimides</hi> and <hi>Iordan</hi> with other, writing of waightes, affirme, that there is proportion betwene waight and waight, and also betwene place &amp; place. Ye see therefore how large is the vse of this fift booke. Wherfore the definitions also thereof are common, although hereof <hi>Euclide</hi> they be accom<g ref="char:EOLhyphen"/>modate and applied onely to Geometry. The first author of this booke was as it is affirmed of many, one <hi>Eudoxus</hi> who was <hi>Platos</hi> scholer,<note place="margin">
                  <hi>The first a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>thor of this booke Eu<g ref="char:EOLhyphen"/>doxus.</hi>
               </note> but it was afterward fra<g ref="char:EOLhyphen"/>med and put in order by <hi>Euclide.</hi>
            </p>
            <div type="part">
               <head>Definitions.</head>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The first defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>A parte is a lesse magnitude in respect of a greater magni<g ref="char:EOLhyphen"/>tude, when the lesse measureth the greater.</head>
                  <p>As in the other bookes before, so in this, the author first setteth orderly the definitions and declarations of such termes and wordes which are necessarily re<g ref="char:EOLhyphen"/>quired to the entreatie of the subiect and matter therof, which is proportion and comparison of proportions or proportionalitie. And first he sheweth what a parte is. Here is to be considered that all the definitions of this fifth booke be general to Geometry and Arithmetique, and are true in both artes, euen as proportion and proportionalitie are common to them both, and chiefly appertayne to number, neither can they aptly be applied to matter of Geometry, but in respect of number and by number. Yet in this booke, and in these definitions here set, <hi>Euclide</hi> semeth to speake of them onely Geometrically, as they are applied to quantitie continu<g ref="char:EOLhyphen"/>all, as to lines, superficieces, and bodies: for that he yet continueth in Geometry. I wil notwithstanding for facilitie and farther helpe of the reader, declare the<g ref="char:cmbAbbrStroke">̄</g> both by example in number, and also in lynes.</p>
                  <p>For the clearer vnderstandyng of a parte, it is to be noted,<note place="margin">
                        <hi>A part taken two maner of wayes.</hi>
                     </note> that a part is taken in the Mathematicall Sciences two maner of wayes.<note place="margin">
                        <hi>The fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>st way.</hi>
                     </note> One way a part is a lesse quan<g ref="char:EOLhyphen"/>titie
<pb n="126" facs="tcp:7062:156" rendition="simple:additions"/>
in respect of a greater, whether it measure the greater o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> no. The second way,<note place="margin">
                        <hi>The second way.</hi>
                     </note> a part is onely that lesse quantitie in respect of the greater, which measureth the greater. A lesse quantitie is sayd to measure or number a greater quantitie,<note place="margin">
                        <hi>How a lesse quantity is sayd to mea<g ref="char:EOLhyphen"/>sure a greater.</hi>
                     </note> when it, beyng oftentymes taken, maketh precisely the greater quantitie without more or lesse, or beyng as oftentymes taken from the greater as it may, there remayneth nothyng. As suppose the line AB to contayne 3. and the lyne CD to contayne 9. the<g ref="char:cmbAbbrStroke">̄</g> doth the line AB measure the line
<figure/>
CD: for that if it be take certayne times, namely, 3. tymes, it maketh precisely the lyne CD, that is 9. without more or lesse. Agayne if the sayd lesse lyne AB be taken from the greater CD, as often as it may be, namely, 3. tymes, there shall remayne nothing of the greater. So the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber 3. is sayde to measure 12. for that beyng taken certayne tymes, namely, foure tymes, it maketh iust 12. the greater quantitie: and also beyng taken from 12. as of<g ref="char:EOLhyphen"/>ten as it may, namely, 4. tymes, there shall remayne nothyng. And in this meaning and signification doth <hi>Euclide</hi> vndoubtedly here in this define a part:<note place="margin">
                        <hi>In what sig<g ref="char:EOLhyphen"/>nificatio<g ref="char:cmbAbbrStroke">̄</g> Eu<g ref="char:EOLhyphen"/>clide here ta<g ref="char:EOLhyphen"/>keth a part.</hi>
                     </note> saying, that it is a lesse magnitude in comparison of a greater, when the lesse measureth the greater. As the lyne AB before set, contayning 3. is a lesse quantitie in compa<g ref="char:EOLhyphen"/>rison of the lyne CD which containeth 9. and also measureth it. For it beyng cer<g ref="char:EOLhyphen"/>tayne tymes taken, namely, 3. tymes, precisely maketh it, or taken from it as often as it may, there remayneth nothyng. Wherfore by this definition the lyne AB is a part of the lyne CD. Likewise in numbers, the number 5. is a part of the number 15. for it is a lesse number or quantitie compared to the greater, and also it measu<g ref="char:EOLhyphen"/>reth the greater: for beyng taken certayne tymes, namely, 3. tymes, it maketh 15. And this kynde of part is called commonly <hi>pars metiens</hi> or <hi>mensurans,</hi>
                     <note place="margin">
                        <hi>Par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> metien<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> or mensuran<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> that is, a mea<g ref="char:EOLunhyphen"/>suryng part: some call it <hi>pars multiplicatina:</hi>
                     <note place="margin">
                        <hi>Pars multipli<g ref="char:EOLhyphen"/>cati<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>a.</hi>
                     </note> and of the barbarous it is called <hi>pars aliquota,</hi>
                     <note place="margin">
                        <hi>Pars aliquota.</hi>
                     </note> that is an aliquote part. And this kynde of parte is commonly vsed in A<g ref="char:EOLhyphen"/>rithmetique.<note place="margin">
                        <hi>This kinde of part co<g ref="char:cmbAbbrStroke">̄</g>monly v<g ref="char:EOLhyphen"/>sed in Arith<g ref="char:EOLhyphen"/>metique.</hi>
                     </note>
                  </p>
                  <p>The other kinde of a part,<note place="margin">
                        <hi>The other kinde of part.</hi>
                     </note> is any lesse quantitie in comparison of a greater, whe<g ref="char:EOLhyphen"/>ther it be in number or magnitude, and whether it measure or no. As suppose the line AB to be 17. and let it be deuided into two partes in the poynt C, namely, in<g ref="char:EOLhyphen"/>to the line AC, &amp; the
<figure/>
line CB, and let the lyne AC the greater part containe 12. and let the line BC the lesse part contayne 5. Now eyther of these lines by this definition is a part of the whole lyne AB. For eyther of them is a lesse magnitude or qua<g ref="char:cmbAbbrStroke">̄</g>tity in co<g ref="char:cmbAbbrStroke">̄</g>pariso<g ref="char:cmbAbbrStroke">̄</g> of the whole lyne AB: but neither of the<g ref="char:cmbAbbrStroke">̄</g> measureth the whole line AB: for the lesse lyne CB contayning 5. taken as ofte<g ref="char:cmbAbbrStroke">̄</g> as ye list, will neuer make precisely AB which contayneth 17. If ye take it 3. tymes it maketh only 15. so lacketh it 2. of 17. which is to litle. If ye take it 4. times, so maketh it 20. the<g ref="char:cmbAbbrStroke">̄</g> are there thre to much, so it neuer maketh precisely 17. but either to much or to litle. Likewise the other part AC measureth not the whole lyne AB: for take<g ref="char:cmbAbbrStroke">̄</g> once, it maketh but 12. which is lesse then 17. and taken twise, it maketh 24. which are more then 17. by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. So it neuer precisely maketh by takyng therof the whole AB, but either more or lesse. And this kynde of part they commonly call <hi>pars constituens,</hi> or <hi>componens</hi>:<note place="margin">
                        <hi>Pars constit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>ens, or compo<g ref="char:EOLhyphen"/>nens.</hi>
                     </note> Because that it with some other part or partes, maketh the whole. As the lyne CB together with the line AC maketh the whole lyne AB. Of the barbarous it is called <hi>pars aliquanta.</hi>
                     <note place="margin">
                        <hi>Pars aliquan<g ref="char:EOLhyphen"/>ta.</hi>
                     </note> In this signification it is taken in <hi>B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rla<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>
                     </hi> in the beginnyng of his booke, in the definition of a part, when he saith: <hi>Euery
<pb facs="tcp:7062:157"/>
lesse number compared to a greater, is sayd to be a part of the greater, whether the lesse mea<g ref="char:EOLhyphen"/>sure the greater, or measure it not.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>Multiplex is a greater magnitude in respect of the lesse, when the lesse measureth the greater.<note place="margin">
                        <hi>The second definition.</hi>
                     </note>
                  </head>
                  <p>As the line CD before set in the first example, is multiplex to the lyne AB. For that CD a lyne contayning 9. is the greater magnitude, and is compared to the lesse, namely, to the lyne AB contayning 3. and also the lesse lyne AB measureth the greater line CD: for taken 3. tymes,
<figure/>
it maketh it, as was aboue sayde. So in numbers 12. is multiplex to 3: for 12 is the greater number, and is compared to the lesse, namely, to 3. which 3. also measu<g ref="char:EOLhyphen"/>reth it:<note place="margin">
                        <hi>Numbers ve<g ref="char:EOLhyphen"/>ry necessary for the vnder<g ref="char:EOLhyphen"/>standing of this booke and the other bookes follo<g ref="char:EOLhyphen"/>wing.</hi>
                     </note> for 3 taken 4 tymes maketh 12. By this worde <hi>multiplex</hi> which is a terme proper to Arithmetike and number, it is easy to consider that there can be no ex<g ref="char:EOLhyphen"/>act knowledge of proportion and proportionalitie, and so of this fifth booke wyth all the other bookes followyng, without the ayde and knowledge of numbers.</p>
               </div>
               <div type="section">
                  <head>Proportion is a certaine respecte of two magnitudes of one kinde,<note place="margin">
                        <hi>The t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ird de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> according to quantitie.</head>
                  <p>
                     <hi>Euclide</hi> as in the first definition, so in this &amp; the other following, and likewise in all his Propositions of this booke, mentioneth onely magnitudes, and geueth his examples and demonstrations of lines: for that hetherto in the 4. bookes before he hath entreated of lines &amp; figures, and so co<g ref="char:cmbAbbrStroke">̄</g>tinueth in his sixth booke following after this, comparing figure to figure, and sides of figures to sides of figures, with<g ref="char:EOLhyphen"/>out mention of number at all. Notwithstanding as it is sayd they are generall to all kinde of quantitie, both discrete and continuall, namely, number and magnitude: and neede for the young reader and studient in these artes to be de<g ref="char:EOLhyphen"/>clared in both. For, the opening of them in numbers (in which they are first and naturally founde) geueth a great and marueilous light to their declaration in magnitudes.<note place="margin">
                        <hi>what propor<g ref="char:EOLhyphen"/>tion is.</hi>
                     </note> 
                     <hi>Proportion</hi> (sayth he) <hi>is a certaine behauiour,</hi> that is, <hi>a certaine respect or comparison of two quantities of one kinde:</hi> as of one line to an other, and one figure generally to an other, and one number to an other, <hi>as touching quantitie,</hi> that is to say, that the quantitie compared,<note place="margin">
                        <hi>Thinges com<g ref="char:EOLhyphen"/>pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ed together this wayes.</hi>
                     </note> is to that wherunto it is compared, eyther equall, or greater, or lesse then it. For after these three maners may thinges be compared the one to the other. But quantities of diuers kindes can not be compared together. A superficies can not be compared to a line: nor number to a body: nor a body to a line or number: for that they are not of one kinde. For example of this definition,<note place="margin">
                        <hi>Example of this definition in magni<g ref="char:EOLhyphen"/>tudes.</hi>
                     </note> take two quantities, namely, two lines AB and CD, and compare the one to the other, namely, AB to
<figure/>
CD according to some certaine respect of greatnes, or lessenes, or equalitie, namely, in this example, let AB be greater then CD, &amp; containe it twise. Now thys comparison, relation, or respect of AB to CD, and generally of any one quantitie to any other, is called proportion. Likewise is
<pb n="127" facs="tcp:7062:157"/>
it, if AB be equall to CD, as in the second example: or if AB be lesse then CD, as in the third example. The like is it also in numbers
<figure/>
comparing 5. to 5. equall to equall,<note place="margin">
                        <hi>Example ther<g ref="char:EOLunhyphen"/>of in numbers.</hi>
                     </note> or 6. to 3. the grea<g ref="char:EOLhyphen"/>ter to the lesse, or 4. to 8. the lesse to the greater. So to the accomplishing of any Proportion there are re<g ref="char:EOLhyphen"/>quired two quantities,<note place="margin">
                        <hi>In proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> two quanti<g ref="char:EOLhyphen"/>ties required, which are cal<g ref="char:EOLhyphen"/>led termes.</hi>
                     </note> and also a comparing or respect of the one to the other. The quantities compared to<g ref="char:EOLhyphen"/>gether are commonly called the <hi>Termes</hi> of the Pro<g ref="char:EOLhyphen"/>portion. And in this booke of <hi>Euclide,</hi> and also in o<g ref="char:EOLhyphen"/>ther writers of <hi>Geometrie,</hi> the first <hi>Terme,</hi> namely, that which is compared is called the antecedent,<note place="margin">
                        <hi>The i. terme called the an<g ref="char:EOLhyphen"/>tecedent.</hi>
                     </note> whether it be equall, greater, or lesse then the other: And the second <hi>Terme,</hi> namely, that wherunto the comparison is made, is called the consequent.<note place="margin">
                        <hi>The ij. terme called the con<g ref="char:EOLhyphen"/>sequent.</hi>
                     </note> As in the former example, the line AB compa<g ref="char:EOLhyphen"/>red to the line CD is antecedent: and the line CD is consequent: And contra<g ref="char:EOLhyphen"/>riwise if the line CD be compared to the line AB, then is the line CD antece<g ref="char:EOLhyphen"/>dent, and the line AB consequent. Albeit in Arithmeticke <hi>Boetius</hi> and others call the terme compared <hi>Dux,</hi>
                     <note place="margin">
                        <hi>Dux.</hi>
                     </note> and the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rme to whom the co<g ref="char:cmbAbbrStroke">̄</g>parison is made they call <hi>Comes.</hi>
                     <note place="margin">
                        <hi>Comes.</hi>
                     </note> This booke hath bene accompted of all men one of the hardest and most intricate of all <hi>Euclides</hi> bookes.<note place="margin">
                        <hi>This booke counted the hardest of all Euclides bookes.</hi>
                     </note> And proportion is a generall knowledge to all learninges, chiefly to the Mathematicalls. Wherefore it shall be very necessary some litle briefe instruction and induction to be here added in the beginning here<g ref="char:EOLhyphen"/>of: of the knowledge and nature of proportions and what they are, and of how many kindes: which thinges are here of <hi>Euclide</hi> supposed to be before knowen, and therefore maketh no mention so distinctly of them.</p>
                  <p>Ye must vnderstand that there are of proportions two generall kindes,<note place="margin">
                        <hi>Proportion of two kindes, rational and irrational.</hi>
                     </note> the one is called rationall, certaine, and knowen, and the other irrationall, vncertaine and vnknowen. Such magnitudes or quantities, which may be expressed by numbre, are called <hi>rationall</hi> magnitudes or quantities<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <note place="margin">
                        <hi>Proportion rational.</hi>
                     </note> As suppose a line, namely, the line AB to containe 5. inches, &amp; compare it to the line CD,
<figure/>
contayning 3. inches: these quantities ye see may be ex<g ref="char:EOLhyphen"/>pressed by numbers, namely, by these numbers 5. and 3: and therefore are <hi>rationall,</hi> and haue the same proporti<g ref="char:EOLhyphen"/>on, that number hath to number, namely, that the number 5. hath to the number 3: and therefore the proportion of the one to the other, is a rationall, certaine, and knowen proportion. And generally when soeuer one number is compared to an other, or two lines or other magnitudes, both which may be expressed by num<g ref="char:EOLhyphen"/>ber, the proportion betwene them is euer rationall, and onely the proportion of such quantities is rationall. So that in Arethmeticke all proportions are rationall,<note place="margin">
                        <hi>In Arithme<g ref="char:EOLhyphen"/>tique all pro<g ref="char:EOLhyphen"/>portions are rational.</hi>
                     </note> for that therein euer one number is compared to an other.</p>
                  <p>There are certaine lines magnitudes or quantities which ca<g ref="char:cmbAbbrStroke">̄</g> not be named and expressed by number, and therefore commonly are called Surd lines or magni<g ref="char:EOLhyphen"/>tudes.<note place="margin">
                        <hi>Proportion irrational.</hi>
                     </note> As suppose the square ABCD to containe 16, then the side or roote ther<g ref="char:EOLhyphen"/>of, namely, the line AB containeth 4, and the diameter of the
<figure/>
same square, namely, the line BC shall be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 32, which is a surd number, and can nor be expressed by any determinate and certaine number, but onely by this maner of circumlocu<g ref="char:EOLhyphen"/>tion Roote square of 32. Now if ye compare the line AB to the line BC, or contrariwise the line BC to the line AB, for that one of them is a surde quantitie, neither can ech of them be expressed by number (and therefore can not haue that
<pb facs="tcp:7062:158"/>
proportion that number hath to number) the proportion betwene them is irratio<g ref="char:EOLhyphen"/>nall, confused, vnknowen, vncertaine, and surd. And this kinde of proportion is found onely in magnitudes, as in lines and figures (and not in numbers) of which he of purpose entreateth in his tenth booke. Wherfore I wil here omit to speake of it, and remit it to his due place. And somewhat will I now say for the elucidation of the first kinde.</p>
                  <p>
                     <note place="margin">
                        <hi>Rational pro<g ref="char:EOLhyphen"/>portion deuided <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>to two kindes.</hi>
                     </note>Proportion rationall is deuided into two kindes, into proportion of equalitie, and into proportion of inequalitie. Proportion of equalitie is, when one quanti<g ref="char:EOLhyphen"/>tie is referred to an other equall vnto it selfe: as if ye compare 5 to 5, or 7 to 7, &amp; so of other. And this proportion hath great vse in the rule of <hi>Cosse.</hi>
                     <note place="margin">
                        <hi>Proportion of equality.</hi>
                     </note> For in it all the rules of equations tende to none other ende but to finde out and bring forth a nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber equall to the number supposed, which is to put the proportion of equalitie.</p>
                  <p>
                     <note place="margin">
                        <hi>Proportion of inequality.</hi>
                     </note>Proportion of inequalitie is, when one vnequall quantity is compared to an o<g ref="char:EOLhyphen"/>ther, as the greater to the lesse, as 8. to 4: or 9. to 3: or the lesse to the greater as 4. to 8: or 3. to 9.</p>
                  <p>
                     <note place="margin">
                        <hi>Proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of the greater to the lesse.</hi>
                     </note>Proportion of the greater to the lesse hath fiue kindes, namely, <hi>Multiplex, Super<g ref="char:EOLhyphen"/>particular, Superpartiens, Multiplex superperticular,</hi> and <hi>Multiplex superpartiens.</hi>
                  </p>
                  <p>Multiplex, is when the antecedent containeth in it selfe the consequent cer<g ref="char:EOLhyphen"/>tayne times without more or lesse:<note place="margin">
                        <hi>Multiplex.</hi>
                     </note> as twice, thrice, foure tymes, and so farther. And this proportion hath vnder it infinite kindes. For if the antecedent contayne the consequent iustly twise, it is called dupla proportion,<note place="margin">
                        <hi>Duple propor<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> as 4 to 2. If thrice tripla,<note place="margin">
                        <hi>Triple, quadru<g ref="char:EOLhyphen"/>ple. Quintuple.</hi>
                     </note> as 9. to 3. If 4. tymes quadrupla as 12. to 3. If 5. tymes quintupla as 15. to 3. And so infinitely after the same maner.</p>
                  <p>
                     <note place="margin">
                        <hi>Superperticu<g ref="char:EOLhyphen"/>lar.</hi>
                     </note>Superperticular is, whe<g ref="char:cmbAbbrStroke">̄</g> the antecede<g ref="char:cmbAbbrStroke">̄</g>t containeth the consequent only once, &amp; moreouer some one part therof as an halfe, a third, or fourth, &amp;c. This kinde also hath vnder it infinite kindes. For if the antecedent containe the consequent once and an halfe, therof it is called <hi>Sesquialtera,</hi>
                     <note place="margin">
                        <hi>Sesquialtera.</hi>
                     </note> as 6. to 4: if once and a third part <hi>Sesqui<g ref="char:EOLhyphen"/>tertia,</hi>
                     <note place="margin">
                        <hi>Sesquitertia.</hi>
                     </note> as 4. to 3: if once and a fourth part <hi>Sesquiquarta,</hi>
                     <note place="margin">
                        <hi>Sesquiquarta.</hi>
                     </note> as 5. to 4. And so in like ma<g ref="char:EOLhyphen"/>ner infinitely.</p>
                  <p>
                     <note place="margin">
                        <hi>Superpartiens.</hi>
                     </note>Superpartiens is, whe<g ref="char:cmbAbbrStroke">̄</g> the antecedent co<g ref="char:cmbAbbrStroke">̄</g>taineth the consequent onely once, &amp; moreouer more partes then one of the same, as two thirdes, three fourthes, foure fifthes and so forth. This also hath infinite kindes vnder it. For if the antecedent containe aboue the consequent two partes, it is called <hi>Superbipartiens,</hi>
                     <note place="margin">
                        <hi>Superbiparti<g ref="char:EOLhyphen"/>ens.</hi>
                     </note> as 7. to 5. If 3. partes <hi>Supertripartiens</hi> as 7. to 4.<note place="margin">
                        <hi>Supertriparti<g ref="char:EOLhyphen"/>ens.</hi>
                     </note> If 4. partes <hi>Superquadripartiens,</hi>
                     <note place="margin">
                        <hi>Superquadri<g ref="char:EOLhyphen"/>partiens.</hi>
                     </note> as 9. to 5. If 5. partes <hi>Superquintipartiens</hi> as 11. to 6.<note place="margin">
                        <hi>Superquinti<g ref="char:EOLhyphen"/>partiens.</hi>
                     </note> And so forth infinitely.</p>
                  <p>
                     <note place="margin">
                        <hi>Multiplex su<g ref="char:EOLhyphen"/>perperticular.</hi>
                     </note>Multiplex Superperticular is when the antecedent containeth the consequent more then once, and moreouer onely one parte of the same. This kinde likewise hath infinite kindes vnder it. For if the antecedent containe the consequent twise and halfe therof, it is called dupla <hi>Sesquialtera,</hi>
                     <note place="margin">
                        <hi>Dupla Sesqui<g ref="char:EOLhyphen"/>altera.</hi>
                     </note> as 5. to 2. If twise and a third <hi>Dupla Sesquitertia</hi> as 7. to 3.<note place="margin">
                        <hi>Dupla sesqui<g ref="char:EOLhyphen"/>tertia.</hi>
                     </note> If thrice and an halfe <hi>Tripla sesquialtera</hi> as 7. to 2.<note place="margin">
                        <hi>Tripla sesqui<g ref="char:EOLhyphen"/>altera.</hi>
                     </note> If foure times and an halfe <hi>Quadr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>pla Sesquialtera,</hi> as 9. to 2. And so goyng on infinitely.</p>
                  <p>
                     <note place="margin">
                        <hi>Multiplex su<g ref="char:EOLhyphen"/>perpartiens.</hi>
                     </note>Multiplex Superpartient, is when the antecedent contayneth the consequent more then once, and also more partes then one of the consequent. And this kinde also hath infinite kindes vnder it. For if the antecedent containe the consequent twise, and two partes ouer, it is called dupla <hi>Superbipartiens</hi> as 8. to 3.<note place="margin">
                        <hi>Dupla superbi<g ref="char:EOLhyphen"/>partiens.</hi>
                     </note> If twice and three partes, dupla <hi>Supertripartiens</hi> as 11. to 4.<note place="margin">
                        <hi>Dupla supertri<g ref="char:EOLhyphen"/>partiens.</hi>
                     </note> If thrice and two partes, it is named Tripla <hi>Superbipartiens</hi> as 11. to 3.<note place="margin">
                        <hi>Tripla superbi<g ref="char:EOLhyphen"/>partiens.</hi>
                     </note> If three tymes and foure partes Treble <hi>Super<g ref="char:EOLhyphen"/>quadripartiens</hi> as 31. to 9.<note place="margin">
                        <hi>Tripla super<g ref="char:EOLhyphen"/>quad<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ipartiens.</hi>
                     </note> And so forth infinitely.</p>
                  <p>Here is to be noted that the denomination of the proportion b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>twene any two numbers, is had by deuiding of the greater by the lesse. For the quotient o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb n="128" facs="tcp:7062:158" rendition="simple:additions"/>
number produced of that diuision is euen the denomination of the proportion.<note place="margin">
                        <hi>How to kno<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the denomi<g ref="char:EOLhyphen"/>nation of any proportion.</hi>
                     </note> Which in the first kinde of proportion, namely, multiplex, is euer a whole number, and in all other kindes of proportion it is a broken number.</p>
                  <p>As if ye will know the denomination of the proportion betwene 9 and 3. De<g ref="char:EOLhyphen"/>uide 9. by 3. so shall ye haue in the quotient 3. which is a whole number, and is the denomination of the proportion: and sheweth that the proportion betwene 9. &amp; 3. is Tripla. So the proportion betwene 12. and 3. is quadrupla, for that 12. beyng deuided by 3. the quotient is 4. and so of others in the kinde of multiplex. And al<g ref="char:EOLhyphen"/>though in this kinde the quotient be euer a whole number, yet properly it is refer<g ref="char:EOLhyphen"/>red to vnitie, and so is represented in maner of a broken number as 5/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and 4/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for v<g ref="char:EOLhyphen"/>nitie is the denomination to a whole number.</p>
                  <p>Likewise the denomination of the proportion betwene 4 and 3 is 1. 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for that 4 deuided by 3. hath in the quotient 1 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> one and a third part, of which third part, it is called <hi>sesquitercia:</hi> so the proportion betwene 7 and 6. is 1 ⅙ one and a sixt, of which fixt part it is called <hi>sesquisexta,</hi> and so of other of that kinde. Also betwene, 7 and 5 the denomination of the proportion is 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> one and two fifthes, which de<g ref="char:EOLhyphen"/>nomination co<g ref="char:cmbAbbrStroke">̄</g>sisteth of two parts, namely, of the munerator and denominator of the quotient of 2. and 5: of which two fifthes it is called <hi>superbipartiens quintas:</hi> for 2 the numerator sheweth the denomination of the number of the partes, and 5. the denominator, sheweth the denominatio<g ref="char:cmbAbbrStroke">̄</g>, what parts they are, &amp; so of others. Also the denomination betwene 5 and 2. is 2 ½ two and a halfe, which consisteth of a whole number and a broken, of 2. the whole number it is dupla, and of the halfe, it is called <hi>sesquialtera,</hi> so is the proportion <hi>dupla sesquialtera.</hi>
                  </p>
                  <p>Agayne the denomination of the proportion betwene 11. and 3. is 3 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> three and two thirdes, consisting also of a whole number and a broken, of 3. the whole numbre it is called tripla, and of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the broken number, it is called <hi>Superbipartiens tertias,</hi> so the proportion is <hi>tripla superbipartiens tertias.</hi> Thus much hetherto tou<g ref="char:EOLhyphen"/>ching proportion of the greater quantitie to the lesse.</p>
                  <p>Proportion of the lesse quantitie to the greater hath as many kindes, as that of the greater to the lesse, which kindes are in the same order:<note place="margin">
                        <hi>Proportion of the lesse in the greater.</hi>
                     </note> and haue also the selfe same names, but that to the names afore put ye must adde here this word <hi>sub.</hi> As comparing the greater to the lesse, it was called <hi>multiplex, superparticular, superpar<g ref="char:EOLhyphen"/>tient, multiplex superparticular,</hi> and <hi>multiplex superpartient,</hi> now comparing the lesse quantitie to the greater, it is called <hi>submultiplex,</hi>
                     <note place="margin">
                        <hi>Submulti<g ref="char:EOLhyphen"/>plex.</hi>
                     </note> 
                     <hi>subsuperparticular,</hi>
                     <note place="margin">
                        <hi>Subsuperpar<g ref="char:EOLhyphen"/>ticular.</hi>
                     </note> 
                     <hi>subsuperpartient,</hi>
                     <note place="margin">
                        <hi>Subsuperper<g ref="char:EOLhyphen"/>tient. &amp;c.</hi>
                     </note> 
                     <hi>submultiplex superparticular,</hi> and <hi>submultiplex superpartient.</hi> And so in like maner to all the inferior kindes of all sortes of proportion ye shall adde that worde <hi>sub.</hi> The examples of the former serue also here, onely transposing the termes of the pro<g ref="char:EOLhyphen"/>portion making the antecedent consequent, and the consequent the antecedent. As 4. to 2. is dupla proportion: so 2. to 4. is subdupla. As 9. to 3. is tripla: so is 3. to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. subtripla. And as 9. to 6. is sesquialtera, so 6. to 9. is subsesquialtera. As 7. to 5. is superbipartiens quintas, so is 5. to 7. subsuperbipartiens quintas. As 5. to 2. is dupla sesquialtera, so is 2. to 5. subdupla sesquialtera. And also as 8. to 3. is dupla superbi<g ref="char:EOLhyphen"/>partiens tertias, so is 3. to 8. subdupla superbipartiens tertias. And so may ye pro<g ref="char:EOLhyphen"/>cede infinitely in all others. Thus much thought I good in this place for the ease of the beginner to be added touching proportion.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:159"/>
                  <head>
                     <note place="margin">
                        <hi>The fourth definition.</hi>
                     </note>Proportionalitie, is a similitude of proportions.</head>
                  <p>As in proportion are compared together two quantities, and proportion is no<g ref="char:EOLhyphen"/>thing els but the respect and comparison of the one to the other, and these quanti<g ref="char:EOLhyphen"/>ties are the termes of the proportion: so in proportionallitie are compared toge<g ref="char:EOLhyphen"/>ther two proportions. And proportionallitie is nothing els, but the respect &amp; com<g ref="char:EOLhyphen"/>parison of the one of them to the other. And these two proportions are the termes of this proportionallitie. He calleth it the similitude, that is, the likenes or idempti<g ref="char:EOLhyphen"/>tie of proporti<g ref="char:EOLhyphen"/>ons:<note place="margin">
                        <hi>Example of this definition in magni<g ref="char:EOLhyphen"/>tudes.</hi>
                     </note>
                     <figure/>
As if ye wil co<g ref="char:cmbAbbrStroke">̄</g>pare the pro<g ref="char:EOLhyphen"/>portion of the line A contay<g ref="char:EOLhyphen"/>nyng 2. to the line B contayning 1, to the proportion of the line C contayning 6. to the line D contayning 3, either proportion is dupla. This likenes, idemptitie, or equallitie of proportion is called proportionallitie.<note place="margin">
                        <hi>Example thereof in numbers.</hi>
                     </note> So in number 9. to 3. and 21. to 7. either pro<g ref="char:EOLhyphen"/>portion is tripla. Where note that proportions compared together, are sayd to be like the one to the other:<note place="margin">
                        <hi>Note.</hi>
                     </note> but magnitudes compared together, are said to be equall the one to the other.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The fifth defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>Those magnitudes are sayd to haue proportion the one to the other, which being multiplied may exceede the one the other.</head>
                  <p>Before he shewed and defined, what proportion was, now by this definition he declareth betwene what magnitudes proportion falleth, saying: That those quanti<g ref="char:EOLhyphen"/>ties are said to haue proportion the one to the other, which being multiplyed, may excede the one the other.<note place="margin">
                        <hi>An example of this <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>efini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note> As for that the
<figure/>
line A being multiplied by what soeuer multiplication or nu<g ref="char:cmbAbbrStroke">̄</g>ber, as taken twise, thrise, or foure, fiue, or more times, or once and halfe, or once and a third, &amp; so of any other part, or partes, may excede and become greater then the line B or contrariwise, then these two lines are said to haue proportion the one to the other. And so ye may see that betwene any two qua<g ref="char:cmbAbbrStroke">̄</g>tities of one kinde, there is a propor<g ref="char:EOLhyphen"/>tion. For the one remayning vnmultiplied, &amp; the other being certaine times mul<g ref="char:EOLhyphen"/>tiplied, shall be greater then it. As 3. to 24. hath a proportion, for leauing 24. vnmul<g ref="char:EOLhyphen"/>tiplied, and multiplying 3. by 9, ye shall produce 27: which is greater then 24, and excedeth it. Here is to be noted, that <hi>Euclide</hi> in defining what quantities haue pro<g ref="char:EOLhyphen"/>portion,<note place="margin">
                        <hi>Why Euclide in defining of Proportion vsed multi<g ref="char:EOLhyphen"/>plication.</hi>
                     </note> was compelled to vse multiplication, or els should not his definition be generall to either kinde of proportion: namely, to rationall and irrationall: to such proportion I say which may be expressed by number, and to such as cannot be ex<g ref="char:EOLhyphen"/>pressed by any determinate number, but remaineth surd and innominable. In rati<g ref="char:EOLhyphen"/>onall quantities which haue one common measure, the excesse of the one aboue the other is knowen, and by it is knowen the proportion, which may be expressed by some determinate number. But in irrationall quantities which haue no co<g ref="char:cmbAbbrStroke">̄</g>mon measure, it is not so. For in them the excesse of the one to the other is euer vn<g ref="char:EOLhyphen"/>knowen,
<pb n="129" facs="tcp:7062:159"/>
&amp; therefore is surd, and innominable. As betwene the side of a square and the diameter therof<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> there is vndoubtedly a proportion, for that the side certaine times multiplied may excede the diameter. Likewise betwene the diameter of a circle and the circumference therof there is certainlie, by this definition, a propor<g ref="char:EOLhyphen"/>tion, for that the diameter certaine times multiplied may excede the circumference of the circle: although neither of these proportions can be named &amp; expressed by number. For this cause therefore vsed <hi>Euclide</hi> this maner of defining by multipli<g ref="char:EOLhyphen"/>cation.</p>
               </div>
               <div type="section">
                  <head>Magnitudes are sayd to be in one or the selfe same propor<g ref="char:EOLhyphen"/>tion,<note place="margin">
                        <hi>The sixth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> the first to the second, and the third to the fourth, when the equimultiplices of the first and of the third beyng compa<g ref="char:EOLhyphen"/>red with the equimultiplices of the second and of the fourth, according to any multiplication: either together exceede the one the other, or together are equall the one to the other, or together are lesse the one then other.</head>
                  <p>In the definition last going before, he shewed what magnitudes haue propor<g ref="char:EOLhyphen"/>tion the one to the other, &amp; now this diffinition sheweth what magnitudes are in one and the selfe same proportion,<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tiin in magni<g ref="char:EOLhyphen"/>tudes.</hi>
                     </note> and how to know whether they be in one and the self same proportion, or not. It is plaine that euery proportio<g ref="char:cmbAbbrStroke">̄</g> hath two termes, so that when ye compare proportion to proportion, ye must of necessitie, haue 4. termes, that is, an antecedent and a consequent, to either of the proportions. As sup<g ref="char:EOLhyphen"/>pose A, B, C, D, to be foure magnitudes, A the first, B the second, C the third, and
<figure/>
D the fourth now if ye take the equimultiplices of A and C the first &amp; the third, that is, if ye multiply A and C by one and the selfe same number, as let the multi<g ref="char:EOLhyphen"/>plex of A be E, and let the equimultiplex of C be F. Likewise also if ye take the equimultiplices of B and D, the second and the fourth, that is if ye multiply them by any one number, whether it be by that number wherby ye multiplied A &amp; C, or by any other number greater or lesse, as let the multiplex of B be G, and the e<g ref="char:EOLhyphen"/>quimultiplex of D be H: how it the equimultiplices of A and C be both greater hen the eqnimultiplices of B and D, that is if the multiplex of A be greater then the multiplex of B, and the multiplex of C be greater then the multiplex of D, or if they be both lesse then they: or both equall to them,<note place="margin">
                        <hi>An example in numbers.</hi>
                     </note> then are the magnitudes A, B and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, D in one and the selfe same proportion.</p>
                  <p>Likewi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> in numbers 8. to 6. hath a proportion, also 4. to 3. hath a proportion:
<pb facs="tcp:7062:160"/>
now to see whether they be in one and the selfe same proportion or not, set them in order as in the example here written, 8 the first, 6 the second, 4 the third, and 3. the fourth. Now take the equemultiplices of 8 and 4. the first, and the third, that is, multiply them by one and the selfe
<figure/>
same number, suppose it be by 3. so the tri<g ref="char:EOLhyphen"/>ple of 8 is 24. &amp; the triple of 4. is 12: like<g ref="char:EOLhyphen"/>wise take the equimultiplices of 6 and 3. the second and the fourth, multiply<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ng them likewise by one and the selfe same number, suppose it be also by 3 as before ye did, the triple of 6 is 18. and the triple of 3 is 9. Now ye see that the triple of 8 the first, namely, 24. excedeth the triple of 6. the second, namely, 18: likewise the triple of 4 the third number, namely, 12. excedeth the triple of 3. the fourth, namely, 9. Wherefore by the first part of this definition, the numbers 8 to 6. and 4 to 3. are in one and the selfe same proportio<g ref="char:cmbAbbrStroke">̄</g>, because that the equemultiplices of 8 and 4. the first &amp; the third, do both exceede the equimultiplices of 6 and 3. the second and the fourth.</p>
                  <p>
                     <note place="margin">
                        <hi>An other ex<g ref="char:EOLhyphen"/>ample in numbers.</hi>
                     </note>Againe, take the same numbers and try the same after this maner. Take the equimultiplices of 8. and 4. the first and the third, multiplieng eche by 3. as before ye did, so shall ye haue 24 for the tri<g ref="char:EOLhyphen"/>ple
<figure/>
of 8. and 12 for the triple of 4. as ye had before. Then take the equimulti<g ref="char:EOLhyphen"/>plices of 6 and 3. the second and the fourth, multipliyng them by some one number, but not by 3 as before ye did: but by 4. so for the quadruple of 6 the second number, shall ye haue 24. and for the quadruple of 3 the fourth number, ye shall haue 12. And now ye see that the equimultiplices of 8 and 4. the first and the third, namely, 24 and 12. are both equall to the multiplices of 6 and 3. the second and the fourth, namely, to 24 and 12. Wherfore the numbers geuen, are by the second part of this definition in one and the selfe same proportion, because the equimultiplices of 8 and 4 the first and the third, are both equall to the equimultiplices of 6 and 3. the seconde and the fourth.</p>
                  <p>
                     <note place="margin">
                        <hi>An other example in numbers.</hi>
                     </note>Agayne to shew the same, and for the fulnes of the diffinition, take the same numbers 8, 6, 4, 3. and take the equimultiplices of 8 and 4. the first and the thirde, multiplieng eche by 2. so haue ye 16 for the duple of 8, the first number, and 8 for the duple of 4 the third number: then take also the equimultiplices of 6 and 3, the second and the fourth, multipliyng eche by 3. so haue ye 18 for the triple of 6 the second, and 9 for the triple of 3. the
<figure/>
fourth number. And now ye see that the equimultiplices of 8 and 4. the first and the third, namely, 16. and 8 are both lesse then the equimultipli<g ref="char:EOLhyphen"/>ces of 6 and 3. the second &amp; the fourth namely 18 and 9. For 16 are lesse then 18, and 8 are lesse then 9. Wherefore by the third part of this diffinition, the numbers proposed are in one and the selfe same proportion, for that the equimultiplices of 8 and 4 the first and the third are both lesse then the equimultiplices of 4 and 3 the second and the fourth.</p>
                  <p>
                     <pb n="130" facs="tcp:7062:160"/>Farther in this diffinition, this particle (according to any multiplication) is most diligently to be considered,<note place="margin">
                        <hi>Note this par<g ref="char:EOLhyphen"/>ticle according to any multi<g ref="char:EOLhyphen"/>plication.</hi>
                     </note> which signifie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h by any multiplication indiffe<g ref="char:EOLhyphen"/>rently whatsoeuer. For whensoeuer the quantities be in one and the selfe same proportion, then by any multiplication whatsoeuer, the equimultiplices of the f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rst and the third, shall exceede the equimultiplices of the second and the fourth, or shall be equall vnto them, or lesse then them. Yet it may so happen by some one multiplication; that the equimultiplices of the first and the third, do exceede the e<g ref="char:EOLhyphen"/>quimultiplices of the second and the fourth, and yet the quantities geuen shal not be in one and the selfe same proportion. As in this example here set, where the e<g ref="char:EOLhyphen"/>quimultiplices of 6 and 5, the first and the thirde, namely, 18. and 15. doo both exceede the equimultiplices of 4 and
<figure/>
3. the second and the fourth, namely, 8 and 6.<note place="margin">
                        <hi>An example where the equimultipli<g ref="char:EOLhyphen"/>ces of the first and third ex<g ref="char:EOLhyphen"/>ceed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the equi<g ref="char:EOLhyphen"/>multiplices of the second and fourth, and yet the quantities geuen are not in one and the selfe same pro<g ref="char:EOLhyphen"/>portion.</hi>
                     </note> yet are not the numbers ge<g ref="char:EOLhyphen"/>uen in one and the selfe same propor<g ref="char:EOLhyphen"/>tion. For 6 hath not that proportion to 4. which 5. hath to 3. In this exam<g ref="char:EOLhyphen"/>ple 6 and 5 the first and the third were multiplied by 3. which made their equimul<g ref="char:EOLhyphen"/>tiplices 18 and 15. which exceede the equimultiplices of 4 and 3, the second and the fourth beyng multiplied by 2. namely, 8 and 6: but if ye shall multiply 6 and 5 the first and the thirde by 2. ye shall produce 12 and 10 for their equimultipli<g ref="char:EOLhyphen"/>ces, and then if ye multiply 4 and
<figure/>
3. the second and the fourth by 3. so shall ye produce for their equimulti<g ref="char:EOLhyphen"/>plices 12 and 9. Now ye see that by this multiplication the equimultipli<g ref="char:EOLhyphen"/>ces of the first and the thirde doo not both exceede the equimultiplices of the second and fourth: for 12 the multiplex of 6 doth not exceede 12 the multiplex of 4. and therfore the numbers or quanti<g ref="char:EOLhyphen"/>ties ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> not in one and the selfe same proportion, for that it holdeth not in all mul<g ref="char:EOLhyphen"/>tiplications whatsoeuer.</p>
                  <p>And because this diffinition requireth all maner of multiplicatio<g ref="char:cmbAbbrStroke">̄</g>s to bring forth the excesses, equalities, and wantes of the antecedents aboue, to, or vnder the con<g ref="char:EOLhyphen"/>sequents; to auoide the tediousnes and infinite labour therof, I haue set forth a rule much to be made of and estemed, wherby ye may in any rationall proportion pro<g ref="char:EOLhyphen"/>duce equimultiplices of the first and the third equall to the equimultiplices of the second and the fourth. The rule is this,<note place="margin">
                        <hi>A rule to pro<g ref="char:EOLhyphen"/>duce equimul<g ref="char:EOLhyphen"/>tiplices of the first and third equall to the equimultipli<g ref="char:EOLhyphen"/>ces of the se<g ref="char:EOLhyphen"/>cond<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>urth.</hi>
                     </note> take two numbers whatsoeuer in that pro<g ref="char:EOLunhyphen"/>portion in which your quantities are, &amp; by the number which is antecedent mul<g ref="char:EOLhyphen"/>tiply the consequents of your proportions, namely, the second and the fourth: and by the number which is the consequent multiply the antecedentes of your pro<g ref="char:EOLhyphen"/>portions, namely, the first and the third<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then necessarily shalbe produced the equi<g ref="char:EOLhyphen"/>multiplices of the first and the third equall to the equimultiplices of the second &amp; the fourth. As by example, take 6 to 2. and 3 to 1, which are in one &amp; the selfe same proportio<g ref="char:cmbAbbrStroke">̄</g>, &amp; taking these two nu<g ref="char:cmbAbbrStroke">̄</g>bers 9 &amp; 3. which are in the same proportio<g ref="char:cmbAbbrStroke">̄</g>, now by 9 the antecedent multiply the cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>que<g ref="char:cmbAbbrStroke">̄</g>ts 2 &amp; 1. and so shal ye haue 18 &amp; 9 for the equimultiplices of the second &amp; the fourth,<note place="margin">
                        <hi>Example thereof.</hi>
                     </note> then by 3 the consequent multiply the antede<g ref="char:cmbAbbrStroke">̄</g>ts 6 &amp; 3, so shal ye haue 18 &amp; <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for the equimultiplices of the first &amp; the third, which are equal to the former equimultiplices of the seco<g ref="char:cmbAbbrStroke">̄</g>d &amp; fourth. Wher<g ref="char:EOLhyphen"/>of it foloweth that if ye multiply 18 &amp; 9 the equimultiplices of the first and the third by any nu<g ref="char:cmbAbbrStroke">̄</g>ber greater the<g ref="char:cmbAbbrStroke">̄</g> 3. wherby they were now multiplied, they shal both euer exceede the equimultiplices of the second &amp; the fourth: &amp; if ye multiply the<g ref="char:cmbAbbrStroke">̄</g>
                     <pb facs="tcp:7062:161"/>
by any number lesse then 3. they shall euer both want of them. So that whatsoe<g ref="char:EOLhyphen"/>uer multiplication it be, they shall euer both exceede, be equal, or want aboue, to, or from 18. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd 9. the equimultiplices of the second and fourth.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The seuenth definition</hi>
                     </note>Magnitudes which are in one and the selfe same proportion, are called Proportionall.</head>
                  <p>As if the lyne A, haue the same proportion to the line B, that the lyne C hath to the lyne D, then are the
<figure/>
said foure magnitudes A, B, C, D, called proportionall. Also in numbers for that 9. to 3. hath that same proportio<g ref="char:cmbAbbrStroke">̄</g> that 12 hath to 4:<note place="margin">9 12</note> therefore these foure nu<g ref="char:cmbAbbrStroke">̄</g>bers 9.3.12.4.<note place="margin">3 4</note> are said to be proportionall. Here is to be noted that this likenes or idemptitie of proportio<g ref="char:cmbAbbrStroke">̄</g> which is called, as before was said proportionalitie, is of two sortes: the one is continuall, the other is discontinuall.<note place="margin">
                        <hi>Proportiona<g ref="char:EOLhyphen"/>lity of two sortes, conti<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>uall and dis<g ref="char:EOLhyphen"/>continuall.</hi>
                     </note> Continuall proportionalitie is, when the magnitudes set in lyke proportion, are so ioyned together, that the second which is consequent to the first, is antecedent to the third, and the fourth which is consequent to the third, is antecedent to the fift, and so continually forth. So eue<g ref="char:EOLhyphen"/>ry quantitie or terme in this proportionalitie, is both antecedent and consequent (consequent in respect of tha which went before,<note place="margin">
                        <hi>An example of continuall proportionali<g ref="char:EOLhyphen"/>tie in num<g ref="char:EOLhyphen"/>bers.</hi>
                     </note> &amp; antecedent in respect of that which followeth) except the first, which is onely antecedent to that which follow<g ref="char:EOLhyphen"/>eth, and the last which is onely consequent to that which went before. Take an ex<g ref="char:EOLhyphen"/>ample in these numbers, 16.8.4.2.1.<note place="margin">16.8.4.2.1.</note> In what proportion 16. is to 8, in the same is 8. to 4, in the same also is 4. to 2, and likewise 2. to 1. For they all are in duple pro<g ref="char:EOLhyphen"/>portion: 16. the first is antecedent to 8, and 8. is consequent vnto it: and the selfe same 8. is antecedent to 4: which 4 beyng consequent to 8. is antecedent to 2, which 2 likewise is consequent to 4. and antecedent to 1: which because he is the last, is onely consequent, and antecedent to none, as 16. because it was the first, was antecedent onely, and consequent to none. Also in this proportionalitie all the magnitudes must of necessitie be of one kynde,<note place="margin">
                        <hi>In coutinnall proportionali<g ref="char:EOLhyphen"/>tie the quan<g ref="char:EOLhyphen"/>tities cannot be of one kinde.</hi>
                     </note> by reason of the continuation of the proportions in this proportionalitie, because there is no proportion betwene quantities of diuers kyndes. Discontinuall proportionalitie is,<note place="margin">
                        <hi>Discontinuall prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rtiona<g ref="char:EOLhyphen"/>litie.</hi>
                     </note> when the magni<g ref="char:EOLhyphen"/>tudes which are set in lyke proportion, are not continually set, as before they were, hauyng one terme referred both to that which went before, and to that which fo<g ref="char:EOLhyphen"/>loweth, but haue their termes distinct and seuered asonder: as the first is antece<g ref="char:EOLhyphen"/>dent to the second, so is the third antecedent to the fourth. Example in numbers, as 8 is to 4.<note place="margin">
                        <hi>Example of discontinual proportionali<g ref="char:EOLhyphen"/>ty in numbers.</hi>
                     </note> so is 6. to 3. for either proportion is duple. Where ye see, how ech pro<g ref="char:EOLhyphen"/>portion hath hys owne antecedent and consequent distinct from the antecedent and consequent of the other, and no one number is antecedent and consequent in diuers respectes. And by reason of the discontinuaunce of the proportions in this proportionalitie,<note place="margin">
                        <hi>In disco<g ref="char:cmbAbbrStroke">̄</g>tinual proportionali<g ref="char:EOLhyphen"/>tie the propor<g ref="char:EOLhyphen"/>tions may be of diuers kindes.</hi>
                     </note> the quantities compared, may be of diuers kyndes, because the consequent in the first proportion is not the antecedent in the second proportion. So that ye may compare superficies to superficies, or body to body in the selfe same proportion that ye do lyne to lyne.</p>
               </div>
               <div type="section">
                  <pb n="131" facs="tcp:7062:161"/>
                  <head>When the equemultiplices being taken,<note place="margin">
                        <hi>The eight definition.</hi>
                     </note> the multiplex of the first excedeth the multiplex of the second, &amp; the multiplex of the third, excedeth not the multiplex of the fourth: then hath the first to the second a greater proportion, then hath the third to the fourth.</head>
                  <p>In the sixt definition was declared what magnitudes are said to be in <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e and the same proportion: now he sheweth in this definition what magnitudes are said to be in a greater proportion. And here is supposed the same order of multiplicati<g ref="char:EOLhyphen"/>on, that there in that definition was vsed: namely, that the first and the third be e<g ref="char:EOLhyphen"/>qually multiplied, that is, by one &amp; the selfe same nu<g ref="char:cmbAbbrStroke">̄</g>ber: and also that the second and the fourth be equally multiplied by
<figure/>
the same or some other number: and then if the multiplex of the first, excede the multiplex of the second:<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note> &amp; the mul<g ref="char:EOLhyphen"/>tiplex of the third; excede not the mul<g ref="char:EOLhyphen"/>tiplex of the fourth, the first hath a grea<g ref="char:EOLhyphen"/>ter proportion to the second, then hath the third to the fourth. As suppose that there be foure quantities, A, B, C, D: of which let A be the first, B the second, C the third, &amp; D the fourth. And let A the first co<g ref="char:cmbAbbrStroke">̄</g>mine 6. and let B the second con<g ref="char:EOLunhyphen"/>taine 2. &amp; C the third 4. &amp; D the fourth 3: Now take the equimultiplices of A and C the first &amp; the third, which let be E and F, so that how multiplex E is to A so multiplex let F be to C: name<g ref="char:EOLhyphen"/>ly for example sake let either of them be triple: so haue you 18. for the multiplex of A, and 12. for the mul<g ref="char:EOLhyphen"/>tiplex of C. Likewise take the equimultiplices of B &amp; D, the second &amp; the fourth, multiplying them also by one and the self same number, as by 4: so haue ye for the multiplex of B the second 8, namely, the line G, and for the multiplex of D the fourth 12, namely, the line H. Now because the line E multiplex to the first, name<g ref="char:EOLhyphen"/>ly, 18, excedeth the line G multiplex to the second, namely, 8: And the line F mul<g ref="char:EOLhyphen"/>tiplex to the third, namely, 12, excedeth not the line H multiplex to the fourth, namely, 12 (for that they are equall) the proportio<g ref="char:cmbAbbrStroke">̄</g> of A to B the first to the second, is greater then the proportion of C to D the third to the fourth. So likewise in nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers: take 11. to 2. &amp; 7. to 3. and mul<g ref="char:EOLhyphen"/>tiply
<figure/>
11. &amp; 7.<note place="margin">
                        <hi>An example in numbers.</hi>
                     </note> (the first, and the third) by 2, so shall ye haue 22. for the multi<g ref="char:EOLhyphen"/>plex of the first, and 14. for the multi<g ref="char:EOLhyphen"/>plex of the third: and multiply 2. and 3. the second and the fourth by 6: so shall the multiplex of the second be 12. and the multiplex of the fourth be 18:
<pb facs="tcp:7062:162"/>
Now ye see 22. the multiplex of the first, excedeth 12, the multiplex of the second. But 14. the multiplex of the third, excedeth not 18. the multiplex of the fourth: Wherefore the proportion of 11. to 2. the first to the second, is greater then the pro<g ref="char:EOLhyphen"/>portion of 7. to 3, the third to the fourth. And so of all other quantities and num<g ref="char:EOLhyphen"/>bers, which are not in one and the selfe same proportion, ye may know when the first to the second hath a greater proportion then the third to the fourth.</p>
                  <div type="subsection">
                     <head>¶An other example.</head>
                     <p>This example haue I set to declare
<figure/>
that although the proportion of the first to the second be greater then the proportion of the third to the fourth, yet the multiplex of the first excedeth not the multiplex of the seco<g ref="char:cmbAbbrStroke">̄</g>d. Wher<g ref="char:EOLhyphen"/>fore it is diligently to be noted,<note place="margin">
                           <hi>Note.</hi>
                        </note> that it is sufficient to shew that the proporti<g ref="char:EOLhyphen"/>on of the first to the second is greater the<g ref="char:cmbAbbrStroke">̄</g> the proportion of the third to the fourth, if the want or lacke of the multiplex of the first from the multiplex of the second, be lesse then the want or lacke of the multiplex of the third to the multiplex of the fourth. As in this example 16. the multiplex of 8. the first, wanteth of 20. the multiplex of 4. the second, foure: wheras 18. the multiplex of 9, the third, wa<g ref="char:cmbAbbrStroke">̄</g>teth of 45, the multiplex of 9 the fourth, 27. And so of all others wheras (the proportions being diuers) the equimultiplices of the first and the third are both lesse, then the equimultiplices of the second and the fourth. Likewise if the equimultiplices of the first and the third do both excede the equimultiplices of the second &amp; the first, the<g ref="char:cmbAbbrStroke">̄</g> shall the excesse of the multiplex of the first aboue the multiplex of the second, be greater the<g ref="char:cmbAbbrStroke">̄</g> the excesse of the multiplex of the third, aboue the multiplex of the fourth. As in these numbers here set, the equimultiplices of 6. and 4. the first and the third, namely, 12. and 8. do both excede the equimultiplices of 2. and 3. the se<g ref="char:EOLhyphen"/>cond and the fourth, namely, 4. and 6. But 12. the multiplex of the first excedeth 4. the multiplex of the second by 4, and 8. the multiplex of the thyrd excedeth 6. the multiplex of the fourth by 2. but 8. is
<figure/>
more then 2. Howbeit this is general<g ref="char:EOLhyphen"/>ly certaine that when soeuer the pro<g ref="char:EOLhyphen"/>portion of the first to the seco<g ref="char:cmbAbbrStroke">̄</g>d is grea<g ref="char:EOLhyphen"/>ter then the proportion of the third to the fourth, there may be found some multiplication, that whe<g ref="char:cmbAbbrStroke">̄</g> the equimul<g ref="char:EOLhyphen"/>tiplices of the first and the third shall be compared to the equimultiplices of the second and the fourth, the multiplex of the first shall excede the multiplex of the second, &amp; the multiplex of the third shall not excede the multiplex of the fourth, according to the plaine wordes of the de<g ref="char:EOLhyphen"/>finition.</p>
                     <p>In like maner when you haue taken the equimultiplices of the first &amp; the third, and also the equimultiplices of the second and the fourth, if the multiplex of the first excede not the multiplex of the second, and the multiplex of the third excede the multiplex of the fourth: then hath the first to the second a lesse proportion, then hath the third to the fourth. As in the example before, if ye chaunge the termes, and make C the first, D the second, A the third, and B the fourth: then shall
<pb n="132" facs="tcp:7062:162"/>
F, namely, 12. the multiplex of the first not excede H, namely, 12. the multiplex of the second: but E, namely, 18. the multiplex of the third excedeth G, namely, 8. the multiplex of the fourth. Wherefore the proportion of C to D, the first to the se<g ref="char:EOLhyphen"/>cond, is lesse then the proportion of A to B, the third to the fourth.</p>
                     <p>Euen so in numbers. As in this ex<g ref="char:EOLhyphen"/>ample,
<figure/>
5. to 4. and 7. to 3. If ye multiply 5. and 7. the first and the third eche by 3, ye shall for the multiplex of 5. the first haue 15. and for the multiplex of 7. the third shall ye haue 21: againe if ye multiply 4. and 3. the second &amp; the fourth by 6, for the multiplex of 4. the second ye shall haue 24, and for the multiplex of 3. the fourth, ye shall haue 18. So ye see that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>5. the multiplex of the first, is lesse then 24, the multiplex of the second: and 21. the multiplex of the third is greater then 18. the multiplex of the fourth. Wherefore the proportion of 5. to 4. the first to the second is lesse then the propor<g ref="char:EOLhyphen"/>tion of 7. to 3. the third to the fourth.</p>
                  </div>
               </div>
               <div type="section">
                  <head>Proportionallitie consisteth at the lest in three termes.<note place="margin">
                        <hi>The ninth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>
                  </head>
                  <p>Before it was sayd, that proportionalitie is a likenesse or an idemptitie of pro<g ref="char:EOLhyphen"/>portions. Wherfore of necessitie in proportionalitie, there must be two proporti<g ref="char:EOLhyphen"/>ons, and euery proportion hath two termes, namely, his antecedent and conse<g ref="char:EOLhyphen"/>quent. Therfore in euery proportionalitie th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>re are foure termes. But for that som<g ref="char:EOLhyphen"/>tyme, one terme supplieth by diuers relations, the roume of two, for in respect to the first it is consequent, and in respect to that which followeth, it is antecedent: therfore three termes at least and not vnder may suffice in proportionalitie, which three are in power foure, and occupy the rome of foure, as is sayd.<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note> As suppose that A hath to B that proportion, that B
<figure/>
hath to C: then are these thre quan<g ref="char:EOLhyphen"/>tities A, B, C, set in the lest number of proportionality.<note place="margin">
                        <hi>Example <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n numbers.</hi>
                     </note> Likewise in num<g ref="char:EOLhyphen"/>bers, as 8. 4. 2. and 9. 6. 4.</p>
               </div>
               <div type="section">
                  <head>When there are three magnitudes in proportion,<note place="margin">
                        <hi>The tenth definition.</hi>
                     </note> the first shall be vnto the third in double proportion that it is to the second. But when there are foure magnitudes in proportion the first shall be vnto the fourth in treble proportion that it is to the se<g ref="char:EOLhyphen"/>cond. And so alwaies in order one more, as the proportion shall be extended.</head>
                  <p>This definition is also vnderstand in continuall proportionalitie.<note place="margin">
                        <hi>Example of this definition in magni<g ref="char:EOLhyphen"/>tudes.</hi>
                     </note> As if the thre magnitudes A, B, C, bee proportionall: then shall the proportion of A the
<pb facs="tcp:7062:163"/>
first to C the thirde, bee
<figure/>
double to the proportion which is betwene A &amp; B the first and the seconde, that is the proportio<g ref="char:cmbAbbrStroke">̄</g> of A to B taken twise, or added to it self (which is all one) shall make the proportio<g ref="char:cmbAbbrStroke">̄</g> of A to C. For the easier vndersta<g ref="char:cmbAbbrStroke">̄</g>dyng of this &amp; the pra<g ref="char:EOLhyphen"/>ctise therof, it shall be much necessary somwhat to instruct the rude beginner how proportions may be added one to an other. Which is done by this rule.</p>
                  <p>
                     <note place="margin">
                        <hi>A rule to adde proporti<g ref="char:EOLhyphen"/>ons to propor<g ref="char:EOLhyphen"/>tions.</hi>
                     </note>Multiply the antecedent of the one proportion by the antecede<g ref="char:cmbAbbrStroke">̄</g>t of the other, and the number produced shall be the antecedent of the proportion which con<g ref="char:EOLhyphen"/>tayneth them both. Likewyse multiply the consequent of the one proportion by the consequent of the other, and the number produced shall be consequent of the proportion which shall contayne them both.</p>
                  <p>
                     <note place="margin">
                        <hi>Example thereof.</hi>
                     </note>
                     <hi>An example.</hi> If ye will adde the proportion which is betwene 4 and 2. (which is dupla) to the proportion which is betwene 9 and 3. (which is tripla) multiply 9. the antecedent of the first proportion by 4. the antecedent of the
<figure/>
second proportion, and ye shall produce 36. which reserue and kepe for the antecedent of the proportion which ye seeke for. Likewise multiply 3 the consequent of the first proportion, by 2 the conse<g ref="char:EOLhyphen"/>quent of the second, so shall ye haue 6. which 6. shall be consequent to the former antecede<g ref="char:cmbAbbrStroke">̄</g>t, namely, to 36. so shal the proportio<g ref="char:cmbAbbrStroke">̄</g> which is betwene 36 and 6. namely, sextupla, contayne in it the two proportions geuen, namely, tripla, and dupla. And by this meanes are they added together, &amp; brought into one. And by this may ye adde all other kyndes of proportions whatsoeuer they be. Now for that the diffinition sayth, that if there be three quantities in pro<g ref="char:EOLhyphen"/>portio<g ref="char:cmbAbbrStroke">̄</g>, that is, what proportio<g ref="char:cmbAbbrStroke">̄</g> the first hath to the seco<g ref="char:cmbAbbrStroke">̄</g>d, the same hath the second to the third, which for example let be tr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ple, as in these nu<g ref="char:cmbAbbrStroke">̄</g>bers, 27. 9. 3. adde triple to triple by the rule abouesaid. And forasmuch as it is easier to worke in small nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers then in great, reduce these proportio<g ref="char:cmbAbbrStroke">̄</g>s to theyr least denomination:<note place="margin">27. 9. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</note> So 27. to 9. reduced to the lest termes in that proportion, is as much as 3. to 1. Likewise 9 to 3 reduced to theyr lest termes are also as much as 3 to 1.<note place="margin">9</note> now adde together these two triple proportions thus reduced, multipliyng 3 by 3.<note place="margin">3—3</note> the one antecedent by the other, so shall ye produce 9 for a new antecedent, then multiply 1 by 1.<note place="margin">1—1</note> the one consequent by the other, so shall you produce 1. which let be consequent to 9. your antecedent, so the proportion betwene 9 and 1. which is noncuple contay<g ref="char:EOLhyphen"/>neth both the two triple proportions. And because they were equal the one to the other, it is duple to eche of them. Ye see <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>lso that the proportion of 27 to 3. the first to the third, is also noncuple. Wherfore according to the definition, the proporti<g ref="char:EOLhyphen"/>on of the first to the third, is double to the proportion of the first to the second, as 9 to 1. beyng noncuple, is double 3 to 1. which is triple, because it contayneth it twise.</p>
                  <p>So if there be 4. quantities in continuall proportion, the proportion of the first to the fourth, shall be triple to the proportion which is betwene the first and the second, that is, it shall contayne it three tymes. As for example, Take 4. numbers in continual proportion 8. 4. 2. 1.<note place="margin">8. 4. 2. 1.</note> Ye see that the proportio<g ref="char:cmbAbbrStroke">̄</g> of 8 to 1.<note place="margin">2 2 2</note> the first to the fourth,<note place="margin">1 1 1</note> is octupla: the proportion of 8 to 4. the first to the second is dupla, now treble dupla proportion, that is, adde 3. dupla proportions together, by the rule before geuen, as ye see in the example. Multiply all the antecedentes together 2.
<pb n="133" facs="tcp:7062:163"/>
the antecedent of the first proportion, by 2. the antecedent of the second, so haue ye 4: which 4. multiply by 2 the antecedent of the third proportio<g ref="char:cmbAbbrStroke">̄</g>, so shal ye haue 8 for a new antecedent. In lyke maner multiply all the consequentes together, 1. the consequent of the first proportion by 1. the consequent of the second propor<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>, so shal ye haue 1, which 1. multiply agayne by 1. the co<g ref="char:cmbAbbrStroke">̄</g>sequent of the third pro<g ref="char:EOLhyphen"/>portion, so shall ye haue agayne 1: which 1. let be consequent to your former ante<g ref="char:EOLhyphen"/>cedent 8: so haue ye 8 to 1. which is octupla, which was also the proportion of the first to the fourth, which octupla is also brought fourth of the addition of thre du<g ref="char:EOLhyphen"/>pla proportions together, and contayneth it three tymes, wherefore octupla is tri<g ref="char:EOLhyphen"/>pla to dupla, and therfore as the diffinition sayth: the proportion of the first to the fourth is tripla to the proportion of the first to the second. And so consequently forth as long as the proportionalitie continueth accordyng to the sentence of the diffinition, the termes of the proportions exceding the number of proportions by one. As if ye haue 5. termes in proportion, the proportio<g ref="char:cmbAbbrStroke">̄</g> of the first to the fifth shal be quadrupla to the proportion of the first to the second, and if there be 6. termes, it shall be quintupla and so in order.</p>
               </div>
               <div type="section">
                  <head>Magnitudes of like proportion, are sayd to be antecedents to antecedentes, and consequentes to consequentes.<note place="margin">
                        <hi>The eleuenth definition.</hi>
                     </note>
                  </head>
                  <p>For that before it was sayd, that proportion was a relation or a respect of one quantitie to an other, now sheweth he what magnitudes are sayd to be of like pro<g ref="char:EOLhyphen"/>portion, namely, these whose antecedents haue like respect to their consequentes, and whose consequents receyue
<figure/>
like respectes of their antecede<g ref="char:cmbAbbrStroke">̄</g>ts.<note place="margin">
                        <hi>Example of this definition in magnitud<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> As putting 4. magnitudes A, B, CD. If A antecedent to B, be dou<g ref="char:EOLhyphen"/>ble to B, and C antecedent to D, be double also to D, the<g ref="char:cmbAbbrStroke">̄</g> haue the two antecedentes like respectes to their consequents. Likewise if B the consequent be halfe of A, and also D the consequent be halfe of C, then the two consequentes B and D receiue of their antecedentes like respectes and relations. And by this dif<g ref="char:EOLhyphen"/>finition, are these magnitudes A, B, C, D, of like proportion.</p>
                  <p>Also in numbers, 9. 3. 6. 2: because 9 the antecedent is triple to 3. his conse<g ref="char:EOLhyphen"/>quent, and the antecedent 6. is also triple to 2 his consequent:<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> the
<figure/>
two antecede<g ref="char:cmbAbbrStroke">̄</g>ts 9 and 6 haue like respectes to their consequentes, and because that 3 the consequent is the subtriple or third part of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. his antecedent, and likewise 2 the consequent is the subtriple or third part of 6. his antecedent, the two consequentes 3 and 2 receiue also lyke re<g ref="char:EOLhyphen"/>spectes of their antecedentes, and therfore are numbers of like proportion.</p>
               </div>
               <div type="section">
                  <head>Proportion alternate, or proportion by permutation is,<note place="margin">
                        <hi>The twelf<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>h definition.</hi>
                     </note> when the antecedent is compared to the antecedent, and the conse<g ref="char:EOLhyphen"/>quent to the consequent.</head>
                  <p>The vnderstanding of this definition &amp; of all the definitions following, depen<g ref="char:EOLhyphen"/>deth of the definition going before, and vse it for a generall supposition, namely, to
<pb facs="tcp:7062:164" rendition="simple:additions"/>
haue foure quantities in proportion. Suppose foure magnitudes A, B, C, D, to be in proportion,<note place="margin">
                        <hi>Example of this de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inition in magnituds.</hi>
                     </note> namely, as A is to B, so let C be to D. Now if ye compare A the ante<g ref="char:EOLhyphen"/>cedent of the first proportion to C the antecedent of the second as to his conse<g ref="char:EOLhyphen"/>quent, &amp; likewise if ye compare
<figure/>
B the co<g ref="char:cmbAbbrStroke">̄</g>sequent of the first pro<g ref="char:EOLhyphen"/>portion as an antecedent to D the consequent of the second as to his consequent: then shall ye haue the magnitudes in this sort: as A to C, ante<g ref="char:EOLhyphen"/>cedent to antecedent,<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> so B to D, consequent to consequent, &amp; this is called permu<g ref="char:EOLhyphen"/>tate proportion or alternate. In numbers as 12. to 6, so
<figure/>
8. to 4. either is dupla. Wherefore by permutation of proportion, as 12. to 8. antecedent to antecedent, so is 6. to 4. consequent to consequent, for either is sesquialtera.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The thirtenth definition.</hi>
                     </note>Conuerse proportion, or propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tion by conuersion is, when the consequent is taken as the antecedent, and so is compared to the antecedent as to the consequent.</head>
                  <p>Suppose as before foure magnitudes in proportion, A, B, C, D, as A to B, so C to D:<note place="margin">
                        <hi>Example of this definition in magnituds.</hi>
                     </note> if ye referre B the consequent of the first proportion, as antecedent, to A the antecedent of the first, as to his consequent: and likewise if ye referre D the conse<g ref="char:EOLhyphen"/>que<g ref="char:cmbAbbrStroke">̄</g>t of the second proportion as an<g ref="char:EOLhyphen"/>tecede<g ref="char:cmbAbbrStroke">̄</g>t
<figure/>
to C the antecede<g ref="char:cmbAbbrStroke">̄</g>t of the se<g ref="char:EOLhyphen"/>cond proportio<g ref="char:cmbAbbrStroke">̄</g>, as to his co<g ref="char:cmbAbbrStroke">̄</g>sequent: the<g ref="char:cmbAbbrStroke">̄</g> shall ye haue the magnitudes in thys order. As B to A co<g ref="char:cmbAbbrStroke">̄</g>sequent to antecede<g ref="char:cmbAbbrStroke">̄</g>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> so D to C consequent to antecede<g ref="char:cmbAbbrStroke">̄</g>t.<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> And thys is called co<g ref="char:cmbAbbrStroke">̄</g>uerse proportion. So also in numbers, 9. to 3, as 6. to 2, eyther is tripla, wher<g ref="char:EOLhyphen"/>fore
<figure/>
comparing 3. to 9, the consequent of the first to hys antecedent 9, and also 2. the consequent of the second to hys antecedent 6, by conuerse proporti<g ref="char:EOLhyphen"/>on it commeth to passe as 3. to 9, so 2. to 6: For ei<g ref="char:EOLhyphen"/>ther is subtripla.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The fourtenth definition.</hi>
                     </note>Proportion composed, or composition of proportion is, when the antecedent and the consequent are both as one compared vnto the consequent.</head>
                  <p>Suppose that in the former foure
<figure/>
magnituds in proportio<g ref="char:cmbAbbrStroke">̄</g>, A, B, C, D, as A is to B, so is <hi>C</hi> to D:<note place="margin">
                        <hi>Example of this definition in magnituds.</hi>
                     </note> if ye adde A and B the antecedent and the consequent of the first proportion together, and compare them so added as one antecedent to B the consequent of the first proportion as to hys consequent: and likewise if ye adde together
<pb n="134" facs="tcp:7062:164"/>
C and D the antecedent and the consequent of the second proportion, and so ad<g ref="char:EOLhyphen"/>ded, compare them as one antecedent to D the consequent of the second propor<g ref="char:EOLhyphen"/>tion, as to his consequent: then shall ye haue the magnitudes in this order. As AB to B, so CD to D, for either of them is tripla. And this is called composed propor<g ref="char:EOLhyphen"/>tion, or composition of proportion. And so also in numbers.<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> As 8. to 4, so 6. to 3: 8. and 4, the antecedent and consequen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the first pro<g ref="char:EOLhyphen"/>portion
<figure/>
added together, make 12: which 12. as antecedent co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>pare to 4. the consequent of the first proportion as to his con<g ref="char:EOLhyphen"/>sequent: so adde together 6. and 3, the antecedent and con<g ref="char:EOLhyphen"/>sequent of the second proportion, they make 9: which 9. as antecedent compare to 3. the consequent of the second proportion, as to his con<g ref="char:EOLhyphen"/>sequent: so shall ye haue by composition of proportion, as 12. to 4, so 9. to 3, for ei<g ref="char:EOLhyphen"/>ther of them is tripla.</p>
               </div>
               <div type="section">
                  <head>Proportion deuided, or diuision of proportio<g ref="char:cmbAbbrStroke">̄</g> is,<note place="margin">
                        <hi>The fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ne definition.</hi>
                     </note> when the ex<g ref="char:EOLhyphen"/>cesse wherein the antecedent excedeth the consequent, is com<g ref="char:EOLhyphen"/>pared to the consequent.</head>
                  <p>Thys definition is the conuerse of the definition going next before:<note place="margin">
                        <hi>This is the co<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>uerse of the former defini<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> in it was vsed composition, and in thys is vsed diuision. As before so now suppose foure magnitudes in proportion AB the first, B the second, CD the third, and D the fourth: as AB to B: so CD to D:
<figure/>
AB,<note place="margin">
                        <hi>Example in magnitudes.</hi>
                     </note> the antecedent of the first pro<g ref="char:EOLhyphen"/>portion excedeth B the consequent of the first proportion by the magni<g ref="char:EOLhyphen"/>tude A, wherfore A is the excesse of the antecedent AB aboue the consequent B: so likewise CD the antecedent of the second proportion, excedeth D the conse<g ref="char:EOLhyphen"/>quent of the same proportion, by the quantitie C, wherefore C is the excesse of the antecedent CD aboue the consequent D. Now if ye compare A the excesse of AB the first antecedent, aboue the consequent B, as antecedent to B the con<g ref="char:EOLhyphen"/>sequent, as to his consequent: also if ye compare D the excesse of the second an<g ref="char:EOLhyphen"/>tecedent CD, aboue the consequent D, as antecedent to D the consequent, as to his consequent: then shall your magnitudes be in this order. As A to B, so is C to D: which is called diuision of proportion, or proportion deuided.<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note>
                  </p>
                  <p>And so in numbers, as 9. to 6, so 12. to 8, either proportion
<figure/>
is sesquialtera: the excesse of 9. the antecedent of the first proportion aboue 6. the consequent of the same is 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: the ex<g ref="char:EOLhyphen"/>cesse of 12. the antecedent of the second proportion aboue 8, the consequent of the same, is 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then if ye compare 3. the ex<g ref="char:EOLhyphen"/>cesse of 9. the first antecedent aboue the consequent, as antecedent to 6, the conse<g ref="char:EOLhyphen"/>quent, as to hys consequent: and also if ye compare .4 the excesse of 12. the second antecedent aboue the consequent, as antecedent, to 8. the consequent, as to hys consequent, ye shall haue your numbers after this maner by diuision of proporti<g ref="char:EOLhyphen"/>on, as 3. to 6: so 4. to 8: for either proportion is subdupla.</p>
               </div>
               <div type="section">
                  <head>Conuersio<g ref="char:cmbAbbrStroke">̄</g> of proportion (which of the elders is commonly cal<g ref="char:EOLhyphen"/>led euerse proportion,<note place="margin">
                        <hi>The sixtene definition.</hi>
                     </note> or euersio<g ref="char:cmbAbbrStroke">̄</g> of proportion) is, whe<g ref="char:cmbAbbrStroke">̄</g> the an<g ref="char:EOLhyphen"/>tecedent
<pb facs="tcp:7062:165"/>
is compared to the excesse, wherein the antecedent excedeth the consequent.</head>
                  <p>
                     <note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note>Foure magnitudes supposed as before, AB the first, B the second, CD the third, and D the fourth. As AB to A, so CD to C: AB the antecedent of the first proportion excedeth B the consequent of the same by the magnitude A, where<g ref="char:EOLhyphen"/>fore A is the excesse of the antece<g ref="char:EOLhyphen"/>dent
<figure/>
AB aboue the consequent B: so also the magnitude C is the ex<g ref="char:EOLhyphen"/>cesse of CD the antecedent of the second proportion aboue D the consequent of the same: now if ye referre AB the antecedent of the first proportion, as antece<g ref="char:EOLhyphen"/>dent, to A the excesse therof aboue the consequent B, as to his consequent: if ye compare also CD the antecedent of the second proportion as antecedent to C the excesse therof aboue the consequent D, as to his consequent: then shall your magnitudes come to thys order. As AB to A, so CD to C, and thys is called conuersion of proportion, and of some euersion of proportion. Likewyse in num<g ref="char:EOLhyphen"/>bers, as 9. to 6, so 12. to 8.<note place="margin">
                        <hi>An example in numbers.</hi>
                     </note> eyther proportion is sesquialtera:
<figure/>
the excesse of 9. the antecedent of the first proportion aboue 6. the consequent of the same is 3: the excesse of 12. the ante<g ref="char:EOLhyphen"/>cedent of the second proportion aboue 8. the consequent of the same, is 4: now co<g ref="char:cmbAbbrStroke">̄</g>pare the antecedent of the first propor<g ref="char:EOLhyphen"/>tion 9. as antecede<g ref="char:cmbAbbrStroke">̄</g>t to 3. the excesse therof aboue 6. the conseque<g ref="char:cmbAbbrStroke">̄</g>t, as to his con<g ref="char:EOLhyphen"/>sequent, likewise compare 12. the antecedent of the second proportion as antece<g ref="char:EOLhyphen"/>dent to 4. the excesse therof aboue 8. the consequent, as to his consequent: so shall your numbers be in thys order by conuersion of proportion: as 9. to 3: so 12. to 4: for either proportion is triple.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The seue<g ref="char:cmbAbbrStroke">̄</g>tenth definition.</hi>
                     </note>Proportion of equalitie is, when there are taken a number of magnitudes in one order, and also as many other magnitudes in an other order, comparing two to two beyng in the same pro<g ref="char:EOLunhyphen"/>portion, it commeth to passe, that as in the first order of mag<g ref="char:EOLhyphen"/>nitudes, the first is to the last, so in the second order of magni<g ref="char:EOLhyphen"/>tudes is the first to the last. Or otherwise it is a compari<g ref="char:EOLhyphen"/>son of extremes together, the middle magnitudes being taken away.</head>
                  <p>To the declaration of thys definition are required two orders of magnitudes equall in number, and in lyke proportion:<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note> As if there be taken in some deter<g ref="char:EOLhyphen"/>minate number certayne magnitudes, namely, foure, A, B, C, D. And also in the same number be taken other quantities, namely, foure, E, F, G, H: then take the equall proportions by two and two: as A to B, so E to F: as B to C,
<pb n="135" facs="tcp:7062:165"/>
so F to G: and as C to D, so G
<figure/>
to H. Now according to the first definition, if A the first magnitude of the first order be to D the last mag<g ref="char:EOLhyphen"/>nitude of the same order, as E the first magnitude of the second order is to H the last magnitude the same, then it is called proportion of equa<g ref="char:EOLhyphen"/>litie, or equall proportion.</p>
                  <p>By the second definition, which is all one in substance with the first, ye leaue the meane magnitudes in eyther order, namely, B, C, on the one side, and F, G on the other side, and onely compare the extremes of ech side together, which by thys de<g ref="char:EOLhyphen"/>finition shall be in lyke proportion, as A is to D, so is E to H.</p>
                  <p>Euen so in numbers, take for example these two orders,
<figure/>
27. 9. 12. 24. 25. and 9. 3. 4. 8. 5. there are in eche order as ye see,<note place="margin">
                        <hi>An example in numbers.</hi>
                     </note> fiue numbers, then see that all the proportions taken by two &amp; two be like: betwene 27 &amp; 9, numbers of the first or<g ref="char:EOLhyphen"/>der, and betwene 9. and 3, num<g ref="char:EOLhyphen"/>bers of the second order, there is one and the self same propor<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>, namely, tripla: also betwene <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, and 12, numbers of the first order, and 3. and 4, numbers of the second order, is like pro<g ref="char:EOLhyphen"/>portion, namely, subsesquitertia proportion: so betwene 12. and 24, numbers of the first order, and 4. and 8, numbers of the second order, is also lyke proportion, namely, subdupla: Last of all, betwene 24. and 15, numbers of the first rowe, and betwene 8. and 5, numbers of the second rowe, the proportion is one, namely, su<g ref="char:EOLhyphen"/>per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ripar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iens quintas. Wherefore by this definition, leauing out all the meane numbers of eche side, ye may compare together onely the e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tremes, and conclude that as 27. of the first row is to 15. the last of the same row, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o is 9. the first of the second rowe to 5, the last of the same rowe: for the proportion of ech is superqua<g ref="char:EOLhyphen"/>dripartiens quintas.</p>
                  <p>Here is to be considered,<note place="margin">
                        <hi>Note.</hi>
                     </note> that it is not of necessitie that all the proportione in eche rowe of numbers be set in like order, as in the one so in the other: but it shall be sufficient that the proportions be the same and in equall number in eche rowe. Whether it be in the selfe same order, or in contrary, or inue<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ed order, it maketh no matter. As in these numbers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 12. 6. 2. in the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rst row<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>. 8<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 4.
<figure/>
in the second. As 12. is to 6, the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the second of the first row, so is 8. to 4. the second to the third of the second row: either i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> duple proportion. And as 6<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the second to the third in the first
<pb facs="tcp:7062:166"/>
order: so is 24. to 8. the first to the second in the second order. Where ye see that the proportions are not placed in one and the selfe same order, and course, and yet notwithstanding ye may conclude by equalitie of proportion, leauing the meanes 6. and 8: as 12. to 2. the first to the last of the first or<g ref="char:EOLhyphen"/>der,
<figure/>
so 24. to 4. the first and last of the second order. And so of others whatsoeuer and how soeuer they be placed.</p>
               </div>
               <div type="section">
                  <head>An ordinate proportionality is, when as the antecedent is to the consequent,<note place="margin">
                        <hi>The eight<g ref="char:EOLhyphen"/>tenth defini<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> so is the antecedent to the consequent, and as the consequent is to another, so is the consequent to an other.</head>
                  <p>For the declaration of this definition are also required two orders of magni<g ref="char:EOLhyphen"/>tudes.<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in mag<g ref="char:EOLhyphen"/>nitudes.</hi>
                     </note> Suppose in the first order, that the antecedent A, to his consequent B, haue the same proportion that the antecedent D, hath to his co<g ref="char:cmbAbbrStroke">̄</g>sequent E in the second order: and make the consequent B
<figure/>
antecede<g ref="char:cmbAbbrStroke">̄</g>t to some other quantitie, as to C. Also make the consequent E antecede<g ref="char:cmbAbbrStroke">̄</g>t to an other qua<g ref="char:cmbAbbrStroke">̄</g>titie, as to F, so that there be the same pro<g ref="char:EOLhyphen"/>portion of B to C, which is of E to to F. And thys disposition of proportions is called ordinate proporti<g ref="char:EOLhyphen"/>onalitie.
<figure/>
Likewise in numbers, 18.9.3 and 6.3.1.<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> As 18. to 9. antece<g ref="char:EOLhyphen"/>dent to consequent, so is 6. to 3. antecedent to consequent: either is dupla proportion: and as 9. the consequent is to an other, namely, to the number 3, so is the consequent 3. to an other, namely, to vnity. And this ordinate proportionalitie may be exte<g ref="char:cmbAbbrStroke">̄</g>ded as farre as ye li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t, as ye may see in the example of numbers in the definition next before.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The nintenth definition.</hi>
                     </note>An inordinate proportionality is, when as the antecedent is to the consequent, so is the antecedent to the consequent: and as the consequent is to an other, so is an other to the ante<g ref="char:EOLhyphen"/>cedent.</head>
                  <p>This definition also as the other before, requireth two orders of magnitudes, Suppose in the first order that the antecede<g ref="char:cmbAbbrStroke">̄</g>t A be to the co<g ref="char:cmbAbbrStroke">̄</g>seque<g ref="char:cmbAbbrStroke">̄</g>t B, as the antece<g ref="char:EOLhyphen"/>de<g ref="char:cmbAbbrStroke">̄</g>t C,<note place="margin">
                        <hi>An example of this defini<g ref="char:EOLhyphen"/>tion in magni<g ref="char:EOLhyphen"/>tudes.</hi>
                     </note> in the second
<figure/>
order is to the conse<g ref="char:EOLhyphen"/>quent D, &amp; let B the conseque<g ref="char:cmbAbbrStroke">̄</g>t of the first proportio<g ref="char:cmbAbbrStroke">̄</g> be to some other, namely, to the magnitude E, as some other, namely, the magnitude F, is to the antecedent C of the second proportio<g ref="char:cmbAbbrStroke">̄</g>:<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> this kinde of proportionalitie is called inordinate or perturbate.
<figure/>
Take also an example in numbers, as 9 to 6. the antecedent to the consequent, so is 3 to 2 the antecedent to the consequent, ei<g ref="char:EOLhyphen"/>ther proportio<g ref="char:cmbAbbrStroke">̄</g> is s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ul tera, and as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the consequent of the first
<pb n="136" facs="tcp:7062:166"/>
proportion, is to an other, namely, to the number 3. so is another namely, the num<g ref="char:EOLhyphen"/>ber 6. to 3. the antecedent of the second proportion, for eyther is dupla propor<g ref="char:EOLhyphen"/>tion.</p>
               </div>
               <div type="section">
                  <head>An extended proportionality is, when as the antecedent is to the consequent, so is the antecedent to the consequent,<note place="margin">
                        <hi>The</hi> 20. <hi>definition.</hi>
                     </note> and as the consequent is to an other, so is the consequent to an other. Apertu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>bate proportionalitie is, when, thre magnitudes be<g ref="char:EOLhyphen"/>ing compared to three other magnitudes,<note place="margin">
                        <hi>The</hi> 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. <hi>defi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>i<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> it co<g ref="char:cmbAbbrStroke">̄</g>meth to passe, that as in the first magnitudes the antecedent is to the conse<g ref="char:EOLhyphen"/>quent, so in the second is the antecedent to the consequent, &amp; as in the first magnitudes the consequent is to an other mag<g ref="char:EOLhyphen"/>nitude, so in the second magnitudes is an other magnitude to the antecedent.</head>
                  <p>These two last definitions here put by Zamberte seeme all one with the other two last before set.<note place="margin">
                        <hi>These two last definitions not found in the greeke ex<g ref="char:EOLhyphen"/>amplers.</hi>
                     </note> Wherfore it is not lyke that they were written and set here by <hi>Euclide,</hi> for that they seeme no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> necessary, but rather superfluous, neither are they found in the Greeke examples commonly set forth in print, nor mentioned of a<g ref="char:EOLhyphen"/>ny that hath written commentaries vpon <hi>Euclide,</hi> olde or new: Not of <hi>Campane, S<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>lius, Pellitarius, Orontius,</hi> nor <hi>Fl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ssates:</hi> wherfore it is not of necessitie to adde vnto them any explanation or example either in magnitudes or in numbers. The examples of the two last definitions set before, may likewise serue for them also.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Theoreme. The <hi>1.</hi> Proposition. If there be a number of magnitudes how many soeuer eque<g ref="char:EOLhyphen"/>multiplices to a like number of magnitudes ech to ech: how multiplex on magnitude is to one, so multiplices are all the magnitudes to all.</head>
               <p>SVppose that there be a number of magnitudes, namely, AB, and DC equimultiplices. to a like number of mag<g ref="char:EOLhyphen"/>nitudes E and F ech to ech. Then I say, that how multi<g ref="char:EOLhyphen"/>plex AB is to E, so multiplices are AB and DC to E and F. For forasmuch as how multiplex AB is to E, so multiplex is DC to F, therefore how many magnitudes there are in AB equall vnto E so in any are there in DC
<pb facs="tcp:7062:167"/>
equall vnto F.<note place="margin">Construction.</note> Deuide AB into the magnitudes
<figure/>
which are equall vnto E, that is, into AG and GB: and likewise DC into the magnitudes which are e<g ref="char:EOLhyphen"/>quall vnto F, that is, into DH and HC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now then the multitude of these DH &amp; HC, is equall vnto the multitude of these AG &amp; GB. And forasmuch as AG is equall vnto E, and DH vnto F: therfore AG and DH, are equall vnto E and F: and by the same reason forasmuch as GB is equall vnto E, and HC vnto F, GB also and HC are equall vnto E and F. Wherefore how many magnitudes there are in AB equall vnto E, so many are there in AB and DC equall vnto E &amp; F: Wherefore how multiplex AB is to E, so multiplices are AB and DC to E and F. If therefore there be a number of magnitudes how many soeuer equemultipli<g ref="char:EOLhyphen"/>ces to a like number of magnitudes ech to ech, how multiplex one magnitude is to one, so multiplices are all the magnitudes to all: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Theoreme. The <hi>2.</hi> Proposition. If the first be equemultiplex to the second as the third is to the fourth, and if the fifth also be equemultiplex to the second as the sixt is to the fourth: then shall the first and the fifth compo<g ref="char:EOLhyphen"/>sed together be equemultiplex to the second, as the third and the sixt composed together is to the fourth.</head>
               <p>SVppose that there be sixe quantities, of which let AB be the first, C the second, DE the third, F the fourth, BG the fifth, &amp; EH the sixt: and suppose that the first, AB, be equemultiplex vnto the second, C, as the third, DE, is to the fourth, F: and let the fift, BG, be equemultiplex vnto the second, C, as the sixt, EH, is to the fourth, F. Then I say, that the first and the fifth composed together, which let be AG, is equemultiplex vnto the second, C, as the third and sixt composed together, which let be DH, is to the fourth, F. For forasmuch as AB is equemultiplex to C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> as DE is
<figure/>
to F, therefore how many magnitudes there are in AB equall vnto C, so many magnitudes are there in DE equall vnto F: and by the same reason how ma<g ref="char:EOLhyphen"/>ny there are in BG equall vnto C, so many also are there in EH equall vnto F. Wherefore how many there are in the whole AG equall vnto C, so many are there in the whole DH equall vnto F. Wherefore how multiplex AG is vnto C, so multiplex is DH vnto F. Wherefore the first and the fifth composed
<pb n="137" facs="tcp:7062:167"/>
together, namely, AG is equemultiplex vnto the second C, as the third and the sixte composed together, namely, DH, is to the fourth F. If therfore the first be equemultiplex to the second as the third is to the fourth, and if the fifth also be equemultiplex to the second as the sixt is to the fourth: then shall the first &amp; the fifth co<g ref="char:cmbAbbrStroke">̄</g>posed together, be equemultiplex to the second, as the third and the sixt composed together<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the fourth: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Theoreme. The <hi>3.</hi> Proposition. If the first be equemultiplex to the second, as the third is to the fourth, and if there be taken equemultiplices to the first &amp; to the third: they shall be equemultiplices to them which were first taken, the one to the second, the other to the fourth.</head>
               <p>SVppose that there be foure magnitudes, of which let A be the first, B the second, C the third, and D the fourth. And let the first, A, be equemulti<g ref="char:EOLhyphen"/>plex to the second, B, as y<hi rend="sup">e</hi> third, C, is to the fourth, D.<note place="margin">Construction.</note> And vnto A and C take equemultiplices, which let be EF and GH, so that how multiplex EF is to A, so multiplex let HG be to C. Then I say, that EF is equemultiplex vn<g ref="char:EOLhyphen"/>to B, as GH is vnto D. For forasmuch as EF is equemultiplex vnto A, as GH is vnto C, therefore how many magnitudes
<figure/>
there are in EF equall vnto A, so many magnitudes also are there in GH equall vnto C. Let EF be deuided into the mag<g ref="char:EOLhyphen"/>nitudes that are equall vnto A, that is, into EK and KF. And likewise GH into the magnitudes equall vnto C, that is, into GL and L H. Now then the multitude of these magnitudes EK and KF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> is equall vnto y<hi rend="sup">e</hi> multitude of these magnitudes GL and LH. And forasmuch as A is equemul<g ref="char:EOLhyphen"/>tiplex to B, as C is to D: but EK is equall vnto A, and GL vnto C, therefore EK is equemultiplex vnto B, as GL is vnto D. And by the same reason KF is equemulti<g ref="char:EOLhyphen"/>plex vnto B, as L H is to D. Now then there are sixe magnitudes whereof EK is the first: B y<hi rend="sup">e</hi> second: GL y<hi rend="sup">•</hi> third: D y<hi rend="sup">•</hi> fourth: KF the fifth: LH the sixt. And forasmuch as the first EK is equemul<g ref="char:EOLhyphen"/>tiplex to the second B, as the third GL is to the fourth D: and the fift KF is equemultiplex to the second B, as the sixt LH is to the fourth D: therefore (by
<pb facs="tcp:7062:168"/>
the second of the fift) the first is the fift composed together, namely, EF is eque<g ref="char:EOLhyphen"/>multiplex vnto the second B, as the third and the sixt composed together, name<g ref="char:EOLhyphen"/>ly, GH is to the fourth D. If therefore the first be equemultiplex to the second, as the third is to the fourth; and if there be taken the equemultiplices to the first and to the third, they shall be equemultiplices to them which were first taken, the one to the second, then other to the fourth: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Theoreme. The <hi>4.</hi> Proposition. If the first be vnto the second in the same proportion that the third is to the fourth: then also the equemultiplices of the first and of the third, vnto the equemultiplices of the second and of the fourth, accordyng to any mnltiplication, shall haue the same proportion beyng compared together.</head>
               <p>SVppose that there be foure magnitudes, of which let A be the first, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the second, C the third, and D the fourth. And let the first A be vnto the second B in the fame proportion, that the third C is vnto the fourth D. And to A and C take equemultiplices E, and F, and likewise to B and D,<note place="margin">Construction.</note> any other equimultiplices G &amp; H. The<g ref="char:cmbAbbrStroke">̄</g> I say y<hi rend="sup">t</hi> as E is to G, so is F to H. Vnto E &amp; F take equemultiplices K
<figure/>
&amp; L, &amp; vnto G and H any other equemultipli<g ref="char:EOLhyphen"/>ces, that is, M and N. And forasmuch as E is equemultiplex vnto A as F is vnto C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and vnto E and F be taken the e<g ref="char:EOLhyphen"/>quemultiplices K &amp; L, therfore (by the <hi>3.</hi> of the fifth) K is equemulti<g ref="char:EOLhyphen"/>plex to A, as L is to C: and by the same reason also M is equemultiplex vnto B, as N is to D. And seing that as A is to B, so is C to D, and of A and C are taken eque<g ref="char:EOLhyphen"/>multiplices K and L, and likewise of B &amp; D are taken other equemultiplices, namely, M and N, therfore if K exceede M, L
<pb n="138" facs="tcp:7062:168"/>
also excedeth N: and if it be equall, it is equall, and if it be lesse<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t is lesse (by the conuerse of the <hi>6.</hi> definition of the fifth). And K and L are equemultiplices to E and F: and M and N are other equemultiplices to G and H. Wherefore as E is to G, so is F to H by the said sixt definition. If therfore the first be vnto the second in the same proportion that the third is to the fourth: then also the eque<g ref="char:EOLhyphen"/>multiplice<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the first and of the third, vnto the equemultiplices of <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> second &amp; of the fourth, according to any multiplication, shall haue the same proportion be<g ref="char:EOLhyphen"/>yng compared to<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>her: which was required to <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> proued.</p>
               <p>An Assumpt.<note place="margin">ALemmae, or an assumpt.</note> Wherfore seing it hath bene proued that if K exceede M, L also excedeth N, and if it be equall it is equall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and if it be lesse, it is lesse: it is manifest that if M exceede K, N also excedeth L: and if it be equall it is e<g ref="char:EOLhyphen"/>quall: and if it be lesse it is lesse: and by this reason as G is to E, so is H to F.</p>
               <div type="section">
                  <head>A Coroll<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>y.</head>
                  <p>Hereby it is manifest that if there be foure magnitudes proportional, they shal also by conuersion be proportionall: that is, if the first be vnto the second,<note place="margin">
                        <hi>A Corollary. Conuerse pro<g ref="char:EOLhyphen"/>portion.</hi>
                     </note> as the thide is to the fourth: then by conuersion as the seconde is to the first, so is the fourth to the third.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Theoreme. The <hi>5.</hi> Proposition. If a magnitude be equemultiplex to a magnitude, as a parte taken away of the one, is to a part taken away from the other: the residue also of the one, to the residue of the other, shal be e<g ref="char:EOLhyphen"/>quemultiplex, as the whole is to the whole.</head>
               <p>SVppose that the whole magnitude AB be vnto the whole magnitude CD equemultiplex, as the part taken away of the one, namely, AE, is to the part taken away of the other, namely, CF. Then I say that the resi<g ref="char:EOLhyphen"/>due<g ref="char:cmbAbbrStroke">̄</g> of the one, namely, EB, is to the residue of the other, namely, DF equemul<g ref="char:EOLhyphen"/>tiplex as the whole AB is to the whole CD. How multiplex AE is to CF, so multiplex make EB to CG.<note place="margin">Construction.</note> And forasmuch as (by y<hi rend="sup">e</hi> first of the fifth) AE is to GF equemultiplex, as AB is to GF:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> but AE is to CF equemulti<g ref="char:EOLhyphen"/>plex, as AB is to CD. Wherfore AB is equemultiplex to either
<figure/>
of these GF and CD. Wherfore GF is equall vnto CD. Take a<g ref="char:EOLhyphen"/>way CF which is common to them both. Wherfore that which re<g ref="char:EOLhyphen"/>mayneth namely, GC, is equall vnto that which remayneth name<g ref="char:EOLhyphen"/>ly, DF. And forasmuch as AE is to CF equemultiplex as EB is to GC, but GC is equall vnto DE, therefore AE is to CF eque<g ref="char:EOLhyphen"/>multiplex as EB is to FD. But AE is put to be equemultiplex to CF, as AB is to CD, wherfore EB is to FD equemultiplex, as AB is to CD. Wherfore the residue EB is to the residue FD e<g ref="char:EOLhyphen"/>quemultiplex, as the whole AB is to the whole CD. If therfore a
<pb facs="tcp:7062:169"/>
magnitude be equemultiplex to a magnitude, as a part taken away of the one is to a parte taken away of the other: the residue of the one also to the residue of the other, shalbe equemultiplex as the whole is to the whole: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Theoreme. The <hi>6.</hi> Proposition. If two magnitudes be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quemultiplices to two magnitudes, &amp; any par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es taken away of them also, be aequemultiplices to the same magnitudes: the residues also of them shal vnto the same magnitudes be either equall, or equemultiplices.</head>
               <p>SVppose that there be two magnitudes AB and CD equemultiplices to two magnitudes E and F, and let the partes take<g ref="char:cmbAbbrStroke">̄</g> away of the mag<g ref="char:EOLhyphen"/>nitudes AB and CD, namely, AG and CH be equemultiplices to the same magnitudes E and F.<note place="margin">Two cases in this Propo<g ref="char:EOLhyphen"/>tion.</note> Then I say that the residues GB and HD, are vnto the selfe same magnitudes E and F either equall, or els eque<g ref="char:EOLhyphen"/>multiplices.</p>
               <p>
                  <note place="margin">The first.</note>Suppose first that GB be equall vnto E. Then I say that
<figure/>
HD is equall vnto F.<note place="margin">Construction.</note> Vnto F put an equall magnitude CK.
And forasmuch as AG is equemultiplex vnto E, as CH is vnto F:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> but GB is equall vnto E, &amp; KC vnto F: therfore AB is equemultiplex to E, as KH is to F. But AB is put eque<g ref="char:EOLunhyphen"/>multiplex vnto E, as CD is to F. Wherfore KH is equimul<g ref="char:EOLunhyphen"/>tiplex vnto F, as CD is to F. And forasmuch as either of these KH and CD are equimultiplices vnto F, therfore (by the 1 common sentence) KH is equall vnto CD. Take away CH which is common to them both. Wherefore the residue KC is equal vnto the residue HD. But KC is equal vnto F, wher<g ref="char:EOLhyphen"/>fore HD is equall vnto F. Wherfore if GB be equal vnto E, DH also shall be equall vnto F.</p>
               <p>And in like sort may we proue, y<hi rend="sup">t</hi> if GB be multiplex to E, HD also shal be so multiplex vnto F. If therfore there be two magnitudes equemultiplices to two magnitudes,<note place="margin">The second.</note> and any parts taken away of them be also equemultiplices to y<hi rend="sup">e</hi> same magnitudes: the residues also of them shall vnto the same magnitudes be either equall, or equemultiplices: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Theoreme. The <hi>7.</hi> Proposition. Equall magnitudes haue to one &amp; the selfe same magnitude,
<pb n="139" facs="tcp:7062:169"/>
one and the same proportion. And one and the same magni<g ref="char:EOLhyphen"/>tude hath to equall magnitudes one and the selfe same pro<g ref="char:EOLhyphen"/>portion.</head>
               <p>SVppose that A and B be equall magnitudes, and take any other mag<g ref="char:EOLhyphen"/>nitude, namely, C. Then I say, that either of these A and B haue vn<g ref="char:EOLhyphen"/>to C one and the same proportion: and that C also hath to either of these A and B one and the same proportion.</p>
               <p>Take the equemultiplices of A and B, and let the same
<figure/>
be D and E:<note place="margin">The first part of this Propo<g ref="char:EOLhyphen"/>sition demon<g ref="char:EOLhyphen"/>strated.</note> and likewise of C, take any other multiplex, and let the same be F. Now forasmuch as D is vnto A e<g ref="char:EOLhyphen"/>quemultiplex as E is to B, but A is equall vnto B, therfore (by the first common sentence) D is equall vnto E. And of C there is taken any other multiplex F. Wherefore if D exceede F, E also excedeth F: and if it be equall it is equal, and if it be lesse it is lesse. But D and E are the equemul<g ref="char:EOLhyphen"/>tiplices of A &amp; B, and F is of C an other multiplex. Wher<g ref="char:EOLhyphen"/>fore as A is to C, so is B to C.</p>
               <p>I say moreouer, that C hath to either of these A and B one and the same proportion.<note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note> For the same order of con<g ref="char:EOLhyphen"/>structio<g ref="char:cmbAbbrStroke">̄</g> remaining, we may in like sort proue, y<hi rend="sup">t</hi> D is equal vnto E, &amp; there is taken an other multiplex to C, name<g ref="char:EOLhyphen"/>ly, F. Wherefore if F exceede D, it also excedeth E: and if it be equall it is equall: and if it be lesse it is lesse. But F is multiplex to C: and D &amp; E are other equemultipli<g ref="char:EOLhyphen"/>ces to A and B. Wherfore as C is to A, so is C to B. Wher<g ref="char:EOLhyphen"/>fore equall magnitudes haue to one and the same magnitude, one and the same proportion: and one and the same magnitude hath to equall magnitudes one and the selfe same proportion: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Theoreme. The <hi>8.</hi> Proposition. Vnequall magnitudes beyng taken, the greater hath to one and the same magnitude a greater proportion then hath the lesse. And that one and the same magnitude hath to the lesse a greater proportion then it hath to the greater.</head>
               <p>SVppose that AB and C be vnequall magnitudes, of which let AB be the greater, and C y<hi rend="sup">e</hi> lesse. And let there be an other magnitude what<g ref="char:EOLhyphen"/>soeuer, namely, D. Then I say that AB hath vnto D a greater pro<g ref="char:EOLhyphen"/>portion then hath C to D: and also that D hath to C a greater propor<g ref="char:EOLhyphen"/>tion,
<pb facs="tcp:7062:170"/>
then it hath to AB. For forasmuch as AB is greater then C, let there be ta<g ref="char:EOLhyphen"/>ken a magnitude equall vnto C, namely, BE.</p>
               <p>Now then the lesse of these two magnitudes AE and EB being multiplied will at the length be greater then D.<note place="margin">The first part of this Propo<g ref="char:EOLhyphen"/>sition demon<g ref="char:EOLhyphen"/>strated.</note>
               </p>
               <p>First let AE be lesse then EB:
<figure/>
and let AE be so oft multiplied vn<g ref="char:EOLhyphen"/>till that which is produced be greater then D,<note place="margin">The first case of the same.</note> and let that multiplex be FG which is greater then D. And how multiplex FG is to AE, so multiplex let GH be to EB, &amp; K to C. Then double D, and let the same be L. And againe treble it &amp; let the same be M: &amp; so forth encreasing by one till such tyme as the multiplex of D so taken be first greater then K, that is, that multiplex which amongst al the mul<g ref="char:EOLhyphen"/>tiplices of D doth first exceede K, &amp; let the same be N, which here is qua<g ref="char:EOLhyphen"/>druple to D, and the first multiplex greater then K. Now forasmuch as K is y<hi rend="sup">e</hi> first multipl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>x lesse then N, therfore K is not lesse then M. And for that FG i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to AE equemultiplex as GH is to EB, therfore (by the first of the fifth) FH is to AB equemultiplex as FG is to AE: but FG is equemultiplex to AE as K is to C. Wherfore FH and K are equemultiplices to AB and C. Againe foras<g ref="char:EOLhyphen"/>much as GH is equemultiplex to EB as K is to C, but EB is equall vnto C, therfore GH is also equall vnto K. But K is not lesse then M: wherfore ney<g ref="char:EOLhyphen"/>ther also is GH lesse then M. But FG is greater then D. Wherefore the whole FH is greater then both these D and M. But both these D and M are equall vnto N, for M is triple to D, and M and D together are quadruple to D, and N also is quadruple to D: wherefore both these M and D are equall vnto N; but FH is greater then M and D, wherfore FH excedeth N, (that is, y<hi rend="sup">e</hi> mul<g ref="char:EOLhyphen"/>tiplex of the first, namely, of AB excedeth the multiplex of the second, namely, of D.) But K excedeth not N (that is, the multiplex of the third, namely, of C, excedeth not the multiplex of the fourth, namely, of the same D:) and FH and K are equemultiplices to AB, and C, and N, is a certayne other multiplex to D. Wherfore (by the <hi>8.</hi> definition) AB hath to D a greater proportion then hath C to D.</p>
               <p>
                  <note place="margin">The second part of the proposition de<g ref="char:EOLhyphen"/>monstrated.</note>I say moreouer that D hath to C a greater proportion then D hath to AB. For the same order of construction still remayning, we may in like sort proue that N is greater then K, and that it is not greater then FH. And N is multiplex to D, and FH and K are certayne other equemultiplices to AB and C. Wher<g ref="char:EOLhyphen"/>fore D hath to C a greater proportion then D hath to AB.</p>
               <p>
                  <pb n="134" facs="tcp:7062:170"/>But now suppose that AE be greater then EB.<note place="margin">The second case of the f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> pa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</note> Now then EB beyng the lesse, and being oftentimes multiplied, will at the length be greater then D. Let it be so multiplied, and let th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> multiplex of EB, namely, GH, be great<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r then D. And how multiplex GH is to EB
<figure/>
so multiplex let FG be to AE and K to C. And by the former reason may we proue, that FH and K are equemulti<g ref="char:EOLhyphen"/>plices to AB and C. Likewise also let N be multiplex of D, &amp; also let it be y<hi rend="sup">e</hi> first multiplex which is greater then FG: wherfore agayn FG is not lesse then M. But GH is greater then D: wherefore the whole FH, excedeth D &amp; M, that is, N: but K excedeth not N. For FG which is greater then GH, that is, then K exceedeth not N. And so followyng the same order we did before, we shall performe the demonstration. Wherefore vnequall magnitudes beyng taken, the greater hath to one and the same magni<g ref="char:EOLhyphen"/>tude a greater proportion then hath the lesse: and that one and the same magni<g ref="char:EOLhyphen"/>tude hath to the lesse a greater proportion then it hath to the greater: which was required to be proued.</p>
               <div type="section">
                  <head>¶ For that Orontius seemeth to demonstrate this more plainly therefore I thought it not amisse here to set it.</head>
                  <p>Suppose that there be two vnequall magnitudes, of which le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the gre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ter, and <hi>C</hi> the lesse: and let there be a certaine other magnitude, namely, <hi>D.</hi> Then I say first, that <hi>AB</hi> hath to <hi>D</hi> a greater proportion then hath <hi>C</hi> to <hi>D.</hi> For forasmuch as by sup<g ref="char:EOLhyphen"/>position <hi>AB</hi> is greater then the magnitude <hi>C</hi>: therefore the magnitude <hi>AB</hi> contay<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>th the same magnitude <hi>C,</hi> and an other
<figure/>
magnitude besides. Let <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> be equall vnto <hi>C</hi>: and let <hi>AE</hi> be the part remayning of the same magnitude.<note place="margin">
                        <hi>First differ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of the first part.</hi>
                     </note> Now <hi>AE</hi> and <hi>EB</hi> are eyther vnequall or equall the one to the o<g ref="char:EOLhyphen"/>ther. First let them be vnequall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>AE</hi> be lesse then <hi>EB.</hi> And vnto <hi>AE</hi> the lesse take any multiplex whatso<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r, so that it be greater the<g ref="char:cmbAbbrStroke">̄</g> the magnitude <hi>D:</hi> and let the same be <hi>FG.</hi> And how multiplex <hi>FG</hi> is to <hi>AE,</hi> so multiplex let <hi>GH</hi> be to <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <hi>K</hi> to <hi>C.</hi> Agayne take the duple of <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which let be <hi>L,</hi> and then the triple, and le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the same be <hi>M.</hi> And so forward, alway<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s adding one: vntill there be produced su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h a multiplex to <hi>D</hi> which shall be n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>xt grea<g ref="char:EOLhyphen"/>ter then <hi>GH</hi> (that is, which amongest the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by the con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all additi<g ref="char:EOLhyphen"/>on
<pb facs="tcp:7062:171"/>
of one, doth first beginne to exce<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>de <hi>GH</hi>) and let th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>me be <hi>N</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which let <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e qua<g ref="char:EOLhyphen"/>druple to <hi>D.</hi> Now then the multiplex <hi>GH</hi> is the next multiplex lesse then <hi>N</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>fore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s not lesse then <hi>M,</hi> that is, is either equall vnto it or greater then it.<note place="margin">
                        <hi>Demonstrati<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e same first differe<g ref="char:cmbAbbrStroke">̄</g>ce.</hi>
                     </note> And foras<g ref="char:EOLhyphen"/>much <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s <hi>FG</hi> is equemultiplex to <hi>AE,</hi> as <hi>GH</hi> is to <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,</hi> therefore how multiplex <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is to <hi>AE,</hi> so multiplex is <hi>FH</hi> to <hi>AB</hi> (by the first of the fift). But how multiplex <hi>FG</hi> is to <hi>AE,</hi> so multiplex is <hi>K</hi> to <hi>C,</hi> therefore how multiplex <hi>FH</hi> is to <hi>AB,</hi> so multiplex is <hi>K</hi> to <hi>C.</hi> Moreouer forasmuch as <hi>GH</hi> and <hi>K</hi> are equemultiplices vnto <hi>EB</hi> and <hi>C</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <hi>EB</hi> is by construction equall vnto <hi>C,</hi> therfore (by the common sentence) <hi>GH</hi> is equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll vnto <hi>K.</hi> But <hi>GH</hi> is not lesse then <hi>M,</hi> as hath before bene shewed, and <hi>FG</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> was put to be greater then <hi>D.</hi> Wherefore the whole <hi>FH</hi> is greater then these two <hi>D</hi> and <hi>M.</hi> But <hi>D</hi> and <hi>M</hi> are equall vnto <hi>N.</hi> For <hi>N</hi> is quadruple to <hi>D.</hi> And <hi>M</hi> being triple to <hi>D,</hi> doth together with <hi>D</hi> make quadruple vnto <hi>D.</hi> Wherefore <hi>FH</hi> is greater then <hi>N.</hi> Farther, <hi>K</hi> is proued to be equall to <hi>GH.</hi> Wherefore <hi>K</hi> is lesse then <hi>N.</hi> But <hi>FH</hi> and <hi>K</hi> are e<g ref="char:EOLhyphen"/>quemultiplices vnto <hi>AB</hi> and <hi>C,</hi> vnto the first magnitude, I say, and the third: and <hi>N</hi> is a certaine other multiplex vnto <hi>D,</hi> which representeth the second &amp; the fourth mag<g ref="char:EOLhyphen"/>nitude. And the multiplex of the first excedeth the multiplex of the second: but the multiplex of the third excedeth not the multiplex of the fourth. Wherefore <hi>AB</hi> the first hath vnto <hi>D</hi> the second a greater proportion, then hath <hi>C</hi> the third to <hi>D</hi> the fourth (by the 8. definition of thys booke).</p>
                  <p>But if <hi>AE</hi> be greater then <hi>EB,</hi> let <hi>EB</hi> the lesse be multiplied vntill there be pro<g ref="char:EOLhyphen"/>duced a multiplex greater then the magnitude <hi>D:</hi> which let be <hi>GH.</hi> And how multi<g ref="char:EOLhyphen"/>plex <hi>GH</hi> is to <hi>EB,</hi>
                     <note place="margin">
                        <hi>Second diffe<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ence.</hi>
                     </note> so multiplex let <hi>FG</hi> be to <hi>AE,</hi> and <hi>K</hi> also to <hi>C.</hi> Then take vnto <hi>D</hi> such a multiplex as is next greater then <hi>FG:</hi> and againe let the same be <hi>N,</hi> which let be quadruple to <hi>D.</hi> And in like sort as be<g ref="char:EOLhyphen"/>fore
<figure/>
may we proue, that the whole <hi>FH</hi> is vnto <hi>AB</hi> equemultiplex as <hi>GH</hi> is to <hi>EB:</hi> and also that <hi>FH</hi> &amp; <hi>K</hi> are equemultiplices vnto <hi>AB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp; <hi>C</hi>: and finally that <hi>GH</hi> is equall vnto <hi>K.</hi> And forasmuch as the multiplex <hi>N</hi> is next greater then <hi>FG</hi>: therefore <hi>FG</hi> is not lesse then <hi>M.</hi> But <hi>GH</hi> is greater then <hi>D</hi> by construction. Wherefore the whole <hi>FH</hi> is greater then <hi>D</hi> and <hi>M</hi>: and so consequent<g ref="char:EOLhyphen"/>ly is greater then <hi>N.</hi> But <hi>K</hi> excedeth not <hi>N</hi>: for <hi>K</hi> is equall to <hi>GH</hi>: for how multiplex <hi>K</hi> is to <hi>EB</hi> the lesse, so multiplex is <hi>FG</hi> to <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> the greater. B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t those magnitudes which are equemultiplice<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> vnto vnequall magnitud<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s are according to the same proportion vnequall. Wherefore <hi>K</hi> is lesse then <hi>FG,</hi> and therefore i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> much lesse then <hi>N.</hi> Wherefore againe the multiplex of the first exceedeth the multiplex of the second: but the multiplex of the third excedeth not th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> multiplex of the fourth. Wherefore (by the 8. definition of the fift) <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> the first, hath to <hi>D</hi> the second, a greater proportion, then hath <hi>C</hi> the third to <hi>D</hi> the fourth.</p>
                  <p>
                     <note place="margin">
                        <hi>Third di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>e<g ref="char:EOLhyphen"/>rence.</hi>
                     </note>But now if <hi>AE</hi> be equall vnto <hi>EB,</hi> eyther
<figure/>
of them shall be equall vnto <hi>C.</hi> Wherfore vnto either of thos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> three magnitudes take eque<g ref="char:EOLhyphen"/>multiplices greater then <hi>D.</hi> So that let <hi>FG</hi> be multiplex to <hi>AE,</hi> and <hi>GH</hi> vnto <hi>EB,</hi> and <hi>K</hi> a<g ref="char:EOLhyphen"/>gayne to <hi>C</hi>: which (by the 6. co<g ref="char:cmbAbbrStroke">̄</g>mon sentence) shall be equall the one to the other. Let <hi>N</hi> al<g ref="char:EOLhyphen"/>so be multiplex to <hi>D,</hi> and be next greater then euery one of them, namely, let it be q<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>adrupl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to <hi>D.</hi> This co<g ref="char:cmbAbbrStroke">̄</g>struction finished, we may again proue that <hi>FH</hi> and <hi>K</hi> are equemultiplices to <hi>AB</hi> and <hi>C</hi>: and that <hi>FH</hi> the multiplex of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> first magnitude exceedeth <hi>N</hi> the multiplex of the second magnitude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>K</hi> t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ulti<g ref="char:EOLhyphen"/>plex
<pb n="141" facs="tcp:7062:171"/>
of the third excedeth not the multiplex of the fourth. Wherfore we may conclude that <hi>AB</hi> hath vnto <hi>D</hi> a greater proportion, then hath <hi>C</hi> to <hi>D.</hi>
                  </p>
                  <p>Now also I say, that the self same magnitude <hi>D</hi> hath vnto the lesse magnitude <hi>C</hi> a grea<g ref="char:EOLhyphen"/>ter proportion, the<g ref="char:cmbAbbrStroke">̄</g> it hath to the greater <hi>AB.</hi>
                     <note place="margin">
                        <hi>The second part <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>f this proposition.</hi>
                     </note> And this may plainly be gathered by the foresayd discourse, without chaunging the order of the magnitudes &amp; of the equemul<g ref="char:EOLhyphen"/>tiplices. For seing that euery way it is before proued, that <hi>FH</hi> excedeth <hi>N,</hi> and <hi>K</hi> is ex<g ref="char:EOLhyphen"/>ceeded of the selfe same <hi>N</hi>: therefore conuersedly <hi>N</hi> excedeth <hi>K,</hi> but doth not excede <hi>FH.</hi> But <hi>N</hi> is multiplex to <hi>D</hi> that is to the first and third magnitude: and <hi>K</hi> is mul<g ref="char:EOLhyphen"/>tiplex to the second, namely, to <hi>C</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <hi>FH</hi> is multiplex to the fourth, namely, to <hi>AB</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wherefore the multiplex of the first excedeth the multiplex of the second: but the mul<g ref="char:EOLhyphen"/>tiplex of the third excedeth not the multiplex of the fourth. Wherefore (by the 8. defi<g ref="char:EOLhyphen"/>nition of this fift booke) <hi>D</hi> the first hath vnto <hi>C</hi> the second a greater proportion, then hath <hi>D</hi> the third to <hi>AB</hi> the fourth: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Theoreme. The <hi>9.</hi> Proposition. Magnitudes which haue to one and the same magnitude one and the same proportion: are equall the one to the other. And those magnitudes vnto whome one and the same mag<g ref="char:EOLhyphen"/>nitude hath one and the same proportion: are also equall.</head>
               <p>SVppose that either of these two magnitudes A and B haue
<figure/>
to C one and the same proportion.<note place="margin">The first par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this Propo<g ref="char:EOLhyphen"/>sition demon<g ref="char:EOLhyphen"/>strated.</note> Then I say that A is e<g ref="char:EOLhyphen"/>quall vnto B. For if it be not, then either of these A and B should not haue to C one &amp; the same proportio<g ref="char:cmbAbbrStroke">̄</g> (by the <hi>8.</hi> of y<hi rend="sup">e</hi> fifth) but by supposition they haue, wherefore A is equall vnto B.</p>
               <p>Againe,<note place="margin">The second part proued.</note> suppose that the magnitude C haue to either of these magnitudes A and B one and the same proportion. Then I say that A is equall vnto B. For if it be not, C should not haue to either of these A and B one and the same proportion (by the former proposi<g ref="char:EOLhyphen"/>tion): but by supposition it hath, wherfore A is equall vnto B. Wher<g ref="char:EOLunhyphen"/>fore magnitudes which haue to one and the same magnitude one and the same proportion, are equall the one to the other. And thos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> magnitudes vnto whome one and the same magnitude hath one and the same proportion, are also equall: which was required to be pro<g ref="char:EOLhyphen"/>ued.</p>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Theoreme. The <hi>10.</hi> Proposition. Of magnitudes compared to one and the same magnitude, that which hath the greater proportion, is the greater. And that magnitude wherunto one and the same magnitude hath the greater proportion, is the lesse.</head>
               <p>
                  <pb facs="tcp:7062:172"/>SVppose that A haue to C a greater proportion then B hath to C. Then I say that A is greater then B.<note place="margin">The first part of this proposi<g ref="char:EOLhyphen"/>tion proued.</note> For if it be not, then either A is equall vnto B or lesse then it. But A cannot be equal vnto B, for then either of these A and B should haue vnto C one and the same proportion (by the <hi>7</hi>
                  <figure/>
of the fifth): but by supposition they haue not, wherfore A is not e<g ref="char:EOLhyphen"/>quall vnto B. Neither also is A lesse then B, for the<g ref="char:cmbAbbrStroke">̄</g> should A haue to C a lesse proportion, then hath B to C (by the <hi>8.</hi> of the fifth) but by supposition it hath not. Wherfore A is not lesse then B. And it is also proued that it is not equall, wherfore A is greater then B.</p>
               <p>
                  <note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note>Agayne suppose that C haue to B a greater proportion then C hath to A. Then I say that B is lesse then A. For if it be not, then is it either equall vnto it or els greater, but B cannot be equall vnto A, for then should C haue to either of these A and B one and the same proportion (by the <hi>7.</hi> of the fifth) but by supposition it hath not, wherfore B is not equall vnto A. Neither also is B greater then A, for then should C haue to B a lesse proportion then it hath to A (by the <hi>8.</hi> of the fifth) but by supposition it hath not: where<g ref="char:EOLhyphen"/>fore B is not greater then A. And it was proued that it is not e<g ref="char:EOLhyphen"/>quall vnto A, wherfore B is lesse then A. Wherfore of magnitudes compared to one and the same magnitude, that which hath y<hi rend="sup">•</hi> grea<g ref="char:EOLhyphen"/>ter proportion, is the greater. And that magnitude wherunto one and the same magnitude hath the greater proportion, is the lesse. Which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>11.</hi> Theoreme. The <hi>11.</hi> Proposition. Proportions which are one and the selfe same to any one pro<g ref="char:EOLhyphen"/>portion, are also the selfe same the one to the other.</head>
               <p>SVpppose that as A is to B, so is C to D, and as C is to D, so is E to F. Then I say that as A is to B, so is E to F.<note place="margin">Construction.</note> Take equemultiplices to A, C and E, which let be G, H, K. And likewise to B, D and F take a<g ref="char:EOLhyphen"/>ny other equemultiplices, which let be L, M, and N. And because as A is to B, so is C to D: and to A and G are taken equemultiplices G &amp; H; &amp; to B and D are take certaine other equemultiplices L &amp; M.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> If therfore G exceede L, then also H excedeth M, and if it be equall it is equall, and if it be lesse it is lesse (by the conuerse of the <hi>6<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> definition of the fifth). Agayne because that as C is to D, so is E to F: and to C and E are taken <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>em<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ltiplices H <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>d K: and likewise to D &amp; F are take<g ref="char:cmbAbbrStroke">̄</g> certaine other equemultiplices M &amp; N. If therfore H exceede M, then also K excedeth N: and if it be equall, it is equall, and if it be lesse, it is lesse (by the same conuerse) But if K exceede M, then also G exce<g ref="char:EOLhyphen"/>deth
<pb n="142" facs="tcp:7062:172"/>
L: and
<figure/>
if it be equal it is equall, and if it be lesse, it is lesse (by the same con<g ref="char:EOLhyphen"/>uerse) Wher<g ref="char:EOLunhyphen"/>fore if G ex<g ref="char:EOLhyphen"/>cede L, then K also ex<g ref="char:EOLhyphen"/>cedeth N, and if it be equal it is e<g ref="char:EOLhyphen"/>quall, and if it be lesse, it is lesse. But G &amp; K are equemultipli<g ref="char:EOLunhyphen"/>ces of A &amp; E. And L &amp; N are certaine other equemultiplices of B &amp; F. Wher<g ref="char:EOLhyphen"/>fore (by the <hi>6.</hi> definition) as A is to B, so is E to F. Proportions therfore which are one and the selfe same to any one proportion, are also the selfe same one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Theoreme. The <hi>12.</hi> Proposition. If there be a number of magnitudes how many soe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>r propor<g ref="char:EOLhyphen"/>tionall: as one of the antecedentes is to one of the co<g ref="char:cmbAbbrStroke">̄</g>sequentes, so are all the antecedentes to all the consequentes.</head>
               <p>SVppose that there be a number of magnitudes how many soeuer, namely, A, B, C, D, E, F, in proportion: so that as A is to B, so let C be to D, and E to F. Then I say, that as A is to B, so <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> A, C, E, to B, D, F. Take equemultiplices to A, C, and E.<note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ction.</note> And let the same be G, H, K. And likewise to B, D, and F, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ake any other equemultiplices, which to be L, M, N. And because that <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> A is to B, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> C to D, and E to F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> And to A, C, E, are taken <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quemulti<g ref="char:EOLhyphen"/>plices G, H, K: and likewise to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, D, F, are taken certaine other equem<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tipli<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s L, M, N. If therefore G exceede L, H also exceedeth M, and KN: and if it be equall, it is equall, and if it be lesse, it is lesse (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y the conuerse of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> defini<g ref="char:EOLhyphen"/>tion
<pb facs="tcp:7062:173"/>
of the fift). Wherfore if G exceede L, then G, H, K, also exceede L, M, N: and if they be equall, they are equall: and if they be lesse, they are lesse (by the same). But G, and G, H, K, are equemultiplices to the magnitude A, and to the
<figure/>
magnitudes A, C, E. For (by the first of the fift) (if there be a number of mag<g ref="char:EOLhyphen"/>nitudes equemultiplices to a like number of magnitudes ech to ech, how multiplex one magnitude, is to one, so multiplices are all the magnitudes, to all). And by the same reason also L, and L, M, N, are equemultiplices to the magnitude B, and to the magnitudes B, D, F: Wherefore as A is to B, so is A, C, E, to B, D, F (by the sixt definition of the fift). If therefore there be a number of magnitudes how many soeuer, proportionall: as one of the antecedentes is to one of the conse<g ref="char:EOLhyphen"/>quentes, so are all the antecedentes to all the consequentes: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Theoreme. The <hi>13.</hi> Proposition. If the first haue vnto the second the self same proportion that the third hath to the fourth, and if the third haue vnto the fourth a greater proportio<g ref="char:cmbAbbrStroke">̄</g> the<g ref="char:cmbAbbrStroke">̄</g> the fifth hath to the sixth: the<g ref="char:cmbAbbrStroke">̄</g> shall the first also haue vnto the second a greater proportion then hath the fifth to the sixth.</head>
               <p>SVppose that there be sixe magnitudes, of which let A be the first, <hi>B</hi> the second, C the third, D the fourth, E the fifth, and F the sixth. Suppose that A the first haue vnto B the second, the self same propor<g ref="char:EOLhyphen"/>tion that C the third hath to D the fourth. And let C the third haue vnto D the fourth, a greater proportion then hath E the fifth to F the sixth.
<pb n="143" facs="tcp:7062:173"/>
Then I say that A the first hath to B the second a greater proportion, then hath E the fifth to F the sixt.<note place="margin">Construction.</note> For forasmuch as C hath to D a greater proportion then hath E to F, therfore there are certaine equemultiplices to C and E, and likewise any other equemultiplices whatsoeuer to D and F, which being compared toge<g ref="char:EOLhyphen"/>ther, the multiplex to C shall exceede the multiplex to D, but the multiplex to E shall not exceede the multiplex to F (by the conuerse of the eight definition of
<figure/>
this booke). Let those multiplices be taken, and suppose that the equemultipli<g ref="char:EOLhyphen"/>ces to C and E, be G and H: and likewise to D and F take any other equemulti<g ref="char:EOLhyphen"/>plices whatsoeuer, and let the same be K and L, so that let G exceede K but let not H exceede L. And how multiplex G is to C, so multiplex let M be to A. And how multiplex K is to D, so multiplex also let N be to B. And because that as A is to B, so is C to D: and to A and C are taken equemultiplices M and G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And likewise to B and D are taken certayne other equemultiplices N &amp; K: if ther<g ref="char:EOLhyphen"/>fore M exceede N, G also excedeth K: and if it be equall it is equall, and if it be lesse it is lesse (by the conuersion of the sixt definition of the fifth.) But by con<g ref="char:EOLhyphen"/>struction G excedet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> K, wherfore M also excedeth N, but H excedeth not L. But M &amp; H are equemultiplices to A &amp; E: and N &amp; L are certaine other e<g ref="char:EOLhyphen"/>quemultiplices<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> whatsoeuer, to B and F. Wherfore A hath vnto B a greater pro<g ref="char:EOLhyphen"/>portion then E hath to F (by the <hi>8.</hi> definition.) If therefore the first haue vnto the second the selfe same proportion that the third hath to the fourth, and if the third haue vnto the fourth a greater proportion then the fifth hath to the sixth, then shall the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> also haue vnto the second a greater proportion then hath the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> to the sixth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Which was required to be proued.</p>
               <div type="section">
                  <head>¶ An addition of Campane.</head>
                  <p>
                     <pb facs="tcp:7062:174"/>
                     <hi>If there be foure quantities,</hi>
                     <note place="margin">
                        <hi>An addition of Campane</hi>
                     </note> 
                     <hi>and if the first haue vnto the second a greater proportion the<g ref="char:cmbAbbrStroke">̄</g> hath the third to the fourth: then shall there be some equemultiplices of the first and the third, which beyng compared to some equemultiplices of the second and the fourth, the mul<g ref="char:EOLhyphen"/>tiplex of the first shall be greater then the multiplex of the second, but the multiplex of the third shall not be greater then the multiplex of the fourth.</hi>
                  </p>
                  <p>Which is thus proued. Suppose that AB haue vnto C a greater proportion the<g ref="char:cmbAbbrStroke">̄</g> hath D to E. And let AF be to C as D is to E. Now then by this proposition &amp; the te<g ref="char:cmbAbbrStroke">̄</g>th, AF is lesse then AB. Let it be lesse then AB by the quantitie FB. And multiply FB vntil there be produced a quantitie greater then C, which let be GH: which also must be such a multiplex, as D be<g ref="char:EOLunhyphen"/>yng
<figure/>
so oftentymes multiplied, maye produce a quanti<g ref="char:EOLhyphen"/>tie not lesse the<g ref="char:cmbAbbrStroke">̄</g> E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> whiche multiplex let be K. And let LG be so multiplex to AF, as GH is to FB, or K to <hi>D.</hi> Now then by the first of this booke LH is equemultiplex to AB as K is to D. And let M be to E the first multiplex greater then K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp; let N be equemultiplex to C as M is to E. Now then N is the first multiplex to C greater then LG: For for that as D is to E, so is AF to C, and K is equemultiplex to D as GL is to AF, also M is equemultiplex to E, as N is to C: therfore (by the 4. of this booke) as K is to M, so is GL to N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> but K is to M the first multiplex lesse then M: wherfore al<g ref="char:EOLhyphen"/>so GL is the first multiplex lesse then N: and GL by supposition is not lesse the<g ref="char:cmbAbbrStroke">̄</g> C. Wher<g ref="char:EOLunhyphen"/>fore take the greatest multiplex of C vnder N: or a multiplex equall to N, if peraduen<g ref="char:EOLhyphen"/>ture N be the first of the multiplices of C, which let be O. Now then then N shall consist of O and C. Wherfore forasmuch as LG is not lesse then O, and GH is greater then C, therfore LH shall be greater then N. And forasmuch as K is lesse then M, therfore that which was required to be proued, is manifest.</p>
                  <p>Although this proposition here put by Campane nedeth no demonstration for that it is but the conuerse of the 8. definition of this booke, yet thought I it not worthy to be omitted, for that it reacheth the way to finde out such equemultipli<g ref="char:EOLhyphen"/>ces, that the multiplex of the first shall excede the multiplex of the second, but the multiplex of the third shall not exceede the multiplex of the fourth.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Theoreme. The <hi>14.</hi> Proposition. If the first haue vnto the second the selfe same proportion that the third hath vnto the fourth: and if the first be greater then the third, the second also is greater then the fourth: and if it be equall it is equall: and if it be lesse it is lesse.</head>
               <p>
                  <pb n="144" facs="tcp:7062:174"/>SVppose that there be foure magnitudes, of which let A be the first, B the second, C the third, and D the fourth: and let A the first, haue vn<g ref="char:EOLhyphen"/>to B the second, the self same proportion that C the third hath vnto D the fourth. And let A be greater then C. Then I say, that B also is greater then D. For forasmuch as A is greater then C, and there is a certaine other magnitude,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> namely,
<figure/>
B, therefore (by the 8. of the fift) A hath vnto B a greater proportion then C hath to B. And as A is to B, so is C to D. Wherfore C also hath vnto D a grea<g ref="char:EOLhyphen"/>ter proportion then C hath to B. But that magnitude wherunto one and the same magnitude hath y<hi rend="sup">e</hi> grea<g ref="char:EOLhyphen"/>ter proportion, is the lesse (by the <hi>10.</hi> of the fift). Wherfore D is lesse then B, &amp; therefore B is grea<g ref="char:EOLhyphen"/>ter then D. And in like sort may we proue, that if A be equall vnto C, D shall also be equall vnto D: and if A be lesse then C, B shall also be lesse then D. If therefore the first haue vnto the second the self same proportion that the third hath vnto the fourth, and if the first be greater then the third, the second also is greater then the fourth, and if it be equall it is equall, and if it be lesse it is lesse: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>15.</hi> Theoreme. The <hi>15.</hi> Proposition. Like partes of multiplices, and also their multiplices compa<g ref="char:EOLhyphen"/>red together, haue one and the same proportion.</head>
               <p>SVppose that AB be equemultiplex to C,
<figure/>
as DE is to F. Then I say, that as C is to F, so is AB to DE. For forasmuch as how multiplex AB is to C,<note place="margin">Construction.</note> so multiplex is DE to F: therfore how many magnitudes there are in AB equall vnto C, so many are there in DE equall vnto F. Deuide AB into the magnitudes equall vnto C, that is, into AG, GH, and HB: and likewise DE into the magnitudes equall vnto E, that is, into DK KL, and LE. Now then the multitude of these AG,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> GH, and HB, is equall to y<hi rend="sup">e</hi> multitude of these DK KL, and LE. And forasmuch as AG, GH, and HB, are equall the one to y<hi rend="sup">e</hi> other and likewise DK, KL, and LE, are also equall the one to y<hi rend="sup">e</hi> other: ther<g ref="char:EOLhyphen"/>fore as AG is to DK, so is GH to KL, and HB
<pb facs="tcp:7062:175"/>
to LE. Wherefore (by the <hi>12.</hi> of the fift) as one of y<hi rend="sup">e</hi> antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherfore as AG is to DK, so is AB to DE. But AG is equall vnto C, and likewise DK to F. Wherefore as C is to F, so is AB to DE. Like partes therefore of multiplices and also their multiplices compared together, haue one and the same proportion that their equemultiplices haue: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>16.</hi> Theoreme. The <hi>16.</hi> Proposition. If foure magnitudes be proportionall: then alternately also they are proportionall.</head>
               <p>SVppose y<hi rend="sup">t</hi> there be foure magnitudes proportionall, namely, A, B, C, D, so that as A is to B, so let C be to D. Then I say, that alternately also they shall be in proportion,<note place="margin">Demo<g ref="char:cmbAbbrStroke">̄</g>stration of alternate proportion.</note> that is, as A is to C, so is B to D. Take eque<g ref="char:EOLhyphen"/>multiplices vnto A &amp; B, and let
<figure/>
the same be E &amp; F. And likewise to C and D take any other eque<g ref="char:EOLhyphen"/>multiplices what soeuer,<note place="margin">Construction.</note> and let the same be G and H.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And foras<g ref="char:EOLhyphen"/>much as how multiplex E is to A, so multiplex is F to B, but like partes of multiplices &amp; also their multiplices haue one and the selfe same proportion the one to the o<g ref="char:EOLhyphen"/>ther (by the former Proposition). Wherefore as A is to B, so is E to F. But as A is to B, so is C to D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore (by the <hi>11.</hi> of the fift) as C is to D, so is E to F. Againe, forasmuch as G and H are equemultiplices to C and D, but like partes of mul<g ref="char:EOLhyphen"/>tiplices and also their multiplices haue the one to the other one and the selfe same proportion (by the <hi>15.</hi> of the fift). Wherefore as C is to D, so is G to H. But as C is to D, so is E to F. Wherefore as E is to F, so is G to H (by the <hi>11.</hi> of the fift). But if there be foure magnitudes in proportion, and if the first be greater then the third, the second also is greater then the fourth: and if it be equall it is equall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and if it be lesse it is lesse (by the <hi>14.</hi> of the fift). If therfore E excede G, F shall also exceede H: and if it be equall, it is equall: and if it be lesse, it is lesse. But E and F are equemultiplices to A and B: and G and H are certaine other eque<g ref="char:EOLhyphen"/>multiplices to C &amp; D. Wherfore (by the <hi>6.</hi> definition of the fift) as A is to C, so is B to D. If therefore there be foure magnitudes proportionall, then alternately also they are proportionall: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb n="145" facs="tcp:7062:175"/>
               <head>The <hi>17.</hi> Theoreme. The <hi>17.</hi> Proposition. If magnitudes composed be proportionall, then also deuided they shall be proportionall.</head>
               <p>SVppose that the magnitudes composed being proportionall, be AB, BE, CD, DF, so that as AB is to BE, so is CD to DF. Then I say,<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of proportion by diuision.</note> that deuided also they shall be proportionall, as AE is to BE, so is CF to DF. Take equemultiplices vnto AE, EB, CF, FD, and let y<hi rend="sup">e</hi> same be GH, HK, LM, and MN.<note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ction<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> And likewise to EB, and FD, take any other eque<g ref="char:EOLhyphen"/>multiplices what soeuer, and let the same be KO, and NP. And forasmuch as how multiplex GH is to AE,<note place="margin">Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ra<g ref="char:EOLhyphen"/>tion.</note> so multiplex is KH to EB, therefore how mul<g ref="char:EOLhyphen"/>tiplex GH is to AE, so multiplex is GK to AB (by the first of the fifth). But how multiplex GH is to AE, so multiplex is LM to CF: Wherfore how mul<g ref="char:EOLhyphen"/>tiplex GK is to AB, so multiplex is LM to CF (by the <hi>11.</hi> of y<hi rend="sup">e</hi> same). Againe forasmuch as how multiplex LM is to CF, so multiplex is MN to DF, therfore how multiplex LM is to CF, so multiplex is LN to CD (by the first of the self same). But how multiplex LM is to CF so mul<g ref="char:EOLhyphen"/>tiplex
<figure/>
is GK to AB. Wherfore how multiplex GK is to AB, so multiplex is LN to CD. Wherefore GK and LN, are equemultiplices to<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AB &amp; CD. Againe forasmuch as how multiplex HK the first is to EB the second, so multiplex is MN the third to FD the fourth. And how multiplex KO the fift is to EB the second, so multiplex is NP the sixt to FD the fourth. Wherfore (by the second of the same) how multiplex HO composed of the first and fift is to EB, so multiplex is MP composed of the third and sixt to FD. And for that as AB is to BE, so is CD to DF: and to AB &amp; CD are taken equemultiplices GK and LN, and likewise to EB and FD are ta<g ref="char:EOLhyphen"/>ken certaine other equemultiplices, that is, HO, and MP. If therefore GK exceede HO, then LN also exceedeth MP: and if it be equall, it is equall: and if it be lesse, it is lesse (by the conuersion of the sixt defi<g ref="char:EOLhyphen"/>nition of the fift). Let GK exceede HO: Wherefore KH common to them both being taken away, the residue GH, shall exceede the residue KO. But if GK exceede HO, then doth LN exceede MP: Wherefore let LN excede MP: and MN which is co<g ref="char:cmbAbbrStroke">̄</g>mon to the both being take<g ref="char:cmbAbbrStroke">̄</g> away, y<hi rend="sup">•</hi> residue LM shall exceede the residue NP. Wherefore if GH exceede KO, then shall LM ex<g ref="char:EOLhyphen"/>ceede NP. And in like sort may we proue, that if GH be equall vnto KO, then LM shall be equall vnto NP: and if it be lesse, it shall be lesse: but GH
<pb facs="tcp:7062:176"/>
and LM are equemultiplices to AE and GF: and likewise KO and NP are certayne other equemultiplices to EB, and FD. Wherfore as AE is to EB, so <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>CF to FD (by the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> definition of the fift). If composed magnitudes there<g ref="char:EOLhyphen"/>fore be proportionall, then also deuided they shall be proportionall: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>18.</hi> Theoreme. The <hi>18.</hi> Proposition. If magnitudes deuided be proportionall: then also composed they shall be proportionall.</head>
               <p>SVppose that the magnitudes deuided being proportionall, be AE, EB, CF, &amp; FD, so that as AE is to EB, so let CF be to FD.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of propor<g ref="char:EOLhyphen"/>tion by com<g ref="char:EOLhyphen"/>position.</note> Then I say, that composed also they shall be proportionall, that is, as AB is to BE, so is CD to DF. For if AB be not vnto BE, as CD is to FD,
<figure/>
then shall AB be vnto BE,<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLunhyphen"/>uerse of the former.</note> as CD is either vnto a magnitude lesse then FD, on vnto a magnitude greater. Let it first be vnto a lesse, namely, to DG.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>aing to an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>mpossibility.</note> And forasmuch as, as AB is to BE, so is CD to DG: the composed magnitudes therefore are proportio<g ref="char:EOLhyphen"/>nall, wherefore deuided also they shall be proportionall (by y<hi rend="sup">e</hi> 
                  <hi>17.</hi> of the first). Wherefore as AE is to EB, so is CG to GD. But by supposition as AE is to EB, so is CF to FD. Wherefore (by the <hi>11.</hi> of the fift) as CG is to GD, so is CF to FD. Now then there are foure magnitudes, CG, GD, CF, and FD: of which the first CG is greater then the third CF. Wherefore (by the <hi>14.</hi> of the fift) the second GD is greater then the fourth FD. But it is also put to be lesse then it: which is impossible. Wherfore it can not be that as AB is to BE, so is CD to a magnitude lesse then FD. In like sort may we proue, that it can not be so to a magni<g ref="char:EOLhyphen"/>tude greater then FD. For by the same order of demonstration, it would follow that FD is greater then the sayd greater magnitude: which is impossible. Wher<g ref="char:EOLhyphen"/>fore it must be to the selfe same. If therefore magnitudes deuided be proportio<g ref="char:EOLhyphen"/>nall, then also composed they shall be proportionall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>19.</hi> Theoreme. The <hi>19.</hi> Proposition. If the whole be to the whole, as the part taken away is to the part taken away: then shall the residue be vnto the residue, as the whole is to the whole.</head>
               <p>
                  <pb n="146" facs="tcp:7062:176"/>SVppose that as the whole AB is to the whole CD,<note place="margin">That which the fift of this booke proued only touching multiplices, this proueth generally of all magnitudes.</note> so is the part taken away AE to the part taken away CF. Then I say, that the residue EB shall be vn<g ref="char:EOLhyphen"/>to the residue FD, as the whole AB is to the whole CD. For for that as the whole AB is to the whole CD, so is AE to CE, therfore alternately also (by the <hi>16.</hi> of the fift) as AB is to AE, so is CD to CF. A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d for that when magnitudes composed are proportionall, the same deuided also are proportionall (by the <hi>17.</hi> of y<hi rend="sup">e</hi> fift)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore as BE is to EA, so is DF to FC, Wherfore alternately also (by the <hi>16.</hi> of the fift) as BE is to DF, so is EA to FC. But as AE is to CF, so (by supposition) is the whole AB to the whole CD. Wherefore the residue EB shall be vnto the residue FD, as the whole AB is to the whole CD. If ther<g ref="char:EOLhyphen"/>fore the whole be to the whole, as the part taken away is to the part taken away, then shall the residue be vnto the residue; as the whole is to the whole: which was required to be proued.</p>
               <div type="section">
                  <head>¶ ALemma or Assumpt.</head>
                  <p>And forasmuch as by supposition as AB is to CD,<note place="margin">ALemma.</note> so is AE to CF: and al<g ref="char:EOLhyphen"/>ternately as AB is to AE, so is CD to CF. And now it is proued, that as, AB is to CD, so is EB to FD. Wherefore againe alternately, as AB is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>B, so is CD to FD. Wherefore it followeth, that as AB is to AE, so is CD to CF: and againe, as the same AB is to EB, so is the same CD to DF.</p>
               </div>
               <div type="section">
                  <head>Corollary.</head>
                  <p>And hereby it is manifest,<note place="margin">A Corollary.</note> that if magnitudes composed be proportionall, then also by conuersion of Proportion (which of some is called Proportion by Eversion,<note place="margin">Conuersion of proportion.</note> and which is, as before it was defined, whe<g ref="char:cmbAbbrStroke">̄</g> the antecedent is compared to the excesse, wherein the antecedent exceedeth the consequent) they shall be proportionall.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>20.</hi> Theoreme. The <hi>20.</hi> Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two and two in eche order, are in one and the same proportion, and if of e<g ref="char:EOLhyphen"/>qualitie
<pb facs="tcp:7062:177"/>
in the first order the first be greater then the third, then in the second order the first also shall be greater then the third: and if it be equall, it shall be equall: and if it be lesse, it shall be lesse.</head>
               <p>SVppose that there be three magnitudes in one order,<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion pertai<g ref="char:EOLhyphen"/>neth to Pro<g ref="char:EOLhyphen"/>portion of e<g ref="char:EOLhyphen"/>quality inor<g ref="char:EOLhyphen"/>dinate propor<g ref="char:EOLhyphen"/>tionality.</note> namely, A, B, C, &amp; let there be as many magnitudes in an other order, which let be D, E, F, which being taken two and two in ech order, let be in one and the same proportion, that is, as A is to B, so let D be to E, and as B is to C, so let E be to F.</p>
               <p>And first of equa<g ref="char:EOLhyphen"/>litie
<figure/>
let A be greater then C.<note place="margin">The first dif<g ref="char:EOLhyphen"/>ference.</note> Then I say, that D also is greater then F: and if it be e<g ref="char:EOLhyphen"/>quall, it is equall, and if it be lesse, it is lesse. For forasmuch as A is greater then C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and there is a certaine o<g ref="char:EOLhyphen"/>ther magnitude, namely, B, but the greater hath to one and the same magni<g ref="char:EOLhyphen"/>tude a greater pro<g ref="char:EOLhyphen"/>portion then hath the lesse (by the <hi>8.</hi> of the fift). Wherefore A hath vnto B a greater proportion then C hath to B. But as A is to B, so is D to E: and as C is to B, so is F to E (by the Cor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>llary of the fourth of the fift). Wherefore D hath vnto E a greater proportion, then F hath to E. But of magnitudes compared to one and the same magnitude that which hath the greater proportion is the grea<g ref="char:EOLhyphen"/>ter (by the <hi>10.</hi> of the fift). Wherefore D is greater then F.</p>
               <p>But now if A be equall vnto C, D also shall be equall vnto F. For then A and C haue vnto B one and the same proportion (by the first part of the seuenth of this booke).<note place="margin">The second difference.</note> And for that as A is to B, so is D to E, and as C is to B, so is F to E: therefore D and F haue vnto E one and the same proportion. Wherefore by the first part of the <hi>9:</hi> of this booke D is equall vnto F.</p>
               <p>But now suppose that A be lesse then C. Then also shall D be lesse then F. For by the <hi>8.</hi> of this booke C shall haue vnto B a greater proportion then hath A to B.<note place="margin">The third difference.</note> But as A is to B, so is D to E by supposition, and as C is to B, so haue we proued is F to E. Wherefore F hath vnto E a greater proportion then hath D to E. Wherefore by the first part of the <hi>10.</hi> of this booke F is greater then D.
<pb n="147" facs="tcp:7062:177"/>
If therefore there be three magnitudes in one order, and as many other magn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>tudes in an other order, which being taken two and two in ech order, are in one and the same proportion, and if of equalitie in the first order the first be greater then the third, then in the second order also the first shall be greater then y<hi rend="sup">e</hi> third and if it be equall, it shall be equall: and if it be lesse, it shall be lesse: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>21.</hi> Theoreme. The <hi>21.</hi> Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which being taken two and two in eche order are in one and the same proportion, and their proportion is perturbate: if of equalitie in the first order the first be greater then the third, the<g ref="char:cmbAbbrStroke">̄</g> in the second order the first also shall be greater then the third: and if it be equall it shall be equall: and if it be lesse it shall be lesse.</head>
               <p>SVppose y<hi rend="sup">t</hi> there be three magnitudes in one order,<note place="margin">Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r proposi<g ref="char:EOLhyphen"/>tion pertai<g ref="char:EOLhyphen"/>neth to Pro<g ref="char:EOLhyphen"/>portion of e<g ref="char:EOLhyphen"/>quality in perturbate proportiona<g ref="char:EOLhyphen"/>lity.</note> namely, A, B, C, and let there be as many other magnitudes in an other, which let be D, E, F: which being taken two &amp; two in ech order, let be in one and the same proportion, and let their proportion be perturbate. So that as A is to B, so let E be to F, &amp; as B is to C, so let D be to E, and of equalitie let A be greater then G. Then I say, that D also is greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n E: and if it be equall it is equall: and if it be lesse it is lesse.</p>
               <p>First let: A be grea<g ref="char:EOLhyphen"/>ter
<figure/>
then C.<note place="margin">The first dif<g ref="char:EOLhyphen"/>ference.</note> And for as<g ref="char:EOLhyphen"/>much as A is greater then C, &amp; there is a cer<g ref="char:EOLhyphen"/>taine other magnitude, namely, B: therfore (by the <hi>8.</hi> of y<hi rend="sup">e</hi> fift) A hath a greater proportio<g ref="char:cmbAbbrStroke">̄</g> vn<g ref="char:EOLhyphen"/>to B, then C hath to B. But as A is to B, so is E to F: and againe as C is to B, so is E to D (by the Corellary of y<hi rend="sup">•</hi> fourth of the fift) Wherfore E hath vnto F a greater proportion, then E hath to D. But that magnitude wherin to one and the same magnitude hath the greater proportion, is the lesse (by the <hi>10.</hi> of the fift) Wher<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:178"/>
F is lesse then D. Wherefore D is greater then F.</p>
               <p>But now if A be e<g ref="char:EOLhyphen"/>quall
<figure/>
vnto C,<note place="margin">The second difference.</note> we may also proue that D is e<g ref="char:EOLhyphen"/>quall vnto F. For then (by the first part of the <hi>7.</hi> of this booke) A and C shall haue vnto B one and the same proporti<g ref="char:EOLhyphen"/>on. And for that as A is to B, so is E to F, &amp; as C is to B, so is E to D: therefore E hath to ei<g ref="char:EOLhyphen"/>ther of these D and F, one and y<hi rend="sup">e</hi> same propor<g ref="char:EOLhyphen"/>tion. Wherefore by the second part of the <hi>9.</hi> of the fift D is equall vnto F.</p>
               <p>
                  <note place="margin">The third difference.</note>Likewise, if A be lesse then C, D also is lesse then F. For then C shall haue vnto B a greater proportion, then hath A to B (by the <hi>8.</hi> of the fift. Wherefore E also <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ath vnto D a greater proportion then it <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ath to F. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore by the second part of the <hi>10.</hi> of this booke D is lesse then F. If therefore there be three magnitudes in one order, &amp; as many other magnitudes in an other order, which being taken two and two in ech order are in one &amp; the same proportion, &amp; their proportion is perturbate, and if of equalitie in the first order the first be greater then the third, then in the second order the first also shall be greater then y<hi rend="sup">e</hi> third<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and if it be equall it shall be equall: and if it be lesse it shall be lesse: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>22.</hi> Theoreme. The <hi>22.</hi> Proposition. If there be a number of magnitudes, how many soeuer in one order, and as many other magnitudes in an other order, which being taken two and two in ech order are in one and the same proportion, they shall also of equalitie be in one and the same proportion.</head>
               <p>
                  <note place="margin">Proportion of equality in or<g ref="char:EOLhyphen"/>dinate propor<g ref="char:EOLhyphen"/>tionality.</note>SVppose that there be a certaine number of magnitudes in one order. As for example: A, B, C, and let there be as many other magnitudes in an other order, which let be D, E, F, which being taken two and two let be in one and the same proportion. So that as A is to B, so let D be to E, and as B is to C, so let E be to F. Then I say, that of equalitie
<pb n="148" facs="tcp:7062:178"/>
they shall be in the same proportion, that is, as A is to C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is D to F.<note place="margin">Construction.</note> Take vnto A and D equemultiplices G &amp; H, and likewise to B &amp; E take any other eque<g ref="char:EOLhyphen"/>multiplices whatsoeuer, namely, K and L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and moreouer vnto C and F take any other equemultiplices also what soeuer, namely, M and N. And forasmuch as,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> as A is to B, so is D to E: and vnto A and D are taken equemultiplices G and H: and likewise vnto B and E are taken certaine other equem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ltiplices K and L: therfore
<figure/>
(by the <hi>4.</hi> of the fift) as G is to K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is H to L. And (by the same reason) as K is to M, so is L to N. Seing therefore y<hi rend="sup">t</hi> there be in order three magnitudes G, K, M, &amp; as many other magnitudes in an other order, namely, H, L, N, which being compared two to two are in one and the same proportion, therefore of equalitie (by the <hi>20.</hi> of the fift) if N exceede M, then shall H exceede G: and if it be equall it shall be equall: and if it be lesse it shall be lesse. But G and H are equemultiplices vnto A and D, and M and N are certaine other equemultiplices vnto C and F: ther<g ref="char:EOLhyphen"/>fore (by the <hi>6.</hi> definition of the fift) as A is to C, so is D to F.</p>
               <p>So also if there be more magnitudes then three in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ither order,<note place="margin">When there are more then three magni<g ref="char:EOLhyphen"/>tudes in either order.</note> the first of the one order shall be to the last, as the first of the other order is to the last. As if there were foure in one order, namely, ABCD, and other foure in the other order, namely, EFGH,<note place="margin">A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>CDE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>GH</note> we may with three magnitudes A, B, C, and E, F, G, proue that as A is to C, so is E to G: And then leauing out in either order the second and taking the fourth, as leauing out B and F, and taking D and H, we may proue by these three and three A, C, D, and E, G, H, that as A is to D, so is E to H. And obseruing this order, thys demonstration will serue how many soeuer the magnitudes be in either order. If therefore there be a number of magnitudes how many soeuer in one order, and as many other magnitudes in an other order, which being taken two and two in eche order are in one and the same proportion, they shall also of equalitie be in one and the same proportion: which was requi<g ref="char:EOLhyphen"/>red to be demonstrated.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:179"/>
               <head>The <hi>23.</hi> Theoreme. The <hi>23.</hi> Proposition. If there be three magnitudes in one order, and as many other magnitudes in an other order, which beyng taken two &amp; two in eche order are in one and the same proportion, and if also their proportion be perturbate: then of equalitie they shall be in one and the same proportion.</head>
               <p>
                  <note place="margin">
                     <hi>Proportion of equality in perturbate proprotiona<g ref="char:EOLhyphen"/>litie.</hi>
                  </note>
                  <hi>SVppose y<hi rend="sup">t</hi> there be in one order three magnitudes, namely, A, B, C, &amp; let be take<g ref="char:cmbAbbrStroke">̄</g> in an other order as many other magnituds, which let be D, E, F which being taken two and two, in eche order let be in one and the same proportion:
<figure/>
and suppose that their pro<g ref="char:EOLunhyphen"/>portio<g ref="char:cmbAbbrStroke">̄</g> be per<g ref="char:EOLhyphen"/>turbate. So that as A is to B, so let E be F, and as B is to C, so let D be to E. The<g ref="char:cmbAbbrStroke">̄</g> I say that as A is to C so is D to F. Take vnto A, B, D,</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>e<g ref="char:EOLhyphen"/>quemultipli<g ref="char:EOLhyphen"/>ces, and let the same be GHK: and likewise vnto C, E, F, take any other equemultiplices whatsoeuer, and let y<hi rend="sup">e</hi> same be LMN. And forasmuch as G and H are equemultiplices vnto A and B, but the partes of equemultiplices are in the same proportion that their equemultipli<g ref="char:EOLhyphen"/>ces are (by y<hi rend="sup">e</hi>
                  </hi> 15. <hi>of the fift) wherfore as A is to B, so is G to H.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>
                  <q>
                     <hi>And by y<hi rend="sup">e</hi> same reason also as E is to F, so is M to N. But as A is to B, so is E to F. Wherfore (by y<hi rend="sup">•</hi>
                     </hi> 11. <hi>of y<hi rend="sup">e</hi> fift) as G is to H, so is M to N. And forasmuch as, as B is to C, so is D to E, and vnto B &amp; D are taken equemultiplices H &amp; K: and likewise vnto C and E are taken certayne other equemultiplices L and M: therfore (by the</hi> 4. <hi>of the fifth) as H is to L, so is K to M, and alternately also (by the</hi> 16. <hi>of the fift) as B is to D, so is C to E. And forasmuch as H and K are the equemultiplices of B and D, but the partes of equemultiplices are in the same proportion that
<pb n="149" facs="tcp:7062:179"/>
their equemultiplices are (by the</hi> 15. <hi>of the fift) therfore as B is to D, so is H to K. But as B is to D, so is C to E, therfore (by the</hi> 11. <hi>of the first) as H is to K, so is C to E. Agayne forasmuch as L and M are the equemultiplices of C and E, therfore as C is to E, so is L to M. But as C is to E, so is H to K: therfore as H it to K, so is L to M, and alternately (by the</hi> 16. <hi>of the fift) as H is to L, so is K to M.*</hi>
                  </q>
               </p>
               <p>(But as <hi>A</hi> is to <hi>B</hi> so is <hi>E</hi> to <hi>F</hi> (by supposition) wherfore as <hi>G</hi> is to <hi>H,</hi> so is E <hi>F,</hi> (by the 11. of the fift.) Agayne for as much as <hi>M</hi> and <hi>N</hi> are equemultiplices vnto <hi>E</hi> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d <hi>F,</hi> th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>efore agayne (by the 15. of the fifth) as <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> is to <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>,</hi> so <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <hi>M</hi> to <hi>N.</hi> But as <hi>E</hi> is to <hi>F,</hi> so haue we proued is <hi>G</hi> to <hi>H,</hi> wherfore as <hi>G</hi> is to <hi>H,</hi> so is <hi>M</hi> to <hi>N</hi> (by the 11. of the fift) And for that <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> is to <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>,</hi> so is <hi>D</hi> to <hi>E</hi> (by supposition). And vnto <hi>B</hi> and <hi>D</hi> are taken e<g ref="char:EOLhyphen"/>quemultiplices <hi>H</hi> and <hi>K</hi>: and vnto <hi>C</hi> and <hi>E</hi> are taken certayne other equemultiplices <hi>L</hi> and <hi>M,</hi> therefore as <hi>H</hi> is to <hi>L</hi> so is <hi>K</hi> to <hi>M,</hi> by the 4 of this booke).</p>
               <p>
                  <hi>But it is proued that as G is to H, so is M to N. Seyng therefore that there are in one orde three<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> magnitudes, namely, G, H, L, and as many other magnitudes in an other order, namely, K, M, N, whiche being ta<g ref="char:EOLhyphen"/>ken two and two in eche order, are in one and the same proportion, and their proportion is perturbate, therfore of equalitie (by the</hi> 21. <hi>of the fifth) if G excede L, then shall K exceede N, and if it be equall, it shall be equall: and if it be lesse it shall be lesse. But G and K are equemultiplices vnto A and D, and L and N are certayne other equemultiplices vnto C and F. Wherfore as A is to C, so is D to F (by the</hi> 6. <hi>definition of the fifth). If therefore there be three magnitudes in one order, and as many other magnitudes in an other order, which beyng taken two and two in eche order, are in one and the same proportion, and if also their proportion be perturbate: then of equalitie they shall be in one and the same pro<g ref="char:EOLhyphen"/>portion: which was required to be proued.</hi>
               </p>
               <p>From this marke * first to the same marke agayne, you may if you will in stede of <hi>The<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>
                  </hi> argumente<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which seeme somewhat intricate, read those argumentes following printed with an other letter, which are very perspicuous and briefe, and followed of the most interpre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ers.</p>
               <p>This Proposition is also true if there be
<figure/>
more then thre magnitudes in either order<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> As for example,<note place="margin">
                     <hi>An addition of Campa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e.</hi>
                  </note> let there be foure. So that as A is to B, so let E be to F, and as B is to C, so let D be to E, &amp; as C is to P, so let Q be to D. Then I say that as A is to P, so is Q to F. For forasmuch as it is before proued, that A is to C, as D is to F taking away. B &amp; E there shall be three m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>gnitudes A, C, P, in one or<g ref="char:EOLhyphen"/>der, and as many other in an other order, namely, Q<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> D, F, which being take<g ref="char:cmbAbbrStroke">̄</g> in ech or<g ref="char:EOLhyphen"/>der two and two, are in one and the same pro<g ref="char:EOLunhyphen"/>portion, and their proportio<g ref="char:cmbAbbrStroke">̄</g> is perturbate, wherfore by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> former propositi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> to P as Q is to F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
               <p>And euen as by the demonstration in three magnitudes is taken the proofe in foure magnitudes by leauing out one of<note place="margin">
                     <hi>Note.</hi>
                  </note>
                  <pb facs="tcp:7062:180"/>
the meanes: so by the demonstration in foure magnitudes is taken the proofe in fiue magnitudes by leauyng out two of the meanes: and by the demonstration in fiue, the proofe in sixe, by leauing out three meanes. And so forward co<g ref="char:cmbAbbrStroke">̄</g>tinually, which is also to be vnderstanded in the former kynde of proportion of equalitie, which is in ordinate proportion.</p>
            </div>
            <div type="part">
               <head>The <hi>24.</hi> Theoreme. The <hi>24.</hi> Proposition. If the first haue vnto the second the same proportion that the third hath to the fourth, and if the fift haue vnto the second the same proportion that the sixt hath to the fourth: then al<g ref="char:EOLhyphen"/>so the first and fift composed together shall haue vnto the se<g ref="char:EOLhyphen"/>cond the same proportion that the third and sixt composed to<g ref="char:EOLhyphen"/>gether haue vnto the fourth.</head>
               <p>SVppose that there be sixe magnitudes AB, C, DE, F, BG, &amp; EH: of which let AB be the first, C the second, DE the third, F y<hi rend="sup">e</hi> fourth, BG the fift, and EH the sixt. And suppose that AB the first, haue vnto C the second, the same proportion that DE the third hath to F the fourth,<note place="margin">That which the second pro<g ref="char:EOLunhyphen"/>positio<g ref="char:cmbAbbrStroke">̄</g> of this booke proued only touching multiplices, is here proued generally tou<g ref="char:EOLhyphen"/>ching magni<g ref="char:EOLhyphen"/>tudes.</note> and let BG the fift haue vnto C the second the same proportion that EH the sixt hath vnto F the fourth. Then I say, that the first and fift compo<g ref="char:EOLhyphen"/>sed together, namely, AG, hath vnto C the second the
<figure/>
same proportion, that the third and sixt composed toge<g ref="char:EOLhyphen"/>ther, namely, DH, hath vnto F the fourth. For, for that as BG is to C, so is EH to F: then also by conuer<g ref="char:EOLhyphen"/>sion (by y<hi rend="sup">e</hi> Corollary of the <hi>4.</hi> of the fift) as C is to BG, so is F to EH. And for that as AB is to C, so is DE to F, but as C is to GB, so is F to EH: therefore of e<g ref="char:EOLhyphen"/>qualitie (by the <hi>22.</hi> of the fift) as AB is to BG, so is DE to EH. And forasmuch as when magnitudes de<g ref="char:EOLhyphen"/>uided are proportionall, they also composed are propor<g ref="char:EOLhyphen"/>tionall (by y<hi rend="sup">e</hi> 
                  <hi>18.</hi> of the fift): therfore as AG is to GB, so is DH to HE: but as BG is to C, so is EH to F: Wherefore againe of equalitie (by the <hi>22.</hi> of the fift) as AG is to C, so is DH to F. If therefore the first haue vnto the second the same proportion that the third hath to the fourth, and if the fift haue vnto the second the same proportion that y<hi rend="sup">e</hi> sixt hath to the fourth: then also the first and fift composed together shall haue vnto the second the same proportion that the third and sixt composed together haue vnto the fourth: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb n="150" facs="tcp:7062:180"/>
               <head>The <hi>25.</hi> Theoreme. The <hi>25.</hi> Proposition. If there be foure magnitudes proportionall: the greatest and the least of them, shall be greater then the other remayning.</head>
               <p>SVppose that there be foure magnitudes proportionall AB, CD, E and F. So that as AB is to CD, so let E be to F. And let the greatest of them be AB, &amp; the lest of them be F. Then I say, that these two magnitudes AB and F, are greater then the two magnitudes CD &amp; E. Forasmuch as AB is supposed to be the greatest of all foure, therefore it is greater then E. There<g ref="char:EOLhyphen"/>fore from the greater AB cut of (by the <hi>3.</hi> of the first) vnto E an equall mag<g ref="char:EOLhyphen"/>nitude AG. and likewise (by the same) from CD cut of vnto F an equall mag<g ref="char:EOLhyphen"/>nitude CH. (Which may be done, for that the magnitude CD is greater then the magnitude E: for that as AB is to CD, so is E to F, therefore alternately as AB is to E, so is CD to F (by the <hi>16.</hi> of the fift). But
<figure/>
AB is greater then E: Wherfore also CD is greater then F: Which thing may also be proued by y<hi rend="sup">e</hi> 
                  <hi>14.</hi> of the same.) Now for that as AB is to CD, so is E to F: but E is e<g ref="char:EOLhyphen"/>quall vnto AG, and F is equall vnto CH: therefore as AB is to CD, so is AG to CH: and forasmuch, as the whole AB is to the whole CD, so is the part taken a<g ref="char:EOLhyphen"/>way AG, to the part taken away CH: therefore the residue GB (by the <hi>1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> of the fift) is vnto the residue HD, as the whole AB is to the whole CD. But AB the first is greater then CD the third: Wherfore GB the second is greater then HD the fourth (by the <hi>14.</hi> of the fift). And forasmuch as AG is equall vnto E, &amp; CH is equall vnto F: therefore AG and F are equall vnto CH and E. And forasmuch as if vnto thinges vnequall be added thinges equall, all shall be vnequall (by the fourth common sentence): therefore seing that GB and DH are vnequall, and GB is the greater, if vnto GB be added AG and F: and likewise if vnto HD be added CH &amp; E, there shall be produced AB and F greater then CD &amp; E. If therefore there be foure magnitudes proportionall, the greatest and the least of them, shall be greater the<g ref="char:cmbAbbrStroke">̄</g> the other remayning: which was required to be demonstrated.</p>
               <p>
                  <hi>Here follow certayne propositions added by</hi> Campane, <hi>which are not to be con<g ref="char:EOLhyphen"/>temned, and are cited euen of the best learned, namely, of</hi> Iohannes Regio montanus, <hi>in the Epitome which he writeth vpon</hi> Ptolome.</p>
               <div type="section">
                  <head>¶ The first Proposition.</head>
                  <p>
                     <hi>If there be foure quantities, and if the proportion of the first to the second, be greater
<pb facs="tcp:7062:181"/>
then the proportion of the third to the fourth: then contrariwise by conuersion, the propor<g ref="char:EOLhyphen"/>tion of the second to the first, shall be lesse then the proportion of the fourth to the third.</hi>
                  </p>
                  <p>Suppose that the proportion of <hi>A</hi> to <hi>B</hi> be greater then the proportion of <hi>C</hi> to <hi>D.</hi> Then I say contrariwise by conuersion,<note place="margin">
                        <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibilitie.</hi>
                     </note> the proportion of <hi>B</hi> to <hi>A</hi> is lesse then the pro<g ref="char:EOLhyphen"/>portion of <hi>D</hi> to <hi>C.</hi> For if the proportion of <hi>B</hi> to <hi>A</hi> be one and the same with the pro<g ref="char:EOLhyphen"/>portion of <hi>D</hi> to <hi>C,</hi> then conuersedly the proportion of <hi>A</hi> to <hi>B</hi> is one and the same with the proportion of
<figure/>
                     <hi>C</hi> to <hi>D,</hi> which is contrary to suppo<g ref="char:EOLhyphen"/>sition. But if the proportion of <hi>B</hi> to <hi>A</hi> be greater then the proportion of <hi>D</hi> to <hi>C.</hi> Then let <hi>E</hi> be vnto <hi>A,</hi> as <hi>D</hi> is to <hi>C.</hi> And by the 13. of this booke, the proportion of <hi>E</hi> to <hi>A</hi> shall be lesse then the pro<g ref="char:EOLhyphen"/>portion of <hi>B</hi> to <hi>A.</hi> Wherfore (by the first part of the tenth of the same) <hi>E</hi> is lesse the<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>B.</hi> And therfore by the seco<g ref="char:cmbAbbrStroke">̄</g>d part of the 8. of the same, <hi>A</hi> hath vnto <hi>E</hi> a greater proportio<g ref="char:cmbAbbrStroke">̄</g> then <hi>A</hi> hath to <hi>B.</hi> And forasmuch as by conuerse proportionalitie <hi>A</hi> is vnto <hi>E</hi> as <hi>C</hi> is to <hi>D,</hi> therfore by the 13. of the same, <hi>C</hi> hath to <hi>D</hi> a greater proportion then hath <hi>A</hi> to <hi>B,</hi> which is contrary to the supposition. For <hi>A</hi> was supposed to haue vnto <hi>B</hi> a greater proportion then hath <hi>C</hi> to <hi>D.</hi> Wherefore the proportion of <hi>B</hi> to <hi>A</hi> is neither one and the same with the proportion of <hi>C</hi> to <hi>D,</hi> nor greater then it. Wherfore it is lesse: which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the same affirmatiuely.</hi>
                     </note>It may also be demonstrated directly. For let <hi>E</hi> be vnto <hi>B</hi> as <hi>C</hi> is to <hi>D.</hi> Then co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>uersedly <hi>B</hi> is to <hi>E</hi> as <hi>D</hi> is to <hi>C.</hi> And forasmuch as <hi>A</hi> is greater then <hi>E</hi> by the first part of the tenth of this booke, therfore by the second part of the 8 of the same <hi>B</hi> hath vnto <hi>A</hi> a lesse proportion then hath <hi>B</hi> to <hi>E.</hi> Wherfore by the 13. of the same <hi>B</hi> hath vnto <hi>A</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lesse proportion then hath <hi>D</hi> to <hi>C</hi>: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The second Proposition.</head>
                  <p>
                     <hi>If there be foure quantities, and if the proportion of the first to the second be greater then the pro<g ref="char:EOLhyphen"/>portion of the third to the fourth, then alternately the proportion of the first to the third, shall be grea<g ref="char:EOLhyphen"/>ter then the proportion of the second to the fourth.</hi>
                  </p>
                  <p>Let <hi>A</hi> haue vnto <hi>B</hi> a greater proportion then hath <hi>C</hi> to <hi>D.</hi> Then I say alternately <hi>A</hi> hath to <hi>C</hi> a greater proportion then hath <hi>B</hi> to <hi>D.</hi> For one and the same proportion it can not haue:<note place="margin">
                        <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> impossibilitie.</hi>
                     </note> for then alternately <hi>A</hi> should be to <hi>B</hi> as <hi>C</hi> is to <hi>D,</hi> which is contrary to the suppositio<g ref="char:cmbAbbrStroke">̄</g>. But if it haue a lesse proportio<g ref="char:cmbAbbrStroke">̄</g>, let <hi>E</hi> be vnto <hi>C</hi> as <hi>B</hi> is to <hi>D.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> by the 13. of this booke
<figure/>
                     <hi>E</hi> hath vnto <hi>C</hi> a greater proportion then hath <hi>A</hi> to <hi>C.</hi> Wherefore (by the first part of the tenth of the same) <hi>E</hi> is greater the<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>A.</hi> Wherefore by the first part of the 8. of the same, E hath to B a greater pro<g ref="char:EOLunhyphen"/>portion then hath <hi>A</hi> to <hi>B.</hi> And forasmuch as by suppositio<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>E</hi> is vnto <hi>C,</hi> as <hi>B</hi> is to <hi>D,</hi> therfore alternately <hi>E</hi> is to <hi>B</hi> as <hi>C</hi> is to <hi>D.</hi> Wherfore by the 13. of the same, <hi>C</hi> hath to <hi>D</hi> a greater proportion then hath <hi>A</hi> to <hi>B,</hi> which is contrary to the suppositio<g ref="char:cmbAbbrStroke">̄</g>. Wherfore the proportion of <hi>A</hi> to <hi>C</hi> is neither one and the same with the proportion of <hi>B</hi> to <hi>D,</hi> no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lesse then it, wherefore it is greater. Which was required to be proued.</p>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the same affirmatiuely.</hi>
                     </note>This may also be demonstrated affirmatiuely, let <hi>E</hi> be vnto <hi>B</hi> as <hi>C</hi> is to <hi>D.</hi> Now the<g ref="char:cmbAbbrStroke">̄</g> by the first part of the tenth of this booke, <hi>E</hi> is lesse then <hi>A</hi>: wherfore by the first parte of the 8. of the same, the proportion of <hi>A</hi> to <hi>C</hi> is greater then the proportion of <hi>E</hi> to <hi>C.</hi> But alternately <hi>E</hi> is to <hi>C</hi> as <hi>B</hi> is to <hi>D.</hi> Wherfore (by the 13. of the same) <hi>A</hi> hath to <hi>C</hi> a
<pb n="152" facs="tcp:7062:181"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                  </p>
               </div>
               <div type="section">
                  <head>¶ The third Proposition.</head>
                  <p>
                     <hi>If there be foure quantities, and if the proportion of the first t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the second be greater, then the proportion of the third to the fourth: then by composition also the proportio<g ref="char:cmbAbbrStroke">̄</g> of the, f<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>th and second to the second, shall be greater then the proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the third and fourth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to the fourth.</hi>
                  </p>
                  <p>Suppose that the proportion of <hi>A</hi> to <hi>B</hi> be greater then the proportion of <hi>C</hi> to <hi>D.</hi> Then I say that the proportion of <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> 
                     <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> to <hi>B</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> greater then the proportion of the whole <hi>CD</hi> to <hi>D.</hi> For the proportion of <hi>AB</hi> to <hi>B</hi> can not be one and the same with the proportion of <hi>CD</hi> to <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <note place="margin">
                        <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibilitie.</hi>
                     </note> for then by diuision also <hi>A</hi> should be vnto <hi>B</hi> as <hi>C</hi> is <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which is contrary to the supposition. Neither
<figure/>
also can it be lesse. For if it be possible let it be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp; let <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nto <hi>B</hi> as <hi>CD</hi> is to <hi>D.</hi> No<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then (by the 12. of this boke <hi>EB</hi> hath vnto <hi>B</hi> a greater propor<g ref="char:EOLhyphen"/>tion then hath <hi>AB</hi> to <hi>B.</hi> Wherefore by the first part of the 10. of the same <hi>EB</hi> is greater then the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <hi>AB.</hi> And by the common sentence <hi>E</hi> is greater then <hi>A.</hi> Wherefore by the first part of the 8. of this booke <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> hath to <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> a grea<g ref="char:EOLhyphen"/>ter proportio<g ref="char:cmbAbbrStroke">̄</g> then hath <hi>A</hi> to <hi>B.</hi> But <hi>E</hi> is to <hi>B</hi> as <hi>C</hi> is to <hi>D</hi> by diuisio<g ref="char:cmbAbbrStroke">̄</g> of proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for <hi>EB</hi> is to <hi>B</hi> as <hi>CD</hi> is to <hi>D.</hi> Wherfore (by the 12. of the same) <hi>C</hi> hath to <hi>D</hi> a greater pro<g ref="char:EOLhyphen"/>portio<g ref="char:cmbAbbrStroke">̄</g> the<g ref="char:cmbAbbrStroke">̄</g> hath <hi>A</hi> to <hi>B,</hi> which is co<g ref="char:cmbAbbrStroke">̄</g>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rary to the supposition. Wherfore the proportion of <hi>AB</hi> to <hi>B</hi> is not one and the same with the proportion of <hi>CD</hi> to <hi>D</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ither also is it lesse then it. Wherfore it is greater: which was required to be proued.</p>
                  <p>This may also be demonstrated a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>firmat<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ely.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the same.</hi>
                     </note> Forasmuch as the proportion of <hi>A</hi> to <hi>B</hi> is greater then the proportion of <hi>C</hi> to <hi>D</hi>: let <hi>E</hi> be vnto <hi>B</hi> as <hi>C</hi> is to <hi>D.</hi> And so by the first part of the 10. of this booke, <hi>E</hi> shall be lesse then <hi>A.</hi> And therfore by the com<g ref="char:EOLhyphen"/>mon sentence <hi>EB</hi> shall be lesse then <hi>AB.</hi> Wherfore by the first part of the 8. of the same <hi>AB</hi> hath vnto <hi>B</hi> a greater proportion then hath <hi>EB</hi> to <hi>B.</hi> But by composition <hi>EB</hi> is to <hi>B</hi> as <hi>CD</hi> is to <hi>D.</hi> For by supposition <hi>E</hi> is vnto <hi>B</hi> as is to <hi>D.</hi> Wherfore (by the 12. of this booke) <hi>AB</hi> hath to <hi>B</hi> a greater proportion then hath <hi>CD</hi> to <hi>D</hi>: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The fourth Proposition.</head>
                  <p>
                     <hi>If there be foure quantities, and if the proportion of the first and the second to the second be greater then the proportion of the third and fourth to the fourth: then by diuision also the proportion of the first to the second, shall be greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n the proportion of the thirde to the fourth.</hi>
                  </p>
                  <p>Suppose that the proportion of <hi>AB</hi> to <hi>B</hi> be greater then the proportion of <hi>CD</hi> to <hi>D.</hi>
                     <note place="margin">
                        <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibilitie.</hi>
                     </note> Then I say that by diuision also the proportion of <hi>A</hi> to <hi>B</hi> is greater then the pro<g ref="char:EOLhyphen"/>portion of <hi>C</hi> to <hi>D.</hi> For it cannot be the same. For then by composition <hi>AB</hi> should be to <hi>B</hi> as <hi>CD</hi> is to <hi>D.</hi> Neither also can it be lesse:
<figure/>
for if the proportion of <hi>C</hi> to <hi>D</hi> be greater then the proportion of <hi>A</hi> to <hi>B,</hi> then by the former proposition, the proportion of <hi>CD</hi> to <hi>D</hi> should be greater then the proportion of <hi>AB</hi> to <hi>B</hi>: which is contrary also to the suppositio<g ref="char:cmbAbbrStroke">̄</g>. Wher<g ref="char:EOLhyphen"/>fore the proportion of <hi>A</hi> to <hi>B</hi> is neither one and the same with the proportion of <hi>C</hi> to <hi>D,</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> it: Wherefore it is greater then it: which was required to be proued.</p>
                  <p>
                     <pb facs="tcp:7062:182"/>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration of the same affirmatiuely.</hi>
                     </note>The same may also be proued affirmatiuely. Suppose that <hi>EB</hi> be vnto <hi>B</hi> as <hi>CD</hi> is to <hi>D.</hi> Now then (by the first part of the 10. of
<figure/>
the fifth) <hi>EB</hi> shall be lesse then <hi>AB:</hi> and there<g ref="char:EOLhyphen"/>fore by the common sentence, <hi>E</hi> is lesse then <hi>A,</hi> wherfore by the first part of the 8. of this booke, the proportion of <hi>E</hi> to <hi>B,</hi> is lesse then the propor<g ref="char:EOLhyphen"/>tion of <hi>A</hi> to <hi>B,</hi> but as <hi>E</hi> is to <hi>B,</hi> so is <hi>C</hi> to <hi>D:</hi> wher<g ref="char:EOLhyphen"/>fore the proportion of <hi>C</hi> to <hi>D,</hi> is lesse then the proportion of <hi>A</hi> to <hi>B.</hi> Wherfore the pro<g ref="char:EOLunhyphen"/>portion of <hi>A</hi> to <hi>B</hi> is greater then the proportion of <hi>C</hi> to <hi>D</hi>: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The fifth Proposition.</head>
                  <p>
                     <hi>If there be foure quantities, and if the proportion of the first and the second to the second be greater then the proportion of the third and the fourth to the fourth: then by euersion the proportion of the first and second to the first, shall be lesse then the proportion of the third and fourth to the third.</hi>
                  </p>
                  <p>Suppose that the proportion of <hi>AB</hi> to <hi>B</hi> be greater then the proportion of <hi>CD</hi> to <hi>D.</hi> Then I say that by euersion the proportion of <hi>AB</hi> to <hi>A</hi> is lesse then the proporti<g ref="char:EOLhyphen"/>on of <hi>CD</hi> to <hi>C.</hi>
                     <note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> For by diuision by the former
<figure/>
proposition the proportion of <hi>A</hi> to <hi>B</hi> is greater then the proportion of <hi>C</hi> to <hi>D.</hi> Wherefore by the first of these propositions conuersedly, <hi>B</hi> hath vnto <hi>A</hi> a lesse proportio<g ref="char:cmbAbbrStroke">̄</g> the<g ref="char:cmbAbbrStroke">̄</g> hath <hi>D</hi> to <hi>C.</hi> Wher<g ref="char:EOLhyphen"/>fore by the 3. of the same by composition, the proportion of <hi>AB</hi> to <hi>A</hi> is lesse the<g ref="char:cmbAbbrStroke">̄</g> the propor<g ref="char:EOLhyphen"/>tion of <hi>CD</hi> to <hi>C:</hi> which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The sixt Proposition.</head>
                  <p>
                     <hi>If there be taken three quantities in one order, and as many in an other order, and if the proportion of the first to the second in the first order, be greater then the proportion of the first to the second in the latter order: then also the proportion of the first to the third in the first order, shall be greater then the proportion of the first to the third in the latter order.</hi>
                  </p>
                  <p>Suppose that there be three qua<g ref="char:cmbAbbrStroke">̄</g>tities in one order <hi>A, B, C,</hi> &amp; as many other qua<g ref="char:cmbAbbrStroke">̄</g>tities in an other order <hi>D, E, F.</hi> And let the proportion of <hi>A</hi> to <hi>B</hi> in the first order be greater then the proportion of <hi>D</hi> to <hi>E</hi> in the second order, and let also the proportion of <hi>B</hi> to <hi>C</hi> in the first order, be greater then the proportion of <hi>E</hi> to <hi>F</hi> in the second order. Then I say that
<figure/>
the pro<g ref="char:EOLhyphen"/>portion of <hi>A</hi> to <hi>C</hi> in the first or<g ref="char:EOLhyphen"/>der, is greater the<g ref="char:cmbAbbrStroke">̄</g> the propor<g ref="char:EOLhyphen"/>tion of <hi>D</hi> to <hi>F</hi> in the second order.<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </note> For let <hi>G</hi> be vnto <hi>C</hi> as <hi>E</hi> is to <hi>F.</hi> Now then by the first part of the 10 of this booke <hi>G</hi> shall be lesse then <hi>B.</hi> And therefore by the second parte of the 8. of the
<pb n="152" facs="tcp:7062:182"/>
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>me, the propor<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>on of <hi>A</hi> to <hi>G</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n the proport<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>on of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rfore the proportion of <hi>A</hi> to <hi>G</hi> is muche greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n the proportion of <hi>D</hi> to <hi>E.</hi> Now then let <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> be <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> 
                     <hi>D</hi> is to <hi>E.</hi> Wherfore by the first part of the 10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the same, <hi>A</hi> is great<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r the<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>H.</hi> And therfore by the first part of the 8. of the same, the proportion of <hi>A</hi> to <hi>C</hi> is greater then the proportion of <hi>H</hi> to <hi>C.</hi> But by proportion of equality <hi>H</hi> is to <hi>C</hi> as <hi>D</hi> is to <hi>F</hi> (for <hi>H</hi> is to <hi>G</hi> as <hi>D</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to <hi>E,</hi> and <hi>G</hi> is to <hi>C</hi> as <hi>E</hi> is to <hi>F.</hi> Wherfore by the 12. of the same <hi>A</hi> hath to <hi>C</hi> a greater proportion then hath <hi>D</hi> to <hi>F</hi>: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The seuenth Proposition.</head>
                  <p>
                     <hi>If there be taken three quantities in one order, and as many other in an other order, and if the proportion of the second to the third in the first order be greater then the proportion of the first to the second in the latter order, if also the proportion of the first to the second in the first order be greater then the proportion of the second to the third in the latter order: then shall the proportion of the first to the third in the first order be greater, then the proportion of the first to the third in the latter order.</hi>
                  </p>
                  <p>Suppose that there be three qua<g ref="char:cmbAbbrStroke">̄</g>tities in one order <hi>A, B, C,</hi> and as many other in an other order <hi>D, E, F.</hi> And let the proportion of <hi>B</hi> to <hi>C</hi> in the first order, be greater then the proportion of <hi>D</hi> to <hi>E</hi> in the second order, and let also the proportion of <hi>A</hi> to <hi>B</hi> in the first order, be greater then the proportion of <hi>E</hi> to <hi>F</hi> in the second order. Then I say that <hi>A</hi> hath to <hi>C</hi>
                     <figure/>
a greater propor<g ref="char:EOLunhyphen"/>tion then hath <hi>D</hi> to <hi>F.</hi> This per<g ref="char:EOLhyphen"/>taineth to pro<g ref="char:EOLhyphen"/>portion of equa<g ref="char:EOLhyphen"/>litie. For let <hi>G</hi> be vnto <hi>C,</hi> as <hi>D</hi> is to <hi>E.</hi> And by the first part of the 10. of this boke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>G</hi> shal be lesse the<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> And therfore by the second part of the 8. of the same, the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>A</hi> to <hi>G</hi> is grea<g ref="char:EOLhyphen"/>ter then the proportion of <hi>A</hi> to <hi>B.</hi> Wherfore <hi>A</hi> hath vnto <hi>G</hi> a much greater proportio<g ref="char:cmbAbbrStroke">̄</g> then hath <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> to <hi>F.</hi> Now then let <hi>H</hi> be vnto <hi>G</hi> as <hi>E</hi> is to <hi>F.</hi> And by the first part of the 10. of the same, <hi>A</hi> shalbe greater then <hi>H.</hi> And by the first part of the 8. of the same, the pro<g ref="char:EOLhyphen"/>portion of <hi>A</hi> to <hi>C</hi> is greater then the proportion of <hi>H</hi> to <hi>C.</hi> But by the 23. of the same the proportion of <hi>H</hi> to <hi>C</hi> is as the proportion of <hi>D</hi> to <hi>F</hi> (for <hi>G</hi> is to <hi>C</hi> as <hi>D</hi> is to <hi>E,</hi> and <hi>H</hi> is to <hi>G</hi> as <hi>E</hi> is to <hi>F.</hi>) Wherfore (by the 12. of the same) the proportion of <hi>A</hi> to <hi>C</hi> is greater then the proportion of <hi>D</hi> to <hi>F,</hi> which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The eight Proposition.</head>
                  <p>
                     <hi>If the proportion of the whole to the whole, be greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n the proportion of a part taken away, to a part taken away: the<g ref="char:cmbAbbrStroke">̄</g> shall the proportion of the residue vnto the residue be greater then the proportion of the whole to the whole.</hi>
                  </p>
                  <p>Suppose that there be two quantities <hi>AB</hi> &amp; <hi>C
<figure/>
D</hi>: from which let there be cutte of these magni<g ref="char:EOLhyphen"/>tudes <hi>AE</hi> and <hi>CF:</hi> and let the residue be <hi>EB</hi> and <hi>FD.</hi> And let the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>AB</hi> to <hi>CD</hi> be grea<g ref="char:EOLunhyphen"/>ter then the proportion of <hi>AE</hi> to <hi>CF.</hi> Then I say that the proportion of <hi>EB</hi> to <hi>FD</hi> is greater then the proportion of <hi>AB</hi> to <hi>CD.</hi> For (by the second
<pb facs="tcp:7062:183"/>
of these propositions now added) alternately the proportion of <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> to <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> greater then the proportion of <hi>CD</hi> to <hi>CF.</hi> And therfore by euersion of proportion (by the 5. of the same) the proportion of <hi>AB</hi> to <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is lesse then the proportion of <hi>CD</hi> to <hi>FD.</hi> Wherfore agayne alternately the proportion of <hi>AB</hi> to <hi>CD</hi> is lesse then the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>EB</hi> to <hi>FD</hi>: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The ninth Proposition.</head>
                  <p>
                     <hi>If quantities how many soeuer in one order be compared to as many other in an other or<g ref="char:EOLhyphen"/>der, and if there be a greater proportion of euery one that goeth before to that wherunto it is referred, then of any that followeth to that wherunto it is referred: the proportion of them all taken together vnto all the other taken together, shall be greater, then the proportion of any that followeth to that wherunto it is compared, and also then the proportion of all them taken together to all the other taken together, but shall be lesse then the proportion of the first to the first.</hi>
                  </p>
                  <p>Suppose that there be three quantities in one order, <hi>A, B, C,</hi> &amp; as many other in an o<g ref="char:EOLhyphen"/>ther order <hi>D, E, F.</hi> And let the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>A</hi> to <hi>D</hi> be greater the<g ref="char:cmbAbbrStroke">̄</g> the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>B</hi> to <hi>E,</hi> let also the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>B</hi> to <hi>E,</hi> be greater then the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>C</hi> to <hi>F.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say that the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>ABC</hi> take<g ref="char:cmbAbbrStroke">̄</g> al together, to <hi>DEF</hi> take<g ref="char:cmbAbbrStroke">̄</g> altogether, is greater the<g ref="char:cmbAbbrStroke">̄</g> the pro<g ref="char:EOLhyphen"/>portion of <hi>B</hi>
                     <figure/>
to <hi>E,</hi> and also then the pro<g ref="char:EOLhyphen"/>portion of <hi>C</hi> to <hi>F,</hi> &amp; more ouer the<g ref="char:cmbAbbrStroke">̄</g> the proportion of <hi>B</hi> &amp; <hi>C</hi> take<g ref="char:cmbAbbrStroke">̄</g> together, to <hi>EF</hi> take<g ref="char:cmbAbbrStroke">̄</g> together, but is lesse then the proportio<g ref="char:cmbAbbrStroke">̄</g> of <hi>A</hi> to <hi>D</hi>:<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> For forasmuch as <hi>A</hi> hath to <hi>D</hi> a greater proportio<g ref="char:cmbAbbrStroke">̄</g> the<g ref="char:cmbAbbrStroke">̄</g> hath <hi>B</hi> to <hi>E,</hi> therfore alternately <hi>A</hi> hath to <hi>B</hi> a greater proportion then hath <hi>D</hi> to <hi>E</hi>: wherfore by co<g ref="char:cmbAbbrStroke">̄</g>position <hi>AB</hi> hath to <hi>B</hi> a greater proportio<g ref="char:cmbAbbrStroke">̄</g> the<g ref="char:cmbAbbrStroke">̄</g> hath <hi>DE</hi> to <hi>E.</hi> And againe alternately <hi>AB</hi> hath to <hi>DE</hi> a greater propor<g ref="char:EOLhyphen"/>tion then hath <hi>B</hi> to <hi>E.</hi> Wherefore by the former proposition <hi>A</hi> hath to <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> a greater proportion then hath <hi>AB</hi> to <hi>DE.</hi> And by the same reason may it be proued that hath to <hi>E</hi> a greater proportion then hath <hi>BC</hi> to <hi>EF.</hi> Wherefore <hi>A</hi> hath to <hi>D</hi> a greater proportion then hath <hi>BC</hi> to <hi>EF.</hi> Wherfore alternately <hi>A</hi> hath to <hi>BC</hi> a grea<g ref="char:EOLhyphen"/>ter proportion then hath <hi>D</hi> to <hi>EF,</hi> wherfore by composition <hi>ABC</hi> hath to <hi>BC</hi> a greater proportion then hath <hi>DEF</hi> to <hi>EF.</hi> Wherfore agayne alternately <hi>ABC</hi> hath to <hi>DEF</hi> a greater proportion then hath <hi>BC</hi> to <hi>EF.</hi> Wherefore (by the former proposition) the proportion of <hi>A</hi> to <hi>D</hi> is greater then the proportion of <hi>ABC</hi> to <hi>DEF</hi>: Which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               </div>
            </div>
            <trailer>The end of the fifth booke of Euclides Elementes.</trailer>
         </div>
         <div n="6" type="book">
            <pb n="153" facs="tcp:7062:183"/>
            <head>¶The sixth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>THIS SIXTH BOOKE, is for vse and practise,<note place="margin">
                  <hi>The argument of this sixth booke.</hi>
               </note> a most speciall booke. In it are taught the proporti<g ref="char:EOLhyphen"/>ons of one figure to an other figure, &amp; of their sides the one to the other, and of the sides of one to the sides of an other, likewise of the angles of the one to the angles of the other. Moreouer it teacheth the description of figures like to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>igures geuen, and mar<g ref="char:EOLhyphen"/>ueilous applications of figures to lines, euenly, or with decrease or excesse, with many other Theo<g ref="char:EOLhyphen"/>remes, not onely of the propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>tions of right lined fi<g ref="char:EOLhyphen"/>gures, but also of sectors of circles, with their angles. On the Theoremes and Problemes of this Booke depend for the most part, the compositions of all instrumentes of measuring length, breadth, or de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>es, and also the reason of the vse of the same instrumentes, as of the Geometricall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>quar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>,<note place="margin">
                  <hi>This booke necessary for the vse of in<g ref="char:EOLhyphen"/>strumentes of Geometry.</hi>
               </note> the Scale of the Astrolabe, the quadrant, the staffe, and such other. The vse of which instrumentes, besides all other mechanicall instrumentes of raysing vp, of mouing, and drawing huge thinges incredible to the ignorant, and infinite other ginnes (which likewise haue their groundes out of this Booke) are of wonderfull and vnspeakeable profite, besides the inestimable pleasure which is in them.</p>
            <div type="part">
               <head>Definitions.</head>
               <div type="section">
                  <head>
                     <hi>1.</hi> Like rectiline figures are such,<note place="margin">
                        <hi>The first de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> whose angles are equall the one to the other, and whose sides about the equall angles are proportionall.</head>
                  <p>As if ye take any
<figure/>
two rectiline figures. As for example, two triangles ABC, and DEF: <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> of the one triangle be e<g ref="char:EOLhyphen"/>quall to the angles of the other, namely, if the angle A be equall to the angle D, and the angle B equall to the angle E, &amp; also the an<g ref="char:EOLhyphen"/>gle C equall to the an<g ref="char:EOLhyphen"/>gle F. And moreouer, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the sides which con<g ref="char:EOLhyphen"/>taine the equall angles be proportionall. As if the side AB haue that proportion to
<pb facs="tcp:7062:184"/>
the side BC, wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ch the side DE hath to the side EF, and also if the side BC be vnto the side CA, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he side EF is to the side FD, and mor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ouer if the side CA be to the side AB, as the side FD is to the side DE, then are these two triangles sayd to be like: and so iudge ye of any other kinde of figures. As if in the paralle<g ref="char:EOLhyphen"/>logrammes ABCD and EFGH, the angle A be equall to the angle E, and the angle B equall to the angle F, and the angle C equall to the angle G, and the an<g ref="char:EOLhyphen"/>gle D equall to the angle H. And farthermore, if the side AC haue that propor<g ref="char:EOLhyphen"/>tion to the side CD which the side EG hath to the side GH, and if also the side CD be to the side DB as the side GH is to the side HF, and moreouer, if the side DB be to the side BA as the side HF is to the side FE, and finally, if the side BA be to the side AC as the side FE is to the side EG, then are these parallelo<g ref="char:EOLhyphen"/>grammes like.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The second de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inition.</hi>
                     </note>
                     <hi>2.</hi> Reciprocall figures are those, when the terme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of proporti<g ref="char:EOLhyphen"/>on are both antecedentes and consequentes in either figure.</head>
                  <p>As if ye haue two parallelogrammes
<figure/>
ABCD and EFGH. If the side AB to the side EF, an antecedent of the first figure to a consequent of the second figure, haue mutually the same proportion, which the side EG hath to the side AC an antece<g ref="char:EOLhyphen"/>dent of the second figure to a consequent of the first figure: then are these two figures Reciprocal. They are called of some, fi<g ref="char:EOLhyphen"/>gures of mutuall sides, and that vndoubted<g ref="char:EOLhyphen"/>ly not amisse nor vnaptly.<note place="margin">
                        <hi>Reciprocall figures called mutuall fi<g ref="char:EOLhyphen"/>gures.</hi>
                     </note> And to make thys definition more plaine, <hi>Campane</hi> and <hi>Pestitarius,</hi> and others<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> thus put it: <hi>Reciprocall figures, are when the sides of other <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> mutually proportionall, as in the example and declaration before geuen.</hi> Among the barbarous they are called <hi>Mutekesia,</hi> reseruing still the Ara<g ref="char:EOLhyphen"/>bike worde.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The third de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>
                     <hi>3.</hi> A right line is sayd to be deuided by an extreme and meane proportion, when the whole is to the greater part, as the grea<g ref="char:EOLhyphen"/>ter part is to the lesse.</head>
                  <p>As if the line AB, be so deuided in the point
<figure/>
C, that the whole line AB haue the same pro<g ref="char:EOLhyphen"/>portion, to the greater part thereof, namely, to AC, which the same greater part AC hath to the lesse part therof, namely, to CB, then is the line AB deuided by an extreme and meane proportion. Commonly it is called <hi>a line deuided by proportion ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ing a meane and two extremes.</hi> How to deuide a line in such sort was taught in the 11. Proposition of the second Booke, but not vnder this forme of proportion.</p>
               </div>
               <div type="section">
                  <pb n="154" facs="tcp:7062:184"/>
                  <head>
                     <hi>4.</hi> The alitude of a figure is a perpendicular line drawen from the toppe to the base.<note place="margin">
                        <hi>The fourth definition.</hi>
                     </note>
                  </head>
                  <p>As the altitude or hight of the triangle ABC, is the line AD being drawen perpendicularly from the poynt A, being the toppe or highest part of the triangle to the base therof BC. So likewise in other figures as ye see in the examples here set. That
<figure/>
which here <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ee calleth the alti<g ref="char:EOLhyphen"/>tude or height of a fi<g ref="char:EOLhyphen"/>gure, in the first booke in the 35. Proposi<g ref="char:EOLhyphen"/>tion and certaine other following, he taught to be contayned within two equidi<g ref="char:EOLhyphen"/>stant lines: so that figures to haue one altitude and to be contayned within two e<g ref="char:EOLhyphen"/>quidistant lines, is all one. So in all these examples, if from the highest point of the figure ye draw an equidistant line to the base therof, and then fro<g ref="char:cmbAbbrStroke">̄</g> that poynt draw a perpendicular to the same base that perpendicular is the altitude of the figure.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>5.</hi> A Proportion is said to be made of two proportions or more, when the quantities of the proportions multiplied the one into the other, produce an other quantitie.<note place="margin">
                        <hi>The fifth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>
                  </head>
                  <p>Of addition of proportions, hath bene somewhat sayd in the declaration of the 10. definition of the fift booke: which in substance is all one with that which is here taught by <hi>Euclide.</hi> By the name of quantities of proportions, he vnderstan<g ref="char:EOLhyphen"/>deth the denominations of proportions. So that to adde two proportions toge<g ref="char:EOLhyphen"/>ther, or more, and to make one of them all, is nothyng els, but to multiply their quantities together, that is to multiply euer the denominator of the one by the denominator of the other.<note place="margin">
                        <hi>By the name of quantities is vnderstan<g ref="char:EOLhyphen"/>ded the deno<g ref="char:EOLhyphen"/>minations of proportions.</hi>
                     </note> Thys is true in all kindes of proportion, whether it be of equalitie, or of the greater inequalitie, when the greater quantitie is referred to the lesse: or of the lesse inequalitie, when the lesse quantitie is referred to the grea<g ref="char:EOLhyphen"/>ter: or of them mixed together. If the proportions be like, to adde two together is to double the one, to adde 3. like is to triple the one, and so forth in like propor<g ref="char:EOLhyphen"/>tions, as was sufficiently declared in the declaration of the 10. and 11. definitions of the fift Booke. Where it was shewed, that if there be 3. quantities in like pro<g ref="char:EOLhyphen"/>portion, the proportion of the first to the thyrd, is the proportion of the first to the second doubled: and if there be foure quantities in like proportion; the proportion of the f<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>st <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o the fourth shall be the proportion of the first to the second <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ipled<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which thing how to do was there taught likewyse in proportions vnlike, the proportion of the first extreme to the last is made of all the meane proportions set
<pb facs="tcp:7062:185"/>
betwene them. Suppose three quantities A, B, C, so that let A haue to B sesqui<g ref="char:EOLhyphen"/>altera proportion, namely, 6. to 4. And let B to C haue sesquitertia proportion, namely, 4. to 3.<note place="margin">
                        <hi>Example of this definitio<g ref="char:cmbAbbrStroke">̄</g>.</hi>
                     </note> Now the proportion of A to
<figure/>
C, the first to the thyrd, is made of the propor<g ref="char:EOLhyphen"/>tion of A to B, and of the proportion of B to to C added together. If ye will adde them to<g ref="char:EOLhyphen"/>gether, ye must by this definition multiply the quantitie or denominator of the one, by the qua<g ref="char:cmbAbbrStroke">̄</g>titie or denominator of the other. Ye must first therefore seeke the denominators of these proportions, by the rule before geuen in the declaration of the definitions of the fift Booke. As if ye deuide A by B, namely, 6. by 4, so shall ye haue in the quotient 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for the denominator of the proportion of A to B: likewise if ye deuide B by C, namely, 4. by 3. ye shall haue in the quotient 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for the denominator of the proportion of B to C, now multiply these two denominators 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the one into the other, by the rule before taught, namely, by multiplying the numerator of the one into the nu<g ref="char:EOLhyphen"/>merator of the other, and also the denominator of the one into the denominator of the other: the numerator of 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which is all one, is 3, the denomi<g ref="char:EOLhyphen"/>nator is 2: the numerator of 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which reduced are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is 4,
<figure/>
the denominator is 3: then multiply 3. by 4, numerator by nu<g ref="char:EOLhyphen"/>merator, so haue ye 12. for a new numerator: likewise multiply 2. by 3. denominator by denominator, ye shall produce 6. for a new denominator: so haue you produced 12. and 6, betwene which there is dupla proportion. Which proportion is also be<g ref="char:EOLhyphen"/>twene A and C, namely, 6. to 3, the first quantitie to the third. Wherfore the pro<g ref="char:EOLhyphen"/>portion of A to C is sayd to be made of the proportion of A to B and of the pro<g ref="char:EOLhyphen"/>portion of B to C, for that it is produced of the multiplication of the quantitie or denominator of the one,<note place="margin">
                        <hi>Example in numbers.</hi>
                     </note> into the quantitie or denominator of the other. And so of all others be they neuer so many. As in these examples in numbers here set 2. 3. 15. 18.<note place="margin">2. 3. 15. 18.</note> In this example the lesse numbers are compared to the greater, as in the former the greater were compared to the lesse: the denominator of the proporti<g ref="char:EOLhyphen"/>on of 2. to 3. is 2/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that is, subsesquialtera, the denomination betwene 3. and 15. is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which is all one, that is, subquintupla, betwene 15. and 18. the denomi<g ref="char:EOLhyphen"/>nation of the proportion is 1/6 that is, subsesquiquinta, multiply all these denomi<g ref="char:EOLhyphen"/>nations together: first the numerators: 2. into 1. produce 2, then 2. into 5. produce 10: which shall be a new numerator. Then the de<g ref="char:EOLhyphen"/>nominators:
<figure/>
3. into 5. produce 15: and 15. into 6. produce 90: which shall be a new denominator. So haue you brought forth 10/9<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or 1/9 which is pro<g ref="char:EOLhyphen"/>portion subnoncupla: which is also the proportion of 2. to 18. Wherefore the proportion of 2. to 18. that is, of the extremes, namely, subnoncupla, is made of the proportions of 2. to 3: of 3. to 15: and of 15. to 18: namely, of subsesquialtera, subquintupla, and subses<g ref="char:EOLhyphen"/>quiquinta.</p>
                  <p>An other example, where the greater inequalitie and the lesse inequalitie are mixed together 6. 4. 2. 3. the denomination of the proportion of 6. to 4, is 1 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>,<note place="margin">
                        <hi>An other example.</hi>
                     </note> of 4. to 2, is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, and of 2. to 3, is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: now if ye multiply as you ought, all these denominations together, ye shall produce 12. to 6, namely, dupla proportion.</p>
                  <p>Forasmuch as so much hath hetherto bene spoken of addition of proportions
<pb n="155" facs="tcp:7062:185"/>
it shall not be vnnecessary somewhat also to say of substraction of them.<note place="margin">
                        <hi>Of substracti<g ref="char:EOLhyphen"/>on of propor<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> Where it is to be noted, that as addition of them, is made by multiplicatio<g ref="char:cmbAbbrStroke">̄</g> of their denomi<g ref="char:EOLhyphen"/>nations the one into the other: so is the substraction of the one from the other done, by diuision of the denomination of the one by the denomination of the o<g ref="char:EOLhyphen"/>ther. As if ye will from sextupla proportion subtrahe dupla proportion, take the denominations of them both. The denomination of sextupla proportion, is 6, the denomination of dupla proportion, is 2. Now deuide 6. the denomination of the one by 2. the denomination of the other: the quotient shall be 3: which is the denomination of a new proportion, namely, tripla: so that when dupla pro<g ref="char:EOLhyphen"/>portion is subtrahed from sextupla, there shall remayne tripla proportion. And thus may ye do in all others.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>6.</hi> A Parallelogramme applied to a right line, is sayd to want in forme by a parallelogramme like to one geuen: whe<g ref="char:cmbAbbrStroke">̄</g> the pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>me applied wanteth to the filling of the whole line, by a parallelogramme like to one geuen:<note place="margin">
                        <hi>The sixth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note> and then is it sayd to exceede, when it exceedeth the line by a parallelogramme like to that which was geuen.</head>
                  <p>As let E be a Parallelogra<g ref="char:cmbAbbrStroke">̄</g>me
<figure/>
geuen, and let AB be a right line, to whom is applied the pa<g ref="char:EOLhyphen"/>rallelogramme ACDF. Now if it want of the filling of the line AB, by the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DFGB being like to the pa<g ref="char:EOLhyphen"/>rallelogramme geuen E, then is the parallelogramme sayd to want in forme by a parallelo<g ref="char:EOLhyphen"/>gramme like vnto a parallelogramme geuen.</p>
                  <p>Likewise if it exceede, as the parallelogramme ACGD applyed to the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> if it exceede it by the
<figure/>
parallelogramme FGBD being like to the parallelo<g ref="char:EOLhyphen"/>gramme F which was ge<g ref="char:EOLhyphen"/>uen, then is the parallelo<g ref="char:EOLhyphen"/>gramme ABGD, sayd to exceede in forme by a pa<g ref="char:EOLhyphen"/>rallelogramme like to a pa<g ref="char:EOLhyphen"/>rallelogramme geuen.</p>
                  <p>This definition is added by <hi>Flussates</hi> as it seemeth, it is not in any co<g ref="char:cmbAbbrStroke">̄</g>mon Greke booke abroad, nor in any Commentary. It is for many Theoremes following very necessary.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:186"/>
               <head>The <hi>1.</hi> Theoreme. The <hi>1.</hi> Proposition. Triangles &amp; parallelogrammes which are vnder one &amp; the self same altitude: are in proportion as the base of the one is to the base of the other.</head>
               <p>SVppose that there be two triangles ABC and ACD, &amp; two paralleiogrammes EC and CF. Which let be set vn<g ref="char:EOLhyphen"/>der one and the selfe same altitude, or perpendicular line drawen from the toppe A to the base BD. Then I say that as the base BC is to the base CD so is the triangle ABC to the triangle ACD: and the parallelogramme EC to the parallelogramme CF. For forasmuch as the parallelogrammes.<note place="margin">Construction.</note> E
<figure/>
C and CF are vnder one and the selfe same altitude, therfore the lines EA and AF make both one right line, and so also do the lines BC and CD: and therefore the lyne EF is a parallel vnto the line BD. Produce the right line DB on <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>che side directly to the pointes H, L (by the <hi>2.</hi> peticion of the first) And vnto the base BC (by the <hi>2.</hi> of the first) put as many equall lines as you will, as for example, two, namely, BG, and GH and vnto the base CD on the other side put as many equall as you did to the other base, which let be DK, and KL. Then draw these right lines AG, AH, AK, and AL.</p>
               <p>And forasmuch as the lines CB, BG, and GH are equall the one to the o<g ref="char:EOLhyphen"/>ther,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</note> therfore the triangles also AHG, AGB and ABC, are (by the <hi>38.</hi> of the first) equall the one to the other. Wherfore how multiplex the base HC is to the base BC, so multiplex also is the triangle AHC to the triangle ABC. And by the same reason also, how multiplex the base LC is to the base DC, so multi<g ref="char:EOLhyphen"/>plex also is the triangle ALC to the triangle ADC. Wherfore if the base HC be equall vnto the base CL, then (by the <hi>38.</hi> of the first) the triangle AHC is equall vnto the triangle ACL. And if the base HC exceede the base CL, then also the triangle AHC excedeth the triangle ACL, and if the base be lesse, the tria<g ref="char:cmbAbbrStroke">̄</g>gle also shall be lesse. Now then there are foure magnitudes, namely, the two bas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s BC and CD, and the two triangles ABC, and ACD, and to the base BC, and to the triangle ABC, namely, to the first and the third, are taken eque<g ref="char:EOLhyphen"/>mul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iplices,
<pb n="156" facs="tcp:7062:186"/>
namely, the base HC and the triangle AHC, and likewise to y<hi rend="sup">e</hi> base CD and to the triangle ADC, namely, to the second and the fourth, are taken certaine other equemultiplices, that is, the base CL, and the triangle ALC. And it hath bene proued that if the multiplex of the first magnitude, that is, the base HC, do exceede the multiplex of the second, that is, the base CL, the multiplex also of the third, that is, the triangle AHC excedeth the multiplex of y<hi rend="sup">e</hi> fourth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that is, the triangle ALC, and if the said base HC be equall to the said ba<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> CL, the triangle also AHC is equall to the triangle ALC, and if it be lesse it i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> lesse. Wherfore by the sixt defini<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ion of y<hi rend="sup">e</hi> fifth, as the first of the foresaid magni<g ref="char:EOLhyphen"/>tudes is to the second, so is the third to the fourth. Wherfore as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.</p>
               <p>And because (by the <hi>41.</hi> of the first) the parallelogramme EC is double to the triangle ABC,<note place="margin">Demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>a<g ref="char:EOLhyphen"/>tion of the second part.</note> and (by the same) the parallelogramme FC is double to the triangle ACD, therfore the parallelogrammes EC and FC are equemultipli<g ref="char:EOLhyphen"/>ces vnto the triangles ABC and ACD. But the partes of equemultiplices (by the <hi>15.</hi> of the fifth) haue one and the same proportion with thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equemultiplices. Wherfore as the triangle ABC is to the triangle ACD, so is the parallelograme EC to the parallelogramme FC. And forasmuch as it hath bene demonstrated, that as the base BC is to the base CD, so is the triangle ABC, to the triangle ACD, and as the triangle ABC is to the triangle ACD so is the parallelo<g ref="char:EOLhyphen"/>gramme EC to the parallelogramme FC. Wherefore (by the <hi>11.</hi> of the fifth) as the base BC is to the base CD, so is the parallelogramme EC to the parallelo<g ref="char:EOLhyphen"/>gramme FC. The parallelogrammes may also be demonstrated a part by them<g ref="char:EOLhyphen"/>selues as the triangles are, if we describe vpon the bases BG, GH, and DK &amp; KL parallelogrammes vnder the self same altitude that the parallelogramme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> geuen are. Wherfore triangles and parallelogrammes which are vnder one and the selfe same altitude, are in proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, as the base of the one is to y<hi rend="sup">e</hi> base of the other: which was required to be demonstrated.</p>
               <div type="section">
                  <head>Here Flussates addeth this Corollary.</head>
                  <p>
                     <hi>If two right lines being geuen, the one of them be deuided how so euer:</hi>
                     <note place="margin">
                        <hi>A Corollary added by Flussates.</hi>
                     </note> 
                     <hi>the rectangle fi<g ref="char:EOLhyphen"/>gures contayned vnder the whole line vndeuided, and eche of the segmentes of the line deui<g ref="char:EOLhyphen"/>ded, are in proportion the one to the other, as the segmentes are the one to the other.</hi> For i<g ref="char:EOLhyphen"/>maginyng the figures BA and AD in the former description, to be rectangled, the rectangle figures contayned vnder the whole right lyne AC, and the segments of the right line BD, which is cu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in the poynt C, namely, the parallelogrammes BA and AD, are in proportion the one to the other, as the segme<g ref="char:cmbAbbrStroke">̄</g>tes BC and CD are.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Theoreme. The <hi>2.</hi> Proposition. If to any one of the sides of a triangle be drawen a parallel
<pb facs="tcp:7062:187"/>
right line, it shall cut the sides of the same triangle proportio<g ref="char:EOLhyphen"/>nally. And if the sides of a triangle be cut proportionally, a right lyne drawn from section to section is a parallel to the o<g ref="char:EOLhyphen"/>ther side of the triangle.</head>
               <p>SVppose that there be a triangle ABC, vnto one of the sides whereof, namely, vnto BC, let there be drawen a parallel line DE cuttyng the sides AC and AB in the pointes E and D. Then I say first that as BD is to DA, so is CE to EA.<note place="margin">The first part of this Theo<g ref="char:EOLhyphen"/>reme.</note> Draw a line from B to E, &amp; also from C to D. Wher<g ref="char:EOLunhyphen"/>fore (by the <hi>37.</hi> of the first) the triangle BDE is equall vnto the triangle CDE: for they are set vpon one and the same base DE, and are contained within the selfe same parallels DE and BC. Consider
<figure/>
also a certaine other triangle ADE. Now thinges equall (by the <hi>7.</hi> of the fifth) haue to one selfe thing one and the same proportion. Wherfore as y<hi rend="sup">e</hi> triangle BDE is to y<hi rend="sup">e</hi> triangle ADE, so is y<hi rend="sup">e</hi> triangle CDE to the triangle ADE. But as y<hi rend="sup">e</hi> triangle BDE is to y<hi rend="sup">e</hi> trian<g ref="char:EOLhyphen"/>gle ADE, so is y<hi rend="sup">e</hi> base BD to y<hi rend="sup">e</hi> base DA (by the first of this booke.) For they are vnder one and the selfe same toppe, namely, E, and ther<g ref="char:EOLhyphen"/>fore are vnder one and the same altitude. And by the same reason as the triangle CDE is to the triangle ADE, so is the lyne CE to the lyne EA. Wherfore (by the <hi>11.</hi> of the fifth) as the line BD is to the line DA, so is the line CE to the line EA.</p>
               <p>But now suppose that in y<hi rend="sup">e</hi> triangle ABC the sides AB &amp; AC be cut propor<g ref="char:EOLhyphen"/>tionally so y<hi rend="sup">t</hi> as BD is to DA, so let CE be to EA, &amp; draw a line from D to E. Then secondly I say y<hi rend="sup">t</hi> the line DE is a parallel to y<hi rend="sup">e</hi> lyne BC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part.</note> For the same order of construction being kept, for y<hi rend="sup">t</hi> as BD is to DA, so is CE to EA, but as BD is to DA, so is y<hi rend="sup">e</hi> triangle BDE to y<hi rend="sup">e</hi> triangle ADE (by the <hi>1.</hi> of the sixt) &amp; as CE is to EA, so (by y<hi rend="sup">e</hi> same) is the triangle CDE to y<hi rend="sup">e</hi> triangle ADE: therfore (by the <hi>11.</hi> of the fifth) as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. Wherfore either of these triangles BDE and CDE haue to the triangle ADE one and the same proportion. Wherefore (by the <hi>9.</hi> of the fifth) the triangle BDE is equall vnto the triangle CDE, and they are vpon one and the selfe base, namely, DE. But triangles equall and set vpon one base, are also contained within the same parallel lines (by the <hi>39.</hi> of the first.) Wherfore the line DE is vnto the line BC a parallel. If therfore to a<g ref="char:EOLhyphen"/>ny one of the sides of a triangle be drawn a parallel line, it cutteth the other sides of the same triangle proportionally. And if the sides of a triangle be cut propor<g ref="char:EOLhyphen"/>tionally, a right lyne drawen from section to section, is parallel to the other side of the triangle: which thing was required to be demonstrated.</p>
               <div type="section">
                  <pb n="157" facs="tcp:7062:187"/>
                  <head>¶ Here also Flussates addeth a Corollary.</head>
                  <p>
                     <hi>If a line parallel to one of the sides of a triangle do cut the triangle, it shall cut of from the whole triangle a triangle like to the whole triangle.</hi>
                     <note place="margin">
                        <hi>A Corollary added by Flussates.</hi>
                     </note> For as it hath bene proued it deui<g ref="char:EOLhyphen"/>deth the sides proportionally. So that as EC is to EA, so is BD to DA, wherfore by the 18. of the fifth, as AC is to AE, so is AB to AD. Wherfore alternately by the 16. of the fifth as AC is to AB, so is AE to AD: wherefore in the two trian<g ref="char:EOLhyphen"/>gles EAD and CAB the sides about the common angle A are proportional. The sayd triangles also are equiangle. For forasmuch as the right lynes AEC and ADB do fall vpon the parallel lynes ED and CB, therefore by the 29. of the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> they make the angles AED and ADE in the triangle ADE equall to the angles ACB and ABC in the triangle ACB. Wherefore by the first definition of this booke the whole triangle ABC is like vnto the triangle cut of ADE.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Theoreme. The <hi>3.</hi> Proposition. If an angle of a triangle be deuided into two equall partes, and if the right line which deuideth the angle deuide also the base: the segmentes of the base shall be in the same proportion the one to the other, that the other sides of the triangle are. And if the segme<g ref="char:cmbAbbrStroke">̄</g>tes of the base be in the same proportion that the other sides of the sayd triangle are: a right drawen from the toppe of the triangle vnto the section, shall deuide the an<g ref="char:EOLhyphen"/>gle of the triangle into two equall partes.</head>
               <p>SVppose that there be a triangle ABC, and (by the <hi>9.</hi> of the first) let the angle BAC be deuided into two equall partes by the right lyne AD, which let cut also the base BC in the point D. Then I say that as the segment BD is to y<hi rend="sup">e</hi> segment DC, so is the side BA to the side AC.<note place="margin">Construction.</note> For by the point C (by the <hi>31.</hi> of the first) draw
<figure/>
vnto the line DA a parallel line CE and extende the line BA till it concurre with the line CE in the point E, and do make the triangle BEC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</note> But the lyne BA shall concurre with the line CE (by the <hi>5.</hi> peticion) for that the angles EBC, and BCE are lesse then two right angles. For the angle ECB is equall to the outwarde and opposite angle ADB (by the <hi>29.</hi> of the first.) And the two angles ADB and DBA of the triangle BAD are lesse then two right angles (by the <hi>17.</hi> of the first) Now for<g ref="char:EOLhyphen"/>asmuch as vpon the parallels AD and EC falleth the right line AC, therefore
<pb facs="tcp:7062:188"/>
by the <hi>29.</hi> of the first) the angle ACE is equall vnto the angle CAD. But vn<g ref="char:EOLhyphen"/>to the angle CAD is the angle BAD supposed to be equall. Wherfore the an<g ref="char:EOLhyphen"/>gle BAD is also equall vnto the angle ACE. Againe because vpon the paral<g ref="char:EOLhyphen"/>lels AD and EC falleth the right line BAE, the outward angle BAD (by the <hi>28.</hi> of the first) is equall vnto the inward angle AEC. But before it was pro<g ref="char:EOLhyphen"/>uell that y<hi rend="sup">e</hi> angle ACE is equall vnto y<hi rend="sup">e</hi> angle BAD, wherfore y<hi rend="sup">e</hi> angle ACE is equall vnto y<hi rend="sup">e</hi> angle AEC. Wherefore (by y<hi rend="sup">e</hi> 6. of y<hi rend="sup">e</hi> first) y<hi rend="sup">e</hi> side AE is equall vnto the side AC. And because to one of y<hi rend="sup">e</hi> sides of y<hi rend="sup">e</hi> triangle BCE, namely, to EC is drawen a parallel line AD, therfore (by y<hi rend="sup">e</hi> 
                  <hi>2.</hi> of y<hi rend="sup">e</hi> sixt) as BD is to DC, so is BA to AE. But AE is equall vnto AC, therfore as BD is to DC, so is BA to AC. But now suppose that as the segment BD is to the
<figure/>
segment DC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n of the second part, which is the conuerse of the first.</note> so is the side BA to the side AC, &amp; draw a line from A to D. Then I say that the an<g ref="char:EOLhyphen"/>gle BAC is by y<hi rend="sup">e</hi> right line AD deuided into two equall partes. For the same order of construction re<g ref="char:EOLhyphen"/>mayning, for that as BD is to DC, so is BA to AC, but as BD is to DC, so is BA to AE (by the <hi>2.</hi> of the sixt) for vnto one of the sides of the triangle BCE, namely, vnto the side EC is drawn a parallel line AD. Wherefore also as BA is to AC, so is BA to AE (by the <hi>11.</hi> of the fifth) Wher<g ref="char:EOLhyphen"/>fore (by the <hi>9.</hi> of the fifth) AC is equall vnto AE. Wherfore also (by the <hi>5.</hi> of the first) the angle AEC is equall vnto the angle ACE, but the angle AEC (by the <hi>29.</hi> of the first) is equall vnto the outward angle BAD: and the angle ACE is equall vnto the angle CAD which is alternate vnto him: wherefore the angle BAD is equall vnto the angle CAD. Wherfore the angle BAC is by the right line AD deuided into two equall partes. Wherefore if an angle of a triangle be deuided into two equall partes, and if the right line which deuideth the angle cut also the base, the segmentes of the base shall be in the same propor<g ref="char:EOLhyphen"/>tion the one to the other, that the other sides of the said triangle are. And if the segmentes of the base be in the same proportion that the other sides of the sayd triangle are, a right line drawen from the toppe of the triangle vnto the section deuideth the angle of the triangle into two equall partes.</p>
               <p>
                  <hi>This construction is the halfe part of that Gnomical figure described in the 43. proposition of the first booke, which Gnomical figure is of great vse in a maner in all Geometrical demonstrations.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Theoreme. The <hi>4.</hi> Proposition. In equiangle triangles, the sides which co<g ref="char:cmbAbbrStroke">̄</g>taine the equall an<g ref="char:EOLhyphen"/>gles are proportionall, and the sides which are subtended vn<g ref="char:EOLhyphen"/>der the equall angles are of like proportion.</head>
               <p>
                  <pb n="158" facs="tcp:7062:188"/>SVppose that there be two equiangle triangles ABC and DCE: and let the angle ABC of the one triangle, be equall vnto y<hi rend="sup">e</hi> angle DCE of the other triangle, and the angle BAC equall vnto y<hi rend="sup">e</hi> angle CDE, and moreouer, the angle ACB equall vnto the angle DEC. Then I say, that those sides of y<hi rend="sup">e</hi> triangles ABC, &amp; DCE, which include the equall angles, are proportionall, and the side which are subtended vnder the equall an<g ref="char:EOLhyphen"/>gles are of like proportion.<note place="margin">Construction.</note> For let two sides of the sayd triangles, namely, two of those sides which are subtended vnder equall angles: as for example the sides BC and CE, be so set that they both make one right line. And because the an<g ref="char:EOLhyphen"/>gles ABC &amp; ACB are lesse then two right angles (by the <hi>17.</hi> of the first): but the angle ACB is e<g ref="char:EOLhyphen"/>quall
<figure/>
vnto the angle DEC: therfore y<hi rend="sup">e</hi> an<g ref="char:EOLhyphen"/>gles ABC &amp; DEC are lesse the<g ref="char:cmbAbbrStroke">̄</g> two right angles. Wherefore the lines BA &amp; ED be<g ref="char:EOLhyphen"/>ing produced, will at the length meete toge<g ref="char:EOLhyphen"/>ther. Let them meete and ioyne together in the poynt F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And because by supposition the angle DCE is equall vnto the an<g ref="char:EOLhyphen"/>gle ABC, therfore the line BF is (by the 28. of the first) a parallell vnto t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e line CD. And forasmuch as by supposition the angle ACB is equall vnto the angle DEC, therefore againe (by the <hi>28.</hi> of the first) the line AC is a parallell vnto the line FE. Wherefore FADC is a parallelogramme. Wherfore the side FA is equall vnto the side DC: and the side AC vnto the side FD (by the 34. of the first). And because vnto one of the sides of the triangle BFE, namely, to FE is drawen a parallell line AC, therefore as BA is to AF, so is BC to CE (by the <hi>2.</hi> of the sixt). But AF is equall vnto CD. Wherfore (by the <hi>11.</hi> of the fift) as BA is to CD, so is BC to CE, which are sides subtended vnder equall angles. Wherefore alternately (by the <hi>16.</hi> of the fift) as AB is to BC, so is DC to CE. Againe forasmuch as CD is a parallell vnto BF, therefore againe (by the <hi>2.</hi> of the sixt) as BC is to CE, so is FD to DE. But FD is equall vnto AC. Wherefore as BC is, to CE, so is AC to DE, which are also sides sub<g ref="char:EOLhyphen"/>tended vnder equall angles. Wherfore alternately (by the <hi>16.</hi> of the fift) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s BC is to CA, so is CE to ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore forasmuch as it hath bene demonstrated, that as AB is vnto BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is DC vnto CE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> but as DC is vnto CA, so is CE vnto ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> it followeth of equalitie (by the <hi>22.</hi> of the fift) that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s BA is vnto AC so is CD vnto DE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore in eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iangle triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, y<hi rend="sup">e</hi> sides which include y<hi rend="sup">e</hi> equall angles are proportionall: and y<hi rend="sup">e</hi> sides which are subt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nded vnder the equall angles are of like proportion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hich was required to be demonstrated.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:189"/>
               <head>The <hi>5.</hi> Theoreme. The <hi>5.</hi> Proposition. If two triangles haue their sides proportionall, the triang<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s are equiangle, and those angles in the<g ref="char:cmbAbbrStroke">̄</g> are equall, vnder which are subtended sides of like proportion.</head>
               <p>SVppose that there be two triangles ABC, &amp; DEF, hauing their sides proportionall, as AB is to BC, so let DE be to EF:<note place="margin">This is the conuerse of the former proposition.</note> &amp; as BC is to AC, so let EF be to DF: and moreouer, as BA is to AC, so let ED be to DF. Then I say, that the triangle ABC is equiangle vnto the triangle DEF: and those angles in them are equall vnder which are subtended sides of like pro<g ref="char:EOLhyphen"/>portion, that is, the angle ABC is equall vnto the angle DEF: and the angle BCA vnto the angle EFD: and moreouer, the angle BAC to y<hi rend="sup">e</hi> angle EDF. Vpon the right line EF,<note place="margin">Construction</note> and vnto the pointes in it E &amp; F, describe (by the 23. of the first) angles equall vnto the angles ABC &amp; ACB, which let be FEG and EFG, namely, let the angle FEG be equall vnto the angle ABC, and let the angle EFG be equall to the angle ACB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the angles ABC and ACB are lesse then two right angles (by the 17. of the first): therefore also the angles FEG and EFG are lesse then two right angles. Wherefore (by the 5. petition of y<hi rend="sup">e</hi> first) y<hi rend="sup">e</hi> right lines EG &amp; FG shall at y<hi rend="sup">e</hi> length concurre. Let the<g ref="char:cmbAbbrStroke">̄</g> concurre in the poynt G. Wherefore EFG is a triangle. Wherefore the angle remayning BAC is equall vnto the angle remay<g ref="char:EOLhyphen"/>ning
<figure/>
EGF (by the first Corollary of the 32. of the first). Wherfore the triangle ABC is equiangle vn<g ref="char:EOLhyphen"/>to the triangle GEF. Wherefore in the triangles ABC and EGF the sides, which include the equall angles (by the 4. of the sixt) are proportionall, and the sides which are subtended vnder the equall an<g ref="char:EOLhyphen"/>gles are of like proportion. Wherefore as AB is to BC, so is GE to EF. But as AB is to BC, so by supposition is DE to EF. Wherefore as DE is to EF, so is GE to EF (by the 11. of the fift). Where<g ref="char:EOLhyphen"/>fore either of these DE and EG haue to EF one and the same proportion. Wherefore (by the 9. of the fift) DE is equall vnto EG. And by the same rea<g ref="char:EOLhyphen"/>son also DF is equall vnto FG. Now forasmuch as DE is equall to EG and EF is common vnto them both, therefore these two sides DE &amp; EF, are equall vnto these two sides GE and EF, and y<hi rend="sup">e</hi> base DF is equall vnto the base FG. Wherefore the angle DEF (by the 8. of the first) is equall vnto the angle GEF: and the triangle DEF (by the <hi>4.</hi> of the first) is equall vnto the triangle GEF: and the rest of the angles of the one triangle
<pb n="159" facs="tcp:7062:189"/>
are equall vnto the rest of the angles of the other triangle the one to y<hi rend="sup">e</hi> other, vn<g ref="char:EOLhyphen"/>der which are subtended equall sides. Wherefore the angle DFE is equall vn<g ref="char:EOLhyphen"/>to the angle GFE: and the angle EDF vnto the angle EGF. And because<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the angle FED is equall vnto the angle GEF: but the angle GEF is equall vnto the angle ABC: therefore the angle ABC is also equall vnto the angle FED. And by the same reason the angle ACB is equall vnto y<hi rend="sup">e</hi> angle DFE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and moreouer, the angle BAC vnto the angle EDF. Wherefore the triangle ABC is equiangle vnto the triangle DEF. If two triangles therefore haue their sides proportionall, the triangles shall be equiangle, &amp; those angles in them shall be equall, vnder which are subtended sides of like proportion: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Theoreme. The <hi>6.</hi> Proposition. If there be two triangles wherof the one hath one angle equall to one angle of the other, &amp; the sides including the equall an<g ref="char:EOLhyphen"/>gles be proportionall: the triangles shall be equiangle, and those angles in them shall be equall, vnder which are subten<g ref="char:EOLhyphen"/>ded sides of like proportion.</head>
               <p>SVppose that there be two triangles ABC, and DEF, which let haue the angle BAC of the one triangle equall vnto the angle EDF of the other triangle, and let the sides including the equall angles be proportio<g ref="char:EOLhyphen"/>nall, that is, as BA is to AC, so let ED be to DF. Then I say, that y<hi rend="sup">e</hi> triangle ABC is equiangle vnto the triangle DEF: and the angle ABC is equall vn<g ref="char:EOLhyphen"/>to the angle DEF, and the angle ACB equall vnto the angle DFE, which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ngles are subte<g ref="char:cmbAbbrStroke">̄</g>ded to sides of like pro<g ref="char:EOLhyphen"/>portion.
<figure/>
Vnto the right line DF, and to the poynt in it D (by the 23. of the first) describe vnto either of y<hi rend="sup">e</hi> angles BAC and EDF,<note place="margin">Constructio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> an equall angle FDG. And vnto the right line DF, and vnto the point in it F (by y<hi rend="sup">e</hi> same) describe vnto y<hi rend="sup">e</hi> angle AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> an equall angle DFG. And forasmuch as the two angles BAC and ACB, are (by the 17. of the first) lesse then two right angles: therefore also the two angles EDG and DFG, are lesse then two right angles. Wherfore y<hi rend="sup">e</hi> lines DG &amp; FG being produced, shall co<g ref="char:cmbAbbrStroke">̄</g>curre (by the 5. petition). Let the<g ref="char:cmbAbbrStroke">̄</g> concurre in the point G. Wherefore DFG is a triangle.
<pb facs="tcp:7062:190"/>
Wherefore the angle remaining ABC is equall vnto the angle remaining DGF (by the 32. of the first). Wherefore the triangle ABC is equiangle vnto the tri<g ref="char:EOLhyphen"/>angle DGF. Wherefore as BA is in proportion to AC, so is GD to DF (by the 4. of the sixt). But it is supposed, that as BA is to AC, so is ED to DF. Wherefore (by the 11. of the fift) as ED is to DF, so is GD to DF. Where<g ref="char:EOLhyphen"/>fore (by the 9. of the fift) ED is equall vnto DG. And DF is common vnto them both. Now then there are two
<figure/>
sides ED and DF equall vnto two sides GD and DF: and the angle EDF (by supposition) is equall vnto the angle GDF. Wherefore (by the 4. of the first) the base EF is equall vnto the base GF, and the triangle DEF is (by the same) equall vnto the triangle GDF, and the other an<g ref="char:EOLhyphen"/>gles remayning in them are equall the one to the other, vnder which are sub<g ref="char:EOLhyphen"/>tended equall sides. Wherefore the an<g ref="char:EOLhyphen"/>gle DFG is equall vnto the angle DFE: and the angle DGF vnto the angle DEF. But the angle DFG is (by construction) equall vnto y<hi rend="sup">e</hi> angle ACB. And the angle DGF is as it hath bene proued, equall to y<hi rend="sup">e</hi> angle ABC. Wherfore also y<hi rend="sup">e</hi> angle ACB is equall vnto the angle DFE. And y<hi rend="sup">e</hi> angle ABC is equall to the angle DEF. But by sup<g ref="char:EOLhyphen"/>position the angle BAC is equall vnto the angle EDF. Wherefore the tri<g ref="char:EOLhyphen"/>angle ABC is equiangle vnto the triangle DEF. If therefore there be two triangles, whereof the one hath one angle equall to one angle of the other, and if also the sides including the equall angles be proportionall: then shall the triangles also be equiangle, and those angles in them shalbe equall, vnder which are subtended sides of like proportion: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Theoreme. The <hi>7.</hi> Proposition. If there be two tria<g ref="char:cmbAbbrStroke">̄</g>gles, wherof the one hath one angle equal to one angle of the other, and the sides which include the other angles, be proportionall, and if either of the other angles re<g ref="char:EOLhyphen"/>mayning be either lesse or not lesse then a right angle: the<g ref="char:cmbAbbrStroke">̄</g> shal the triangles be equiangle, and those angles in them shall be equall, which are contayned vnder the sides proportionall.</head>
               <p>
                  <pb n="160" facs="tcp:7062:190"/>SVppose that there be two triangles ABC and DEF, which let haue one angle of the one, equall to one angle of the other, namely, the angle BAC equall vnto the angle EDF. And let the sides which include the other angles, namely, the angles ABC and DEF be proportio<g ref="char:EOLhyphen"/>nall, so that as AB is to BC, so let DE be to EF. And let the other angles re<g ref="char:EOLhyphen"/>mayning, namely, ACD and DFE be first either of them lesse then a right angle. Then I say that the triangle ABC is equiangle vnto the triangle DEF. And that the angle ABC is equall vnto the an<g ref="char:EOLhyphen"/>gle
<figure/>
DEF,<note place="margin">The first part of this proposition.</note> namely, the angles which are contai<g ref="char:EOLhyphen"/>ned vnder the sides proportionall, and that the angle remayning, namely, y<hi rend="sup">e</hi> angle C is equall vn<g ref="char:EOLhyphen"/>to the angle remayning, namely, to y<hi rend="sup">e</hi> angle F. For first the angle ABC is either equall to the an<g ref="char:EOLhyphen"/>gle DEF, or els vnequall. If the angle ABC be equall to the angle DEF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> then the angle remai<g ref="char:EOLhyphen"/>ning, namely, ACB, shall be equall to the angle remayning DFE (by the corollary of the 32. of the first) And therfore the triangles ABC and DEF are equiangle. But if the angle ABC be vnequall vnto the angle DEF, then is the one of them grea<g ref="char:EOLunhyphen"/>ter then the other. Let the angle ABC be the greater, and vnto the right line AB and vnto the point in it B (by the 23. of the first) describe vnto the angle DEF an equall angle ABG. And forasmuch as the angle A is equall vnto the angle D, and the angle ABG is equall vnto the angle DEF, therfore the an<g ref="char:EOLhyphen"/>gle remayning AGB is equall vnto the angle remayning DFE (by the corol<g ref="char:EOLhyphen"/>lary of the 32. of the first. Wherfore the triangle ABG is equiangle vnto the tri<g ref="char:EOLhyphen"/>angle DEF. Wherfore (by the 4. of the sixth) as the side AB is to the side BG, so is the side DE to the side EF. But by suppposition the side DE is to the side EF, as the side AB is to the side BC. Wherfore (by the 11. of the fifth) as the side AB is to the side BC so is the same side AB to the side BG. Wher<g ref="char:EOLhyphen"/>fore AB hath to either of these BC and BG one and the same proportion, and therfore (by the 9. of the fifth) BC is equall vnto BG. Wherefore (by the 5. of the first) y<hi rend="sup">e</hi> angle BGC is equall vnto y<hi rend="sup">e</hi> angle BCG: but by supposition y<hi rend="sup">e</hi> angle BCG is lesse then a right angle. Wherfore the angle BGC is also lesse then a right angle. Wherfore (by the 13. of the first) the side angle vnto it, namely, AGB is greater then a right angle, and it is already proued that the same angle is equall vnto the angle F. Wherfore the angle F is also greater then a right angle. But it is supposed to be lesse which is absurde. Wherefore the angle ABC is not vnequall vnto the angle DEF, wherfore it is equall vnto it. And the angle A is equall vnto the angle D by supposition. Wherfore the angle remayning, name<g ref="char:EOLhyphen"/>ly, C, is equall vnto the angle remayning, namely, to F (by the corollary of the 32. of the first) Wherfore the triangle ABC is equiangle vnto the triangle DEF.<note place="margin">The second part of this proposition.</note>
               </p>
               <p>But now suppose that either of the angles ACB and DFE be not lesse then
<pb facs="tcp:7062:191"/>
a right angle. That is, let either of them be a right angle, or either of them grea<g ref="char:EOLhyphen"/>ter then a right angle. Then I say againe that in that case also the triangle ABC is equiangle vnto the triangle DEF. For if either of them be a right angle, forasmuch as all right angles are (by the 4. peticion) equall the one to the other, straight way will follow the intent of the proposition. But if either of them be greater then a right angle, then the same order of construction that was before being kept, we may in like sort proue that the side BC is equall vnto the side BG. Wherfore also the angle BCG is equall vnto the angle BGC. But the angle BCG is greater then a right angle. Wherfore also the angle BGC is greater the<g ref="char:cmbAbbrStroke">̄</g> a right angle. Wherfore two angles of the triangle BGC are greater then two right angles: which (by the 17. of the first) is impossible. Wherfore the angle ABC is not vnequall vnto the angle DEF. And therfore is it equall: but the angle A is equall vnto the angle D (by supposition) Wher<g ref="char:EOLhyphen"/>fore
<figure/>
the angle remayning, namely, C is equal vnto the angle remayning, namely, to F (by the corollary of the 32. of the first). Wherfore the triangle ABC is equi<g ref="char:EOLhyphen"/>angle vnto the triangle DEF. If there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore there be two triangles whero<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the one hath one angle equall to one angle o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the other, and the sides which include the other angle be proportionall, and if either of the other angles remaining be either lesse or not lesse then a right angle, the triangle shall be equiangle, and those angles in them shall be equall which are contained vnder sides proportionall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>8.</hi> Theoreme. The <hi>8.</hi> Proposition. If in a rectangle triangle be drawen from the right angle vn<g ref="char:EOLhyphen"/>to the base a perpendicular line, the perpendicular line shall deuide the triangle into two triangles like vnto the whole, and also like the one to the other.</head>
               <p>
                  <q>SVppose that there be a rectangle triangle ABC, whose right angle let be BAC: and (by the 12. of the first) from the point A to the line BC let there be drawen a perpendicular line AD,<note place="margin">Construction.</note> which perpendi<g ref="char:EOLhyphen"/>cular line let deuide the whole triangle ABC into these two trian<g ref="char:EOLhyphen"/>gles ABD and ADC. (Note that this perpendicular line AD, drawen from the right angle to the base, must needes fall within the triangle ABC, &amp; so deuide the triangle into two triangles. For if it should fall without, then pro<g ref="char:EOLhyphen"/>ducing the side BC vnto the perpendicular line, there should be made a triangle, whose outward angle being an acute angle, should be lesse then the inward and opposite angle which is a right angle: which is co<g ref="char:cmbAbbrStroke">̄</g>trary to the 16. of the first. Nei<g ref="char:EOLhyphen"/>ther
<pb n="161" facs="tcp:7062:191"/>
can it fall vpon any of the sides AB or AC for th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n two angles of one and the selfe same triangle should not be lesse than <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wo right angles, contrary to the selfe same 17. of the first. Wherefore it fall<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>h within the triangle ABC)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </q>
Then I say, that either of these trian<g ref="char:EOLhyphen"/>gles
<figure/>
ABD and ADC, are like vnto the whole triangle ABC, and more<g ref="char:EOLhyphen"/>ouer, that they are like the one to the o<g ref="char:EOLhyphen"/>ther. First that the triangle ABD is like vnto the whole triangle ABC is thus proued.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Forasmuch as (by the 4. petition) the angle BAC is equall vn<g ref="char:EOLhyphen"/>to the angle ADB, for either of them is a right angle. And in the two triangles ABC and ABD the angle B is common. Wherefore the angle remayning, namely, ACB is (by the Corollary of the 32. of the first) equall vnto the angle remayning, namely, to BAD. Wher<g ref="char:EOLhyphen"/>fore the triangle ABC is equiangle vnto the triangle ABD. Wherefore the sides which containe the equall angles, are (by the 4. of the sixt) proportionall. Wherfore as the side CB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which subtendeth y<hi rend="sup">e</hi> right angle of the triangle ABC, is vnto the side BA which subtendeth the right angle of the triangle ABD, so is the same side AB which subtendeth the angle C of the triangle ABC, vn<g ref="char:EOLhyphen"/>to the side BD which subtendeth y<hi rend="sup">e</hi> angle BAD of the triangle ABD, which is equall vnto the angle C: and moreouer, the side AC vnto the side AD which subtend the angle B common to both the triangles. Wherfore the triangle ABC is like vnto the triangle ABD (by the 1. definition of the sixt). In like maner also may we proue, that the triangle ADC is like vnto the triangle ABC. For the right angle ADC is equall to the right angle BAC, &amp; the angle at the point C is common to either of those triangles. Wherefore the angle remayning, namely, DAC is equall to the angle remaining<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> namely, to ABC (by y<hi rend="sup">e</hi> Corolla<g ref="char:EOLhyphen"/>ry of the 32. of the first). Wherefore the triangles ABC &amp; ADC are equian<g ref="char:EOLhyphen"/>gle. And therefore (by the 4. of the sixt) the sides which are about the equall an<g ref="char:EOLhyphen"/>gles are proportionall. Wherefore as in the triangle ABC the side BC is to the side CA, so in the triangle ADC is the side AC to the side DC: and againe, as in the triangle ABC the side CA is to the side AB, so in the triangle ADC is the side CD to the side DA. And moreouer, as in the triangle ABC the side CB is to the side BA, so in the triangle ADC is the side CA to the side AD. Wherefore the triangle ADC is like vnto the whole triangle ABC. Wherfore either of these triangles ABD, &amp; ADC is like vnto y<hi rend="sup">e</hi> whole triangle ABC.</p>
               <p>I say also, that the triangles ABD and ADC are like the one to the other. For forasmuch as the right angle BDA is equall vnto y<hi rend="sup">e</hi> right angle ADC (by the 4. petition) and as it hath already bene proued, the angle BAD is equall vn<g ref="char:EOLhyphen"/>to the angle C: therefore the angle remayning, namely, B is equall vnto the an<g ref="char:EOLhyphen"/>gle remayning, namely, to DAC (by the Corollary of the 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first). Wher<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:192"/>
the triangle ABD is equiangle vnto the triangle ADC. Wherefore as the side BD which in the triangle ABD subtendeth the angle BAD is vnto the side DA which in the triangle ADC subtendeth the angle C which is e<g ref="char:EOLhyphen"/>quall vnto the angle BAD, so is the side AD which in y<hi rend="sup">e</hi> triangle ABD sub<g ref="char:EOLhyphen"/>tendeth the angle B, vnto the side DC which in the triangle ADC subtendeth the angle DAC which is equall vnto the angle B: and moreouer, so is the side BA vnto the side AC which subtende the right angles. Wherefore the triangle ABD is like vnto the triangle ADC. If therefore in a rectangle triangle be drawen from the right angle vnto the base a perpendicular line, the perpendicu<g ref="char:EOLhyphen"/>lar line shall deuide the triangle into two triangles like vnto the whole, and also like the one to the other: which was required to be proued.</p>
               <div type="section">
                  <head>Corollary.</head>
                  <p>Hereby it is manifest, that if in a rectangle triangle be draw<g ref="char:EOLhyphen"/>en from the right angle vnto the base a perpendicular line, the same line drawen is a meane proportionall betwene the secti<g ref="char:EOLhyphen"/>ons of the base: and moreouer, betwene the whole base and ei<g ref="char:EOLhyphen"/>ther of the sections, the side annext to the sayd section is the meane proportionall. For is was proued, that as CD is to DA, so is DA to DB: and moreouer, as CB is to BA, so is BA to BD: and finally, as BC is to CA, so is CA to CD.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>1.</hi> Probleme. The <hi>9.</hi> Proposition. A right line being geuen, to cut of fro<g ref="char:cmbAbbrStroke">̄</g> it any part appointed.</head>
               <p>LEt the right line geuen be AB. It is requi<g ref="char:EOLhyphen"/>red
<figure/>
that from the same line AB be cut of any part appoynted. Suppose that a thyrd part be appoynted to be cut of.<note place="margin">Construction.</note> From y<hi rend="sup">e</hi> point A draw a right line AC making with the line AB an angle: and in the line AC take a poynt at all aduentures, which let be D. And beginning at D put vnto AD two equall lines DE &amp; EC (by the 2. of the first). And draw a right line from B to C: and by the point D (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>1. of the first) draw vnto BC a pa<g ref="char:EOLhyphen"/>rallell line DF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as vnto one of y<hi rend="sup">•</hi> sides of the triangle ABC, namely, vnto y<hi rend="sup">e</hi> side BC is draw<g ref="char:EOLhyphen"/>en a parallell line FD, it followeth by y<hi rend="sup">e</hi> 2. of this booke,
<pb n="162" facs="tcp:7062:192"/>
that as CD is in proportion vnto DA, so is BF to FA. But (by construction) CD is double to DA. Wherefore y<hi rend="sup">•</hi> line BF is also double to the line FA. Wher<g ref="char:EOLhyphen"/>fore the line BA is treble vnto the line AF. Wherfore from the right line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> geuen AB, is cut of a third part appoynted, namely, AF: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>2.</hi> Probleme. The <hi>10.</hi> Proposition. To deuide a right line geue<g ref="char:cmbAbbrStroke">̄</g> not deuided, like vnto a right line geuen beyng deuided.</head>
               <p>SVppose that the right line geuen not deuided be AB, and the right lyne geuen being deuided, let be AC. It is required to deuide y<hi rend="sup">e</hi> line AB which is not deuided like vnto the line AC which is deuided.<note place="margin">Construction.</note> Suppose the lyne AC be deuided in the pointes D and E, &amp; let y<hi rend="sup">•</hi> lines AB &amp; AC so be put, that they make an angle at all aduentures, and draw a line from B to C, and by the pointes D and E draw vnto the line BC (by the 31. of the first) two parallel lines DF and EG: and by the point D vnto
<figure/>
the line AB (by the same) draw a parallel line DHK. Wherfore either of these figure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> FH and HB are parallelogrammes. Wherfore the line DH is equall vnto the line FG, and the line HK is equall vnto the line GB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And be<g ref="char:EOLhyphen"/>cause to one of the sides of the triangle DKC, namely, to the side KC is drawn a paral<g ref="char:EOLhyphen"/>lel line HE, therefore the line CE (by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 2. of the sixt) is in proportion vnto the line ED as the line KH is to the line HD: but the line KH is equall vnto the line BG, and the line HD is equal vnto the line GF. Wher<g ref="char:EOLunhyphen"/>fore (by the 11. of the fift) as CE is vnto ED, so is BG to GF. Agayne because to one of the sides of the triangle AGE, namely, to GE is drawn a parallel lyne FD, therfore the line ED (by the 2. of the sixth) is in proportion vnto the lyne DA, as the line GF is to the line FA. And it is already proued that as CE is to ED, so is BG to GF. VVherfore as CE is to ED, so is BG to GF, and as ED is to DA, so is GF to FA. VVherfore the right line geuen not deuided, name<g ref="char:EOLunhyphen"/>ly, AB is deuided like vnto the right line geuen being deuided, which is AC: which was required to be done.</p>
               <div type="section">
                  <head>¶ A Corollary out of Flussates.<note place="margin">
                        <hi>A Corollary out of Flussates.</hi>
                     </note>
                  </head>
                  <p>By this Proposition we may deuide any right line geuen, accordyng to the pro<g ref="char:EOLhyphen"/>portion
<pb facs="tcp:7062:193"/>
of any right lynes geuen. For let those right lynes hauyng proportion be ioyned together directly, that they may make all one right lyne, and then ioyne them to the lyne geuen anglewise. And so proceede as in the proposition, where you see that the right line geuen AB is deuided into the right lynes AF, FG and GB which haue the selfe same proportion that the right lines AD, DE, and EC haue.</p>
                  <p>By this and the former proposition also may a right line geuen be easily deui<g ref="char:EOLhyphen"/>ded into what partes so euer you will name.<note place="margin">
                        <hi>By this and the former propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ition may a right line be deui<g ref="char:EOLhyphen"/>ded into what partes soeuer you will.</hi>
                     </note> As if you will deuide the line AB in<g ref="char:EOLhyphen"/>to three equall partes, let the lyne DE be made equall to the lyne AD, and the lyne EC made equall to the same by the third of the first. And then vsing the selfe same maner of construction that was before: the lyne AB shall be deuided into three equall partes. And so of any kynde of partes whatsoeuer.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>3.</hi> Probleme. The <hi>11.</hi> Proposition. Vnto two right lines geuen, to finde a third in proportion with them.</head>
               <p>SVppose that there be two right lines geuen BA and AC, and let them be so put that they comprehend an angle howsoeuer it be. It is required to finde vnto BA and vnto AC a third line in proportion.<note place="margin">Construction.</note> Produce y<hi rend="sup">e</hi> lynes AB and AC vnto the pointes D and E. And vnto the
<figure/>
line AC (by the 2. of the first) put an equall line BD, and draw a lyne from B to C. And by the pointe D (by the 31. of the first) draw vnto the lyne BC a parallel lyne DE, which let concurre with the line AC in the point E.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as vnto one of the sides of the triangle ADE, namely, to DE is drawne a parallel line BC: therfore as AB is in proportion vnto BD, so (by the 2. of the sixt) is AC vnto CE. But the lyne BD is equall vnto the line AC. VVherfore as the lyne AB is to the line AC, so is the line AC to the line CE. VVherfore vnto the two right lines geuen AB and AC is found a third line CE in proportio<g ref="char:cmbAbbrStroke">̄</g> with them: which was required to be done.</p>
               <div type="section">
                  <head>¶ An other way after Pelitarius.</head>
                  <p>Let the lines AB and BC be set directly in such sort that they both make one right line.<note place="margin">
                        <hi>An other way after Pelita<g ref="char:EOLunhyphen"/>rius.</hi>
                     </note> Then fro<g ref="char:cmbAbbrStroke">̄</g> the point A erect the lyne AD makyng with the lyne AB an angle at all aduentures. And put the lyne AD equall to the lyne BC. And draw a right line from D to B which produce beyond the poynt B vnto the point E. And by the point C draw vnto the lyne DA a parallel lyne CE concurring with the lyne DE in the point E. Then I say that the line CE is the third line proportionall with the lines AB and BC. For for<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asmuch
<pb n="163" facs="tcp:7062:193"/>
as by the 15. of the first, the angle B of the
<figure/>
triangle ABD is equall to the angle B of the tri<g ref="char:EOLhyphen"/>angle CBE, and by the 29. of the same, the angle A is equall to the angle C, and the angle D to the angle E: therefore by the 4. of this booke AB is to DA, as BC is to CE. Wherfore (by the 11. of the fifth) AB is to BC as BC is to CE: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>¶ An other way also after Pelitarius.</head>
                  <p>Let the lines AB and BC be so ioyned toge<g ref="char:EOLhyphen"/>ther,<note place="margin">
                        <hi>An ot<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> way after Pelitarius.</hi>
                     </note>
                     <figure/>
that they may make a right angle, namely, ABC. And drawe a line from A to C, and from the point C drawe vnto the line AC a perpendicular CD (by the 11. of the first) And produce the lyne CD till it concurre with the line AB produced vn<g ref="char:EOLhyphen"/>to the pointe D. Then I say that the line BD is a third lyne proportionall with the lines AB and BC: which thing is manifest by the corollary of the 8. of this booke.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>4.</hi> Probleme. The <hi>12.</hi> Proposition. Vnto three right lines geuen to finde a fourth in proportion with them.</head>
               <p>SVppose that the three right lines geuen be A, B, C. It is required to finde vnto A, B, C, a fourth line in proportio<g ref="char:cmbAbbrStroke">̄</g> with them. Let there be taken two right lines DE &amp; DF comprehending an angle as it shall happen, namely, EDF.<note place="margin">Construction.</note> And (by the 2. of the first) vnto the line A put an equall line DG. And vn<g ref="char:EOLhyphen"/>to the line B (by the same)
<figure/>
put an equall line GE. And moreouer, vnto y<hi rend="sup">e</hi> line C put an equall line DH. Then draw a line fro<g ref="char:cmbAbbrStroke">̄</g> G to H. And by the poynt E (by the 31. of the first) draw vnto the line GH a parallell line EF. Now forasmuch as vnto one of the sides of the triangle DEF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> namely, vnto y<hi rend="sup">•</hi> side EF is drawen a parallell line GH: therefore (by the 2. of the sixt) as the line DG is to the line GE, so is the line DH to the line HF.
<pb facs="tcp:7062:194"/>
But the line DG is equall vnto the line A, and the line GE is equall vnto the line B, and the line DH vnto the line C. Wherfore as the line A is vnto the line B, so is the line C vnto the line HF. Wherfore vnto the three right lines geuen A, B, C, is found a fourth line HF in proportion with them: which was requi<g ref="char:EOLhyphen"/>red to be done.</p>
               <div type="section">
                  <head>¶ An other way after Campane.</head>
                  <p>Suppose that there be three right lines AB, BC, and BD. It is required to adde vn<g ref="char:EOLhyphen"/>to them a fourth line in proportion with them.<note place="margin">
                        <hi>An other way after Campane.</hi>
                     </note> Ioyne AB the first, with BD the third, in such sort that they both make one right line, namely, AD. And vpon the said lyne AB erect from the point B the second
<figure/>
line BC making an angle at all aduen<g ref="char:EOLhyphen"/>tures. And draw a line from A to C. Then by the point D draw the lyne DE parallel to the line AC, which pro<g ref="char:EOLhyphen"/>duce vntill it concurre in the point E, with the line CB being likewise pro<g ref="char:EOLhyphen"/>duced to the point E. Then I say that the line BE is the fourth line in pro<g ref="char:EOLhyphen"/>portion with the lines AB, BC, and BD: so that as AB is to BC, so is BD to BE. For forasmuch as by the 15 and 29. of the first the two triangles ABC and DBE are equiangle, therfore (by the 4. of this booke) AB is to BC, as BD is to BE: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>5.</hi> Probleme. The <hi>13.</hi> Proposition. Vnto two right lines geuen, to finde out a meane proportionall.</head>
               <p>SVppose the two right lines geuen to be AB and BC. It is required be<g ref="char:EOLhyphen"/>twene th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se two lines AB and BC to finde out a meane line proportio<g ref="char:EOLhyphen"/>nall. Let the lines AB and BC be so ioyned together that they both make one right line, namely, AC.<note place="margin">Construction.</note> And vp<g ref="char:EOLhyphen"/>on
<figure/>
the line AC describe a semicircle ADC, and from the poynt B raise vp vnto the line AC (by the 11. of the first) a perpendicular line BD, cut<g ref="char:EOLhyphen"/>ting the circumference in the point D: and draw a line from A to D, and an other from D to C. Now forasmuch as (by the 31. of the third) the angle in the semicircle ADC is a right angle,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> and for that in the rectangle triangle ADC is drawen from the right angle vnto the base a perpendicular line DB: therefore (by the Corollary of the 8. of the sixt) the line DB is a meane proportionall betwene the segme<g ref="char:cmbAbbrStroke">̄</g>tes of the base AB &amp; BC. Wherefore betwene the two right lines geuen, AB &amp; BC, is found a meane proportionall DB: which was required to be done.</p>
               <div type="section">
                  <pb n="164" facs="tcp:7062:194"/>
                  <head>¶ A Proposition added by Pelitarius.</head>
                  <p>
                     <hi>A meane proportionall beyng geuen, to finde out in a line geuen the two extremes.</hi>
                     <note place="margin">
                        <hi>A proposition added by Pe<g ref="char:EOLhyphen"/>litarius.</hi>
                     </note> 
                     <hi>Now it behoueth that the meane geuen be not greater then the halfe of the lyne geuen.</hi>
                  </p>
                  <p>Suppose that the meane geuen be <hi>AB,</hi> and let the right line geuen be <hi>BC.</hi> It is re<g ref="char:EOLhyphen"/>quired in the line <hi>BC</hi> to finde out two extremes, betwene which <hi>AB</hi> shal be the mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportionall. So that yet the lyne AB be not greater then the halfe part of the line <hi>BC.</hi> For so could it not be a meane. Ioyne the lines <hi>AB</hi> and <hi>BC</hi> directly in such sort, that they both make one right line, name<g ref="char:EOLunhyphen"/>ly,
<figure/>
                     <hi>AC.</hi> Then vppon the line <hi>BC</hi> de<g ref="char:EOLhyphen"/>scribe the semicircle <hi>BEC.</hi> And from the point <hi>A</hi> erect vnto the lyne <hi>AC</hi> a perpendicular line <hi>AD</hi>: which lyne <hi>AD</hi> put equal vnto the line <hi>AB.</hi> And by the point <hi>D</hi> draw vnto the line <hi>AC</hi> a parallel line <hi>DE,</hi> which vndoub<g ref="char:EOLhyphen"/>tedly shall either cut or touch the se<g ref="char:EOLhyphen"/>micircle, as in the point E, for that the line <hi>AD</hi> is not greater then the semidiameter. Then from the point <hi>E</hi> draw vnto the line <hi>BC</hi> a perpendicular line <hi>EF</hi> (by the 12. of the first) Then I say that the line <hi>BC</hi> is so deuided in the point <hi>F,</hi> that the lyne <hi>AB</hi> is a meane proportionall betwene the lines <hi>BF</hi> and <hi>FC.</hi> Which thing is manifest (by the 31. of the third) &amp; corollary of the 8. of this booke. For the line <hi>FE</hi> is equal to the line <hi>AD</hi> by the 34. of the first, and so is equall to the line <hi>AB</hi>: then if we draw the ryght lines <hi>BE</hi> and <hi>CE,</hi> there shall be made a rectangle triangle <hi>BEC.</hi> And so by the sayd co<g ref="char:EOLhyphen"/>rollary, the line <hi>BF</hi> shall be to the lyne <hi>FE</hi> (and therfore to the line <hi>AB</hi>) as the line <hi>FE</hi> is to the lyne <hi>FC</hi>: which was required to be done.</p>
                  <p>Flussates putteth this Proposition added by Pelitarius as a corollary following of this 13. proposition.</p>
               </div>
            </div>
            <div type="part">
               <head>The 9. Theoreme. The <hi>14.</hi> Proposition. In equall parallelogrammes which haue one angle of the one equall vnto one angle of the other, the sides shall be recipro<g ref="char:EOLhyphen"/>kall, namely, those sides which containe the equall angles. And if parallelogrammes which hauing one angle of the one, equal vnto one angle of the other, haue also their sides reciprokal, namely, those which contayne the equall angles, they shall al<g ref="char:EOLhyphen"/>so be equall.</head>
               <p>SVppose that there be two equall Parallelogrammes AB and BC, hauing the angle B of the one equall vnto the angle B of the other.<note place="margin">The first part of this propo<g ref="char:EOLhyphen"/>sition.</note> And let the lines DB and DE be set directly in such sort that they both make one right line, namely, DE. And then (by the <hi>14.</hi> of the first) shall the lines FB and BG be so set that they shall make also one right line, namely, GF. Then I say, that the sides of the parallelogrammes AB and BC, which containe the equall
<pb facs="tcp:7062:195"/>
angles, are reciprocally proportionall: that is, as BD is to BE, so is GB to BF. Make complete the parallelogramme FE by producing the sides AF and CE, till they concurre in the poynt H. Now forasmuch as the parallelogramme AB is (by supposition) equall vnto the parallelogramme BC, and there is a certaine other parallelogramme FE:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the of the same.</note> therfore (by the <hi>7.</hi> of the fift) as the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AB is to the parallelogramme FE, so
<figure/>
is the parallelogramme BC to the pa<g ref="char:EOLhyphen"/>rallelogramme FE. But as the paralle<g ref="char:EOLhyphen"/>logramme AB is to y<hi rend="sup">e</hi> parallelogramme FE, so is the side DB to the side BE (by the first of this booke). And (by the same) as the parallelogramme BC is to the parallelogramme FE, so is the side GB to the side BF. Wherefore also (by the 11. of the fift) as the side DB is to the side BE, so is the side GB to the side BF. Wherefore in the parallelogrammes AB and BC the sides which con<g ref="char:EOLhyphen"/>taine the equall angles, are reciprokally proportionall: which was first required to be proued.</p>
               <p>But now suppose that the sides about the equall angles be reciprokally propor<g ref="char:EOLhyphen"/>tionall so that as the side DB is to the side BE,<note place="margin">The second part which is the conuerse of the first.</note> so let the side GB be to the side BF. Then I say, y<hi rend="sup">t</hi> the parallelogramme AB is equall vnto y<hi rend="sup">e</hi> parallelogramme BC. For, for that as the side DB is to the side BE, so is the side GB to the side BF: but as the side DB is to the side BE, so (by the 1. of the sixt) is the paralle<g ref="char:EOLhyphen"/>logramme AB to the parallelogramme FE: and as the side GB is to the side BF, so is the parallelogramme BC to the parallelogramme FE. Wherefore also (by the 11. of the fift) as the parallelogramme AB is to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me FE, so is the parallelogramme BC to the same parallelogramme FE. Wherefore the parallelogramme AB is equall vnto the parallelogramme BC (by the 9. of the fift). Wherefore in equall and equiangle parallelogrammes the sides which con<g ref="char:EOLhyphen"/>taine the equall angles are reciprokall: and if in equiangle parallelogrammes the sides which containe the equall angles be reciprokall, the parallelogrammes also shall be equall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Theoreme. The <hi>15.</hi> Proposition. In equal triangles which haue one angle of the one equall vn<g ref="char:EOLhyphen"/>to one angle of the other , those sides are reciprokal, which in<g ref="char:EOLhyphen"/>clude the equall angles. And those tria<g ref="char:cmbAbbrStroke">̄</g>gles which hauyng one angle of the one equall vnto one angle of the other, haue also
<pb n="165" facs="tcp:7062:195"/>
their sides which include the equall angles reciprokal, are al<g ref="char:EOLhyphen"/>so equall.</head>
               <p>SVppose that there be two equall triangles ABC, and ADE hauing one angle of the one equall vnto one angle of the other, namely, the angle BAC equall vnto the angle DAE. Then I say that in those triangles ABC and ADE, the sides which include y<hi rend="sup">e</hi> equal angles, are reciprokallie propor<g ref="char:EOLhyphen"/>tionall,<note place="margin">The first par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this propo<g ref="char:EOLhyphen"/>sition.</note> that is, as the side CA is to the side AD, so is the side EA to the side AB. For let y<hi rend="sup">e</hi> lines CA and AD be so put, y<hi rend="sup">t</hi> they both make directly one right line. And so also the lines EA and AB
<figure/>
shal both make one right line (by the 14. of the first) And draw a line from B to D. Now forasmuch as (by supposition) the triangle ABC is equall vnto the triangle ADE.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the same.</note> And there is a certaine other tri<g ref="char:EOLhyphen"/>angle BAD, vnto which the two equall triangles being compared, it will follow by the 7. of the fifth, that as the triangle ABC is vnto the triangle BAD, so is y<hi rend="sup">e</hi> trian<g ref="char:EOLhyphen"/>gle EAD to the same triangle BAD. But as the triangle ABC is to the tria<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>gle BAD, so by the 1. of the sixth, is the base CA to the base AD: and as the triangle EAD is to the triangle BAD, so (by the same) is the base EA to the base AB. Wherfore (by the 11. of the fifth) as the side CA is to the side AD, so is the side EA to the side AB. Wherefore in the triangles ABC and ADE the sides which include the equall angles are reciprokally proportionall.</p>
               <p>But now suppose that in the triangles ABC and ADE, the sides which in<g ref="char:EOLhyphen"/>clude the equall angles, be reciprokally proportionall,<note place="margin">The second part which is the conuerse of the first.</note> so that as the side CA is to the side AD, so let the side EA be to the side AB. Then I say that the triangle ABC is equall vnto the triangle ADE. For agayne draw a line from B to D. And for that as the line CA is to the line AD, so is the line EA to the line AB, but as the line CA is to the line AD, so is the triangle ABC to the triangle BAD, and as the line EA is to the line AB so is the triangle EAD to the tri<g ref="char:EOLhyphen"/>angle BAD. Wherfore as the triangle ABC is to the triangle BAD so is the triangle EAD to y<hi rend="sup">e</hi> same triangle BAD. Wherfore either of these triangles ABC and EAD haue vnto y<hi rend="sup">•</hi> triangle BAD one and y<hi rend="sup">•</hi> selfe same proportion. Wher<g ref="char:EOLunhyphen"/>fore (by the 9. of the fifth) the triangle ABC is equal vnto the triangle EAD. If therfore there be taken equall triangles hauyng one angle of the one equal vn<g ref="char:EOLhyphen"/>to one angle of the other, those sides in them shal be reciprokal, which include the equal angles. and those triangles which hauing one angle of the one equall vnto one angle of the other, haue also their sides which include the equall angles reci<g ref="char:EOLhyphen"/>prokal, shal also be equall: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:196"/>
               <head>The <hi>11.</hi> Theoreme. The <hi>16.</hi> Proposition. If there be foure right lines in proportion, the rectangle figure comprehended vnder the extremes: is equall to the rectangle figure contayned vnder the meanes. And if the rectangle fi<g ref="char:EOLhyphen"/>gure which is contained vnder the extremes, be equall vnto the rectangle figure which is contayned vnder the meanes: then are those foure lines in proportion.</head>
               <p>SVppose that there b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> foure right lines in proportio<g ref="char:cmbAbbrStroke">̄</g>, namely, AB, CD, E, and F: so that as the line AB is to the line CD, so let the line E be to the line F. Then I say, that the rectangle figure comprehended vn<g ref="char:EOLhyphen"/>der the extremes AB and F,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</note> is equall vnto the rectangle figure con<g ref="char:EOLhyphen"/>tayned vnder the meanes CD and E. From the poynt A (by the 11. of the first) raise vp vnto the right line AB a perpendicular line AG. And (by the same) from the point C vnto the right line CD raise vp a perpendicular line CH. And (by the 2. of the first) put the line AG equall vnto the line F, and put also y<hi rend="sup">e</hi> line CH equall vnto the line E, and make complete the parallelogrammes GB and HD. Now for that by supposition as the line AB is to the line CD, so is the line E to the line F. But the line E is equall vnto the line CH, &amp; the line F vn<g ref="char:EOLhyphen"/>to the line AG, therefore as the line AB is to the line CD, so is the line CH to the line AG. Wherefore in the
<figure/>
parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes BG and DH the sides which include y<hi rend="sup">e</hi> equall angles, are reciprokally proportionall. But equiangle parallelogrammes whose sides which include the equall an<g ref="char:EOLhyphen"/>gles, are reciprokall, are also equall (by the 14. of the sixt). Wherefore the parallelogramme BG is equall vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DH. But the parallelogramme BG is that which is contayned vnder the lines AB and F, for the line AG is put equall vnto the line F. And the parallelogramme DH is that which is contained vnder the lines CD and E, for the line CH is put equall vnto the line E. Where<g ref="char:EOLunhyphen"/>fore the rectangle figure contained vnder the lines AB and F, is equall vnto the rectangle figure contayned vnder the lines CD and E.<note place="margin">The second part which is the conuerse of the first.</note>
               </p>
               <p>But now suppose that the rectangle figure comprehended vnder the lines AB and F, be equall vnto the rectangle figure co<g ref="char:cmbAbbrStroke">̄</g>prehended vnder y<hi rend="sup">e</hi> lines CD &amp; E.
<pb n="166" facs="tcp:7062:196"/>
Then I say, that the foure right lines AB, CD, E and F, are proportionall, that is, as the line AB is to the line CD, so is the line E to the line F. The same or<g ref="char:EOLhyphen"/>der of construction that was before being kept, forasmuch as that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AB and F is equall vnto that which is contained vnder the lines CD and E, but that which is contayned vnder the lines AB and F is the parallelogramme BG, for the line AG is equall vnto the line F. And that also which is contained vnder the lines CD &amp; E is the parallelogramme DH, for the line CH is equall vnto the line E. Wherefore the parallelogramme BG is equall vnto the parallelogramme DH, &amp; they are also equiangle. But in pa<g ref="char:EOLhyphen"/>rallelogrammes equall &amp; equiangle the sides which include the equall angles are reciprokall (by the 14. of the sixt). Wherfore as the line AB is to the line CD, so is the line CH to the line AG, but the line CH is equall vnto the line E, and the line AG is equall vnto y<hi rend="sup">e</hi> line F. Wherefore as the line AB is to the line CD, so is the line E to the line F. If therefore there be foure right lines in proportion, the rectangle figure comprehended vnder the extremes, is equall to the rectangle figure contayned vnder the meanes. And if the rectangle figure which is contai<g ref="char:EOLhyphen"/>ned vnder the extremes, be equall vnto the rectangle figure which is contained vnder the meanes, then are those foure lines in proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>12.</hi> Theoreme. The <hi>17.</hi> Proposition. If there be three right lines in proportion, the rectangle figure comprehended vnder the extremes, is equall vnto the square that is made of the meane. And if the rectangle figure which is made of the extremes, be equal vnto the square made of the meane, then are those three right lines proportional.</head>
               <p>SVppose that there be three lines in proportion A, B, C, so that as A is to B, so let B be to C. Then I say that the rectangle figure comprehended vnder y<hi rend="sup">e</hi> lines A and C is equall vnto y<hi rend="sup">e</hi> square made of the line B.<note place="margin">The first part of th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> Theoreme.</note> Vnto the line B (by the 2. of the
<figure/>
first) put an equall line D. And because by supposition) as A is to B, so is B to C, but B is equall vnto D, where<g ref="char:EOLhyphen"/>fore (by the 7. of the fifth) as A is to B, so is D to C, but if there be foure right lines pro<g ref="char:EOLunhyphen"/>portionall, the rectangle fi<g ref="char:EOLhyphen"/>gure comprehended vnder
<pb facs="tcp:7062:197"/>
the extremes is equall vnto the rectangle figure comprehended vnder the meanes (by the 16. of the sixt). Wherfore that which is contained vnder the lines A and C is equall vnto that which is comprehended vnder the lines B and D. But that which is contained vnder the lines B and D, is the square of the line B, for the line B is equall vnto the line D. Wherfore the rectangle figure comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vn<g ref="char:EOLhyphen"/>der the lines A and C is equall vnto the square made of the lyne B.</p>
               <p>
                  <note place="margin">The second part which is the conuerse of the first.</note>But now suppose that that
<figure/>
which is comprehended vn<g ref="char:EOLhyphen"/>der the lines A &amp; C be equal vnto the square made of the line B. Then also I say, that as the line A is to the line B, so is the line B to the lyne C. The same order of constru<g ref="char:EOLhyphen"/>ction that was before, beyng kept, forasmuch as y<hi rend="sup">t</hi> which is contained vnder the lynes A and C is equall vnto the square which is made of the line B. But the square which is made of the line B is that which is contained vnder y<hi rend="sup">e</hi> lines B &amp; D, for the line B is put equall vnto the line D. Wherefore that which is contayned vn<g ref="char:EOLhyphen"/>der the lines A and C is equall vnto that which is contayned vnder the lines B and D. But if the rectangle figure comprehended vnder the extremes, be equall vnto the rectangle figure comprehended vnder the meane lynes, the foure right lines shall be proportionall (by the 16. of the sixth) Wherfore as the line A is to the line B, so is the line D to the line C. But the line B is equall vnto the lyne D. Wherfore as the line A is to the lyne B, so is B to the line C. If therefore there be three right lynes in proportion, the rectangle figure comprehended vnder the ex<g ref="char:EOLhyphen"/>tremes, is equall vnto the square that is made of the meane. And if the rectangle figure which is contayned vnder the extremes, be equall vnto the square made of the meane, then are those three right lines proportional: which was required to be demonstrated.</p>
               <div type="section">
                  <head>
                     <note place="margin">A Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ollary.</note>¶ Corollary added by Flussates.</head>
                  <p>Hereby we gather that euery right lyne is a meane proportionall betwene euery two right lines which make a rectangle figure equall to the square of the same right lyne.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>6.</hi> Probleme. The <hi>18.</hi> Proposition. Vpon a right line geuen, to describe a rectiline figure like, and in like sort situate vnto a rectiline figure geuen.</head>
               <p>
                  <pb n="167" facs="tcp:7062:197"/>SVppose that the ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ht line geuen be AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let the rectiline figure ge<g ref="char:EOLhyphen"/>uen be EG. It is required vpon the right line geuen AB to describe a rectiline figure like, and in like sort situate vnto the rectiline figure ge<g ref="char:EOLhyphen"/>uen GE. Drawe a line from H to F, and vnto the right line AB and to y<hi rend="sup">e</hi> p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t in it A,<note place="margin">Description of the rectiline figure r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>qui<g ref="char:EOLhyphen"/>red.</note> make vnto the angle E an equall angle DAB (by the 23. of the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) and vnto the right line AB and vnto the point in it B (by the same) make vn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the angle EFH an equall angle ABD. Wherefore y<hi rend="sup">e</hi> angle remayning EHF is equall vnto the angle remayning ADB. Wherefore the triangle HEF is equiangle vnto the triangle DAB. Wherefore (by the 4. of the sixt) as the side HF is in proportion
<figure/>
to the side DB, so is the side HE to the side DA, and y<hi rend="sup">•</hi> side EF to the side AB. Againe (by the 23. of the first) vnto the right line BD and vnto y<hi rend="sup">e</hi> point in it D, make vnto the angle FHG an equall an<g ref="char:EOLhyphen"/>gle BDC, and (by the same) vnto the right line BD and vnto the point in <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> B, make vnto the angle HFG an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall angle DBC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the angle remayning, namely, G, is equall vnto the angle remayning, namely, to C. Where<g ref="char:EOLhyphen"/>fore the triangle HFG is equiangle vnto the triangle DBC. Wherefore (by the 4. of the sixt) as the side HF is in proportion to the side DB, so is the side HG to the side DC, and the side GF to the side CB. And it is already proued that as HF is to DB, so is HE to DA, and EF to AB. Wherefore (by the 11. of the fift) as EH is to AD, so is EF to AB, and HG to DC, and more<g ref="char:EOLhyphen"/>ouer, GF to CB. And forasmuch as the angle EHF is equall vnto the angle ADB, and the angle FHG is equall vnto the angle BDC: therefore the whole angle EHG is equall vnto the whole angle ADC, and by the same rea<g ref="char:EOLhyphen"/>son the angle EFG is equall vnto the angle ABC. But (by construction) the angle E is equall vnto the angle A, and the angle G is proued equall vnto the an<g ref="char:EOLhyphen"/>gle C. Wherefore the figure AC is equiangle, vnto the figure EG, and those sides which in it include the equall angles are proportionall, as we haue before proued. Wherefore the rectiline figure AC is (by the first definition of the sixt) like vnto the re<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>line figure geuen EG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore vpon the right line geuen AB is described a rectiline figure AC like &amp; in like sort <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>uat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>to the recti<g ref="char:EOLhyphen"/>line figure geuen EG: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>13.</hi> Theoreme. The <hi>19.</hi> Proposition.
<pb facs="tcp:7062:198"/>
Like triangles are one to the other in double proportion that the sides of lyke proportion are.</head>
               <p>SVppose the triangles like to be ABC and DEF, hauing the angle B of the one triangle, equal vnto the angle E of the other triangle, &amp; as AB is to BC, so let DE be to EF, so that let BC &amp; EF be sides of like proportion. Then I say that the proportion of the triangle ABC vnto the triangle DEF is double to the proportion of the side BC to the side EF. Vnto the two lines BC and EF (by the <hi>10.</hi> of the sixth) make a third lyne in proportion BG, so that as BC is to EF, so let EF be to BG, and draw a lyne from A to G. Now forasmuch as AB is to BC, as DE is to EF, therfore alter<g ref="char:EOLhyphen"/>nately (by the <hi>16.</hi> of the fifth) as AB is to DE, so is BC to EF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> But as BC is to EF, so is EF to BG, wherfore also (by the
<figure/>
                  <hi>11.</hi> of the fifth) as AB is to DE, so is EF to BG. Wherfore the sides of the triangles ABG &amp; DEF, which include y<hi rend="sup">e</hi> equal angles are reciprokally proportionall. But if in triangles hauing one angle of the one equall to one angle of y<hi rend="sup">e</hi> other, the sides which include y<hi rend="sup">e</hi> equall an<g ref="char:EOLhyphen"/>gles, be reciprokal, the triangles also (by y<hi rend="sup">e</hi> 15. the sixth) shal be equall. Wherfore the triangle ABG is equall vnto y<hi rend="sup">e</hi> triangle DEF. And for y<hi rend="sup">t</hi> as y<hi rend="sup">•</hi> line BC is to y<hi rend="sup">e</hi> line EF, so is the line EF to y<hi rend="sup">e</hi> line BG: but if there be three lines in proportion, the first shall haue to the third double proportion that it hath to the second (by the <hi>10.</hi> de<g ref="char:EOLhyphen"/>finition of the fifth) therfore the line BC hath vnto the line BG double propor<g ref="char:EOLhyphen"/>tion that it hath to the line EF. But as BC is to BG, so (by the <hi>1.</hi> of the sixth) is the triangle ABC to the triangle ABG. Wherfore the tiangle ABC is vnto the triangle ABG in double proportion that the side BC is to the side EF. But the triangle ABG is equall to the triangle DEF. Wherfore also the triangle ABC is vnto the triangle DEF in double proportion that the side BC is to the side EF. Wherfore lyke triangles are one to the other in double proportion that the sides of like proportion are: which was required to be proued.</p>
               <div type="section">
                  <head>Corollary.</head>
                  <p>
                     <note place="margin">A Corollary.</note>Hereby it is manifest that if there be three right lines in pro<g ref="char:EOLhyphen"/>portion, as the first is to the third, so is the triangle described vpon the first, vnto the triangle described vpon the second, so that the sayd triangles be like, and in lyke <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ort described, for it hath bene proued that as the lyne CB is to the line BG, so
<pb n="168" facs="tcp:7062:198"/>
is the triangle ABC to the triangle DEF: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>14.</hi> Theoreme. The <hi>20.</hi> Proposition. Like Poligonon figures, are deuided into like triangles and equall in number, and of like proportion to the whole. And the one Poligonon fig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re is to the other Poligonon figure in double proportion that one of the sides of like proportion is to one of the sides of like proportion.</head>
               <p>SVppose y<hi rend="sup">t</hi> the like Poligonon figures be ABCDE, &amp; FGHKL, hauing the angle at the point F equall to the angle at the point A, and the angle at the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equall to the angle at the point B, and the an<g ref="char:EOLhyphen"/>gle at the point H equall to y<hi rend="sup">•</hi> angle at the point C: and so of the rest. And moreouer, as the side AB is to the side BC, so let the side FG be to the side GH, and as the side BC is to the side CD, so let the side GH be to the side HK and so forth. And let the sides AB &amp; FG be sides of like proportion. Then I
<figure/>
say first,<note place="margin">The first par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this Theoreme.</note> that these Poligonon figures ABCD<g ref="char:EOLhyphen"/>E &amp; FG<g ref="char:EOLhyphen"/>HKL, are deui<g ref="char:EOLhyphen"/>ded into like trian<g ref="char:EOLhyphen"/>gles and equall in number. For draw these right lines, AC, AD, FH, &amp; FK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And forasmuch as (by supposition, that is, by reason the figure ABCDE is like vnto the figure FGHKL) the angle B is equall vnto the angle G, and as the side AB is to the side BC, so is the side FG to the side GH, it followeth that the two triangles ABC and FGH haue one angle of the one equall to one angle of the other, and haue also the sides about the equall angles proportionall. Wherefore (by the <hi>6.</hi> of the sixt) the triangle ABC is equiangle vnto the trian<g ref="char:EOLhyphen"/>gle FGH. And those angles in the<g ref="char:cmbAbbrStroke">̄</g> are equall, vnder which are subtended sides of like proportion: namely, the angle BAC is equall to the angle GFH, and the angle BCA to the angle GHF. Wherefore (by the 4. of the sixt) the sides which are about the equall angles are proportionall: and the sides which are sub<g ref="char:EOLhyphen"/>tended
<pb facs="tcp:7062:199"/>
vnder the equall angles are of like proportion. Wherfore as AC is to BC, so is FH to GH. But by supposition as BC is to CD, so is GH to HK. Wherefore of equalitie (by the 22. of the fift) as AC is to CD, so is FH to HK. And forasmuch as by supposition the whole angle BCD is equall to the whole angle GHK, and it is proued that the angle BCA is e<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>all to the an<g ref="char:EOLhyphen"/>gle GHF: therefore the angle remayning ACD is equall to the angle remay<g ref="char:EOLhyphen"/>ning FHK (by the 3. common sentence). Wherefore the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ria<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>gl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s ACD, and FHK, haue againe one angle of the one equall to one angle of the other, and the sides which are about the equall sides are proportionall. Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame sixt of this booke) the triangles ACD &amp; FHK are equiangle. And (by the 4. of this booke) the sides, which are about the equall angles <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> proportio<g ref="char:EOLhyphen"/>nall. And by the same reason may we proue that the triangle AD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is equian<g ref="char:EOLhyphen"/>gle vnto the triangle FKL. And that the sides which are about the equall an<g ref="char:EOLhyphen"/>gles are proportionall. Wherefore the triangle ABC is like to y<hi rend="sup">e</hi> triangle FGH, and the triangle ACD to the triangle FHK, and also the triangle ADE to the triangle FKL (by the first definition of this sixt booke). Wherfore the Po<g ref="char:EOLhyphen"/>ligonon figures geuen ABCDE, and FGHKL, are deuided into triangles like and equall in number.</p>
               <p>
                  <note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note>I say moreouer, that the triangles are the one to the other, and to the whole Poligonon figures proportionall: that is, as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK, and y<hi rend="sup">e</hi> triangle ADE to the triangle FKL: and as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. For for<g ref="char:EOLhyphen"/>asmuch as the triangle ABC is like to the triangle FGH, and AC and FH are sides of like proportion, therfore the proportion of the triangle ABC to the triangle
<figure/>
FGH is double to the pro<g ref="char:EOLhyphen"/>portion of the side AC to the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> FH (by y<hi rend="sup">e</hi> for<g ref="char:EOLhyphen"/>mer Pro<g ref="char:EOLhyphen"/>position). And therefore also the proportion of the triangle ACD to the triangle FKH, is double to the proportion that the same side AC hath to the side FH. Where<g ref="char:EOLhyphen"/>fore (by the 11. of the fift) as the triangle ABC is to the triangle FGH, so is the triangle ACD to the triangle FHK. Againe, forasmuch as the triangle ACD is like to the triangle FHK, and the sides AD &amp; FK are of like pro<g ref="char:EOLhyphen"/>portion: therefore the proportion of the triangle ACD to the triangle FHK
<pb n="169" facs="tcp:7062:199"/>
is double to the proportion of the side AD to the side FK (by the foresayd 19. of the sixt). And by the same reason the proportion of the triangle ADE to y<hi rend="sup">e</hi> tri<g ref="char:EOLhyphen"/>angle FKL, is double to the proportion of the same side AD to the side FK. Wherfore (by the 11. of the fift) as the triangle ACD is to the triangle FHK, so is the triangle ADE to the triangle FKL. But as the triangle ACD is to the triangle FHK, so is it proued that the triangle ABC is to the triangle FGH. Wherefore also (by the <hi>11.</hi> of the fift) as the triangle ABC is to the tri<g ref="char:EOLhyphen"/>angle FGH, so is the triangle ADE to the triangle FKL. Wherefore the foresayd triangles are proportionall: namely, as ABC is to FGH, so is ACD to FHK, and ADE to FKL. Wherefore (by the <hi>12.</hi> of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as the triangle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. Wher<g ref="char:EOLhyphen"/>fore the triangles are proportionall both the one to the other, &amp; also to the whole Poligonon figures.</p>
               <p>Lastly I say,<note place="margin">The third part.</note> that the Poligonon figure ABCDE hath to the Poligonon fi<g ref="char:EOLhyphen"/>gure FGHKL a double proportion to that which the side AB hath to the side FG: which are sides of like proportion. For it is proued, that as the trian<g ref="char:EOLhyphen"/>gle ABC is to the triangle FGH, so is the Poligonon figure ABCDE to the Poligonon figure FGHKL. But the triangle ABC hath to the triangle FGH a double proportion to that which the side AB hath to the side FG (by the former 19. Proposition of this booke): for it is proued, that y<hi rend="sup">e</hi> triangle ABC is like to the triangle FGH. Wherefore the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to the proportion of the side AB to the side FG: which are sides of like proportion. Wherefore like Poligonon figures are deuided. &amp;c. as before<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
               <div type="section">
                  <head>The first Corollary.</head>
                  <p>Hereby it is manifest, that all like rectiline figures what soe<g ref="char:EOLhyphen"/>uer, are the one to the other in double proportion that the sides of like proportion are.<note place="margin">The first Co<g ref="char:EOLhyphen"/>rollary.</note> For any like rectiline figures whatsoeuer are by this Proposition deuided into like triangles and equall in number.</p>
               </div>
               <div type="section">
                  <head>The second Corollary.</head>
                  <p>Hereby also it is manifest, that if there be three right lines proportionall,<note place="margin">The second Corollary.</note> as the first is to the third, so is the figure descri<g ref="char:EOLhyphen"/>bed vppon the first to the figure described vpon the second, so that the sayd figures be like and in like sort described.
<pb facs="tcp:7062:200"/>
For it is proued, that the proportion of the Poligonon figure ABCDE to the Poligonon figure FGHKL is double to y<hi rend="sup">e</hi> proportion of the side AB to the side FG. And if (by the 11. of the sixt) vnto the lines AB and FG we take a third line in proportion, namely, MN, the first line, namely, AB shall haue vnto the third line, namely, to MN, double proportion that it hath to the second line, namely, to FG (by the 10. definition of the fift). Wherfore as y<hi rend="sup">e</hi> line AB is to the line MN, so is the rectiline figure ABC to the rectiline figure FGH, the sayd rectiline figures being like &amp; in like sort described.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>15.</hi> Theoreme. The <hi>21.</hi> Proposition. Rectiline figures which are like vnto one and the same recti<g ref="char:EOLhyphen"/>line figure, are also like the one to the other.</head>
               <p>SVppose there be two rectiline figures A and B like vnto the rectiline figure C. Then I say that the figure A is also like vnto the figure B. For forasmuch as the figure A is like vnto the figure C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> it is also equi<g ref="char:EOLhyphen"/>angle vnto it (by the conuersion of the first definition of the sixth) &amp; the sides including the equall angles
<figure/>
shall be proportionall. Agayne foras<g ref="char:EOLhyphen"/>much as the figure B is like vnto the figure C, it is also (by the same defini<g ref="char:EOLhyphen"/>tion) equiangle vnto it, and the sides about the equall angles are proportio<g ref="char:EOLhyphen"/>nall. Wherfore both these figures A and B are equiangle vnto the figure C, and the sides about the equall angles are proportionall. Wherfore (by the first common sentence) the figure A is equiangle vnto the figure B, and the sides a<g ref="char:EOLhyphen"/>bout the equall angles are proportionall, wherfore the figure B is like vnto the figure A, which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The <hi>16.</hi> Theoreme. The <hi>22.</hi> Proposition. If there be foure right lines proportionall, the rectiline figures also described vpon them beyng lyke, and in like sorte situate, shall be proportional. And if the rectiline figures vppon them described be proportional, those right lynes also shall be pro<g ref="char:EOLhyphen"/>portionall.</head>
               <p>
                  <pb n="170" facs="tcp:7062:200"/>SVppose there be foure right lines AB, CD, EF, and GH, and as AB is to CD, so let EF be to GH. And vpon the lines AB and CD (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the sixth) let there be described two rectiline figures KAB, and LCD like the one to the other, and in like sort situate. And vpon the lynes EF and GH (by the same) let there be described also two rectiline figures MF and NH like the one to the other, and in like sorte situate.<note place="margin">The first part of this proposition.</note> Then I say that as the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure KAB is
<figure/>
to the figure LCD, so is the figure MF to the figure NH. Vnto the lines AB and CD (by the 11. of the sixth) make a third lyne in propor<g ref="char:EOLhyphen"/>tion, namely, O: and vnto the lines EF and GH in like sort make a third lyne in a line proportion, namely, P. And for that as the line AB is to the line CD, so is the line EF to the line GH, but as the line CD is to the line O, so is the line GH to the lyne P. Wherfore of equality (by the 22. of the fifth) as the lyne AB is vnto the line O, so is the lyne EF to the line P. But as the line AB is to the line O, so is the figure KAB to the figure LCD (by the second corollary of the 20. of the sixth). And as the line EF is to the lyne P, so is the figure M F to the figure NH. Where<g ref="char:EOLhyphen"/>fore (by the 11. of the fifth) as the figure KAB is to the figure LCD, so is the fi<g ref="char:EOLhyphen"/>gure M F to the figure NH.</p>
               <p>But now suppose that as the figure KAB is to the figure LCD,<note place="margin">The second part which is the conuerse of the first.</note> so is the figure M F to the figure NH, then I say that as the line AB is to the line CD, so is the line EF to the line GH. As the line AB is to the lyne CD, so (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the sixth) let the lyne EF be to the lyne QR, and vpon the lyne QR (by the 18. of the sixth) describe vnto either of these figures MF and NH a like fi<g ref="char:EOLhyphen"/>gure, and in like sort situate SR. Now forasmuch as the lyne AB is to the lyne CD, so is the lyne EF to the line QR, and vpon the lines AB and CD are de<g ref="char:EOLhyphen"/>scribed two figures lyke, and in like sort situate KAB and LCD, and vpon the lines EF and QR are described also two figures like, and in like sort situate MF and SR, therfore as the figure KAB is to the figure LCD, so is the figure MF to the figure SR: wherfore also (by the 11. of the fifth) as the figure MF is to the figure SR, so is the figure MF to the figure NH, wherfore the figure
<pb facs="tcp:7062:201"/>
M F hath to either of these figures NH, and SR one and the same proportion, wherfore by the 9. of the fifth, the figure NH is equal vnto the figure SR. And it is vnto it like, and in like sort situate. <note n="*" place="margin">Note that this is proued in the assumpt following.</note> But in like and equall rectiline figures beyng in like sort situate, the sides of like proportion on which they are described are equall. Wherfore y<hi rend="sup">e</hi> line GH is equall vnto the line QR. And because as the lyne AB is to the line CD, so is the line EF to the line QR, but the line QR is equall vnto the line GH, therfore as the line AB is to he line CD, so is the line EF to the line GH.
<figure/>
If therefore there be foure right lines proportionall, the rectiline figures al<g ref="char:EOLhyphen"/>so described vpon them beyng like and in lyke sort situate shall be proportionall And if the rectiline figures vpon them described beyng like and in like sort situ<g ref="char:EOLhyphen"/>ate be proportionall, those right lines also shall be proportional: which was required to be proued.</p>
               <div type="section">
                  <head>An Assumpt.</head>
                  <p>And now that in like and equall figures, being in like sort situate, the sides of like proportion are also equall (which thing was before in this proposition taken as graunted) may thus be proued.<note place="margin">An Assumpt.</note> Suppose y<hi rend="sup">t</hi> the rectiline figures NH and SR be equall and like, and as HG is to GN, so let RQ be to QS, and let GH and QR be sides of like proportion. Then I say that the side RQ is equall vnto the side GH. For if they be vnequall, the one of them is greater then the other, let the side RQ be greater then the side HG. And for that as the line RQ is to the line QS, so is the line HG to the line GN, and alternately also (by the 16. of the fifth) as the line RQ is to the line HG, so is the line QS, to the lyne GN, but the line RQ is greater then the line HG. Wherfore also the line QS is grea<g ref="char:EOLunhyphen"/>ter then y<hi rend="sup">e</hi> line GN. Wherefore also y<hi rend="sup">e</hi> figure RS is greater then the figure HN but (by supposition) it is equall vnto it, which is impossible. Wherfore y<hi rend="sup">e</hi> line QR is not greater then y<hi rend="sup">e</hi> line GH. In like sorte also may we proue that it is not lesse then it, wherfore it is equall vnto it: which was required to be proued.</p>
                  <p>
                     <pb n="171" facs="tcp:7062:201"/>Flussates <hi>demonstrateth this second part more briefly,</hi>
                     <note place="margin">An other demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stration of the second part af<g ref="char:EOLhyphen"/>ter Flussates.</note> 
                     <hi>by the first corollary of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>0. of this boke, thus. Forasmuch as the rectiline figures are by supposition in one and the same proportion, and the same pro<g ref="char:EOLhyphen"/>portion is double to the proportion of the sides AB to CD, and EF to GH (by the foresaid corollary) the proportion also of the sides shall be one and the selfe same (by the 7. common sentence) namely, the line AB shall be vnto the line CD as the line EF is to the line GH.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>17.</hi> Theoreme. The <hi>23.</hi> Proposition. Equiangle Parallelogrammes haue the one to the other that proportion which is composed of the sides.</head>
               <p>SVppose the equiangle Parallelogrammes to be AC and CF, hauing the angle BCD of the one equall to the angle ECG of the other. Then I say, that the parallelogramme AC is vnto the parallelogramme CF in that proportion which is composed of the proportion of their sides, that is, of that which the side BC hath to the side CG, and of that which the side DC hath to the side CE. Let the lines BC and CG be so put that they both make one right line (by the 14. of the first). Wherefore
<figure/>
(by the same) the lines DC and CE shall make also one right line. Make complete the parallelogramme DG by producing the sides AD and FG till they concurre in the point H, and let there be put a certaine right line K. And as the line BC is to the line CG, so (by the 12. of the sixt) put vnto the line K a line in the same proportion which let be L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and as DC is to CE, so vnto L put a line in the same pro<g ref="char:EOLhyphen"/>portion, namely, M. Wherefore the proportions of the lines K to L, and L to M, are one and y<hi rend="sup">e</hi> same with the proportions of the sides BC to CG, and D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to CE: but the proportion of K to M, is composed of the proportions of K to L, &amp; L to M:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the proportion of K to M, is composed of the proportions of the sides BC to CG, &amp; EC to CD. And for that as the line BC is to the line CG, so is the parallelogramme AC to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me CH (by the 1. of the sixt). But as the line BC is to the line CG, so is the line K to the line L. Wherefore also (by the 11. of the fift) as the line K is to the line L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the parallelogramme AC to the parallelogramme CH. Againe, for that as the line DC is to the line CE, so is the parallelogramme CH to the parallelogramme CF: but as the line DC is to the line CE, so is the line L to the line M: Wherefore also (by y<hi rend="sup">e</hi> same) as the line L is to the line M, so is the parallelogramme CH to the parallelo<g ref="char:EOLhyphen"/>gramme CF. And forasmuch as it is proued, that as the line K is to the line L, so is the parallelogramme AC to the parallelogramme CH, and as the line L is to the line M, so is the parallelogramme CH to the parallelogramme CF: ther<g ref="char:EOLhyphen"/>fore of equalitie (by the 22. of the fift) as the line K is to the line M, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the pa<g ref="char:EOLhyphen"/>rallelogramme AC to the parallelogramme CF. But as it hath before bene pro<g ref="char:EOLhyphen"/>ued, the proportion of the line K to the line M, is composed of the proportions of
<pb facs="tcp:7062:202"/>
the sides BC to CG, and DC to CE. Wherefore also the proportion of the pa<g ref="char:EOLhyphen"/>rallelogramme AC to the parallelogramme CF, is composed of the proportions of the sides BC to CG, and DC to CE. Wherefore equiangle parallelogrammes haue the one to the other that proportion which is composed of the proportions of the sides: which was required to be proued.</p>
               <p>Flussates <hi>demonstrateth this Theoreme without taking of these three lines, K, L, M, after this maner.</hi>
               </p>
               <p>
                  <note place="margin">An other demonstration after flussates.</note>
                  <hi>Forasmuch as (sayth he) it hath bene declared vpon the 10. definition of the fift booke, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ift definition of this booke, that the proportions of the extremes consist of the proportions of the meanes, let vs suppose two equiangle parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes</hi> ABGD, <hi>and</hi> GEZI, <hi>and let the angles at the poynt</hi> G <hi>in eyther be equall. And let the lines</hi> BG <hi>and</hi> GI <hi>be set directly that they both make one
<figure/>
right line, namely,</hi> BGI. <hi>Wherefore</hi> EGD <hi>also shall be one right line by the conuerse of the 15. of the first. Make complete the parallelogramme</hi> GT. <hi>Then I say, that the proportion of the paral<g ref="char:EOLhyphen"/>lelogrammes</hi> AG <hi>&amp;</hi> GZ <hi>is composed of the pro<g ref="char:EOLhyphen"/>portions of the sides</hi> BG <hi>to</hi> GI, <hi>and</hi> DG <hi>to</hi> GE. <hi>For forasmuch as that there are three magni<g ref="char:EOLhyphen"/>tudes,</hi> AG, GT, <hi>and</hi> GZ, <hi>and</hi> GT <hi>is the meane of the sayd magnitudes: and the proportion of the extremes</hi> AG <hi>to</hi> GZ <hi>consisteth of the meane pro<g ref="char:EOLhyphen"/>portions (by the 5. definition of this booke) namely, of the proportion of</hi> AG <hi>to</hi> GT, <hi>and of the proportion</hi> GT <hi>to</hi> GZ: <hi>But the proportion of</hi> AG <hi>to</hi> GT <hi>is one and the selfe same with the proportion of the sides</hi> BG <hi>to</hi> GI <hi>(by the first of this booke). And the proportion also of</hi> GT <hi>to</hi> GZ <hi>is one and the selfe same with the proportion of the other sides, namely,</hi> DG <hi>to</hi> GE <hi>(by the same Proposition). Wherefore the proportion of the parallelogrammes</hi> AG <hi>to</hi> GZ <hi>consisteth of the proportions of the sides</hi> BG <hi>to</hi> GI, <hi>and</hi> DC <hi>to</hi> GE. <hi>Wherefore equiangle parallelogrammes are the one to the other in that proportion which is composed of theyr sides: which was required to be proued.</hi>
               </p>
            </div>
            <div type="part">
               <head>The <hi>18.</hi> Theoreme. The <hi>24.</hi> Proposition. In euery parallelogramme, the parallelogrammes about the dimecient are lyke vnto the whole, and also lyke the one to the other.</head>
               <p>
                  <note place="margin">Demonstration of this propositio<g ref="char:cmbAbbrStroke">̄</g> wherein is first proued that the parallegramme EG is like to the whole parallelo<g ref="char:EOLhyphen"/>gra<g ref="char:cmbAbbrStroke">̄</g>me ABCD.</note>SVppose y<hi rend="sup">t</hi> there be a parallelogramme ABCD, and let the dimecient therof be AC: and let the parallelogrammes about the dimecient AC, be EG and HK. Then I say that either of these parallelogrames EG and HK is like vnto the whole parallelogramme ABCD, and also are lyke the one to the other. For forasmuch as to one of the sides of the trian<g ref="char:EOLhyphen"/>gle ABC, namely, to BC is drawen a parallel lyne EF, therfore as BE is to EA, so (by the 2. of the sixt) is CF to FA. Agayne forasmuch as to one of y<hi rend="sup">e</hi> sides
<pb n="172" facs="tcp:7062:202"/>
of the triangle ADC, namely, to CD is drawen a parallel lyne F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, therefore (by the same) as CF is to FA, so is DG to GA. But as CF is to FA, so is it pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ued that BE is to EA. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore as BE is to EA, so (by the 11. of the fifth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is DG to GA. Wherfore by composition (by the 18. of the fifth) as BA is to AE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is DA to AG. And alternately (by the 16. of the fifth) as BA is to AD, so is EA to AG. Wherfore in the parallelogrammes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> ABCD and EG y<hi rend="sup">e</hi> sides which are about the common angle BAD are proportionall. And because y<hi rend="sup">e</hi> line GF is a parallel vnto the lyne DC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore the angle AGF (by the 29<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> first) is equall vnto y<hi rend="sup">e</hi> angle ADC, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> y<hi rend="sup">e</hi>
                  <figure/>
angle GFA equall vnto y<hi rend="sup">e</hi> angle DCA and the angle DAC is common to the two triangles ADC and AFG: Wher<g ref="char:EOLhyphen"/>fore the triangle DAC is equiangle vnto the triangle AGF. And by the same reason the triangle ABC is equi<g ref="char:EOLhyphen"/>angle vnto the triangle AEF. Wher<g ref="char:EOLhyphen"/>fore the whole parallelogramme ABCD is equiangle vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me EG. Wherfore as AD is in proportion to DC, so (by the 4. of the sixth) is AG to GF, and as DC is to CA, so is GF to FA. And as AC is to CB, so is AF to FE. And moreouer as CB is to BA, so is FE to EA. And forasmuch as it is proued that as D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to CA, so is GF to FA: but as AC is to C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, so is AF to FE. Wherfore of equalitie (by the 22. of the fifth) as DC is to CB, so is GF to FE. Wherefore in the parallelogrammes ABCD and EG, the sides which include the equall angles are proportionall. Wherefore the parallelogramme ABCD is (by the first definition of the sixth) like vnto the parallelogramme EG.</p>
               <p>And by the same reason also the parallelogramme ABCD is like to the pa<g ref="char:EOLhyphen"/>rallelogramme KH:<note place="margin">That the paral<g ref="char:EOLhyphen"/>lelogra<g ref="char:cmbAbbrStroke">̄</g>me KH is like to the whole parallelo<g ref="char:EOLhyphen"/>gramme AB<g ref="char:EOLhyphen"/>CD.</note> wherefore either of these parallelogrammes EG and KH is like vnto the parallelogramme ABCD. But rectiline figures which are like to one and the same rectiline figure are also (by the 21. of the sixth) like the one to the other. Wherefore the parallelogramme EG is like to the parallelo<g ref="char:EOLhyphen"/>gramme HK.<note place="margin">That the paral<g ref="char:EOLhyphen"/>lelogrammes EG and KH are like the one to the other.</note> Wherfore in euery parallelogramme, the parallelogrammes a<g ref="char:EOLhyphen"/>bout the dimecient are like vnto the whole, and also like the one to the other. Which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other more briefe demonstration after Flussates.</head>
                  <p>Suppose that there be a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>ABCD,</hi> whose dime<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ient let b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,</hi>
                     <note place="margin">
                        <hi>An other Demonstra<g ref="char:EOLhyphen"/>tion after Flussates.</hi>
                     </note> about which let consist these parallelogrammes <hi>EK</hi> and <hi>TI,</hi> hauing the angles at the pointes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> with the whole parallelogramme <hi>ABCD.</hi> Then I say, that those pa<g ref="char:EOLhyphen"/>rallelogrammes <hi>EK</hi> and <hi>TI</hi> are like to the whole parallelogramme <hi>DB</hi> and also al<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb facs="tcp:7062:203"/>
like the one to the other. For forasmuch as <hi>BD, EK,</hi>
                     <figure/>
and <hi>TI</hi> are parallelogrammes, therefore the right line <hi>AZG</hi> falling vpon these parallell lines <hi>AEB, KZT,</hi> and <hi>DI G,</hi> or vpon these parallell lines <hi>AKD, EZI,</hi> and <hi>BTG,</hi> maketh these angles equall the one to the other, namely, the angle <hi>EAZ</hi> to the angle <hi>KZA,</hi> &amp; the an<g ref="char:EOLhyphen"/>gle <hi>EZA</hi> to the angle <hi>KAZ,</hi> and the angle <hi>TZG</hi> to the angle <hi>ZGI,</hi> and the angle <hi>TGZ</hi> to the angle <hi>IZG,</hi> and the angle <hi>BAG</hi> to the angle <hi>AGD:</hi> and finally, the angle <hi>BGA</hi> to the angle <hi>DAG.</hi> Wherefore (by the first Corollary of the 32. of the first, and by the 34. of the first) the angles remayning are equall the one to the other, namely, the angle <hi>B</hi> to the angle <hi>D,</hi> and the angle <hi>E</hi> to the angle <hi>K,</hi> and the angle <hi>T</hi> to the angle <hi>I.</hi> Wherefore these triangles are equiangle and therefore like the one to the other, namely, the triangle <hi>ABG</hi> to the triangle <hi>GDA,</hi> and the triangle <hi>AEZ</hi> to the triangle <hi>ZKA,</hi> &amp; the triangle <hi>ZTG</hi> to the triangle <hi>GIZ.</hi> Wherefore as the side <hi>AB</hi> is to the side <hi>BG,</hi> so is the side <hi>AE</hi> to the side <hi>EZ,</hi> and the side <hi>ZT</hi> to the side <hi>TG.</hi> Wherefore the parallelogrammes contayned vnder those right lines, namely, the parallelogrammes <hi>ABGD, EK,</hi> &amp; <hi>TI,</hi> are like the one to the other (by the first definition of this booke). Wherefore in euery parallelogramme the parallelogrammes. &amp;c. as before: which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
               </div>
               <div type="section">
                  <head>¶ A Probleme added by Pelitarius.</head>
                  <p>
                     <q>Two equiangle Parallelogrammes being geuen, so that they be not like, to cut of from one of them a parallelogramme like vnto the other.</q>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>An addition of Pelitarius.</hi>
                     </note>Suppose that the two equiangle parallelogrammes be <hi>ABCD</hi> and <hi>CEFG,</hi> which let not be like the one to the other. It is required from the Parallelogramme <hi>ABCD,</hi> to cut of a parallelogramme like vnto the parallelogramme <hi>CEFG.</hi> Let the angle <hi>C</hi> of the one be equall to the angle <hi>C</hi> of the other. And let the two parallelogrammes be so <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>, that the lines <hi>BC</hi> &amp; <hi>CG</hi> may make both
<figure/>
one right line, namely, <hi>BG.</hi> Wherefore also the right lines <hi>DC</hi> and <hi>CE</hi> shall both make one right line, namely, <hi>DE.</hi> And drawe a line from the poynt <hi>F</hi> to the poynt <hi>C,</hi> and produce the line <hi>FC</hi> till it co<g ref="char:cmbAbbrStroke">̄</g>curre with the line <hi>AD</hi> in the poynt <hi>H.</hi> And draw the line <hi>HK</hi> parallell to the line <hi>CD</hi> (by the 31. of the first). Then I say, that from the paralle<g ref="char:EOLhyphen"/>logramme <hi>AC</hi> is cut of the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>CDHK,</hi> like vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>EG.</hi> Which thing is manifest by thys 24. Propo<g ref="char:EOLhyphen"/>sition. For that both the sayd parallelo<g ref="char:EOLhyphen"/>grammes are described about one &amp; the selfe same dimetient. And to the end it might the more plainly be seene, I haue made complete the Parallelogramme <hi>ABGL.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>¶ An other Probleme added by Pelitarius.</head>
                  <p>
                     <q>Betwene two rectiline Superficieces, to finde out a meane superficies proportionall.</q>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Another ad<g ref="char:EOLhyphen"/>dition of Pelitarius.</hi>
                     </note>Suppose that the two superficieces be <hi>A</hi> and <hi>B,</hi> betwene which it is required to place a meane superficies proportionall. Reduce the sayd two rectiline figures <hi>A</hi> and <hi>B</hi>
                     <pb n="173" facs="tcp:7062:203"/>
vnto two like parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes (by the 18. of this booke) or if you thinke good reduce eyther of them to a square, (by the last of the second). And let the said two parallelo<g ref="char:EOLhyphen"/>grammes like the one to the other and equall to the superficieces <hi>A</hi> and <hi>B,</hi> be <hi>CDEF</hi> and <hi>FGHK.</hi> And let the angles <hi>F</hi> in either of them be equall, which two angles let be placed in such sort, that the two parallelogrammes <hi>ED</hi> and <hi>HG</hi> may be about one and the selfe same dimetient <hi>CK</hi> (which is done by putting the right lines <hi>EF</hi> and <hi>FG</hi> in such sort that they both make one right line, namely,
<figure/>
                     <hi>EG</hi>). And make co<g ref="char:cmbAbbrStroke">̄</g>plete the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>CLK M.</hi> Then I say, that either of the supplements <hi>FL</hi> &amp; <hi>FM</hi> is a meane proportionall betwene the superficieces <hi>CF</hi> &amp; <hi>FK,</hi> that is, betwene the superficieces <hi>A</hi> and <hi>B:</hi> name<g ref="char:EOLhyphen"/>ly, as the superficies <hi>HG</hi> is to the superficies <hi>FL,</hi> so is the same superficies <hi>FL</hi> to the superficies <hi>ED.</hi> For by this 24. Proposition the line <hi>HF</hi> is to the line <hi>FD,</hi> as the line <hi>GF</hi> is to the line <hi>FE.</hi> But (by the first of this booke) as the line <hi>HF</hi> is to the line <hi>FD,</hi> so is the su<g ref="char:EOLhyphen"/>perficies <hi>HG</hi> to the superficies <hi>FL</hi>: and as the line <hi>GF</hi> is to the line <hi>FE,</hi> so also (by the same) is the superficies <hi>FL</hi> to the superficies <hi>ED.</hi> Wherfore (by the 11. of the fift) as the superficies <hi>HG</hi> is to the superficies <hi>FL,</hi> so is the same superficies <hi>FL</hi> to the superficies <hi>ED</hi>: which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>7.</hi> Probleme. The <hi>25.</hi> Proposition. Vnto a rectiline figure geuen to describe an other figure lyke, which shal also be equall vnto an other rectiline figure geuen.</head>
               <p>SVppose y<hi rend="sup">t</hi> the rectiline figure geue<g ref="char:cmbAbbrStroke">̄</g>, wherunto is required an other to be made like be ABC, and let the other rectiline figure whereunto the same is required to be made, equal be D. Now it is required to describe a rectiline figure like vnto the figure ABC, and equall vnto the fi<g ref="char:EOLhyphen"/>gure D.<note place="margin">Construction.</note> Vppon the line BC
<figure/>
describe (by the 44. of the first) a parallelogramme BE equall vnto the triangle ABC, and by the same vpon the line CE, describe the paralle<g ref="char:EOLhyphen"/>logramme C M equall vnto the rectiline figure D, and in the said parallelogramme let the angle FCE, be equall vnto the angle CBL. And forasmuch as the angle FCE is by construction equall to the angle CBL,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> adde the an<g ref="char:EOLunhyphen"/>gle BCE common to them both. Wherefore the angles LBC and BCE are e<g ref="char:EOLhyphen"/>quall
<pb facs="tcp:7062:204"/>
vnto the angles BCE and ECF, but the angles LBC and BCE are e<g ref="char:EOLhyphen"/>quall to two right angles (by the 29. of the first) wherfore also the angles BCE and ECF are equall to two right angles. Wherfore the lines BC and CF (by the 14. of the first) make both one right line, namely, BF, and in like sort do the lines LE and EM make both one right line, namely, LM. Then (by the 13. of the sixth) take the meane proportionall betwene the lines BC and CF, which let be GH. And (by the 18. of the sixth) vpon the line GH, let there be described a rectiline figure KHG like vnto the rectiline figure ABC, and in like sorte si<g ref="char:EOLhyphen"/>tuate. And for that as the line BC, is to the line GH, so is the line GH to y<hi rend="sup">e</hi> line CF:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> but if there be thre right lines proportional, as the first is to the third, so is the figure which is described of the first vnto the figure which is described of the second, the said figures being like and in like sort situate (by the second corrella<g ref="char:EOLhyphen"/>ry of the 20. of the sixth) wherfore as the line BC is to the line CF, so is the tri<g ref="char:EOLhyphen"/>angle ABC to the triangle KGH. But as the line BC is to the lyne CF, so is the parallelogramme BE to the parallelogramme EF (by the 1. of the sixth). Wherfore as the triangle A
<figure/>
BC, is to the triangle KGH so is the parallelogramme BE to the parallelogramme EF. Wherfore alternately also (by the 16. of the fifth) as the triangle ABC is to the pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>me BE, so is the triangle KGH, to the parallelogramme EF: but the triangle ABC is e<g ref="char:EOLhyphen"/>qual vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me BE, wherfore also the trian<g ref="char:EOLhyphen"/>gle KGH is equall vnto the parallelogramme EF: but the parallelogramme FE is equall vnto the rectiline figure D. Wherfore also the rectiline figure KGH is equall vnto the rectiline figure D, and the rectiline figure KGH is by supposition like vnto the recti<g ref="char:EOLhyphen"/>line figure ABC. Wherefore there is described a rectiline figure KGH lyke vnto the rectiline figure geuen ABC, and equall vnto the other rectiline figure geuen D: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>19.</hi> Theoreme. The <hi>26.</hi> Proposition. If from a parallelogramme be taken away a parallelograme like vnto the whole and in like sorte set, hauing also an angle common with it, then is the parallelogramme about one and the selfe same dimecient with the whole.</head>
               <p>
                  <pb n="174" facs="tcp:7062:204"/>SVppose that there be a parallelogramme ABCD, and from the paral<g ref="char:EOLhyphen"/>lelogramme ABCD, take away a parallelogramme AF like vnto the parallelogramme ABCD, and in like sort situate, hauing also the an<g ref="char:EOLhyphen"/>gle DAB common with it. Then I say, that the parallelogrammes ABCD and AF are both about one and the self same <note n="*" place="margin">
                     <p>By the di<g ref="char:EOLhyphen"/>metie<g ref="char:cmbAbbrStroke">̄</g>t is vnderstand here the dime<g ref="char:EOLhyphen"/>tient which is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rawen from the angle which is com<g ref="char:EOLhyphen"/>mon to them both to the op<g ref="char:EOLhyphen"/>posite angle.</p>
                     <p>Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</p>
                  </note> dimecient AFC, that is, that the dimecient AFC of the whole parallelogramme ABCD passeth by the angle F of the parallelogramme AF, and is common to either of the parallelogrammes. For if AC do not passe by the point F, then if it be possible let it passe by some o<g ref="char:EOLhyphen"/>ther point, as AHC doth. Now then the dimetient AHC shall cut eyther the side GF or the side EF of y<hi rend="sup">e</hi> parallelogramme AF. Let it cut y<hi rend="sup">e</hi> side GF in the point H. And (by the 31. of the first) by the point H let there be drawen to ei<g ref="char:EOLhyphen"/>ther of these lines AD and BC a parallel line HK wherfore GK is a paralle<g ref="char:EOLhyphen"/>logramme, and is about one and the selfe same
<figure/>
dimetient with y<hi rend="sup">e</hi> parallelogramme ABCD. And forasmuch as y<hi rend="sup">e</hi> parallelogrammes ABCD and GK are about one and the self same dimecient, therfore (by the 24. of the sixth) the parallelogramme ABCD is like vnto the parallelogramme GK. Wherfore as the line DA is to the line AB so is the line GA to the line AK (by the conuersion of the first definition of the sixth) And for that the pa<g ref="char:EOLhyphen"/>rallelogrammes ABCD, and EG are (by supposition) like, therfore as the line DA is to the lyne AB so is the line GA to the line AE. Wherfore the line GA hath one and the selfe proportion to either of these lines AK and AE. Wherfore (by the 9. of the fifth) the line AK is e<g ref="char:EOLhyphen"/>quall vnto y<hi rend="sup">e</hi> line AE, namely, y<hi rend="sup">e</hi> lesse to y<hi rend="sup">e</hi> greater, which is impossible. The selfe same inconuenience also will follow, if you put the dimetient AC to cut the side FE. Wherfore AC the dimetie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t of the whole parallelogramme ABCD pas<g ref="char:EOLhyphen"/>seth by the angle and poynt F. And therfore the parallelogramme AEFG is a<g ref="char:EOLhyphen"/>bout one and the selfe same dimetient with the whole parallelogramme ABCD. Wherfore if from a parallelogramme be taken away a parallelograme lyke vnto the whole, and in lyke sorte situate, hauing also an angle common with it, then is that parallelogramme about one and the selfe same dimetient with the whole: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other demonstration after Flussates, which proueth this proposition affirmatiuely.</head>
                  <p>From the parallelogramme ABGD let there be taken away the parallelogramme AEZK like and in like sorte situate with the whole parallelogramme ABGD, and ha<g ref="char:EOLhyphen"/>uing also the angle A common with the whole parallelogramme.<note place="margin">
                        <hi>An other way after Flussates.</hi>
                     </note> Then I say that both their diameters, namely, AZ and AZG do make one and the selfe same right line. De<g ref="char:EOLhyphen"/>uide the sides AB and B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> into two equall partes in the pointes C and F (by the 10. of
<pb facs="tcp:7062:205"/>
the first.) And drawe a line from C to F.
<figure/>
Wherefore the line CF is a parallel to the right line AG (by the corollary added by Campane after the 29. of the first) Wher<g ref="char:EOLhyphen"/>fore the angles BAG and BCF are equall (by the 29. of the first): but the angle EA<g ref="char:EOLhyphen"/>Z is equall vnto the angle BAG (by reaso<g ref="char:cmbAbbrStroke">̄</g> the parallelogrammes are supposed to be like) wherefore the same angle EAZ is e<g ref="char:EOLhyphen"/>quall to the angle BCF, namely, the out<g ref="char:EOLhyphen"/>ward angle to the inward and opposite an<g ref="char:EOLhyphen"/>gle. Wherfore (by the 28. of the first) the lines AZ and CF are parallel lynes. Now then the lines AZ and AG being parallels to one and the selfe same lyne, namely, to CF do concurre in the point A. Wherefore they are set directly the one to the other, so that they both make one right line (by that which was added in the ende of the 30. proposition of the first) wherfore the parallelo<g ref="char:EOLhyphen"/>grammes ABGD, and AEZK are about one and the selfe fame dimetient: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>20.</hi> Theoreme. The <hi>27.</hi> Proposition. Of all parallelogrammes applied to a right line wanting in fi<g ref="char:EOLhyphen"/>gure by parallelogrammes like and in like sort situate to that parallelograme which is described of the halfe line: the grea<g ref="char:EOLhyphen"/>test parallelogramme is that which is described of the halfe line being like vnto the want.</head>
               <p>
                  <note place="margin">
                     <p>In this propositio<g ref="char:cmbAbbrStroke">̄</g> are two cases, in the first the pa<g ref="char:EOLhyphen"/>rallelogramme compared to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me described of the halfe line is de<g ref="char:EOLhyphen"/>scribed vpon a line greater the<g ref="char:cmbAbbrStroke">̄</g> the halfe line: In the second vpo<g ref="char:cmbAbbrStroke">̄</g> a line lesse.</p>
                     <p>The first case where the pa<g ref="char:EOLhyphen"/>rellelogramme compared name<g ref="char:EOLunhyphen"/>ly AF is descri<g ref="char:EOLhyphen"/>bed vpon the line AK which is greater then the halfe line AC.</p>
                  </note>LEt there be a right line AB, and (by the 10. of the first) deuide it in two equall partes in the point C. And vnto the right line AB apply a parallelogramme AD wanting in figure by the parallelogramme DB, which let be like and in like sort described vnto the parallelo<g ref="char:EOLhyphen"/>gramme described of halfe the line AB, which is, BC. Then I say, that of all the parallelogrammes which
<figure/>
may be applied vnto the line AB and which wa<g ref="char:cmbAbbrStroke">̄</g>t in figure by parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes like and in like sort situate vnto the pa<g ref="char:EOLhyphen"/>rallelogramme DB the greatest is the parallelogramme AD. For vnto y<hi rend="sup">e</hi> right line AB let there be applied a parallelo<g ref="char:EOLhyphen"/>gramme AF wanting in figure by the parallelogramme FB, which let be like and in like sort situate vnto the paralle<g ref="char:EOLhyphen"/>logramme DB. Then I say, that the parallelogramme AD is greater then the parallelogramme AF. For forasmuch
<pb n="175" facs="tcp:7062:205"/>
as the parallelogramme DB is like vnto the parallelogramme <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>B, therfore (by the 26. of the sixt) they are about one and the selfe same dimetient.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of this case.</note> Let their di<g ref="char:EOLhyphen"/>metient be DB, and make complete the figure. Now forasmuch as (by the 43. of y<hi rend="sup">e</hi> first) the supplement FC is equall vnto the supplement FE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> adde the figure FB common to them both. Wherefore the whole figure CR is equall vnto the whole figure KE. But the figure CR is equall vnto the figure CG (by the 36. of the first) for that the base AC is equall vnto the base CB. Wherefore the fi<g ref="char:EOLhyphen"/>gure GC is equall vnto the figure KE. Adde the figure CF common vnto them both. Wherfore the whole figure AF is equall vnto the whole Gnomon LMN. But the whole parallelogramme DB is greater then the Gnomon LMN (by the 3. common sentence). Wherefore also it is greater then the parallelogramme AF. But the parallelogramme AD is equall vnto the parallelogramme DB (by the 36. of the first). Wherefore the parallelogramme AD is greater then the parallelogramme AF. Wherefore of all parallelogrammes applied to a right line wanting in figure by parallelogrammes like and in like sort situate, to that parallelogramme which is described of y<hi rend="sup">e</hi> halfe line, the greatest parallelogra<g ref="char:cmbAbbrStroke">̄</g>me is that which is described of the halfe of the line, being like vnto the want: which was required to be proued.</p>
               <p>Againe, let AB be deuided into two equall partes in the point C, and let the parallelogramme applied vpon the halfe line be AL,<note place="margin">The second case where the parallelo<g ref="char:EOLhyphen"/>gramme compared namely AE is described vpon the line AD which is lesse then the line AC, Demonstrati<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond case.</note> wanting in figure by the parallelogramme LB, which let be like and in like sort situate vnto the paralle<g ref="char:EOLhyphen"/>logramme AL. Againe vnto the line AB let there be applied an other paralle<g ref="char:EOLhyphen"/>logramme AE wanting in figure by y<hi rend="sup">e</hi> pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>me
<figure/>
EB being like and in like sort situate vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me LB which is described vpon halfe of the line AB. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the parallelogramme AL appli<g ref="char:EOLhyphen"/>ed vnto halfe the line is greater then the parallelogramme AE. For forasmuch as the parallelogramme EB is like vnto the parallelogramme LB, they are (by the 26. of the sixt) about one and the same dimeti<g ref="char:EOLhyphen"/>ent. Let their dimetient be EB, and make complete the whole figure, and for that the figure LF is equall vnto the figure LH (by the 36. of the first) for the base FG is equall vnto the base GH, ther<g ref="char:EOLhyphen"/>fore the figure LF is greater then the figure KE. But the figure LF is equall vnto the figure DL (by the 43. of the first). Wherefore the figure DL is greater then the figure KE: put the figure KD common to them both. Wher<g ref="char:EOLhyphen"/>fore the whole parallelogramme AL is greater then the whole parallelogramme AE: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:206"/>
               <head>The <hi>8.</hi> Probleme. The <hi>28.</hi> Proposition. Vpon a right line geuen, to apply a parallelogramme equall to a rectiline figure geuen, &amp; wanting in figure by a paralle<g ref="char:EOLhyphen"/>logramme like vnto a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me geuen. Now it beho<g ref="char:EOLhyphen"/>ueth that the rectiline figure geuen, whereunto the parallelo<g ref="char:EOLhyphen"/>gra<g ref="char:cmbAbbrStroke">̄</g>me applied must be equall, be not greater the<g ref="char:cmbAbbrStroke">̄</g> that paralle<g ref="char:EOLhyphen"/>logramme, which so is applied vpon the halfe lyne, that the defectes shall be like, namely, the defect of the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me applied vpon the halfe line, and the defect of the parallelo<g ref="char:EOLhyphen"/>gramme to be applied (whose defect is required to be like vn<g ref="char:EOLhyphen"/>to the parallelogramme geuen).</head>
               <p>SVppose the right line geuen to be AB, and let the rectiline figure ge<g ref="char:EOLhyphen"/>uen wherunto is required to apply vpon the right line AB an equall rectiline figure be C, which figure C, let not be greater then that pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>me which is so applied vpon the halfe line, that the defectes shall be like, namely, the defect of the parallelogramme applied vpon the halfe line, and the defect of the parallelogramme to be applied (whose defect is requi<g ref="char:EOLhyphen"/>red to be like vnto the parallelogramme geue<g ref="char:cmbAbbrStroke">̄</g>). And let the figure whereunto the defect or want of the parallelogramme is
<figure/>
required to be like D. Now it is requi<g ref="char:EOLhyphen"/>red vpo<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> right line geue<g ref="char:cmbAbbrStroke">̄</g> AB, to describe vnto the rectiline figure geuen C, an e<g ref="char:EOLhyphen"/>qual parallelogramme wanting in figure by a parallelogramme like vnto D.<note place="margin">Construction.</note> Let the line AB (by the 10. of the first) be deuided into two equall partes in y<hi rend="sup">e</hi> point E. And (by the 18. of the sixth) vppon the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>B describe a rectiline figure EBFG like vnto the parallelogramme D and in like sort situate, which shall also be a parallelograme. And make complete the parallelogramme AG.<note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> Now then the parallelogramme AG is either equal vnto the rectiline figure C, or greater then it by supposition. If the parallelo<g ref="char:EOLhyphen"/>gramme AG be equal vnto the rectiline figure C, then is that done which we <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ought for.<note place="margin">The first case.</note> For then vpo<g ref="char:cmbAbbrStroke">̄</g> the right line AB is described vnto the rectiline figure
<pb n="176" facs="tcp:7062:206"/>
geuen C an equal parallelogramme AG wanting in figure by the parallelograme GB,<note place="margin">The second case.</note> which is like vnto the parallelogramme D. But if AG be not equal vnto C then is AG greater then C, but AG is equall vnto GB (by the first of the sixt). Wherfore also GB is greater then C. Take the excesse of the rectiline figure BG aboue the rectiline figure C (by that which Pelitarius addeth after the 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first) And vnto that excesse (by the 15. of the sixt) describe an equall recti<g ref="char:EOLhyphen"/>line figure KLMN like and in like sort situate vnto the rectiline figure D. But the rectiline figure D is like vnto the rectiline GB, wherfore also the recti<g ref="char:EOLhyphen"/>line figure KLMN is like vnto the rectiline figure GB (by the 25. of the sixt) Now then let the sides KL and GE be sides of like proportion, let also y<hi rend="sup">e</hi> sides LM and GF be sides of like proportion. And forasmuch as the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me GB is equal vnto the figures C and KM, therfore the parallelogramme GB is greater then the parallelogramme KM. Wherefore also the side GE is greater then the side KL, and the side GF is greater then the side LM, vnto the side KL put an equall line GO (by the 2. of the first) and likewise vnto the side LM put an equall line GP. And make perfect the parallelogramme OGPX. Wherfore the parallelogramme GX is equal &amp; like vnto the parallelogramme KM. But the parallelogramme KM is lyke vnto the parallelogramme GB. Wherfore also the parallelogramme GX is like vnto the parallelogramme GB. Wherfore the parallelogrammes GX and GB are (by the 26. of the sixt) about one and the self same dimecient. Let their dimecient be GB, and make complete the figure. Now forasmuch as the parallelogramme BG is equall vnto the recti<g ref="char:EOLhyphen"/>line figure C, and vnto the parallelogramme KM, and the parallelogramme GX, which is part of the parallelogramme GB, is equal vnto KM. Wherfore the Gnomon remayning YQV is equall vnto the rectiline figure remayning, name<g ref="char:EOLhyphen"/>ly, to C. And forasmuch as the supplement PR is equall vnto the supplement OS, put the parallelogramme XB common vnto them both. Wherfore the whole parallelogramme PB is equall vnto the whole parallelograme OB. But the pa<g ref="char:EOLhyphen"/>rallelogramme OB is equal vnto the parallelogramme TE by the 1. of the sixt, (for the side AE is equal vnto the side EB) Wherfore the parallelogramme TE is equal vnto the parallelogramme PB. Put the parallelogramme OS com<g ref="char:EOLhyphen"/>mon to them both. Wherfore the whole parallelogramme TS is equall vnto the whole gnomon YQV. But it is proued that the gnomon YQV is equal vnto the rectiline figure C. Wherfore also the parallelogramme TS is equal vnto the re<g ref="char:EOLhyphen"/>ctiline figure C. Wherfore vpon the right line geuen AB is applied a parallelo<g ref="char:EOLhyphen"/>gramme TS equal vnto the rectiline figure geuen C, and wanting in figure by a parallelogramme XB which is like vnto the parallelogramme geuen D, for the parallelogramme XB is like vnto the parallelogramme GX: which was re<g ref="char:EOLhyphen"/>quired to be done.</p>
               <div type="section">
                  <head>¶ A Corollary added by Flussates.</head>
                  <p>
                     <hi>Hereby it is manifest, that if vpon a right line be applied a parallelogramme wantyng
<pb facs="tcp:7062:207"/>
in figure by a square,</hi>
                     <note place="margin">
                        <hi>A Corollary ad<g ref="char:EOLhyphen"/>ded by Flussa<g ref="char:EOLhyphen"/>tes, and is put of Theon as an assumpt be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ore the</hi> 17. <hi>proposi<g ref="char:EOLhyphen"/>tion of the te<g ref="char:cmbAbbrStroke">̄</g>th booke: which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>or that it follo<g ref="char:EOLhyphen"/>weth of this proposition I thought it not amisse here to place.</hi>
                     </note> 
                     <hi>the parallelogramme ap<g ref="char:EOLhyphen"/>plied
<figure/>
shall be equall to the rectangle figure which is contayned vnder the segments of the line geuen which are made by the application.</hi> For the rest of the line is equall to the o<g ref="char:EOLhyphen"/>ther side of the parallelogramme applied. For that they are sides of one &amp; the selfe same square, as the parallelogramme AG is contained vnder the lines AD and DB, or DG which is equall to DB.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>9.</hi> Probleme. The <hi>29.</hi> Proposition. Vpon a right line geuen to apply a parallelogramme equall vnto a rectiline figure geuen, and exceeding in figure by a pa<g ref="char:EOLhyphen"/>rallelogramme like vnto a parallelogramme geuen.</head>
               <p>SVppose the right line geuen to be AB, and let the rectiline figure geuen whereunto is required vpon the line AB to apply an equall parallelo<g ref="char:EOLhyphen"/>gramme be C: let also the parallelogramme wherunto y<hi rend="sup">e</hi> excesse is requi<g ref="char:EOLhyphen"/>red to be like, be D. Now it is requi<g ref="char:EOLhyphen"/>red
<figure/>
vpon the right line AB to apply in parallelogramme equall vnto the rectiline figure C, and exceeding in fi<g ref="char:EOLhyphen"/>gure by a parallelogramme like vnto the parallelogramme D.<note place="margin">Construction.</note> Let the line AB be (by the 10. of the first) deui<g ref="char:EOLhyphen"/>ded into two equall partes in the point E. And vpon the line EB (by the 18. of the sixt) describe a parallelo<g ref="char:EOLhyphen"/>gramme BF like vnto the figure D, and in like sort situate. And vnto both
<pb n="177" facs="tcp:7062:207"/>
these figures BF and C, describe an equall rectiline figure GH like vnto the figure D and in like sort situate (by the 25. of the sixt). Wherefore the paralle<g ref="char:EOLhyphen"/>logramme GH is (by the 21. of the sixt) like vnto the parallelogramme BF. Let the sides KH and FL be sides of like proportion, and so also let the sides KG and FE be. And forasmuch as the parallelogramme GH is (by constructi<g ref="char:EOLhyphen"/>on) greater then the parallelogramme FB, therefore the line KH is greater then the line FL, and the line KG is greater then the line FE. Extend the lines FL and FE to the pointes M &amp; N, and vnto the line KH put an equall line FLM, and likewise vnto the line KG put an equall line FEN: and make perfect the figure MN. Wherefore the parallelogramme MN is equall and like vnto the parallelogramme GH. But the parallelogramme GH is like vnto the parallelogramme EL. Wherefore also the parallelogramme MN is like vnto the parallelogramme EL. Wherefore the parallelogrammes EL and MN are (by the 26. of the sixt) about one and the same dimetient. Let the sayd dimetient be FO, and make perfect the figure.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as the pa<g ref="char:EOLhyphen"/>rallelogramme GH is equall vnto the figures EL and C. But by construction the parallelogramme GH is equall vnto the parallelogramme MN. Wherfore the parallelogramme MN is equall vnto the figures EL and C. Take away the figure EL which is common to them both. Wherefore the Gnomon remay<g ref="char:EOLhyphen"/>ning, namely, VYX, is equall vnto the rectiline figure C. And forasmuch as the line AE is equall vnto the line EB, therefore the parallelogramme A N is (by the 36. of the first) equall vnto the parallelogramme NB, that is, vnto the parallelogramme LP, which (by the 43. of the first) is equall vnto the pa<g ref="char:EOLhyphen"/>rallelogramme NB. Adde the parallelogramme BO common to them both. Wherefore the whole parallelogramme AO is equall vnto the Gnomon VYX. But the Gnomon VYX is equall vnto the rectiline figure C. Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme AO is equall vnto the rectiline figure C. Wherefore vpon the right line geuen AB is applied the parallelogramme AO equall vnto the recti<g ref="char:EOLhyphen"/>line figure geuen C, and exceeding in figure by the parallelogramme QP which is like vnto the parallelogramme geuen D. For the parallelogramme D is like vnto the parallelogramme BF: and the parallelogramme BF is like vnto the parallelogramme PQ: for they are about one and the selfe same dimetient: which was required to be done.</p>
            </div>
            <div type="part">
               <head>The <hi>10.</hi> Probleme. The <hi>30.</hi> Proposition. To deuide a right line geuen by an extreme and meane pro<g ref="char:EOLhyphen"/>portion.</head>
               <p>SVppose the right line geuen to be AB. It is required to devide the line AB by an extreme and meane proportion.<note place="margin">Construction.</note> Vpon the line AB describe (by the 46. of the first) a square BC. And vpon the line AC (by the
<pb facs="tcp:7062:208"/>
29. of the sixt) applie a parallelogramme CD equall vnto the square BC,<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> and exceding in figure by the figure AD like vnto the figure BC. Now BC is a square. Wherefore also AD is a square. And forasmuch as BC is equall vnto CD, take away the figure CE which is common
<figure/>
to them both. Wherefore the figure remayning, namely, BF, is equall to the figure remayning, namely, to AD, and the angle E of the one is e<g ref="char:EOLhyphen"/>quall vnto the angle E of the other. Wherefore (by the 2. definition of the sixt, and by the 14. of the sixt) the sides of the figures BF and DA, which containe the equall angles, are reciprokall. Wherefore as the side FE is to the side ED, so is the side AE to the side EB. But the side FE is equall vnto the line AC, that is, vnto the line AB, and the side ED is equall vnto the line AE (by the 34. of the first). Wherefore as the line BA is to the line AE, so is the line AE to the line EB. But the line AB is greater then the line AE. Wherfore also the line AE is grea<g ref="char:EOLhyphen"/>ter then the line EB. Wherefore the right line AB is deuided by an extreme and meane proportion in the point E: and the greater segment thereof is AE: which was required to be done.</p>
               <div type="section">
                  <head>An other way.</head>
                  <p>
                     <note place="margin">An other way.</note>Suppose the right line geuen to be AB. It is required to deuide the line AB by an extreme and meane proportion. Deuide the line AB in the point C (by the 11. of the second) in such sort that the rectangle figure comprehended vnder the lines AB and BC may be equall vnto the square de<g ref="char:EOLhyphen"/>scribed
<figure/>
of the line CA. And forasmuch as that which is comprehended vnder the lines AB &amp; BC, is equall vnto the square made of the line AC, therefore as the line BA is to the line AC, so (by the 17. of the sixt) is the line AC to the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> CB. Wherfore the line AB is deuided by an extreme &amp; meane proportion in the point C. which was required to be done.</p>
               </div>
            </div>
            <div type="part">
               <head>The <hi>21.</hi> Theoreme. The <hi>31.</hi> Proposition. In rectangle triangles the figure made of the side subtending the right angle, is equal vnto the figures made of the sides co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>prehending the right angle, so that the sayd thr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e figures b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <pb n="178" facs="tcp:7062:208"/>
b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> like and in like sort described.</head>
               <p>SVppose that there be a triangle ABC, whose angle BAC let be a right angle. The<g ref="char:cmbAbbrStroke">̄</g> I say that the figure which is described of the lyne BC is equall vnto the two figures which are described of the lines BA &amp; AC, the said thre figures being like the one to the other, and in like sort described. From the point A (by the 12. of the first) let there be drawne vn<g ref="char:EOLhyphen"/>to the line BC a perpendiculer line AD.<note place="margin">Construction.</note> Now forasmuch as in y<hi rend="sup">e</hi> rectangle trian<g ref="char:EOLhyphen"/>gle ABC is drawen from the right angle A vnto the base BC a perpendicular line AD,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore the triangles ABD, and ADC set vpon the perpendiculer line, are like vnto the whole triangle ABC, and also like the one to the other (by the 8. of the sixt). And forasmuch as the triangle ABC is like vnto the triangle ABD, therfore as the line CB is to the line BA, so is the line AB to the lyne BD. Now for that there are three right lines proportional, therfore (by the 2. correllary of the 20. of the sixth) as the first is to the third, so is the figure made of the first, to the figure made
<figure/>
of the second, the said figures being like and in like sorte de<g ref="char:EOLhyphen"/>scribed. Wherefore as the lyne BC is to the line BD, so is the figure made of the line BC to the figure made of the line BA, they beyng like and in lyke sort described. And by y<hi rend="sup">e</hi> same reason as the line BC is to the lyne CD, so is the figure made of the line BC, to the figure made of the line CA, they be<g ref="char:EOLhyphen"/>ing like and in like sort descri<g ref="char:EOLhyphen"/>bed. Wherfore as the line BC is to the lines BD and DC, so is the figure made of the line BC to the figures made of the lines BA and AC, they being like and in like sort described. But the line BC is equall vnto the lines BD and DC, wherfore the figure made of the lyne BC is equall vnto the figures made of the lines BA and AC, they being like and in like sort described. Wherfore in rectan<g ref="char:EOLhyphen"/>gle triangles the figure made of the side subtending the right angle is equall vnto the figures made of the sides comprehending the right angle, so that the said thr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> figures be like and in like sort described: which was required to be proued.</p>
               <div type="section">
                  <head>An other way.</head>
                  <p>Forasmuch as (by the first correllary of the 20. of the sixth) like rectiline fi<g ref="char:EOLhyphen"/>gures are in double proportion to that that the sides of like proportion are, ther<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:209"/>
the rectiline figure made of the line BC is vnto t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e rectiline figure <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ade of the line BA in double proportion to that that the line CB is to the line BA, and (by the same) the square also made of the line BC is vnto the square made of the line BA in double proportion
<figure/>
to that that the line CB is vn<g ref="char:EOLhyphen"/>to the line BA. Wherfore also as the rectiline figure made of the line CB is to the rectiline figure made of the line BA, so is the square made of the lyne CB to the square made of the line BA. And by the same rea<g ref="char:EOLunhyphen"/>son also, as the rectiline figure made of the line BC is to the rectiline figure made of y<hi rend="sup">e</hi> line CA, so is the square made of the line BC to the square made of the line CA. Wherfore also as the rectiline figure made of the line BC is to the rectiline figures made of the lines BA and AC, so is the square made of the line BC to the squares made of the lines BA and AC. But the square made of y<hi rend="sup">e</hi> line BC is equall vnto the squares made of the lines BA and AC (by the 47. of the first) Wherefore also the rectiline figure made of the line BC is equall vnto the rectiline figures made of the lines BA and AC, the said three figures beyng lyke and in like sort described.</p>
                  <div type="subsection">
                     <head>The conuerse of this Proposition after Campane.</head>
                     <p>
                        <note place="margin">
                           <hi>The conuerse of the former proposition.</hi>
                        </note>
                        <hi>If the figure described of one of the sides of a triangle be equall to the figures which are described of the two other sides, the sayd figures being like and in like sort described, the tri<g ref="char:EOLhyphen"/>angle shall be a rectangle triangle.</hi>
                     </p>
                     <p>Suppose that ABC be a triangle, and let the fi<g ref="char:EOLhyphen"/>gure
<figure/>
described of the side BC be equall to the two figures described of the sides AB and AC, the said figures being like, and in like sort descri<g ref="char:EOLhyphen"/>bed. Then I say that the angle A is a right angle. Let the angle CAD be a right angle, and put the line AD equall to the line AB, and drawe a lyne from D to C. Now then by this 31. proposition, the figure made of the line CD is equall to the two figures made of the lines AC and AD, the said figures being like and in like sort described. Wherefore also it is equall vnto the figure made of the line BC, which is by supposition e<g ref="char:EOLhyphen"/>qual to the two figures made of the lines AC and AD (for the line AD is put equall to the line AB) wherfore the line DC is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>quall to the line BC. Wherfore (by the 8. of the first) the angle BAC is a right angle, which was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <pb n="179" facs="tcp:7062:209"/>
               <head>The <hi>22.</hi> Theoreme. The <hi>32.</hi> Proposition. If two triangles be set together at one angle, hauing two sides of the one proportionall to two sides of the other, so that their sides of like proportion be also parallels: then the other sides remayning of those triangles shall be in one right line.</head>
               <p>
                  <hi>SVppose the two triangles to be ABC, and DCE, and let two of their sides AC &amp; DC make an angle ACD, and let the said triangles haue two sides of the one, namely, BA and AC proportionall to two sides of the other,</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>namely, to DC and DE, so y<hi rend="sup">t</hi> as AB is to AC, so let DC be to DE. And let AB be a parallell vnto DC, and AC a parallell vnto DE. Then I say, that the lines BC and CE are in one right line. For forasmuch as the line AB is a parallell vnto the line DC, and
<figure/>
vpon the<g ref="char:cmbAbbrStroke">̄</g> lighteth a right line AC:
<figure/>
therefore (by the 29. of the first) the alternate angles BAC and ACD are equall the one to the other. And by the same reason the angle CDE is equall vnto y<hi rend="sup">e</hi> same angle ACD. Wherefore the angle BAC is equall vnto the angle CDE. And foras<g ref="char:EOLhyphen"/>much as there are two triangles ABC and DCE, hauing the angle A of the one equall to the angle D of the other, and the sides about the equall angles are (by supposition) proportionall, that is, as the line BA is to the line AC, so is the line CD to the line DE, ther<g ref="char:EOLhyphen"/>fore the triangle ABC is (by the 6. of the sixt) equiangle vnto the triangl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> DCE. Wherefore the angle ABC is equall vnto the angle DCE. And it is proued, that the angle ACD is equall vnto the angle BAC. Wherefore the whole angle ACE is equall vnto the two angles ABC and BAC. Put the an<g ref="char:EOLhyphen"/>gle ACB common to them both. Wherefore the angles ACE and ACB are e<g ref="char:EOLhyphen"/>quall vnto the angles CAB, ACB, &amp; CBA. But the angles CAB, ACB, and CBA, are (by the 32. of the first) equall vnto two right angles. Wherefore also the angles ACE and ACB are equall to two right angles. Now then vn<g ref="char:EOLhyphen"/>to the right line AC, and vnto the point in it C, are drawen two right lines BC and CE, not on one and the same side, making the side angles ACE &amp; ACB equall to two right angles. Wherefore the lines BC and CE (by the 14. of the first) are set directly and do make one right line. If therefore two triangles be set together at one angle hauing two sides of the one proportionall to two sides of y<hi rend="sup">e</hi> other, so y<hi rend="sup">t</hi> their sides of like proportion be also parallels: then y<hi rend="sup">e</hi> sides remay<g ref="char:EOLhyphen"/>ning of those triangles shall be in one right line: which was required to be proued.</hi>
               </p>
               <p>
                  <pb facs="tcp:7062:210"/>Although <hi>Euclide</hi> doth not distinctly set forth the maner of proportion of like rectiline figures, as he did of lines in the 10. Propositio<g ref="char:cmbAbbrStroke">̄</g> of this Booke, and in the 3. following it, yet as <hi>Flussates</hi> noteth, is that not hard to be done by the 22. of thys Booke. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or two like rectiline figures being geuen to finde out a third proportio<g ref="char:EOLhyphen"/>nall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> also betwene two rectiline superficieces geuen to finde out a meane propor<g ref="char:EOLhyphen"/>tionall (which we before taught to do by <hi>Pelitarius</hi> after the 24. Proposition of this booke): and moreouer three like rectiline figures being geuen to finde out a fourth proportionall like and in like sort described, and such kinde of proportions, are easie to be found out by the proportions of lines. As thus. If vnto two sides of like proportion we should find out a third proportionall by the 11. of this boke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the rectiline figure described vpon that line shall be the third rectiline figure pro<g ref="char:EOLhyphen"/>portionall with the two first figures geuen by the 22. of thys booke. And if be<g ref="char:EOLhyphen"/>twene two sides of like proportion be taken a meane proportionall by the 13. of thys Booke: the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure described vpon the sayd meane shall likewise be a meane proportionall betwene the two rectiline figures geue<g ref="char:cmbAbbrStroke">̄</g> by the same 22. of the sixt. And so if vnto three sides ge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>en be found out the fourth side proportionall (by the 12. of this booke) the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure described vpon the sayd fourth line shall be the fourth rectiline figure proportionall. For if the right lines be proporti<g ref="char:EOLhyphen"/>onall, the rectiline figures described vpon them shall also be proportionall, so that the said rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igures be like &amp; in like sort described by the said 22. of the sixt.</p>
            </div>
            <div type="part">
               <head>The <hi>23.</hi> Theoreme. The <hi>33.</hi> Proposition. In equal circles, the angles haue one and the selfe same pro<g ref="char:EOLhyphen"/>portion that the circumfere<g ref="char:cmbAbbrStroke">̄</g>ces haue, wherin they co<g ref="char:cmbAbbrStroke">̄</g>sist, whe<g ref="char:EOLhyphen"/>ther the angles be set at the centres or at the circumferences. And in like sort are the sectors which are described vppon the centres.</head>
               <p>SVppose the equall circles to be ABC, and DEF, whose centres let be G and H, and let the angles set at their centres G and H, be BGC, and EHF: and let the angles set at their circumferences be BAC and EDF. Then I say that as the circumference BC is to the circumference EF so is the angle BGC to the angle EHF: and the angle BAC to the angle EDF, and moreouer the sector GBC to the sector HEF. Vnto the circumference BC (by the 38. of the third) put as many equall circumferences in order as you will, namely, CK and KL, and vnto the circumference EF put also as many e<g ref="char:EOLhyphen"/>quall circumferences in number as you will, namely, FM, and MN. And drawe these right lines GK,<note place="margin">That the an<g ref="char:EOLhyphen"/>gles at th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> ce<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ter are in pro<g ref="char:EOLhyphen"/>portio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, as the circumfe<g ref="char:EOLhyphen"/>rences wheron they are.</note> GL, HM, and HN. Now forasmuch as the circumfe<g ref="char:EOLhyphen"/>rences BC, CK, and KL are equall the one to the other, the angles also BGC, and CGK, and KGL are (by the 27. of the third) equall the one to the other. Therfore how multiplex the circumference BL is to the circumference BC, so multiplex is y<hi rend="sup">e</hi> angle BGL to the angle BGC. And by y<hi rend="sup">e</hi> same reason also, how multiplex the circumference NE is to the circumference EF, so multiplex is the
<pb n="180" facs="tcp:7062:210"/>
angle NHE to the angle EHF. Wherfore if the circumference BL be equal vnto the circumference EN, the angle BGL is equall vnto the angle EHN, and if the circumference BL be greater then the circumference EN, the angle BGL is greater then the angle NHE, and if the circumference be lesse, the angle is lesse. Now then there are foure magnitudes, namely, the two circumfe<g ref="char:EOLhyphen"/>rences BC and EF, and the two angles that is, BGC, and EHF, and to the circumference BC and to the angle BGC, that is, to the first and third are take<g ref="char:cmbAbbrStroke">̄</g> equemultiplices, namely, the circumference BL, and the angle BGL, and like<g ref="char:EOLhyphen"/>wise to the circumference EF, and to the angle EHF, that is, to the second and fourth, are taken certayne other equemultiplices, namely, the circumference EN and the angle EHN. And it is proued, that if the circumference BL exceede the circumference EN, the angle also BGL exceedeth the angle EHN. And if the circumference be equall, the angle is equall, and if the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce be lesse, the angle also is lesse. Wherfore (by the 6. definition of the fifth) as the circumfe<g ref="char:EOLhyphen"/>rence BC is to the circumference EF, so is the angle BGC to the angle EHF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
               </p>
               <p>But as the angle BGC is to the angle EHF,<note place="margin">That the an<g ref="char:EOLhyphen"/>gles at the cir<g ref="char:EOLhyphen"/>cumferences are so also.</note> so is the angle BAC to the angle EDF, for the angle BGC is double to the angle BAC, and the angle EHF is also double to the angle EDF (by the 20. of the third) Wherfore as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Wherfore in equall circles the angles are in one and the selfe same proportion that their circumferences are, whether the angles be set at the centres, or at the circumferences: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
               <p>I say moreouer that as the circumference BC is to the circumference EF,<note place="margin">That the sec<g ref="char:EOLhyphen"/>tors are so also.</note> so is the sector GBC to the sector HEF. Draw these lines BC and CK. And in the circumferences BC and CK take pointes at all aduentures, namely, P and O. And draw lines from B to P, and from P to C, from C to O, and from O to K. And forasmuch as (by the 15. definition of the first) the two lines BG and GC are equall vnto the two lines CG and GK, and they also comprehend equall an<g ref="char:EOLhyphen"/>gles,
<pb facs="tcp:7062:211"/>
                  <figure/>
therfore (by the 4. of the first) the base BC is equall vnto the base CK, &amp; the triangle GBC is equall vnto the triangle GCK. And seing that the circum<g ref="char:EOLunhyphen"/>ference BC is equall vnto the circumference CK, therefore the circumference remayning of the whole circle ABC, namely, the circumference BLAKOC, is equall vnto the circum<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>erence remayning of the selfe same circle ABC, name<g ref="char:EOLhyphen"/>ly, to the circumference CPBLAK. Wherfore the angle BPC is equall vnto the angle COK (by the 27. of the third) Wherfore (by the 10. definition of the third) the segment BPC is like vnto the segment COK, and they are set vpon equall right lines BC and KC. But like segmentes of circles which consist vpon equall right lines are also equall the one to the other (by the 24. of the third). Wherfore the segment BPC is equall vnto the segment COK. And the trian<g ref="char:EOLhyphen"/>gle GBC is equall vnto the triangle GCK. Wherfore the sector GBC is equall vnto the sector GCK. And by the same reason also, the sector GKL is e<g ref="char:EOLhyphen"/>quall vnto either of the sectors GBC and GCK. Wherfore y<hi rend="sup">e</hi> three sectors GBC and GCK, and GKL, are equall y<hi rend="sup">e</hi> one to y<hi rend="sup">e</hi> other. And by the same reason also the sectors HEF and HFM, and HMN are equall the one to the other. Wherefore how multiplex the circumference BL is to the circumference BC so multiplex is the sector GLB to the sector GBC. And by the same reason how multiplex the circumference NE is to the circumference EF, so multiplex is y<hi rend="sup">e</hi> sector HEN to the sector HEF. If therfore the circumference BL be equall vnto y<hi rend="sup">e</hi> circumference EN, y<hi rend="sup">e</hi> sector also BGL is equall vnto y<hi rend="sup">e</hi> sector EHN. And if the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce BL excede y<hi rend="sup">e</hi> circumference EN, y<hi rend="sup">e</hi> sector also BGL excedeth the sector EHN. And if the circumference be lesse, the sector also is lesse. Now the<g ref="char:cmbAbbrStroke">̄</g> there are foure magnitudes, namely, the two circumferences BC and EF, and the two sectors GBC &amp; HEF, and to the circumference BC, &amp; to the sector GBC, namely, to the first and the third, are taken equemultiplices, that is, the circumference BL, and the sector GBL, and likewise to the circum<g ref="char:EOLhyphen"/>ference EF, and to the sector HEF, namely, to the second and fourth, are ta<g ref="char:EOLhyphen"/>ken certayne other equimultiplices, namely, the circumference EN and the sec<g ref="char:EOLhyphen"/>tor
<pb n="181" facs="tcp:7062:211"/>
HEN. And it is proued that if the circumference BL excede the circumfe<g ref="char:EOLhyphen"/>rence EN, the sector also BGL excedeth the sector EHN. And if the circumference be equall, the segment also is equall, and if the circumference be lesse, the segment also is lesse. Wherfore (by the conuersion of the sixt definition of the fifth) as the circumference BC is to the circumference EF, so is y<hi rend="sup">e</hi> sector GBC vnto the sector HEF: which is all that was required to be proued.</p>
               <div type="section">
                  <head>Corollary. And hereby it is manifest, that as the sector is to the sector, so is angle to angle by the <hi>11.</hi> of the fifth.</head>
                  <p>
                     <hi>Flussates</hi> here addeth fiue Propositions wherof one is a Probleme hauing three Corollaryes following of it, and the rest are Theoremes : which for that they are both witty, &amp; also serue to great vse, as we shall afterward see, I thought not good to omitte, but haue here placed them: but onely that I haue not put them to fol<g ref="char:EOLhyphen"/>lowe in order with the Propositions of <hi>Euclide</hi> as he hath done.</p>
                  <div type="subsection">
                     <head>¶The first Proposition added by Flussates.</head>
                     <p>
                        <q>To describe two rectiline figures equall and like vnto a rectiline figure geuen and in like sort situate, which shall haue also a proportion geuen.</q>
                     </p>
                     <p>Suppose that the rectiline figure geuen be <hi>ABH.</hi>
                        <note place="margin">
                           <hi>Construction of the Pro<g ref="char:EOLhyphen"/>bleme.</hi>
                        </note> And let the proportion geuen be the proportion of the lines <hi>GC</hi> and <hi>CD.</hi> And (by the 10. of this booke) deuide the line <hi>AB</hi> like vnto the line <hi>GD</hi> in the poynt <hi>E</hi> (so that as the line <hi>GC</hi> is to the line <hi>CD,</hi> so let the line <hi>AE</hi> be to the line <hi>EB</hi>). And vpon the line <hi>AB</hi> describe a semicircle <hi>AFB.</hi> And from the poynt <hi>E</hi> erect (by the 11. of the first) vnto the line <hi>AB</hi> a perpendicular line <hi>EF</hi> cutting the circumference in the poynt <hi>F.</hi> And draw these lines <hi>AF</hi> and <hi>FB.</hi> And vpon either of these lines describe rectiline figures like vnto the rectiline figure <hi>AHB</hi> and in like sort situate (by the 18. of the sixt): which let be <hi>AKF,</hi> &amp; <hi>FIB.</hi> Then I say, that the rectiline figures <hi>AKF,</hi> and <hi>FIB,</hi>
                        <figure/>
haue the proportion geue<g ref="char:cmbAbbrStroke">̄</g> (namely, the propor<g ref="char:EOLhyphen"/>tion of the line <hi>GC</hi> to the line <hi>CD</hi>) and are e<g ref="char:EOLhyphen"/>quall to the rectiline figure geuen <hi>ABH</hi> vnto which they are described like and in like sort si<g ref="char:EOLhyphen"/>tuate. For forasmuch as <hi>AFB</hi> is a semicircle, therefore the angle <hi>AFB</hi> is a right angle (by the 31. of the third) and <hi>FE</hi> is a perpendicular line.<note place="margin">Demonstar<g ref="char:EOLhyphen"/>tion of the same.</note> Wherefore (by the 8. of this booke) the triangles <hi>AFE</hi> and <hi>FBE</hi> are like both to the whole triangle <hi>AFB</hi> and also the one to the other. Wherefore (by the 4. of this booke) as the line <hi>AF</hi> is to the line <hi>FB,</hi> so is the line <hi>AE</hi> to the line <hi>EF,</hi> and the line <hi>EF</hi> to the line <hi>EB,</hi> which are sides co<g ref="char:cmbAbbrStroke">̄</g>tayning equall angles. Wher<g ref="char:EOLhyphen"/>fore (by the 22. of this booke) as the rectiline fi<g ref="char:EOLhyphen"/>gure described of the line <hi>AF</hi> is to the rectiline figure described of the line <hi>FB,</hi> so is the recti<g ref="char:EOLhyphen"/>line figure described of the line <hi>AE</hi> to the recti<g ref="char:EOLhyphen"/>line figure described of the line <hi>EF,</hi> the sayd rectiline figures being like and in like sort
<pb facs="tcp:7062:212"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ituate. But as the rectiline figure described of the line <hi>AE</hi> being the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>irst, is to the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described of the line <hi>EF</hi> being the second, so is the line <hi>AE</hi> the first, 10. the line <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>B</hi> the third (by the 2. Corollary of the 20. of thys booke). Wherfore the recti<g ref="char:EOLhyphen"/>line figure described of the line <hi>AF</hi> is to the rectiline figure described of the line <hi>FB,</hi> as the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> is to the line <hi>EB.</hi> But the line <hi>AE</hi> is to, the line <hi>EB</hi> (by construction) as the line <hi>GC</hi> is to the line <hi>CD.</hi> Wherefore (by the 11. of the fift) as the line <hi>GC</hi> is to the line <hi>CD,</hi> so is the r<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described of the line <hi>AF</hi> to the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> the line <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>B,</hi> the sayd rectiline figures being like and in like sort described. But the <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> described o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the lines <hi>AF</hi> and <hi>FB,</hi> are equall to the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure d<gap reason="illegible" resp="#PDCC" extent="5 letters">
                           <desc>•••••</desc>
                        </gap>bed o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the line <hi>AB,</hi> vnto which they are (by construction) described lyke and in like sort situate. Wherefore there are described two rectiline figures <hi>AKF</hi> and <hi>FIB</hi> equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ll and like vnto the rectiline figure geuen <hi>ABH</hi> and in like sort situate, and they ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e also the one to the other the proportion geuen, namely, the proportion of the line <hi>GC</hi> to the line <hi>CD</hi>: which was required to be done.</p>
                  </div>
                  <div type="subsection">
                     <head>¶The first Corollary.</head>
                     <p>
                        <hi>To resolue a rectiline figure geue into two like rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>igures which shall hau<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> also a proporti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> ge<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e<g ref="char:cmbAbbrStroke">̄</g>.</hi> For i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> there be put three right lines in the proportio<g ref="char:cmbAbbrStroke">̄</g> geue<g ref="char:cmbAbbrStroke">̄</g>, and if the line <hi>AB</hi> be cu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> in the same proportion that the first line is to the third,<note place="margin">
                           <hi>The first Co<g ref="char:EOLhyphen"/>rollary.</hi>
                        </note> the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igures described of the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> and <hi>FB</hi> (which figures haue the same proportion that the lines <hi>AE</hi> and <hi>EB</hi> haue) shall be in double proportion to that which the lines <hi>AF</hi> and <hi>FB</hi> are (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>irs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>orollary o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the 20. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> this booke). Wherefore the right lines <hi>AF</hi> and <hi>FB</hi> are the o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e to the other in the same proportion that the first of the three lines put is to the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>or t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> line to the third, namely, the line <hi>AE</hi> to the line <hi>EB</hi> is in dou<g ref="char:EOLhyphen"/>ble propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tion that it is to the second, by the 10. de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>inition of the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ The second Corollary.</head>
                     <p>
                        <hi>Hereby may we learne, how from a rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>igure geuen to take away a part appointed, lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ing, the rest of the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>igure like vnto the whole.</hi>
                        <note place="margin">
                           <hi>The second Corollary.</hi>
                        </note> For if fro<g ref="char:cmbAbbrStroke">̄</g> the right line <hi>AB</hi> be cut of a part appoynted, namely, <hi>EB</hi> (by the 9. of this booke) as the line <hi>AE</hi> is to the line <hi>EB,</hi> so is the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described of the line <hi>AF</hi> to the rectiline figure described of the line <hi>FB</hi> (the sayd <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igures being supposed to be like both the one to the other and also to the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described of the line <hi>AB,</hi> and being also in like sort situate). Wherfore taking away <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rom the rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure described of the line <hi>AB,</hi> the rectiline figure de<g ref="char:EOLhyphen"/>scribed of the line <hi>FB,</hi> the residue, namely, the rectiline figure described of the line <hi>AF</hi> shall be both like vnto the whole rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure geuen described of the line <hi>AB,</hi> and in like sort situate.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ The third Corollary.</head>
                     <p>
                        <hi>To compose two like rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>igures into one rectiline figure like and equall to the same figures.</hi> Let their sides of like proportio<g ref="char:cmbAbbrStroke">̄</g> be set so that they make a right angle,<note place="margin">
                           <hi>The third Corollary.</hi>
                        </note> as the lines <hi>AF</hi> and <hi>FB</hi> are. And vpo<g ref="char:cmbAbbrStroke">̄</g> the line subtending the said angle, namely, the line <hi>AB,</hi> describe a rectiline <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>igure like vnto the rectiline figures geuen and in like sort situate (by the 18. of this booke) and the same shall be equall to the two rectiline figures geuen (by the 31. of this booke).</p>
                  </div>
                  <div type="subsection">
                     <head>¶ The second Proposition.</head>
                     <p>
                        <hi>If two right lines cut the one the other obtuseangled wise, and from the endes of the lines which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ut the one the other be drawen perpendicular lines to either line: the lines which are betwene the endes and the perpendicular lines are cut reciprokally.</hi>
                     </p>
                     <p>
                        <pb n="182" facs="tcp:7062:212"/>Suppose that there be two right lines <hi>AB</hi> and <hi>GD</hi> cutting the one the other in the point <hi>E,</hi> and making an obtuse angle in the section <hi>E.</hi> And from the endes of the lines, namely, <hi>A</hi> and <hi>G,</hi> let there be drawen to either line perpendicular lines, namely, from the point <hi>A</hi> to the line <hi>GD,</hi> which let be <hi>AD,</hi> and from the point <hi>G</hi> to the right line <hi>AB:</hi> which let be <hi>GB.</hi> Then I say, that the
<figure/>
right lines <hi>AB</hi> and <hi>GD</hi> do, betwene the end <hi>A</hi> and the perpendicular <hi>B,</hi> and the end <hi>G</hi> and the perpendicular <hi>D,</hi> cut the one the other reciprokally in the point <hi>E:</hi> so that as the line <hi>AE</hi> is to the line <hi>ED,</hi> so is the line <hi>GE</hi> to the line <hi>EB.</hi>
                        <note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion of this proposition.</hi>
                        </note> For foras<g ref="char:EOLhyphen"/>much as the angles <hi>ADE</hi> and <hi>GBE,</hi> are right angles, therfore they are equall. But the angles <hi>AED</hi> and <hi>GEB</hi> are also e<g ref="char:EOLhyphen"/>quall (by the 15. of the first). Wherefore the angles remayning, namely, <hi>EAD,</hi> &amp; <hi>EGB,</hi> are equall (by the Corollary of the 32. of the first). Wherefore the triangles <hi>AED</hi> and <hi>GEH,</hi> are equiangle. Wherfore the sides about the equall angles shall be proportionall (by the 4. of the sixt). Wherfore as the line <hi>AE</hi> is to the line <hi>ED,</hi> so is the line <hi>GE</hi> to the line <hi>EB.</hi> If therefore two right lines cut the one the other obtuseangled wife. &amp;c: which was required to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ The third Proposition.</head>
                     <p>
                        <q>If two right lines make an acute angle, and from their endes be drawen to ech line perpendicular lines cutting them: the two right lines geuen shall be reciprokally pro<g ref="char:EOLhyphen"/>portionall as the segmentes which are about the angle.</q>
                     </p>
                     <p>Suppose that there be two right lines <hi>AB</hi> and <hi>GB,</hi> making an acute angle <hi>ABG.</hi> And from the poyntes <hi>A</hi> and <hi>G</hi> let there be drawen vnto the lines <hi>AB</hi> and <hi>GB</hi> per<g ref="char:EOLhyphen"/>pendicular lines <hi>AC</hi> and <hi>GE,</hi> cutting the lines <hi>AB</hi> and <hi>GB</hi> in the poyntes <hi>E</hi> and <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi> Then I say, that the lines, namely <hi>AB</hi> to <hi>GB,</hi> are reciprokally proportionall, as the seg<g ref="char:EOLhyphen"/>mentes, namely, <hi>CB</hi> to <hi>EB</hi> which are about the acute
<figure/>
angle <hi>B.</hi>
                        <note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion of this propositions</hi>
                        </note> For forasmuch as th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> right angles <hi>ACB</hi> and <hi>GER</hi> are equall, and the angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <hi>ABG</hi> is common to the triangles <hi>ABC,</hi> and <hi>GBE</hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> therefore the angles re<g ref="char:EOLhyphen"/>mayning <hi>BAC</hi> and <hi>EGB</hi> are equall (by the Corolla<g ref="char:EOLhyphen"/>ry of the 32. of the first). Wherfore the triangles <hi>ABC</hi> and <hi>GBE</hi> are equiangle. Wherefore the side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> about the equall angles are proportionall (by the 4. of the sixe)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> so that, as the line <hi>AB</hi> is to the line <hi>FC,</hi> so is the line <hi>GB</hi> to the line <hi>BE.</hi> Wherefore alternately as the line <hi>AB</hi> is to the line <hi>GB</hi> so is the line <hi>CB</hi> to the line <hi>BE.</hi> If therefore two right lines mak<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>te angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp;c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> which was required to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> The fourth Proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </head>
                     <p>
                        <q>If in a circle be drawen two right lines cutting the one the other, the sections of the one to the sections of the other shall be reciprokally proportionall.</q>
                     </p>
                     <p>In the circle, <hi>AGB</hi> let these two right lines <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> one the other in the poynt <hi>E.</hi>
                        <note place="margin">
                           <hi>Demonstrati<g ref="char:EOLhyphen"/>on of this pro<g ref="char:EOLhyphen"/>position.</hi>
                        </note> Th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> I say, that reciprokally <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> line <hi>AE</hi> is to the line <hi>ED,</hi> so is the line <hi>GE</hi> to the line <hi>EB.</hi> For forasmuch as (by the 35. of the third) the rectangle fi<g ref="char:EOLhyphen"/>gure
<pb facs="tcp:7062:213"/>
contayned vnder the lines <hi>AE</hi> and
<figure/>
                        <hi>EB</hi> is equall to the rectangle figure contay<g ref="char:EOLhyphen"/>ned vnder the lines <hi>GE</hi> and <hi>ED,</hi> but in e<g ref="char:EOLhyphen"/>quall rectangle parallelogrammes the sides about the equall angles are reciprokall (by the 14. of the sixt). Therefore the line <hi>AE</hi> is to the line <hi>ED</hi> reciprokally as the line <hi>GE</hi> is to the line <hi>EB</hi> (by the second defi<g ref="char:EOLhyphen"/>nition of the sixt). If therefore in a circle be drawen two right lines. &amp;c: which was required to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>¶The fift Proposition.</head>
                     <p>
                        <q>If from a poynt geuen be drawen in a plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>icies two right lines to the concaue circumference of a circle: they shall be reci<g ref="char:EOLhyphen"/>prokally proportionall with their partes take<g ref="char:cmbAbbrStroke">̄</g> without the circle. And moreouer a right line drawen from the sayd poynt &amp; touching the circle, shall be a meane proportionall betwene the whole line and the vtter segment.</q>
                     </p>
                     <p>Suppose that there be a circle <hi>ABD,</hi> and without it
<figure/>
take a certayne poynt, namely, <hi>G.</hi> And from the point <hi>G</hi> drawe vnto the concaue circumference two right lines <hi>GB</hi> and <hi>GD,</hi> cutting the circle in the poyntes <hi>C</hi> and <hi>E.</hi> And let the line <hi>GA</hi> touch the circle in the point <hi>A.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say, that the lines, namely, <hi>GB</hi> to <hi>GD</hi> are reciprokally as their parts taken without the circle, namely, as <hi>GC</hi> to <hi>GE.</hi>
                        <note place="margin">
                           <hi>Demonstrati<g ref="char:EOLhyphen"/>on of the first part of this proposition.</hi>
                        </note> For forasmuch as (by the Corollary of the 36. of the third) the rectangle figure contayned vnder the lines <hi>GB</hi> and <hi>GE</hi> is equall to the rectangle figure contayned vn<g ref="char:EOLhyphen"/>der the lines <hi>GD</hi> and <hi>GC,</hi> therefore (by the 14. of the sixt) reciprokally as the line <hi>GB</hi> is to the line <hi>GD,</hi> so is the line <hi>GC</hi> to the line <hi>GE,</hi> for they are sides contayning equall angles. I say moreouer, that betwene the lines <hi>GB</hi> and <hi>GE,</hi> or betwene the lines <hi>GD</hi> and <hi>GC</hi> the touch line <hi>GA</hi> is a meane proportionall.<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the se<g ref="char:EOLhyphen"/>cond part.</hi>
                        </note> For forasmuch as the re<g ref="char:EOLhyphen"/>ctangle figure comprehended vnder the lines <hi>GB</hi> and <hi>GE</hi> is equall to the square made of the line <hi>AG</hi> (by the 36. of the third) it followeth that the touch line <hi>GA</hi> is a meane propor<g ref="char:EOLhyphen"/>tionall betwene the extremes <hi>GB</hi> and <hi>GE</hi> (by the second part of the 17. of the sixt) for that by that Proposition the lines <hi>GB, GA,</hi> and <hi>GE</hi> are proportionall. And by the same reason may it be proued that the line <hi>GA</hi> is a meane proportionall betwene the lines <hi>GD</hi> and <hi>GC,</hi> and so of all others. If therefore from a poynt geuen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp;c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> which was required to be demonstrated.</p>
                  </div>
               </div>
            </div>
            <trailer>The end of the sixth booke of Euclides Elementes.</trailer>
         </div>
         <div n="7" type="book">
            <pb n="183" facs="tcp:7062:213"/>
            <head>¶The seuenth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>HETHERTO IN THE SIXE bookes before hath <hi>Euclide</hi> passed through, and entreated of the Elementes of Geome<g ref="char:EOLhyphen"/>trie without the ayde and succor of nomber. But the matters which remayne to be taught and to be spoken of in these his Geometri<g ref="char:EOLhyphen"/>call bookes which follow as in the tenth, eleuenth, and so forth, he could by no meanes fully and clearely make plaine &amp; demonstrate, without the helpe and ayd of nombers. In the tenth is entreated of lines irrationall and vncertaine, and that of many &amp; sondry kindes: and in the eleuenth &amp; the other following he teacheth the natures of bodyes, and compareth theyr sides and lines together. All which for the most part are also irrationall. And as rationall quantites, and the comparisons and proportions of them, cannot be knowen, nor exactly tried, but by the meane of nomber, in which they are first sene and perceiued: euen so likewise cannot irrationall quantities be knowen and found out without nomber. As straightnes is the triall of crokednes, and inequalitie is tried by equalitie: so are quantities irrationall perceiued and knowen by quantities rationall: which are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>irst and chiefely found among nombers. Wherefore in these three bookes following, being as it were in the middest of his Elementes, he is compelled of necessitie to entreate of nombers,<note place="margin">
                  <hi>Why Euclide in the middest of his workes was compelled to adde these three bookes of numbers.</hi>
               </note> although not so fully, as the nature of nombers requireth, yet so much as shall seme to be fit, and sufficiently to serue for his purpose. Wherby is sene the necessitie, that the one Arte, namely, Geometrie, hath of the other, namely, of Arithmeticke. And also of what excelle<g ref="char:cmbAbbrStroke">̄</g>cy and worthines Arithmeticke is aboue Geo<g ref="char:EOLhyphen"/>metrie:<note place="margin">
                  <hi>Arithmetike of more excellency then Geometry.</hi>
               </note> in that, Geometrie boroweth of it principles, ayd, and succor, and is as it were maymed with out it. Whereas Arithmeticke is of it selfe, sufficient and neadeth not at all any ayde of Geometrie, but is absolute and perfit in it selfe, and may well be taught and attayned vnto without it. Agayne the mat<g ref="char:EOLhyphen"/>ter or subiect where about Geometrie is occupied, which are lines, figures, and bodyes, are such as of<g ref="char:EOLhyphen"/>fer them selues to the sences, as triangles, squares, circles, cubes, and other are sene &amp; iudged to be such as they are, by the sight: but nomber, which is the subiect and matter of Arithmeticke, falleth vnder no sence, nor is represented by any shape, forme, or figure: and therefore cannot be iudged by any sence, but only by consideration of the minde, and vnderstanding. Now thinges sensible are farre vnder in de<g ref="char:EOLhyphen"/>gree then are thinges intellectuall:<note place="margin">
                  <hi>Things intellec<g ref="char:EOLhyphen"/>tuall of more worthines the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> things sensible.</hi>
               </note> and are of nature much more grosse then they. Wherefore nomber, as being only intellectuall, is more pure, more immateriall, and more subtile, farre then is magnitude: and exte<g ref="char:cmbAbbrStroke">̄</g>deth it selfe farther. For Arithmeticke, not onely aydeth Geometrie: but ministreth principles, and groundes to many other, nay rather to all other sciences and artes.<note place="margin">
                  <hi>Arithmetike ministreth prin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ciples and groundes in a maner to all sciences.</hi>
               </note> As to musicke, Astronomy, na<g ref="char:EOLhyphen"/>turall philosophy, perspectiue, with others. What other thing is in musicke entreated of, then nomber contracted to sound and voyce? In Astronomie, who without the knowledge of nomber can doo any thing, either in searching out of the motions of the heauens, and their courses, either in iudging and foreshewing the effectes of them? In natural philosophie, it is of no small force. The wisest and best lear<g ref="char:EOLhyphen"/>ned philosophers that haue bene, as <hi>Pithagoras, Timeus, Plato,</hi> and their followers, found out &amp; taught most pithely and purely, the secret and hidden knowledge of the nature and condicion of all thinges, by nombers, and by the proprieties and passions of them. Of what force nomber is in perspectiue, let him declare and iudge, who hath any thing traueled therein. Yea to be short, what can be worthely and with prayse practised in common life of any man of any condition, without the knowledge of nomber. Yea it hath bene taught of the chiefest amongst philosophers, that all naturall thinges are framed, and haue their constitucion of nomber. <hi>Boetius</hi> sayth <hi>Hoc fuit principal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in anim<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ditoris exemplar:</hi>
               <note place="margin">
                  <hi>Boetius: Cap.</hi> 2. <hi>Lib. prim. Arithmeti.</hi>
               </note> Nomber (sayth he) was the principall example and patron in the minde of the creator of the world. Doth not that great philopher <hi>Timaus</hi> in his booke,<note place="margin">
                  <hi>Timaus.</hi>
               </note> &amp; also <hi>Plato</hi> in his <hi>Tim<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>,</hi> following him, shew how the soule is composed of harmonicall nombers, and consonantes of musicke. Nomber compaseth all thinges, and is (after these men) the being and very essence of all thinges. And ministreth ayde and helpe, as to all other knowledges, so also no small to Geometrie. Which thing causeth <hi>Euclid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </hi> in the midest of his booke of Geometrie, to inserte and place these three bookes of Arithmeticke: as without the ayde of which he could not well passe any father.</p>
            <p>In this seuenth booke, he <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>irst placeth the generall principles, and first groundes of Arithmetick<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, and setteth the diffinitions of nomber or multitude and the kinds therof: as in the first boke, he did of magnitude and the kindes and partes thereof.<note place="margin">
                  <hi>The argument of the seuenth booke.</hi>
               </note> After that he entreateth of nombers generally, and of theyr partes: and searcheth and demonstrateth in generall the most common passions and proprieties of the same, and chie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ely of nombers prime or incommensurable, and of nombers composed or com<g ref="char:EOLhyphen"/>mensurable and of their proprieties and partely also of the comparison o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> proportion of one nomber to an other.</p>
            <div type="part">
               <pb facs="tcp:7062:214"/>
               <head>¶Definitions.</head>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The first defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>1 Vnitie is that, whereby euery thing that is, is sayd to be on.</head>
                  <p>As a point in magnitude, is the least thing in magnitude, and no magnitude at all, &amp; yet the ground and beginning of all magnitudes: euen so is vnitie in multitude or nomber, the least thing in nomber, and no nomber at all, and yet the ground and beginning of all nombers. And therefore it is here in this place, of <hi>Euclide</hi> first defined: as in the first booke, for the like reason and cause was a point first defined. <hi>Vnitie,</hi> sayth <hi>Euclide, is that whereby euery thing is sayd to be one:</hi> that is, vnitie is that, whereby euery thing is deuided and seperated from an other, and remayneth on in it selfe pure and distinct fro<g ref="char:cmbAbbrStroke">̄</g> all others. O<g ref="char:EOLhyphen"/>therwise, were not this vnitie,<note place="margin">
                        <hi>Without vnity should be confu<g ref="char:EOLhyphen"/>sion of thinges.</hi>
                     </note> whereby all thinges are seioyned the on from the other, all things should suffer mixtion and be in confusion. And where confusion is, there is no order, nor any thing can be ex<g ref="char:EOLhyphen"/>actly knowen, either what it is, or what is the nature, and what are the properties thereof. Vnitie there<g ref="char:EOLhyphen"/>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e is that which maketh euery thing to be that which it is. <hi>Boetius</hi> sayth very aptly:<note place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>oetius in his booke d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> vnita<g ref="char:EOLhyphen"/>te &amp; vno.</hi>
                     </note> 
                     <hi>Vnum quod<expan>
                           <am>
                              <g ref="char:abque"/>
                           </am>
                           <ex>que</ex>
                        </expan>, idea est, quia vnum numero est,</hi> that is euery thing therfore is (that is, therefore hath his being in nature, and is that it is) for that it is on in nomber. According whereunto <hi>Iordane</hi> (in that most excellent and abso<g ref="char:EOLhyphen"/>lute worke of A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ithmeticke which he wrote) defineth vnitie after this maner.</p>
                  <p>
                     <hi>Vnitas, est res per se discretio:</hi> that is, vnitie is properly, and of it selfe the difference of any thing. That is,<note place="margin">
                        <hi>An other desi<g ref="char:EOLhyphen"/>nition of vnity.</hi>
                     </note> vnitie is that whereby euery thing doth properly and essentially differ, and is an other thing from all others. Certainely a very apt de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inition and it maketh playne the definition here set of <hi>Euclide.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The second de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>2 Number is a multitude composed of vnities.</head>
                  <p>As the number of three, is a multitude composed and made of three vnities. Likewise the number of fiue is nothing ells, but the composition &amp; putting together of fiue vnities. Although as was before sayde, betwene a poynt in magnitude, and vnitie in multitude, there is great agreement and many thi<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap> are com<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>on to them both, (for as a poynt is the beginning of magnitude, so is vnitie the be<g ref="char:EOLhyphen"/>ginning of nomber. And as a poynt in magnitude is indiuisible, so is also vnitie in number indiuisible) yet in this they differ and disagree.<note place="margin">
                        <hi>Differen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e be<g ref="char:EOLhyphen"/>twene a point and vnity.</hi>
                     </note> There is no line or magnitude made of pointes, as of his partes. So that although a point be the beginning of a lyne, yet is it no part therof. But vnitie, as it is the begin<g ref="char:EOLhyphen"/>ning of number, so is it also a part therof, which is somewhat more manifestly set of <hi>Boetius</hi> in an other d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ffinition of number which he geueth in his Arithmetike,<note place="margin">
                        <hi>Boetius.</hi>
                     </note> which is thus.</p>
                  <p>
                     <hi>Numerus, est quantitat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> acernus ex vnitatibus profusus,</hi> that is. Number is a masse or heape of quanti<g ref="char:EOLhyphen"/>ties produced of vnities: which diffinition in substance is all one with the first, wherin is said most plain<g ref="char:EOLunhyphen"/>ly, that the heape or masse,<note place="margin">
                        <hi>An other desi<g ref="char:EOLhyphen"/>nition of num<g ref="char:EOLhyphen"/>ber.</hi>
                     </note> that is, the whole substance of the quantitie of number is produced &amp; made of vnities. So that vnitie is as it were the very matter of number. As foure vnities added together are the matter wherof the number 4. is made, &amp; eche of these vnities is a part of the number foure, name<g ref="char:EOLhyphen"/>ly, a fourth part, or a quarter. Vnto this diffinition agreeth also the definition geuen of <hi>Iordane,</hi>
                     <note place="margin">
                        <hi>Iordane.</hi>
                     </note> which is thus. <hi>Number is a quantitie which gathereth together thinges seuered a sonder.</hi>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition of num<g ref="char:EOLhyphen"/>bers.</hi>
                     </note> As fiue men beyng in them<g ref="char:EOLhyphen"/>selues seuered and distincte, are by the number fiue brought together, as it were into one masse, and so of others. And although vnitie be no number, yet it contayneth in it the vertue and power of all num<g ref="char:EOLhyphen"/>bers, and is set and taken for them.<note place="margin">
                        <hi>Vnity hath in it the vertue and power of all numbers.</hi>
                     </note>
                  </p>
                  <p>In this place (for the Farther elucidation of thinges, partly before set, and chiefly hereafter to be set, because <hi>Euclide</hi> here doth make mention of diuers kyndes of numbers, and also defineth the same) is to be noted, that number may be considered three maner of wayes. First, number may be considered absolutely, without comparyng it to any other number,<note place="margin">
                        <hi>Number consi<g ref="char:EOLhyphen"/>dered three ma<g ref="char:EOLhyphen"/>ner of way<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> or without applieng it to any other thing, one<g ref="char:EOLhyphen"/>ly vewing <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd paysing what it is in it selfe, and in his owne nature onely, and what partes it hath, and what proprieties and passions. As this number sixe, may be considered absolutely in his owne nature, that it is an euen number, and that it is a perfect number, and hath many mo conditions and proprie<g ref="char:EOLhyphen"/>ties. And so conceiue ye of all other numbers, whatsoeuer, of 9. 12. and so forth.</p>
                  <p>An other way, number may be co<g ref="char:cmbAbbrStroke">̄</g>sidered by way of co<g ref="char:cmbAbbrStroke">̄</g>parison, and in respect of some other number either as equall to it selfe, or as greater the<g ref="char:cmbAbbrStroke">̄</g> it selfe, or as lesse the<g ref="char:cmbAbbrStroke">̄</g> it selfe. As 12. may be co<g ref="char:cmbAbbrStroke">̄</g>sidered, as co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>pared to 12. which is equall vnto it, or as to 24. which is greater then it, for 12 is the halfe thereof, of as to 6. which is lesse then it, as beyng the double therof. And of this consideration of numbers ariseth and springeth all kyndes and varieties of proportio<g ref="char:cmbAbbrStroke">̄</g>: as hath before bene declared in the explanation of the principles of the fift booke, so that of that matter it is needelesse any more to be sayd in this place.</p>
                  <p>The third way may numbers be considered as they are applied to formes and figures of Geometry<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And numbers so considered are not represented by figures or charactes of number commonly vsed in Arithmetique, but are signified by certayne pointes or prickes, which represent the vnities which they contayne: which, accordyng to the diuersitie of the disposition and placing of them, may represent di<g ref="char:EOLhyphen"/>uers formes and figures of Geometry<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and accordyng to the nature of the figure which it represen<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eth, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t taketh his name, and is called a trianguler number, a square number, a cube
<figure/>
number or after any other figure. As if the figure of 10. be described by his vni<g ref="char:EOLhyphen"/>ties
<pb n="184" facs="tcp:7062:214"/>
that they be orderly set in a straight course, so that they represent the forme of a line, then is the number 10 called a lineall number. And if the same number to be so described by his vnities,
<figure/>
that it shew forth the forme of a triangle, then it is called a trianguler nu<g ref="char:cmbAbbrStroke">̄</g>ber<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as ye here see. Likewise if 12 be in such sort described by his vnities, that it represe<g ref="char:cmbAbbrStroke">̄</g>teth that form<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or figure which in Geometry is called a figure on the one side longer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then shall the number 12 be called a number hauyng the one side longer, and so may you conceaue of all others.</p>
                  <p>Thus much of this for the declaration of the thinges following.</p>
               </div>
               <div type="section">
                  <head>3 A part is a lesse number in comparison to the greater when the lesse measureth the greater.<note place="margin">
                        <hi>The third defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                  </head>
                  <p>As the number 3 compared to the number 12. is a part. For 3 is a lesse number then is 12. and more<g ref="char:EOLhyphen"/>ouer it measureth 12 the greater number. For 3 taken (or added to it selfe) certayne times (namely, 4 tymes) maketh 12. For 3 foure tymes is 12. Likewise is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a part of 8: 2 is lesse then 8, and taken 4 tymes it maketh 8. For the better vnderstandyng of this diffinition, and how this worde Parte, is diuersly ta<g ref="char:EOLhyphen"/>ken in Arithmetique and in Geometry, read the declaration of the first diffinition of the 5. booke.</p>
               </div>
               <div type="section">
                  <head>4 Partes are a lesse number in respect of the greater, when the lesse measu<g ref="char:EOLhyphen"/>reth not the greater.<note place="margin">
                        <hi>The fourth de<g ref="char:EOLhyphen"/>finition.</hi>
                     </note>
                  </head>
                  <p>As the number 3 compared to 5, is partes of 5 and not a part. For the number 3 is lesse then the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLunhyphen"/>ber 5, and doth not measure 5. For taken once it maketh but 3. once 3 is 3, which is lesse then 5. and 3 taken twise maketh 6, which is more then 5. Wherfore it is no part of 5 but partes, namely, three fifth partes of 5. For in the number 3 there are 3 vnities, and euery vnitie is the fifth part of 5. Wherfore 3 is three fifth partes of 5, and so of others.</p>
               </div>
               <div type="section">
                  <head>5 Multiplex is a greater number in comparison of the lesse, when the lesse measureth the greater.<note place="margin">
                        <hi>The fifth defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                  </head>
                  <p>As 9 compared to 3 is multiplex, the number 9 is greater then the number 3. And moreouer 3 the lesse number measureth 9 the greater number. For 3 taken certaine tymes, namely, 3 tymes maketh 9. three tymes three is 9. For the more ample and full knowledge of this definition, read what is sayd in the explanation of the second definition of the 5 booke, where multiplex is sufficiently entreated of with all his kyndes.</p>
               </div>
               <div type="section">
                  <head>6 An euen number is that, which may be deuided into two equal partes.<note place="margin">
                        <hi>The sixth defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                  </head>
                  <p>As the number 6 may be deuided into 3 and 3 which are his partes, and they are equall, the one not exceding the other. This definition of Euclide is to be vnderstand of two such equall partes, which ioy<g ref="char:EOLhyphen"/>ned together, make the whole number: as 3 and 3 (the equall partes of 6) ioyned together, make 6, for otherwise many numbers both euen and odde may be deuided into many equall partes, as into 4. 5. 6<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or mo, and therfore into 2. As 9 may be deuided into 3 and 3 which are his partes, and are also equall, for the one of them excedeth not the other: yet is not therfore this number <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> an euen number, for that 3 and 3 (these equall partes of 9) added together make not 9 but onely 6. Likewise taking the definiti<g ref="char:EOLhyphen"/>on so generally, euery number whatsoeuer should be an euen number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for in that sort of vndersta<g ref="char:cmbAbbrStroke">̄</g>ding there is no number, but that it may be deuided into two equall partes: as this number 7 may be deui<g ref="char:EOLhyphen"/>ded into 3 partes namely, 3. 1. and 3. of which two, namely, 3 and 3 are equal, yet i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> not 7 an euen num<g ref="char:EOLhyphen"/>ber, because 3 and 3 added together, make not 7. <hi>Boetius</hi> therfore in the first booke of his Arithmetike,<note place="margin">
                        <hi>Boetius.</hi>
                     </note> for the more playnes, after this maner defineth an euen number: <hi>Num<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ru<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> est, qui potest in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>qu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>lia due diuidi, vna medi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>n intercedenta,</hi> that is:</p>
                  <p>An euen number is that which may be deuided into two equall partes, without an vnitie comming betwene them.<note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition of an eue<g ref="char:cmbAbbrStroke">̄</g> number.</hi>
                     </note> As 8 is deuided into 4 and 4 two equall partes without an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> vnitie comming betwene them, which added together, make 8, so that the sence of this definition is, that an euen number is that which is deuided into two such equall partes, which are his two halfe partes.</p>
                  <p>Here is also to be noted,<note place="margin">
                        <hi>Note.</hi>
                     </note> that a part is taken in this de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inition, and in certaine definitions following, not in the signification as it was before de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ined, namely, for such a part as<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> measureth the whole num<g ref="char:EOLhyphen"/>ber, but for any part which helpeth to the making of the whole, and into which the whole may be r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>solued, so are 3 and 5 partes of 8 in this sence but not in the other sence. For neyther 3 not 5 measureth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Pithagoras and his scholers gaue an other definition of an euen number (which definition <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>
                     </hi> al<g ref="char:EOLhyphen"/>so hath) after this maner.<note place="margin">
                        <hi>Pithagora<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note>
                  </p>
                  <p>
                     <pb facs="tcp:7062:215"/>
                     <hi>An euen number is that which in one and the same deuision is deuided into the greatest, and into the least: in<g ref="char:EOLhyphen"/>to the greatest, as touching space, and into the least as touching quantitie.</hi>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note> As 10. is deuided into 5 &amp; 5. which <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> his greatest partes (which greatnes of partes he calleth space) and in the same diuision the number 10 is deuided but into two partes: but into lesse the<g ref="char:cmbAbbrStroke">̄</g> two partes nothing can be deuided, which thing he calleth quantitie, so that 10 deuided into 5 and 5. in that one deuision, is deuided into the greatest, namely, into his halues<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and into the least, namely, into two partes, and no mo.</p>
                  <p>There is also another de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inition more auncient, which is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                     <hi>An euen number is that which may be deuided into two equall partes, and into two vnequall partes, but in ney<g ref="char:EOLhyphen"/>ther diuision (to the constitucion of the whole) to the euen part is added the odde, neither to the odde is added the euen.</hi> As 8 may be deuided diuersly, partly into euen partes, as into 4 and 4. likewise into 6 and 2. and partly into odde partes, as into 5 and 3. also into 7 and 1. In all which deuisions, ye see no odde parte ioyned to an euen, nor an euen part ioyned to an odde: but if the one be euen, the other is euen, and if the one be odde, the other is odde. In the two first deuisions, both partes were euen, and in the two last, both partes were odde. It is to be considered that the two partes added together must make the whole.</p>
                  <p>This diffinition is generall and common to all euen numbers, except to the number 2. which can not be deuided into two vnequall partes, but onely into two vnities, which are equall.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>There is yet geue<g ref="char:cmbAbbrStroke">̄</g> an other diffinition of an euen number, namely, thus. <hi>An euen number is that which onely by an vnitie either aboue it or vnder it, differeth from an odde.</hi> As 8 being an euen number differeth from 9 an odde number, being aboue it but by one. And also from 7. vnder it, it differeth likewise by one, and so o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> others.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The seuenth definition.</hi>
                     </note>7 An odde number is that which cannot be deuided into two equal partes: or that which onely by an vnitie differeth from an euen number.</head>
                  <p>As the number 5 can be by no meanes deuided into two equall partes, namely, such two which ad<g ref="char:EOLhyphen"/>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d together, shall make 5. Or by the second definition, 5 an odde number, differeth from 6 an euen number aboue <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t, by 1. And the same 5 differeth from 4 an euen number vnder it likewise by 1.</p>
                  <p>
                     <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> number after this maner.</hi>
                  </p>
                  <p>
                     <hi>An odde number is that which cannot be deuided into two equall partes, but that an vnitie shall be betwene them</hi> As if ye deuide 5 into 2 and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. which are two equall partes,<note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition of an od number.</hi>
                     </note> there remaineth one or an vnitie be<g ref="char:EOLhyphen"/>twene them to make the whole number 5.</p>
                  <p>There is yet an other definition of an odde number. <hi>An odde number is that, which being deuided into two vnequall partes howsoeuer, the one is euer euen, and the other odde.</hi> As if 9 be deuided into two partes, which added together, maketh the whole, namely, into 4 and 5. which are vnequall:<note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note> ye see the one is euen, namely, 4 and the other is odde, namely, 5. so if ye deuide 9 into 6 and 3. or into 8 and 1. the one part is euer euen, and the other odde.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The eight defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>8 A number euenly euen (called in latine <hi>pariter par</hi>) is that number, which an euen number measureth by an euen number.</head>
                  <p>As 8 is a number euenly euen. For 4 an euen number measureth 8 by 2, which is also an euen num<g ref="char:EOLhyphen"/>ber. This definition hath much troubled many, and seemeth not a true definition, for that there are many numbers which euen numbers do measure, and that by euen numbers, which yet are not euen<g ref="char:EOLhyphen"/>ly eue<g ref="char:cmbAbbrStroke">̄</g> numbers, after most mens minds: as 24. which 6 an eue<g ref="char:cmbAbbrStroke">̄</g> number doth measure by foure, which is also an euen number, and yet as they thinke is not 24 an euenly euen number, for that 8 an euen nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber doth measure also <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>4 by 3. an odde number.<note place="margin">
                        <hi>Campane.</hi>
                     </note> Wherfore <hi>Campane</hi> to make his sentence plaine, after this maner setteth forth this definition. <hi>Pariter par est quem cuncti pares <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>um numerantes, paribus vicibu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> numerane,</hi> that is:</p>
                  <p>
                     <note place="margin">
                        <hi>An other de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>i<g ref="char:EOLhyphen"/>nition of an eue<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>ly euen num<g ref="char:EOLhyphen"/>ber.</hi>
                     </note>An euenly euen number is, when all the euen numbers which measure it, doo measure it by euen times, that is, by euen numbers, as 16. All the euen numbers whith measure 16, as are 8. 4. and 2. do measure it by euen numbers. As 8 by 2, twise 8 is 16:4 by 4, foure times 4 is 16: and 2 by 8, 8 times 2 is 16. Which particle <hi>(all euen number)</hi> added by <hi>Campane</hi> maketh 24 to be no euenly euen number. For that some one euen number measureth it by an odde number as 8 by 3. <hi>Flussates</hi> also is plainly of this minde,<note place="margin">
                        <hi>Flussates.</hi>
                     </note> that <hi>Euclide</hi> gaue not this definition in such maner as it is by <hi>Theon</hi> written, for the largenes &amp; generalitie therof, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or that it extendeth it to in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inite numbers which are not euenly euen as he thin<g ref="char:EOLhyphen"/>keth, for which cause in place therof, he geueth this definition.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                     <hi>A number euenly euen, is that which onely euen numbers do measure.</hi> As 16 is measured of none but of euen numbers, and therfore is euenly euen. There is also of <hi>Boetius</hi> geuen an other definition of more facili<g ref="char:EOLhyphen"/>tie,<note place="margin">
                        <hi>Boetius.</hi>
                     </note> including in it no doubt at all, which is most commonly vsed of all writers, and is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                     <hi>A number euenly euen is that which may be diuided into two euen partes, and that part agayne into two euen partes, and so continually deuiding without stay <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> come to vnitie.</hi> As by example<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 64. may be deuided
<pb n="185" facs="tcp:7062:215"/>
into 31 and. 32. And either of these partes may be deuided into two euen partes, for 32 may be deuided into 16 and 16. Againe, 16 may be deuided into 8 and 8 which are euen partes, and 8 into 4 and 4. A<g ref="char:EOLhyphen"/>gaine 4 into <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, and last of all may <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be deuided into one and one.</p>
               </div>
               <div type="section">
                  <head>9 A number euenly odde <hi>(called in latine pariter impar)</hi> is that which an e<g ref="char:EOLhyphen"/>uen number measureth by an odde number.<note place="margin">
                        <hi>The ninth definition.</hi>
                     </note>
                  </head>
                  <p>As the number 6 which 2 an euen number measureth by 3 an odde number, thre times 2 is 6. Like<g ref="char:EOLhyphen"/>wise 10. which 2. an euen number measureth by 5 an odde number. In this diffinition also is found by all the expositors of <hi>Euclide,</hi> the same want that was found in the diffinition next before. And for that it extendeth it selfe to large, for there are infinite numbers which euen numbers do measure by odde numbers, which yet after their mindes are not eue<g ref="char:cmbAbbrStroke">̄</g>ly odde nu<g ref="char:cmbAbbrStroke">̄</g>bers, as for example 12. For 4 an eue<g ref="char:cmbAbbrStroke">̄</g> nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber, measureth 12 by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> an odde number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> three times 4 is 12. yet is not 12 as they thinke an euenly odde number. Wherfore <hi>Campane</hi> amendeth it after his thinking,<note place="margin">
                        <hi>Campane.</hi>
                     </note> by adding of this worde <hi>all,</hi> as he did in the first, and defineth it after this maner.</p>
                  <p>
                     <hi>A number euenly odde is, when all the euen numbers which measure it, do measure it by vneuen tymes, that is, by an odde number.</hi>
                  </p>
                  <p>As 10. is a number euenly odde,<note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note> for no euen number but onely 2 measureth 10. and that is by 5 an odde number. But not all the euen numbers which measure 12. do measure it by odde numbers. For 6 an euen number measureth 12 by 2 which is also euen. Wherfore 12 is not by this definition a num<g ref="char:EOLhyphen"/>ber euenly odde. <hi>Flussates</hi> also offended with the ouer large generalitie of this definition to make the definition agree with the thing defined putteth it after this maner.<note place="margin">
                        <hi>Flussates.</hi>
                     </note>
                  </p>
                  <p>
                     <hi>A number euenly odde, is that which an odde number doth measure onely by an euen number.</hi>
                  </p>
                  <p>As 14. which 7. an odde number doth measure onely by 2. which is an euen number.<note place="margin">
                        <hi>An other definition.</hi>
                     </note> There is also an other definition of this kinde of number commonly geuen of more plaines, which is this.</p>
                  <p>
                     <hi>A number euenly odde in that which may be deuided into two equall partes, but that part cannot agayne be de<g ref="char:EOLhyphen"/>uided into two equall partes:</hi> as 6. may be deuided into two equall partes into 3. and 3. but neither of them can be deuided into two equall partes: for that 3. is an odde number and suffereth no such diuision.<note place="margin">
                        <hi>An other definition.</hi>
                     </note>
                  </p>
               </div>
               <div type="section">
                  <head>10 A number oddly euen <hi>(called im lattin in pariter par)</hi> is that which an odde number measureth by an euen number.<note place="margin">
                        <hi>The tenth definition.</hi>
                     </note>
                  </head>
                  <p>As the number 12: for 3. an odde number measureth 12. by 4. which is an euen number: three times 4. is 12.</p>
                  <p>This definition is not founde in the greeke neither was it doubtles euer in this maner written by <hi>Euclide</hi>:<note place="margin">
                        <hi>This definition not found in the Greeks.</hi>
                     </note> which thing the slendernes and the imperfection thereof and the absurdities following also of the same declare most manifestly. The definition next before geuen is in substance all one with this, For what number soeuer an euen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> number doth measure by an odde, the selfe same number doth an odde number measure by an euen. As 2. an eue<g ref="char:cmbAbbrStroke">̄</g> number measureth 6. by 3. an odde number. Wherfore 3. an odde number doth also measure the same number 6. by 2. an eue<g ref="char:cmbAbbrStroke">̄</g> nu<g ref="char:cmbAbbrStroke">̄</g>ber. Now if these two definiti<g ref="char:EOLunhyphen"/>ons be definitions of two distinct kindes of numbers, then is this number 6. both euenly euen, and also euenly odde and so is contayned vnder two diuers kindes of numbers. Which is directly agaynst the authoritie of <hi>Euclide</hi> who playnely. p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ouo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h here after in the 9. booke, that euery nomber whose halfe is an odde number, is a number euenly odde onely. Flussates hath here very well noted, that these two euenly odde, and oddely euen, were taken of <hi>Euclide</hi> for on and the selfe same kinde of nomber. But the number which here ought to haue bene placed is called of the best interpreters of <hi>Euclide, numerus pariter par &amp; nupar,</hi> that is a number eue<g ref="char:cmbAbbrStroke">̄</g>ly eue<g ref="char:cmbAbbrStroke">̄</g>, and eue<g ref="char:cmbAbbrStroke">̄</g>ly odde. Ye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and it is so called of <hi>Euclide</hi> him selfe in the 34. proposition of his 9. booke: which kinde of number <hi>Campanus</hi> and <hi>Flussates</hi> in steade of the insufficient and v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>apt definition before geuen, assigne this definition.</p>
                  <p>
                     <hi>A number euenly euen, and euenly <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>dde, is that which an euen number doth measure sometime by an euen number, and sometime by an odde.</hi>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>
                  </p>
                  <p>As the number 12: for 2. an euen number, measureth 12. by 6. an euen number: two times 6. i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 12. Al<g ref="char:EOLhyphen"/>so 4. an euen number measureth the same number 12. by 3. an odde number. Add therefore is 12. a num<g ref="char:EOLhyphen"/>ber euenly euen, and euenly odde, and so of such others.</p>
                  <p>The cause why that <hi>Campanus</hi> and <hi>Flussates</hi> were so scrupelous in amending (as they supposed) the two definitions before,<note place="margin">
                        <hi>Campane and Flussates with other <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap>
                        </hi>
                     </note> namely, of a number eue<g ref="char:cmbAbbrStroke">̄</g>ly euen and of a number euenly odde, the one by ad<g ref="char:EOLhyphen"/>ding this word <hi>all,</hi> and the other by adding this word <hi>onely,</hi> was for that they were offended with the la<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>eue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and generalitie of them<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by them, on and the selfe same number might be compre<g ref="char:EOLhyphen"/>hended vnder either definition. And so, the selfe number should be both euenly euen, and also euenly odde: which they tooke for an absurditie. For that they are two distinct and diuers kindes of numbers. But all things well and iustly conceiued, it shall not be hard nor amisse to thinke, that these definitio<g ref="char:cmbAbbrStroke">̄</g>s were set and written by <hi>Euclide</hi> in such forme and ma<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er; as they are deliuered vnto vs by <hi>Theon:</hi> and
<pb facs="tcp:7062:216"/>
that they neede not these corrections and amendementes by adding these wordes <hi>all</hi> and <hi>onely,</hi> for ad<g ref="char:EOLhyphen"/>mit that they be distinct kinds of numbers, may not contraries be attributed in diuerse respectes to one thing? May not one line be sayd to be great and little, compared to diuers? Great in comparison of a lesse, and lesse in comparison of a greater? Euen so one number in diuers respects may be of diuers and contrary kindes of numbers.<note place="margin">
                        <hi>One number in di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ers respectes may be of di<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> kindes of numbers.</hi>
                     </note> What are more diuers them a square number and a cube number. And yet <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 64. in diuers respectes a number both square and cube. In respect of 8. to be his roote, it is a square number: for 8. times 8. is 64. and in respect of 4. to be his roote, it is a cube number for 4. times 4. fower times is 64: so in diuers respectes it is both, without any absurditie at all. Likewise this number 6. in di<g ref="char:EOLhyphen"/>uers respectes, is a number on the on side longer, and also a triangular number: which yet are diuers and distinct kindes of numbers. For 6. described by his vnities resembling a figure
<figure/>
of leng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he and breadthe hauing two sides, namely 2. and 3. is a plaine or superfici<g ref="char:EOLhyphen"/>all number of one side longer. And if the same 6. be so described by his vnities, that it representeth the figure of a triangle, then is it and beareth it the name of a trian<g ref="char:EOLhyphen"/>gler <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igure: as here ye may see the forme of either. And if ye extende in description therof all his vnities in length onely, so is 6 also a lineall number. So you see 6 in di<g ref="char:EOLhyphen"/>uers respectes is a lineall number, a number on the one side longer, and also a trigo<g ref="char:EOLhyphen"/>nall or trianguler number, and yet therby no inconuenience at all. And why may not likewise one and the selfe number in diuers respectes be accompted a number both euenly euen, and euenly odde? Yea <hi>Euclide</hi> him selfe doth most manifes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ly proue the same, and in the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ame wordes, if it be diligently wayed, in his ninth booke. For he sayth, that all numbers being double in continuall course from the number 2. be euenly euen numbers only: and agayne all numbers whose halues are odde, are euenly odde numbers only: and that number which neither is duple from the number two nor hath to his halfe an odde number is a number euenly euen and a number euenly odde. What in this can be spoke<g ref="char:cmbAbbrStroke">̄</g> more playnely? So that by <hi>Euclide</hi> it is no inconuenience that on num<g ref="char:EOLhyphen"/>ber, as 12, f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r example, in diuers respectes should be both a number euenly euen, and also a number e<g ref="char:EOLhyphen"/>uenly odde. In respect that 6. an euen number measureth 12. by 2. an euen number, 12 is a number euen<g ref="char:EOLhyphen"/>ly euen: and in respect that 4. an euen number measureth 12. by 3. an odde number, 12. is a number e<g ref="char:EOLhyphen"/>uenly odde. And thus iudge ye of all others like.</p>
                  <p>There is also an other definition geuen of this kinde of number by <hi>Boetius</hi> and others commonly which is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>Boe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ius def<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ni<g ref="char:EOLhyphen"/>tion of a num<g ref="char:EOLhyphen"/>ber euenly eue<g ref="char:cmbAbbrStroke">̄</g>, and euenly <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>d</hi>
                     </note>
                     <hi>A number eue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ly euen and euenly odde is that, which may be deuided into two equall partes, and eche of them may a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ayne be deuided into two equall partes: and so forth. But this deuision is at lenghth stayd, and continueth not till it come to vnitie.</hi> As for example 48: which may be deuided into two equall partes, namely, into 24. and 24. Agayne 24. which is on of the partes may be deuided into two equall partes 12. and 12. A<g ref="char:EOLhyphen"/>gayne 12. into 6. and 6. And agayne 6 may be deuided into two equall partes, into 3. and 3: but 3. cannot be deuided into two equall partes. Wherefore the deuision there stayeth: and continueth not till it come to vnitie as it did in these numbers which are euenly euen only.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The eleuenth definition.</hi>
                     </note>11 A number odly odde is that, which an odde number doth measure by an odde number.</head>
                  <p>As 25, which 5. an odde number, measureth by an odde number, namely, by 5. Fiue times fiue is 25: Likewise 21. whom 7. an odde number doth measure by 3, which is likewise an odde number. Three times 7. is 21.</p>
                  <p>
                     <note place="margin">
                        <hi>Flus<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ates.</hi>
                     </note>
                     <hi>Flussatus</hi> geueth this definition following of this kinde of number, which is all one in substance with the former definition.</p>
                  <p>
                     <hi>A number odly odde, it that, which onely an odde number doth measure.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>An other defi<g ref="char:EOLhyphen"/>nition.</hi>
                     </note>As 15. for no number measureth 15. but onely 5. and 3: also 25: none measureth it but onely 5. which is an odde number, and so of others.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The twelfth definition.</hi>
                     </note>12 A prime (or first) number is that, which onely vnitie doth measure.</head>
                  <p>As 5.7.11.13. For no number measureth 5, but onely vnitie. For v. vnities make the number 5. So no number measureth 7, but onely vnitie .2. taken 3. times maketh 6. which is lesse then 7: and 2. taken 4. times is 8, which is more then 7. And so of 11.13. and such others. So that all prime numbers,<note place="margin">
                        <hi>Prime numbers called incompo<g ref="char:EOLhyphen"/>sed numbers.</hi>
                     </note> which also are called first numbers, and numbers vncomposed, haue no part to measure the<g ref="char:cmbAbbrStroke">̄</g>, but onely vnitie.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The thirtenth definition.</hi>
                     </note>13 Numbers prime the one to the other are they, which onely vnitie doth measure, being a common measure to them.</head>
                  <p>As 15. and 22. be numbers prime the one to the other .15. of it selfe is no prime number, for not one<g ref="char:EOLhyphen"/>ly
<pb n="186" facs="tcp:7062:216"/>
vnitie doth measure it, but also the numbe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 5. and 3, for <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. times 5. is lx. Likewise 22. is of it selfe no prime number: for it is measured by 2. and 11, besides vnitie. For 11. twise, or 2. eleuen times, make 22. So that although neither of these two numbers 15. and 22. be a prime or incomposed number, but ey<g ref="char:EOLhyphen"/>ther haue partes of his owne, whereby it may be measured beside vnitie: yet compared together, they are prime the one to the other: for no one number doth as a common measure, measure both of them but onely vnitie, which is a common measure to all numbers. The numbers 5. and 3. which measure 15. will not measure 22: againe, the numbers 2. and 11. which measure 22, do not measure 15.</p>
               </div>
               <div type="section">
                  <head>14 A number composed, is that which some one number measureth.<note place="margin">
                        <hi>The fourtenth defini<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion.</hi>
                     </note>
                  </head>
                  <p>A number composed is not measured onely by vnitie, as was a prime number, but hath some num<g ref="char:EOLhyphen"/>ber which measureth it. As 15: the number 3. measureth 15, namely, taken 5. times. Also the number 5. measureth 15, namely, taken 3. times. 5. times <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, and 3. times 5, is 15. Likewise 18. is a composed num<g ref="char:EOLhyphen"/>ber, it is measured by these numbers 6.3.9.2. and so of others. These numbers are also called common<g ref="char:EOLhyphen"/>ly second numbers, as contrary to prime or first numbers.</p>
               </div>
               <div type="section">
                  <head>15 Numbers composed the one to the other, are they, which some one number, being a common measure to them both, doth measure.<note place="margin">
                        <hi>The fiftenth definition.</hi>
                     </note>
                  </head>
                  <p>As 12. and 8. are two composed numbers the one to the other. For the number 4, is a common mea<g ref="char:EOLhyphen"/>sure to them both: 4. taken three times maketh 12: and the same 4. taken two tymes maketh 8. So are 9. and 15: 3. measureth them both. Also 10. and 25: for 5. measureth both of them: and so infinitely of others. In thys do numbers composed the one to the other or second numbers, differre from numbers prime the one to the other: for that two numbers being composed the one to the other, ech of them seuerally is of necessitie a composed number. As in the examples before 8. and 12. are composed num<g ref="char:EOLhyphen"/>bers: likewise 9. and 15: also 10, and 25: but if they be two numbers prime the one to the other, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>it is not of necessitie, that ech of them seuerally be a prime number. As 9. and 22. are two numbers prime the one to the other: no one number measureth both of them: and yet neither of them in it selfe and in his owne nature is a prime number, but ech of them is a composed number. For 3. measureth 9, and 11. and 2. measure 22.</p>
               </div>
               <div type="section">
                  <head>16 A number is sayd to multiply a number, when the number multiplyed,<note place="margin">
                        <hi>The sixtenth definition.</hi>
                     </note> is so oftentimes added to it selfe, as there are in the number multiplying vni<g ref="char:EOLhyphen"/>ties: and an other number is produced.</head>
                  <p>In multiplication are euer required two numbers, the one is whereby ye multiply, commonly called the multiplier or multiplicant, the other is that which is multiplied.<note place="margin">
                        <hi>Two numbers required in multiplication.</hi>
                     </note> The number by which an other is multiplied, namely, the multiplyer, is sayd to multiply. As if ye will multiply 4. by 3, then is three sayd to multiply 4: therefore according to this definition, because in 3. there are three vnities: adde 4,3. times to it selfe, saying 3. times 4: so shall ye bring forth an other number, namely, 12, which is the summe produced of that multiplication: and so of all other multiplications.</p>
               </div>
               <div type="section">
                  <head>17 When two numbers multiplying them selues the one the other, produce an other:<note place="margin">
                        <hi>The seuententh definition.</hi>
                     </note> the number produced is called a plaine or superficiall number. And the numbers which muliply them selues the one by the other, are the sides of that number.</head>
                  <p>As let these two numbers 3. and 6, multiply the one the other, saying, 3. times 6, or sixe tymes 3, they shall produce 18. Thys number 18. thus produced, is called a plaine number, or a superficiall number. And the two multiplying numbers which produced 12, namely, 3. and 6,
<figure/>
are the sides of the same superficiall or plaine number, that is, the length and breadth thereof. Likewise if 9. multiply 11, or eleuen nine, there shal be produced 99. a plaine number, whose sides are the two numbers 9. and 11 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as the length and breadth of the same. They are called plaine and superficiall numbers,<note place="margin">
                        <hi>Why they are called superfici<g ref="char:EOLhyphen"/>all numbers.</hi>
                     </note> because being described by their vnities on a plaine superficies, they represent some superficiall forme or figure Geo<g ref="char:EOLhyphen"/>metricall, hauing length and breadth. As ye see of this example: and so of others. And all such plaine or superficiall numbers do euer re<g ref="char:EOLhyphen"/>present right angled figures as appeareth in the example.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:217"/>
                  <head>
                     <note place="margin">
                        <hi>The eightenth definition.</hi>
                     </note>18 When three numbers multiplyed together y<hi rend="sup">e</hi> one into the other, produce any number, the number produced, is called a solide number: and the numbers multiplying them selues the one into y<hi rend="sup">e</hi> other, are the sides therof.</head>
                  <p>As taking these three nu<g ref="char:cmbAbbrStroke">̄</g>bers 3.4. 5. multiply the one into the other.
<figure/>
First 4. into 5. saying, foure times 5. is 20: then multiply that number pro<g ref="char:EOLhyphen"/>duced, name<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y, 20. into 3: which is the third number, so shall ye produce 60. which is a solide number: and the three numbers which produced the number, namely, 3.4. and 5. are the sides of the same. And they are called solide numbers,<note place="margin">
                        <hi>Why they are called solid numbers.</hi>
                     </note> because being described by their vnities, they re<g ref="char:EOLhyphen"/>present solide and bodylicke figures of Geometry, which haue length, breadth, and thicknes. As ye see this number 60. expressed here by hys vnities. Whose length is hys side 5, his breadth is 3, and thicknes 4. And thus may ye do of all other three nu<g ref="char:cmbAbbrStroke">̄</g>bers multiplying the one the other.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The ninetenth definition.</hi>
                     </note>19 A square number is that which is equally equall: or that which is contay<g ref="char:EOLhyphen"/>ned vnder two equall numbers.</head>
                  <p>As multiply two equall numbers the one into the other. As 9. by 9. ye
<figure/>
shall produce 81, which is a square number. <hi>Euclide</hi> calleth it a number e<g ref="char:EOLhyphen"/>qually equall, because it is produced of the multiplication of two equall numbers the one into the other. Which numbers are also sayd in the se<g ref="char:EOLhyphen"/>cond definition to contayne a square number. As in the definitions of the second booke two lines are sayd to containe a square or a parallelogramme figure. It is called a square number,<note place="margin">
                        <hi>Why it is called a square num<g ref="char:EOLhyphen"/>ber.</hi>
                     </note> because being described by his vnities it representeth the figure of a square in Geometry. As ye here see doth the number 81. whose sides, that is to say, whose length and breadth, are 9. and 9, equall numbers: which also are sayd to contayne the square number 81: and so of others.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The twenteth definition.</hi>
                     </note>20 A cube number is that which is equally equall equally: or that which is con<g ref="char:EOLhyphen"/>tayned vnder three equall numbers.</head>
                  <p>As multiply three equall numbers the one into
<figure/>
the other, as 9, 9, and 9: first 9. by 9, so shall ye haue 81: which agayne multiply by 9, so shall ye produce 729. which is a cube number. And <hi>Euclide</hi> calleth it a number equally equall equally, because it is pro<g ref="char:EOLhyphen"/>duced of the multiplication of three equall num<g ref="char:EOLhyphen"/>bers the one into the other: which three numbers are sayd in the second definition (wherein he spea<g ref="char:EOLhyphen"/>keth more applying to Geometry) to contayne the cube number. It is therefore called a cube number,<note place="margin">
                        <hi>Why it is called a cube number.</hi>
                     </note> because being described by hys vnities, it represen<g ref="char:EOLhyphen"/>teth the forme of a cube in Geometry, whose sides, that is to say, whose length, breadth, and thicknes, are the three equall numbers 9, 9, and 9, of which he was produced: which three sides also are sayd to containe the cube number 729: beholde here the description therof.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The twenty one definition.</hi>
                     </note>21 Numbers proportionall are, when the first is to the second equemultiplex, as the third is to the fourth, or the selfe same part, or the selfe same partes.</head>
                  <p>Here he defineth which numbers are called proportionall, that is, what numbers haue one and the selfe same proportion. For example 6. to 3: and 4. to 2, are numbers proportionall, and haue one and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> self same proportion: for 6. the first is to 3. the seco<g ref="char:cmbAbbrStroke">̄</g>d equemultiplex, as 4. the third is to 2. the fourth:
<pb n="187" facs="tcp:7062:217"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. is double to 3: and so is 4. double to 2. Likewise these foure numbers are in like proportion 3.9.4.1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. for what part 3. is of 9, such part is 4. of 12 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>3. of 9. is a third part, so is also 4. of 12. a third part. So are these foure numbers also in proportion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.5: 4. 10: what partes 2. are of 5, such partes are 4. of 10: 2, of 5, are two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ift partes, likewise 4. of 10 are two fift partes. Moreouer, these numbers 8.6: 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.9: be in pro<g ref="char:EOLhyphen"/>portion, for what and how many parts 8. are of 6, such &amp; so many parts are 12. of 9: 8. of 6, is foure third partes, for one third part of 6. is 2, which taken foure times maketh 8: so 12. of 9, is also foure thyrd partes: for one third part of 9. is 3, which taken foure times make 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. And so conceaue ye of all other proporti<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>ll numbers.</p>
                  <p>In the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> definitio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the v. booke, <hi>Euclide</hi> gaue a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>arre other definition of magnitudes proportio<g ref="char:EOLhyphen"/>nall, and much vnlike to this which he here geueth of numbers proportionall:<note place="margin">
                        <hi>Why the defi<g ref="char:EOLhyphen"/>nition of pro<g ref="char:EOLhyphen"/>portionall mag<g ref="char:EOLhyphen"/>nitudes is vn<g ref="char:EOLhyphen"/>like to the defi<g ref="char:EOLhyphen"/>nitio of propor<g ref="char:EOLhyphen"/>tionall num<g ref="char:EOLhyphen"/>bers.</hi>
                     </note> the reason is as there al<g ref="char:EOLhyphen"/>so was partly noted, for that there he gaue a definition common to all quantities discrete and continu<g ref="char:EOLhyphen"/>all, rationall, and irrationall: and therefore was constrayned to geue the definition by the excesse, equa<g ref="char:EOLhyphen"/>litie or want of their equemultiplices, and that generally onely: for that irrationall quantities haue no certayne part or common measure to be measured by or knowen, neyther can they be expressed by any certayne numbers. But here in this place because in numbers there are no irrationall quantities, but all are certaynly knowen, so that both they and the proportions betwene them may be expressed by num<g ref="char:EOLhyphen"/>bers certayne and knowen, by reason of their partes certayne, and for that they haue some common measure to measure them (at the lest vnitie which is a common measure to all numbers) he geueth here this definition of proportionall numbers, by that the one is like equemultiplex to the other, or the same part, or the same partes: which definition is much easier then was the other: and is not so large, as is the other, neither extendeth it self generally to all kinde of quantities rationall and irrationall, but con<g ref="char:EOLhyphen"/>tayneth it selfe within the limites and bondes of rationall quantitie and numbers.</p>
               </div>
               <div type="section">
                  <head>22 Like plaine numbers, and like solide numbers, are such,<note place="margin">
                        <hi>The twenty two definition.</hi>
                     </note> which haue their sides proportionall.</head>
                  <p>Before he shewed that a plaine number hath two sides, and a solide number three sides. Now he sheweth by this definition which plaine numbers, and which solide numbers are like the one to the other. The likenes of which numbers dependeth altogether of the proportions of the sides of these numbers. So that if the two sides of one plaine number, haue the same proportion the one to the o<g ref="char:EOLhyphen"/>ther, that the two sides of the other plaine number haue the one to the other, then are such two plaine numbers like. For an example 6 and 24 be two plaine numbers, the sides of 6 be 2 and 3, two tymes 3 make 6: the sides of 24 be 4 and 6, foure times 6 makes 24. Againe the same proportion that is betwene 3 and 2 the sides of 6. is also betwene 6 and 4 the sides of 24. Wherfore 24 and 6 be two like plaine and superficiall numbers. And so of other plaine numbers. After the same manner is it in solide num<g ref="char:EOLhyphen"/>bers. If three sides of the one be in like proportion together, as are the three sides of the other, then is the one solide number like to the other. As 24 and 192 be solide numbers, the sides of 24. are 2. 3. and 4, two tymes there taken 4 times are 24. the sides of 192 are 4.6. and 8: for foure tymes 6. 8 times make 192. Againe the proportion of 4 to 3 is sesquitercia, the proportion of 3 to 2 is sesquialtera, which are the proportions of the sides of the one solide number, namely, of 24: the proportion betwene 8 and 6 is sesqu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ercia, the proportion betwene 6 and 4 is sesquialtera, which are the proportions of the sides of the other solide number, namely, of 191. And they are one and the same with the proportione of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> of the other wherfore th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se two solide numbers 24 and 192 be like, and so of other solide nu<g ref="char:cmbAbbrStroke">̄</g>bers.</p>
               </div>
               <div type="section">
                  <head>23 A perfect number is that, which is equall to all his partes.<note place="margin">
                        <hi>The twenty three definitio<g ref="char:cmbAbbrStroke">̄</g>.</hi>
                     </note>
                  </head>
                  <p>As the partes of 6 are 1. 2. 3. three is the halfe of 6, two the third part, and 1. the sixth part, and mo p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rtes 6 hath not: which thr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tes 1.<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.3. added together, make 6 the whole number, whose partes they are. Wherfore 6 is a perfect number. So likewise is 28 a perfect number, the partes whereof are these numbers 14. 7. 2. and 1: 14 is the halfe therof, 7 is the quarter, 4 is the seuenth part, 2 is a four<g ref="char:EOLhyphen"/>tenth part, and 1 an 28 part, and these are all the partes of 28. all which, namely, 1, 2, 3, 4, 7 and 14 added together, make iustly without more or l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sse 28. Wherfore <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> a perfect number, and so of others the like. This kinde of numbers is very rare and seldome found.<note place="margin">
                        <hi>Perfect num<g ref="char:EOLhyphen"/>bers rare &amp; of great vse in magike &amp; in secret philoso<g ref="char:EOLhyphen"/>phy.</hi>
                     </note> From 1 to 10, there is but one perfect num<g ref="char:EOLunhyphen"/>ber, namely, 6. From 10 to an 100, there is also but one, that is, 28. Also from 100 to 1000 there is but one which is 496. From 1000 to 10000 likewise but one. So that betwene euery s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ay in numbring, which is euer in the tenth place, there is found but one perfect number And for their rarenes and great perfec<g ref="char:EOLhyphen"/>tion, they are of maruelous vse in magike, and in the secret part of philosophy.</p>
                  <p>This kinde of number is called perfect,<note place="margin">
                        <hi>In what respect a number is perfect.</hi>
                     </note> in respect <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> numbers which are i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>perfect. For as the nature of a perfect number standeth in this, that all partes added together are equall to the whole: and make the whole: so in an imperfect nu<g ref="char:cmbAbbrStroke">̄</g>ber all the parts added together are not equal to the whole, nor make the whole, but make either more or lesse. Wherefore of imperfect numbers there are two kindes,<note place="margin">
                        <hi>Two kinds of imperfect numbers.</hi>
                     </note> the one is called <hi>abundan<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> or a building, the other <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>,</hi> or wanting.</p>
                  <p>A number abunding is that whose partes being all added together make more then the w<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ds number whose partes they are, as 12. is an abundant number. For all the parte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of 12. namely, 6.4.3.2.
<pb facs="tcp:7062:218"/>
and 1. added together make 16: which are more then 12. Likewise 18. is a number abunding, all his part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>mely, 9.6.3.2. and 1. added together make 20. which are more then 18: and so of others.</p>
                  <p>
                     <note place="margin">
                        <hi>A <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>mber wan<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </note>A number diminute, or wanting is that whose partes being all added together, make lesse then the whole, or number whose partes they are.</p>
                  <p>As 9. is a diminute, or wanting number, for all his partes, namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 3. and 1. (moe partes he hath not) added together make onely 4: which are lesse then 9. Also 26. is a diminute nu<g ref="char:cmbAbbrStroke">̄</g>ber, all his partes, name<g ref="char:EOLhyphen"/>ly. 13.2.1. added togeth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r make onely 16: which is a number much lesse then 26. And so of such like.</p>
                  <p>
                     <note place="margin">
                        <hi>Common sen<g ref="char:EOLhyphen"/>tences.</hi>
                     </note>
                     <hi>CAmpane</hi> and <hi>Flussates</hi> here adde certayne common sentences, some of which<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for that they are in these three bookes following somtimes alledged, I thought good here to annexe.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>irst common se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tence.</hi>
                     </note>1 The lesse part is that which hath the greater denomination: and the greater part is that, which hath the lesse denomination.</head>
                  <p>As the numbers 6. and 8. are either of them a part of the number 24:6. is a fourth part, 4. times 6. is 24: and 8. is a third part, 3. times 8. is 24. Now forasmuch as 4 (which denominateth what part 6. is of 24) is greater then 3. (which denominateth what part 8. is of 24.) therefore is 6. a lesse part of 24<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then is 8. and so is 8. a greater part of 24. the<g ref="char:cmbAbbrStroke">̄</g> 6. is. And so in others.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>cond <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ommon sentence.</hi>
                     </note>2 Whatsoeuer numbers are equemultiplices to one &amp; the selfe same nu<g ref="char:cmbAbbrStroke">̄</g>ber, or to equall numbers, are also equall the one to the other.</head>
                  <p>As if vnto the number 3 be taken two numbers containing the same number foure times, that is being equemultiplices to the same number three: the sayd two numbers shalbe equall. For 4. times 3. will euer be 12. So also will it be if vnto the two equal num<g ref="char:EOLhyphen"/>bers 3. &amp; 3. be taken two numbers, the one co<g ref="char:cmbAbbrStroke">̄</g>taining the one number 3. foure times, the other containing the other number 3. also foure times, that is, being equemultipli<g ref="char:EOLhyphen"/>ces to the equall numbers 3. and 3.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>Third common sentence.</hi>
                     </note>3 Those numbers to whome one and the selfe same number is equmulti<g ref="char:EOLhyphen"/>plex, or whose equemultiplices are equall: are also equall the on to the other.</head>
                  <p>As if the number 18. be equemultiplex to any two numbers, that is, contayne any two numbers twise, thrise, fower times &amp;c? As for example 3. times: then are the sayd two numbers equall. For 18. deuided by 3. will euer bring forth 6. So that that diuision made twise will bring forth 6. and 6. two equall numbers. So also would it follow if the two numbers had equall equemultiplices, namely, if 18. and 18. which are equall num<g ref="char:EOLhyphen"/>bers contayned any two numbers 3. times.</p>
               </div>
               <div type="section">
                  <head>4 If a number measure the whole, and a part taken away: it shall also mea<g ref="char:EOLhyphen"/>sure the residue.<note place="margin">
                        <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>urth common sentence.</hi>
                     </note>
                  </head>
                  <p>As if from 24. be taken away 9. there remaineth 15. And for as much as the number 3 measureth the whole number 24, &amp; also the number take<g ref="char:cmbAbbrStroke">̄</g> away, namely, 9. it shall al<g ref="char:EOLhyphen"/>so measure the residue, which is 15<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For 3. measureth 15 by fiue, fiue times 3. is 15. And so of others.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>th common sentence.</hi>
                     </note>5 If a number measure any number: it also measureth euery number that the sayd number measureth.</head>
                  <p>As the number 6. measuring the number 12. shall also measure all the numbers that 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. measureth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as the numbers 24.36.48.60. and so forth: which the number 12. doth
<pb n="188" facs="tcp:7062:218"/>
measure by the number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 2.3.4. and 5. And for as much as the number 12. doth measure the numbers 24.36.48. and 60. And the nu<g ref="char:cmbAbbrStroke">̄</g>ber 6, doth measure the number 12. (namely by 2.) It followeth by this commo<g ref="char:cmbAbbrStroke">̄</g> sentence, that the number 6. measureth eche of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se numbers 24. 36.48. and 60. And so of others.</p>
               </div>
               <div type="section">
                  <head>6 If a number measure two numbers,<note place="margin">
                        <hi>Sixth common sentence.</hi>
                     </note> it shall also measure y<hi rend="sup">e</hi> number com<g ref="char:EOLhyphen"/>posed of them.</head>
                  <p>As the number 3 measureth these two numbers 6. and 9<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> it measureth 6. by 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and 9. by 3. And therefore by this common sentence it measureth the number 15. which is composed of the numbers 6. and 9: namely it measureth it by 5.</p>
               </div>
               <div type="section">
                  <head>7 If in numbers there be proportions how manysoeuer equall or the selfe same to one proportion:<note place="margin">
                        <hi>Seuenth com<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>mon sentence.</hi>
                     </note> they shall al<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o be equall or the selfe same the one to the other.</head>
                  <p>As yf the proportion of the number 6. to the number 3. be as the proportion of the number 8. to the number 4, if also the proportion of the number 10. to the num<g ref="char:EOLhyphen"/>ber 5. be as the proportion of the number 8. to the number 4: then shall the proportion of the number 6. to the number 3. be as the proportion of the number 10. is to the number 5: namely, eche proportion is duple. And so of others. <hi>Euclide</hi> in his <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. booke the 11. proposition demonstrated this also in continuall quantitie: which although as touching that kinde of quantitie it might haue bene put also as a principle (as in num<g ref="char:EOLhyphen"/>bers he taketh it) yet for that in all magnitudes theyr proportion can not be expressed, (as hath before bene noted &amp; shalbe afterward in the tenth booke more at large made manifest:) therefore he demonstrateth it there in that place, and proueth that it is true as touching all proportions generally whither they be rational or irrationall.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The first Proposition. The first Theoreme. If there be geuen two vnequall numbers, and if in taking the lesse continu<g ref="char:EOLhyphen"/>ally from the greater, the number remayning do not measure the number going before, vntill it shall come to vnitie: then are those numbers which were at the beginning geuen, prime the one to the other.</head>
               <p>SVppose that there be two vnequal numbers <hi>AB</hi> the greater, and <hi>CD</hi> the lesse, and from <hi>AB</hi> the greater, take away <hi>CD</hi> the lesse as o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ten as you can leauing <hi>FA,</hi>
                  <note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ctio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> and from <hi>CD</hi> take away <hi>FA</hi> as often as you can, leauing the number <hi>GC.</hi> And from <hi>FA</hi> take away <hi>GC</hi> as often as you
<figure/>
can, and so do continually till there remayne onely vnitie, which let be <hi>HA.</hi>
                  <note place="margin">Demonstrati<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> leading to an absurditie.</note> Then I say that no number measureth the numbers <hi>AB</hi> and <hi>CD.</hi> For if it be possible let some number measure them, and let the same be <hi>E.</hi> Now <hi>CD</hi> measuring <hi>AB</hi> leaueth a lesse number then it selfe, which let be <hi>FA</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And <hi>FA</hi> measuring <hi>DC</hi> leaueth also a lesse then it selfe, namely, <hi>GC.</hi> And <hi>GC</hi> meas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ring <hi>FA</hi> leaueth v<g ref="char:EOLhyphen"/>nitie <hi>HA.</hi> And forasmuch as the number <hi>E</hi> measureth <hi>DC,</hi> and the number <hi>CD</hi> mea<g ref="char:EOLhyphen"/>sureth the number <hi>BF,</hi> therfore the number <hi>E</hi> also measureth <hi>BF,</hi> and it measureth the
<pb facs="tcp:7062:219"/>
whole number <hi>BA,</hi> wherfore it also measureth that which remayneth, namely, the number <hi>FA</hi> (by the 4. co<g ref="char:cmbAbbrStroke">̄</g>mon sentence of the seuenth). But the number <hi>AF</hi> measureth the number <hi>DG,</hi> wherfore <hi>E</hi> also measureth <hi>DG.</hi> And it measureth also the whole <hi>DC,</hi> wherfore it also measureth that which remayneth, namely, the number <hi>GC</hi> (by the same common sen<g ref="char:EOLhyphen"/>tence): but <hi>GC</hi> measureth the number <hi>FH,</hi> wherfore also <hi>E</hi> measureth <hi>FH,</hi> and it mea<g ref="char:EOLhyphen"/>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>reth the whole number <hi>FA,</hi> wherfore (by the former common sentence) it also measu<g ref="char:EOLhyphen"/>reth that which remayneth <hi>HA,</hi> which is vnitie, it selfe being a number, which is impossi<g ref="char:EOLhyphen"/>ble. Wherfore no number doth measure the numbers <hi>A</hi>B and <hi>CD,</hi> wherfore the numbers <hi>AB</hi> and <hi>CD</hi> are prime numbers the one to the other: which was required to be proued.</p>
               <div type="section">
                  <head>The conuerse of this proposition after <hi>Campane.</hi>
                  </head>
                  <p>And if the two numbers, namely AB and CD be prime the one to the other. Then the lesse being continually taken from the greater there shalbe no stay of that sustraction, till that you come to vnitie. For if in the continuall substraction ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be a stay before you come to vnitie.<note place="margin">
                        <hi>The conuerse of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>his proposition.</hi>
                     </note>
                     <figure/>
Suppose that HA be the number whereat the stay is made, which also being subtrahed out of GC leaueth nothing. Wherfore HA measureth GC wher<g ref="char:EOLhyphen"/>fore also it measureth FH by the 5. common sentence of the seuenth. And for as much as it also measureth i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> selfe, therefore it also measureth the whole AF by the sixth common sentence of the seuenth, wherfore also it measureth DG by the 5. common sentence. But it is before proued that it measureth GC, wherfore it measureth the whole CD, by the sixth common sentence of the seuenth: wherfore also it measureth BF by the 5. common sentence of the seuenth. And it is also proued that it measu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eth FA, wherfore also it measureth the whole number AB by the sixth common sentence of the seuenth. Now for as much as the number HA measureth the numbers AB and CD, therfore the numbers AB and CD are numbers composed: wherfore they are not prime the one to the othe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: which is contra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y to the supposition.</p>
                  <p>And by this proposition if there be two numbers geuen. It is easy to finde out, whether they be prime the one to the other or no.<note place="margin">
                        <hi>How to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>now whether two numbers geuen be prime the one to the other.</hi>
                     </note> For if by such continual substraction of the lesse <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rom the greater, you come at the length to vnitie. Then are those numbers geuen prime the one to the other. But if there be a stay before you come to vnitie, then are the numbers geuen, numbers composed the one to the other.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 1. Probleme. The 2. Proposition. Two numbers being geuen not prime the one to the other, to finde out their greatest common measure.</head>
               <p>SVppose the two numbers geuen not prime the one to the other to be <hi>AB</hi> and <hi>CD.</hi> It is required to finde out the greatest common measure of the said numbers <hi>AB</hi> and <hi>CD.</hi>
                  <note place="margin">Two cases in this probleme.</note> Now the numbe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>CD</hi> either measureth
<figure/>
the number <hi>AB</hi> or not. If <hi>CD</hi> measure <hi>AB</hi> it also mea<g ref="char:EOLhyphen"/>sureth it selfe.<note place="margin">The first case.</note> Wherefore <hi>CD</hi> is a common measure to the numbers <hi>CD</hi> and <hi>AB.</hi> And it is manifest also that it is the greatest common measure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for there is no number greater then <hi>CD</hi> that will measure <hi>CD.</hi>
               </p>
               <p>
                  <note place="margin">The second case.</note>But if <hi>CD</hi> do not measure <hi>AB,</hi> then if of the numbers <hi>AB</hi>
                  <figure/>
and <hi>CD,</hi> the lesse be continually taken away from the greater, there will before you come to vnitie, be left a number, which will measure the number going before (by the 1. of the seuenth). For if there should not, then should the numbers <hi>AB</hi> and <hi>CD</hi> be prime the one to the other, which is contrary to the supposition. Let the sayd number left by the continuall substraction of the lesse number out of the greater be <hi>FC.</hi> So that le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the number <hi>CD</hi> measuring <hi>AB,</hi> and subtrahed out of it as often as you can le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e a lesse number then it selfe, namely <hi>AE.</hi> And let <hi>AE</hi> measuring <hi>CD,</hi> and subtrahed out of it
<pb n="189" facs="tcp:7062:219"/>
as often as you can leaue a lesse-then it selfe namely, <hi>CF.</hi> And suppos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t <hi>CF</hi> do so mea<g ref="char:EOLhyphen"/>sure <hi>AE</hi> that there remayne nothing. Then I say that <hi>CF</hi> is a common measure to the numbers <hi>AB</hi> and <hi>CD.</hi> For forasmuch as <hi>CF</hi> measureth <hi>AE,</hi> and <hi>AE</hi> measureth <hi>DF,</hi>
                  <note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the second case.</note> therefore <hi>CF</hi> also measureth <hi>DF</hi> (by the fifth common sentence of the seuenth) and it likewise measureth it selfe, wherfore it also measureth the whole <hi>CD</hi> (by the sixth common sentence of the seuenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> but <hi>CD</hi> measureth <hi>BE,</hi> wherefore <hi>CF</hi> also measureth <hi>BE</hi> (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ifte common sentence of the seuenth). And it measureth also <hi>EA</hi>:<note place="margin">That CF is a common mea<g ref="char:EOLhyphen"/>sure to the numbers AB and CD.</note> wherefore it also measureth the whole <hi>BA</hi> (by the sixth common sentence of the seuenth): and it also measu<g ref="char:EOLhyphen"/>reth <hi>CD</hi> as we haue before proued: wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the number <hi>CF</hi> measureth the numbers <hi>AB</hi> &amp; <hi>CD</hi> wherfore the number <hi>CF</hi> is a commo<g ref="char:cmbAbbrStroke">̄</g> measure to the numbers <hi>AB</hi> &amp; <hi>CD.</hi>
               </p>
               <p>I say also that it is the greatest common measure. For if <hi>CF</hi> be not the greatest commo<g ref="char:cmbAbbrStroke">̄</g> measure to <hi>AB</hi> and <hi>CD,</hi>
                  <note place="margin">That CF is the greatest common mea<g ref="char:EOLhyphen"/>sure to AB and CD.</note> let there be a number greater then
<figure/>
                  <hi>CF,</hi> which measureth <hi>AB</hi> and <hi>CD</hi>: which let be <hi>G.</hi> And forasmuch as <hi>G</hi> measureth <hi>CD,</hi> and <hi>CD</hi> measureth <hi>BE,</hi> therefore <hi>G</hi> also measureth <hi>BE</hi> (by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ft common sentence of the seuenth). And it measureth the whole <hi>AB,</hi> where<g ref="char:EOLhyphen"/>fore also it measureth the residue, namely, <hi>AE</hi> (by the 4. common sentence of the se<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth). But <hi>AE</hi> measureth <hi>DF,</hi> wherefore <hi>G</hi> also measureth <hi>DF</hi> (by the foresayd 5. com<g ref="char:EOLhyphen"/>mon sentence of the seuenth). And it measureth the whole <hi>CD.</hi> Wherefore it also measu<g ref="char:EOLhyphen"/>reth the residue <hi>FC</hi>: namely, the greater number the lesse: which is impossible. No number therefore greater then <hi>CF</hi> shall measure those numbers <hi>AB</hi> and <hi>CD</hi>: wherefore <hi>CF</hi> is the greatest common measure to <hi>AB</hi> and <hi>CD</hi>: which was required to be done.</p>
               <div type="section">
                  <head>Corrolary.</head>
                  <p>
                     <hi>Hereby it is manifest, that if a number measure two numbers it shall also measure their greatest common measure.</hi> For if it measure the whole &amp; the part taken away, it shall alwayes measure the residue also, which residue is at the length, the greatest common measure of the two numbers geuen.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 2. Probleme. Th 3. Proposition. Thre numbers being geue<g ref="char:cmbAbbrStroke">̄</g>, not prime the one to the other: to finde out their greatest common measure.</head>
               <p>SVppose the three numbers geuen not prime the one to the other
<figure/>
to be <hi>A, B, C.</hi> Now it is required vnto the sayd numbers <hi>A, B, C</hi> to finde out the greatest common measure. Take the greatest common measure of the two numbers <hi>A</hi> and <hi>B</hi> (by the 2 of the seuenth) which let be <hi>D</hi>: which number <hi>D</hi> either measureth the num<g ref="char:EOLhyphen"/>ber <hi>C</hi> or not.</p>
               <p>First let <hi>D</hi> measure <hi>C.</hi> And it also measureth the numbers <hi>A</hi> and <hi>B,</hi>
                  <note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> wherfore <hi>D</hi> measureth the numbers <hi>A, B, C.</hi> Wherefore <hi>D</hi> is a common measure vnto the numbers <hi>A, B, C.</hi> Then I say also, that it is the greatest common measure vnto them. For if <hi>D</hi> be not the greatest common measure vnto the numbers <hi>A, B, C,</hi>
                  <note place="margin">The first case.</note> let some number greater then <hi>D</hi> measure the numbers <hi>A, B, C.</hi> And let the same number be <hi>E.</hi> Now forasmuch as <hi>E</hi> measu<g ref="char:EOLhyphen"/>reth the numbers <hi>A, B, C,</hi> it measureth also the numbers <hi>A, B.</hi> Wherefore it measureth also
<pb facs="tcp:7062:220"/>
the greatest common measure of the numbers <hi>A, B</hi> (by the Co<g ref="char:EOLhyphen"/>rollary
<figure/>
of the second of the seuenth). B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t the greatest common measure of the numbers <hi>A, B,</hi> is the number <hi>D</hi> (by con<g ref="char:EOLhyphen"/>struction). Wherefore the number <hi>E</hi> measureth the number <hi>D,</hi> namely, the greater the lesse: which is impossible. Where<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore no number greater the<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>D</hi> measureth the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>A, B, C.</hi> Wherefore <hi>D</hi> is the greatest common measure to the numbers <hi>A, B, C.</hi>
               </p>
               <p>
                  <note place="margin">The second case.</note>But now suppose that <hi>D</hi> do not measure <hi>C.</hi> First I say that <hi>D</hi> &amp; <hi>C</hi> are not prime num<g ref="char:EOLhyphen"/>bers the one to the other. For forasmuch as the numbers <hi>A, B, C,</hi> are not prime the one to the other (by supposition) some one number will measure them: but that number that mea<g ref="char:EOLhyphen"/>sureth the numbers <hi>A, B, C,</hi> shall also measure the numbers <hi>A, B,</hi> and shall likewise measure the greatest co<g ref="char:cmbAbbrStroke">̄</g>mon measure of <hi>AB,</hi> namely, <hi>D</hi> (by the Corollary of the second of the seue<g ref="char:cmbAbbrStroke">̄</g>th). And the sayd number measureth also <hi>C.</hi> Wherfore some one number measureth the num<g ref="char:EOLhyphen"/>bers <hi>D</hi> and <hi>C.</hi> Wherefore <hi>D</hi> and <hi>C</hi> are not prime the one to the other.</p>
               <p>Now then let there be taken (by the 2. of the seuenth) the greatest common measure vnto the numbers <hi>D</hi> and <hi>C,</hi> which let be the number <hi>E.</hi> And forasmuch as <hi>E</hi> measureth <hi>D,</hi> and <hi>D</hi> measureth the numbers <hi>A, B,</hi> therefore <hi>E</hi> also measureth the numbers <hi>A, B</hi> (by the sixt common sente<g ref="char:cmbAbbrStroke">̄</g>ce): and it measureth also <hi>C.</hi> Wherfore <hi>E</hi> measureth the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>A, B, C.</hi> Wherefore <hi>E</hi> is a common measure vnto the numbers <hi>A, B, C.</hi> I say also that it is the grea<g ref="char:EOLhyphen"/>test. For if <hi>E</hi> be not the greatest common measure vnto the numbers <hi>A, B, C,</hi> let there be some number greater then <hi>E,</hi> which measureth the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>A, B, C.</hi> And let the same num<g ref="char:EOLhyphen"/>ber be <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</hi> And forasmuch as <hi>F</hi> measureth the numbers <hi>A, B, C</hi>: it measureth also the num<g ref="char:EOLhyphen"/>bers <hi>A, B.</hi> Wherefore also it measureth the greatest common measure of the numbers <hi>A, B</hi> (by the Corollary of the 2. of the seuenth). But the greatest common measure of the numbers <hi>A, B,</hi> is <hi>D.</hi> Wherefore <hi>E</hi> measureth <hi>D.</hi> And it measureth also the number <hi>C.</hi> Wherefore <hi>F</hi> measureth the numbers <hi>D, C.</hi> Wherefore also (by the same Corollary) it measureth the greatest common measure of the numbers <hi>D, C.</hi> But the greatest common measure of the numbers <hi>D, C,</hi> is <hi>E.</hi> Wherfore <hi>F</hi> measureth <hi>E,</hi> namely, the greater number the lesse: which is impossible. Wherefore no number greater then <hi>E</hi> shall measure the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>A, B, C.</hi> Wher<g ref="char:EOLhyphen"/>fore <hi>E</hi> is the greatest common measure to the numbers <hi>A, B, C</hi>: which was required to be done.</p>
               <div type="section">
                  <head>¶Corollary.</head>
                  <p>Wherefore it is manifest, that if a number measure three numbers, it shall also measure their greatest common measure. And in like sort more num<g ref="char:EOLhyphen"/>bers being geue<g ref="char:cmbAbbrStroke">̄</g> not prime the one to the other may be found out their grea<g ref="char:EOLhyphen"/>test common measure, and the Corollary will followe.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 2. Theoreme. The 4. Proposition. Euery lesse number is of euery greater number, either a part, or partes.</head>
               <p>
                  <note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note>SVppose there be two numbers <hi>A</hi> and <hi>BC.</hi> Of which let <hi>BC</hi> be the lesse. Then I say, that <hi>BC</hi> is either a part or partes of <hi>A.</hi> For the numbers <hi>A</hi> and <hi>BC</hi> are ei<g ref="char:EOLhyphen"/>ther prime the one to the other, or not. First let <hi>A</hi> and <hi>BC</hi> be prime the one to the other.<note place="margin">The first case.</note> And deuide the number <hi>BC</hi> into those vnities which are in it. Now euery one of
<pb n="190" facs="tcp:7062:220"/>
the vnities which <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> in <hi>BC</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>me certaine part of <hi>A</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore <hi>BC</hi> are partes of <hi>A.</hi>
               </p>
               <p>But now suppose that the number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>A</hi> and <hi>BC</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> not prime the one to the other. Then <hi>BC</hi> either measureth <hi>A</hi> or not measureth it.<note place="margin">The second case.</note> If <hi>BC</hi> measure:
<figure/>
                  <hi>A,</hi> then is <hi>BC</hi> a part of <hi>A.</hi> But if not<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> take (by the 2. of the seuenth) the greatest common <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of <hi>A</hi> and <hi>BC,</hi> and let the same be <hi>D.</hi> And let <hi>BC</hi> be deuided into as man<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>te<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> as it hath equall vnto <hi>D,</hi> that is into <hi>BE, EF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and <hi>FC.</hi> And forasmuch as <hi>D</hi> measureth <hi>A,</hi> therefore <hi>D</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a part of <hi>A.</hi> But <hi>D</hi> is equall vnto euery one of these partes <hi>BE, EF,</hi> and <hi>FC.</hi> Wherfore also euery one of these partes <hi>BE, EF,</hi> and <hi>FC,</hi> is a part of <hi>A.</hi> Wher<g ref="char:EOLhyphen"/>fore the number <hi>BC</hi> is partes of <hi>A.</hi> Wherefore euery lesse number is of euery greater nu<g ref="char:cmbAbbrStroke">̄</g>ber, either a part or partes: which was required to <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> proued.</p>
            </div>
            <div type="part">
               <head>¶ The 3. Theoreme. The 5. Proposition. If a number be a part of a number, and an other nu<g ref="char:cmbAbbrStroke">̄</g>ber the selfe same part of an other number, then both the numbers added together shall be the selfe same part of both the numbers added together, which one number was of one number.</head>
               <p>SVppose the number <hi>A</hi> to be a part of the number <hi>BC,</hi> and let an other number, namely, <hi>D,</hi> be the selfe same part of an other number, namely, of <hi>EF.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say, <note place="margin">This proposi<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>, and the <hi>6.</hi> proposition in discrete quan<g ref="char:EOLhyphen"/>titie, answer to the first of the fifth in continual quantitie.</note> that the numbers <hi>A</hi> and <hi>D</hi> added together, are the selfe same part of the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>bers <hi>BC</hi> and <hi>EF</hi> added together, that <hi>A</hi> is of <hi>BC.</hi> For forasmuch as what part the number <hi>A</hi> is of the number <hi>BC,</hi> the self<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> same part
<figure/>
is the number <hi>D</hi> of the number <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> F,</hi> therfore h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>w many num<g ref="char:EOLhyphen"/>bers there are in <hi>BC</hi> equall vnto <hi>A,</hi> so many numbers are there in <hi>EF</hi> equall vnto <hi>D.</hi> Deuide I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> say, <hi>BC</hi> into the num<g ref="char:EOLhyphen"/>bers that are equall vnto <hi>A,</hi> that is, into <hi>BG,</hi> and <hi>GC,</hi> and likewise deuide <hi>EF</hi> into the numbers that are equall vnto <hi>D,</hi> that is, into <hi>EH</hi> and <hi>HF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Now then the multitude of these <hi>BG</hi> and <hi>GC</hi> is equall vnto the multitude of these <hi>EH</hi> and <hi>HF.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as <hi>BG</hi> is equall vnto <hi>A,</hi> and <hi>EH,</hi> vnto <hi>D,</hi> therefore <hi>BG</hi> and <hi>EH</hi> are equall vnto <hi>A</hi> &amp; <hi>D.</hi> And by the same reason forasmuch as <hi>GC</hi> is equall vnto <hi>A,</hi> and <hi>HF</hi> vnto <hi>D</hi>: therefore <hi>GC</hi> and <hi>HF</hi> are also equall vnto <hi>A</hi> and <hi>D.</hi> Wherefore how many numbers there are in <hi>BC</hi> equall vnto <hi>A</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so many are there in <hi>BC</hi> and <hi>EF</hi> equall vnto <hi>A</hi> and <hi>D.</hi> Wherefore how multiplex <hi>BC</hi> is to <hi>A,</hi> so multiplex are both the numbers <hi>BC</hi> and <hi>EF,</hi> to both the numbers <hi>A</hi> and <hi>D.</hi> Wherefore what part <hi>A</hi> is of <hi>BC,</hi> the selfe part also are <hi>A</hi> &amp; <hi>D</hi> added together, of <hi>BC</hi> and <hi>EF</hi> added together: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 4. Theoreme. The 6. Proposition. If a number be partes of a number, and an other number the selfe same partes of an other number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> then both numbers added together shall be of both numbers added together the selfe same partes, that one number was of one number.</head>
               <p>
                  <pb facs="tcp:7062:221"/>SVppose that the number <hi>AB</hi> be partes of the number <hi>C,</hi> and let an other number, namely, <hi>DE</hi> be the selfe same partes of an other number, namely, of <hi>F.</hi> Then I say, that the numbers <hi>AB</hi> and <hi>DE</hi> added together, are of the numbers <hi>C</hi> and <hi>F</hi> added together the selfe same partes, that <hi>AB</hi> is of <hi>C.</hi>
                  <note place="margin">Construction.</note> For forasmuch
<figure/>
as what partes <hi>AB</hi> is of <hi>C,</hi> the selfe same partes is <hi>DE</hi> of <hi>F</hi>: therefore how many partes of C there are in <hi>AB,</hi> so many partes of <hi>F</hi> are there in <hi>DE.</hi> Deuide <hi>AB</hi> into the partes of <hi>C,</hi> that is, into <hi>AG</hi> and <hi>GB,</hi> and likewise <hi>DE</hi> into the partes of <hi>F,</hi> that is, into <hi>DH</hi> and <hi>HE.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now then the multitude of these <hi>AG</hi> and <hi>GB</hi> is equall to the multitude of these <hi>DH</hi> and <hi>HE.</hi> And forasmuch as what part <hi>AG</hi> is of <hi>C,</hi> the selfe same part is <hi>DH</hi> of <hi>F</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore what part <hi>AG</hi> is of <hi>C,</hi> the selfe same part is <hi>AG</hi> and <hi>DH</hi> added together of <hi>C</hi> and <hi>F</hi> added together. And by the same reason also what part <hi>GB</hi> is of <hi>C,</hi> the selfe same part is <hi>GB</hi> and <hi>HE</hi> added together of <hi>C</hi> and <hi>F</hi> added together. Wherefore what partes <hi>AB</hi> is of <hi>C,</hi> the selfe same partes are <hi>AB</hi> and <hi>DE</hi> added together of <hi>C</hi> and <hi>F</hi> added toge<g ref="char:EOLhyphen"/>ther: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 5. Theoreme. The 7. Proposition. If a number be the selfe same part of a number, that a part taken away is of a part taken away: then shall the residue be the selfe same part of the resi<g ref="char:EOLhyphen"/>due, that the whole was of the whole.</head>
               <p>
                  <note place="margin">Thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proposi<g ref="char:EOLhyphen"/>tion and the next following in discret qua<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>titie answe<g ref="char:EOLhyphen"/>reth to the fifth propositio<g ref="char:cmbAbbrStroke">̄</g> of the fifth boke in conti<g ref="char:EOLhyphen"/>nuall qua<g ref="char:cmbAbbrStroke">̄</g>tity.</note>SVppose that the number <hi>AB</hi> be of the number <hi>CD</hi> the selfe same part, that the part taken away <hi>AE</hi> is of the part taken away <hi>CF.</hi> Then I say, that the residue <hi>EB</hi> is of the residue <hi>FD</hi> the self same part that the whole <hi>AB</hi> is of the whole <hi>CD.</hi> What part <hi>AE</hi> is of <hi>CF,</hi> the selfe same part let <hi>EB</hi>
                  <figure/>
be of <hi>GC.</hi> And for that what part <hi>AE</hi> is of <hi>CF,</hi> the same part is <hi>EB</hi> of <hi>CG,</hi> therefore what part <hi>AE</hi> is of <hi>CF,</hi> the same part (by the 5. of the seuenth) is <hi>AB</hi> of <hi>FG.</hi> But what part <hi>AE</hi> is of <hi>CF,</hi> the same part (by supposition) is <hi>AB</hi> of <hi>CD.</hi> Wher<g ref="char:EOLhyphen"/>fore what part <hi>AB</hi> is of <hi>FG,</hi> the self same part is <hi>AB</hi> of <hi>CD.</hi>
                  <note place="margin">Construction.</note> Wherefore <hi>AB</hi> is one &amp; the selfe same part of both these numbers <hi>GF</hi> and <hi>CD.</hi> Wherefore <hi>GF</hi> is equall vnto <hi>CD</hi> (by the second common sentence of the seuenth). Take away <hi>CF</hi> which is common to them both. Wherefore the residue <hi>GC</hi> is equall vnto the residue <hi>FD.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as what part <hi>AE</hi> is of <hi>CF,</hi> the same part is <hi>EB</hi> of <hi>GC</hi>: but <hi>GC</hi> is equall vnto <hi>FD</hi>: therefore what part <hi>AE</hi> is of <hi>FC,</hi> the selfe same part is <hi>EB</hi> of <hi>FD.</hi> But what part <hi>AE</hi> is of <hi>CF,</hi> the same part is <hi>AB</hi> of <hi>CD.</hi> Wherefore what part <hi>EB</hi> is of <hi>FD,</hi> the same part is <hi>AB</hi> of <hi>CD.</hi> Wherefore the residue <hi>EB</hi> is of the residue <hi>FD</hi> the selfe same part that the whole <hi>AB</hi> is of the whole <hi>CD</hi>: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 6. Theoreme. The 8. Proposition. If a number be of a number the selfe same partes, that a part taken away is of a part taken away, the residue also shall be of the residue the selfe same partes that the whole is of the whole.</head>
               <p>
                  <pb n="191" facs="tcp:7062:221"/>SVppose that the number <hi>AB</hi> be of the number <hi>CD</hi> the selfe same partes that the part taken away <hi>AE</hi> is of the part taken <hi>CF.</hi> Then I say, that the residue <hi>EB</hi> is of the residue <hi>FD</hi> the selfe same partes that the whole <hi>AB</hi> is of the whole <hi>CD.</hi> Vnto <hi>AB</hi> put an equall nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>GH.</hi>
                  <note place="margin">Constu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ction.</note> Wherefore what partes <hi>GH</hi> is of <hi>CD,</hi> the selfe same partes is <hi>AE</hi> of <hi>CF.</hi> Deuide <hi>GH</hi> into the partes of <hi>CD,</hi> that is, <hi>GK,</hi> and <hi>KH,</hi> and likewise <hi>AE</hi> into the partes of <hi>CF,</hi> that is, into <hi>AL</hi> and <hi>LE.</hi> Now then the multitude of these <hi>GK</hi> and <hi>KH,</hi> is equall vnto the multitude of these <hi>AL</hi> and <hi>LE.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as what part <hi>GK</hi> is of <hi>CD,</hi> the selfe same part is <hi>AL</hi> of <hi>CF</hi>: but <hi>CD</hi> is greater then <hi>CF.</hi> Wherefore <hi>GK</hi> is greater then <hi>AL.</hi> Put vnto <hi>AL</hi> an equall
<figure/>
number <hi>MG.</hi> Wherefore what part <hi>GK</hi> is of <hi>CD,</hi> the same part is <hi>GM</hi> of <hi>CF.</hi> Wherefore the residue <hi>MK</hi> is (by the 7. of the seuenth) of the residue <hi>FD,</hi> the selfe same part that the whole <hi>GK</hi> is of the whole <hi>CD.</hi> Againe, forasmuch as what part <hi>KH</hi> is of <hi>CD,</hi> the selfe same part is <hi>EL</hi> of <hi>CF</hi>: but <hi>CD</hi> is greater then <hi>CF.</hi> Wherefore <hi>HK</hi> is greater then <hi>EL.</hi> Put vnto <hi>EL</hi> an e<g ref="char:EOLhyphen"/>quall number <hi>KN.</hi> Wherefore what part <hi>KH</hi> is of <hi>CD,</hi> the selfe same part is <hi>KN</hi> of <hi>CF.</hi> Wherefore the residue also <hi>NH</hi> is (by the 7. of the seuenth) of the residue <hi>FD,</hi> the selfe same part that the whole <hi>KH</hi> is of the whole <hi>DC.</hi> Wherefore both these <hi>MK</hi> and <hi>NH</hi> added together are (by the 5. of the seuenth) of <hi>DF</hi> the self same partes that the whole <hi>HG</hi> is of the whole <hi>CD.</hi> But both these <hi>MK</hi> and <hi>NH</hi> added together, are equall vnto <hi>EB.</hi> And <hi>HG</hi> is equall vnto <hi>BA.</hi> Wherefore the residue <hi>EB</hi> is of the residue <hi>FD</hi> the selfe same partes that the whole <hi>AB</hi> is of the whole <hi>CD</hi>: which was required to be proued.</p>
               <div type="section">
                  <head>¶An other demonstration after <hi>Flussates.</hi>
                  </head>
                  <p>Suppose that the number AB be of the number CD the selfe same partes that the part taken away AE is of the part taken away CF.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Flussa<g ref="char:EOLhyphen"/>tes.</hi>
                     </note> Then I say, that the residue EB is of the residue FD the selfe same partes that the whole AB is of the whole CD. Let EB be of CI the self same partes that AB is of CD, or AE of C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Now forasmuch as EB is of CI the selfe same partes that AE is of CF: therefore both these AE and EB added together are of both these CF and CI added
<figure/>
together (that is, the whole AB is of the whole FI) the selfe same partes that AE is of CF (by the sixt of this booke). But what partes AE is of CF, the selfe same partes is the number AB of the number CD (by supposition). Wherefore what partes the number AB is of the nu<g ref="char:cmbAbbrStroke">̄</g>ber FI, tho selfe same partes is the same number AB of the number CD. Wherefore the numbers FI and CD are equall. Take a<g ref="char:EOLhyphen"/>way the number CF which is common to them both. Wherefore the numbers remayning CI and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D are equall. Wherefore what partes the number EB is of the number CI, the selfe same partes is the same number EB of the number FD. But what partes EB is of CI, the selfe same partes (by con<g ref="char:EOLhyphen"/>struction) is AB of CD. Wherefore what partes the residue EB is of the residue FD, the selfe same partes is the whole AB of the whole CD: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 7. Theoreme. The 9. Proposition. If a number be a part of a number, and if an other number be the self same part of an other nu<g ref="char:cmbAbbrStroke">̄</g>ber: then alternately what part or partes the first is of the third, the self same part or partes shall the second be of the fourth.</head>
               <p>SVppose that the number A be of the number BC the
<figure/>
selfe same part, that an other number D is of an other number EF. And let A be lesse then D. Then I say, that alternately what part or partes A is of D, the selfe same part or partes is BC of EF.<note place="margin">Construction.</note> For forasmuch as what part A is of BC, the selfe same part is D of EF, therefore how many
<pb facs="tcp:7062:222"/>
numbers there are in BC equall vnto A, so many are there in
<figure/>
EF equall vnto D. Deuide BC into the numbers equall vnto A, that is, into BG &amp; GC: and likewise EF into the num<g ref="char:EOLhyphen"/>bers equall vnto D, that is, into EH and HF. Now then the multitude of these BG and GC, is equall vnto the multitude of these EH &amp; HF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the numbers BG and GC are equall the one to the other, the numbers also EH and HF are equall the one to the other: and the multitude of these BG &amp; GC is equall vnto the multitude of these EH and HF. Wherefore what part or partes BG is of EH, the self same part or partes is GC of HF. Wherefore what part or partes BG is of EH, the selfe same part or partes (by the fift &amp; sixt of the seuenth) are BG and GC added together, of EH and HF added together. But BG is equall vnto A, and EH vnto D. Wherefore what part or partes A is of D, the selfe same part or partes is BG of EF: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 8. Theoreme. The 10. Proposition. If a number be partes of a number, and an other nu<g ref="char:cmbAbbrStroke">̄</g>ber the self same partes of an other number, then alternately what partes or part the first is of the third, the selfe same partes or part is the second of the fourth.</head>
               <p>SVppose that the number <hi>AB</hi> be of the number <hi>C</hi> the selfe same partes, that an other number <hi>DE</hi> is of an other nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>F,</hi> and let <hi>AB</hi> be lesse then <hi>DE.</hi> Then I say, that alternately also what part or partes <hi>AB</hi> is of <hi>DE,</hi> the selfe same partes or part is <hi>C</hi> of <hi>F.</hi> Forasmuch as what partes <hi>AB</hi> is of <hi>C,</hi> the selfe same partes is <hi>DE</hi> of <hi>F</hi>:<note place="margin">Construction.</note> therefore how many partes of <hi>C</hi> there are
<figure/>
in <hi>AB,</hi> so many partes of <hi>F</hi> also are there in <hi>DE.</hi> Deuide <hi>AB</hi> into the partes of <hi>C,</hi> that is, into <hi>AG</hi> and <hi>GB.</hi> And likewise <hi>DE</hi> into the partes of <hi>F,</hi> that is, <hi>DH</hi> and <hi>HE.</hi> Now then the multitude of these <hi>AG</hi> and <hi>GB,</hi> is equall vnto the multi<g ref="char:EOLhyphen"/>tude of these <hi>DH</hi> and <hi>HE.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as what part <hi>AG</hi> is of <hi>C,</hi> the selfe same part is <hi>DH</hi> of <hi>F,</hi> therefore alter<g ref="char:EOLhyphen"/>nately also (by the former) what part or partes <hi>AG</hi> is of <hi>DH,</hi> the selfe same part or partes is <hi>C</hi> of <hi>F.</hi> And by the same reason also what part or partes <hi>GB</hi> is of <hi>HE,</hi> the same part or partes is <hi>C</hi> of <hi>F.</hi> Wherefore what part or partes <hi>AG</hi> is of <hi>DH,</hi> the selfe same part or partes is <hi>AB</hi> of <hi>DE</hi> (by the 6. of the seuenth). But what part or partes <hi>AG</hi> is of <hi>DH,</hi> the selfe same part or partes is it proued that <hi>C</hi> is of <hi>F.</hi> Wherefore what partes or part <hi>AB</hi> is of D<hi>E,</hi> the selfe same partes or part is <hi>C</hi> of <hi>F</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 9. Theoreme. The 11. Proposition. If the whole be to the whole, as a part taken away is to a part taken away: then shall the residue be vnto the residue, as the whole is to the whole.</head>
               <p>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> discret qua<g ref="char:cmbAbbrStroke">̄</g>titie ans<g ref="char:EOLhyphen"/>wereth to the ninth prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>si<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> of the fifth boke in conti<g ref="char:EOLhyphen"/>nual qua<g ref="char:cmbAbbrStroke">̄</g>titie.</note>SVppose that the whole number AB be vnto the whole number CD, as the part take<g ref="char:cmbAbbrStroke">̄</g> away AE, is to the part take<g ref="char:cmbAbbrStroke">̄</g> away CF. The<g ref="char:cmbAbbrStroke">̄</g> I say that the residue EB, is to the resi<g ref="char:EOLhyphen"/>due FD, as the whole AB is to the whole CD. For forasmuch as, AB is to CD, as AE is to CF: therfore what part or partes AB is of CD, the selfe
<figure/>
same part or partes is AE of CF. Wherfore also the residue EB is of the residue FD(by the 8. of the seuenth) the selfe same parte
<pb n="192" facs="tcp:7062:222"/>
o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> partes that AB is of CD.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore also (by the <hi>21.</hi> definition of this booke) as EB is to FD, so is AB to CD: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 10. Theoreme. The 12. Proposition. If there be a multitude of numbers how many soeuer proportionall: as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes.</head>
               <p>SVppose that there be a multitude of nu<g ref="char:cmbAbbrStroke">̄</g>bers how many soeuer proportional, name<g ref="char:EOLhyphen"/>ly, A, B, C, D, so that as A is to B, so let C be to D.<note place="margin">This in discret qua<g ref="char:cmbAbbrStroke">̄</g>tity ans<g ref="char:EOLhyphen"/>wereth to the twelfe propo<g ref="char:EOLhyphen"/>sition of the fifth in conti<g ref="char:EOLhyphen"/>nual qua<g ref="char:cmbAbbrStroke">̄</g>tity.</note> Then I say that as one of the antecedentes, namely, A is to one of the consequentes, namely, to B, or as C is to D, so are all the antecedentes: namely, A and C to all the consequentes,
<figure/>
namely, to B and D. For forasmuch as (by supposition) as A is to B, so is C to D, therfore what parte or partes A is of B, the selfe same part or partes is C of D (by the <hi>21.</hi> definition of this booke) where<g ref="char:EOLhyphen"/>fore alternately what part or partes A is of C the selfe same parte or partes is B of D (by the ninth and tenth of the seuenth) wherefore both these numbers added together,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> A and C, are of both these numbers B and D added to<g ref="char:EOLhyphen"/>gether the selfe same part or partes that A is of B (by the 5. and 6. of the seuenth) wherfore (by the <hi>21.</hi> definition of the seuenth) as one of the antecedents, namely, A, is to one of the con<g ref="char:EOLhyphen"/>sequentes, namely, to B, so are all the antecedentes A and C to all the consequentes B &amp; D. Which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 11. Theoreme. The 13. Proposition. If there be foure numbers proportionall: then alternately also they shall be proportionall.</head>
               <p>
                  <hi>SVppose that there be foure numbers proportional,</hi>
                  <note place="margin">
                     <hi>This in dis<g ref="char:EOLhyphen"/>crete quanti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y answereth to the sixtenth proposition of the fifth booke in continuall quantitie.</hi>
                  </note> 
                  <hi>A, B, C, D, so that as A is to B, so let C be to D. Then I say that alternately also they shalbe proportional, that is. as A is to C, so is B to D. For forasmuch as (by supposition) as A is to B, so is C to D, therfore (by the</hi> 21. <hi>definition of this booke) what part or partes
<figure/>
A is of B the selfe same part or partes is C of D. Therfore alternate<g ref="char:EOLhyphen"/>ly what part or partes A is of C the selfe same part or partes is B of D (by the 9. of the seuenth) &amp; also (by the 10. of the same): wherfore as A is to C, so is B to D (by the</hi> 21. <hi>definition of this booke): which was required to be proued.</hi>
               </p>
               <p>Here is to be noted, that although in the foresayd example and demonstration the number A be supposed to be lesse then the number B, and so the number C is lesse then the number D:<note place="margin">
                     <hi>Note.</hi>
                  </note> yet will the same serue also though A be supposed to be greater then B, wherby also C shall be greater then D, as in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s example here put. For for that (by supposition) as A is to B, so is C to D, and A is supposed to be greater then B, and C greater then D: therefore (by the 21. definition of this
<figure/>
Booke) how multiplex A is to B, so multiplex is C to D, and therefore what part or partes B is of A, the selfe same part or partes is D of C. Wherefore alternately what part or partes B is of D, the selfe same part or partes is A of C, and therefore by the same definition, B is to D, as A is to C. And so must you vnderstand of the former Proposition next going before.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:223"/>
               <head>¶The 12. Theoreme. The 14. Proposition. If there be a multitude of numbers how many soeuer, and also other num<g ref="char:EOLhyphen"/>bers equall vnto them in multitude, which being compared two and two are in one and the same proportion: they shall also of equalitie be in one and the same proportion.</head>
               <p>
                  <note place="margin">
                     <hi>This in dis<g ref="char:EOLhyphen"/>crete quantity an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>wereth to t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> twe<g ref="char:cmbAbbrStroke">̄</g>ty one proposition o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the fifth booke in continuall quantitie.</hi>
                  </note>
                  <hi>SVppose that there be a multitude of numbers how many soeuer: namely, A, B, C, and let the other numbers equall vnto them in multitude be D, E, F: which be<g ref="char:EOLhyphen"/>ing compared two and two, let be in one and the same proportion: that is, as A to B, so let D be to E: and as B is to C, so let E be to F. Then I say, that of equa<g ref="char:EOLhyphen"/>litie, as A is to C, so is D to F. For forasmuch as by suppositi<g ref="char:EOLhyphen"/>on
<figure/>
as A is to B, so is D to E: therefore alternately also (by the 13 of the seuenth) as A is to D, so is B to E. Againe, for that as B is to C, so is E to F:</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>therfore alternately also (by the self same) as B is to E, so is C to F. But as B is to E, so is A to D. VVher<g ref="char:EOLhyphen"/>fore (by the seuenth common sentence of the seuenth) as A is to D, so is C to F. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore alternately (by the 13. of the seuenth) as A is to C, so is D to F: which was required to be demon<g ref="char:EOLhyphen"/>strated.</hi>
               </p>
               <p>After this Proposition, <hi>Campane</hi> demonstrateth in numbers these foure kindes of proportionalitie,<note place="margin">
                     <hi>Certaine ad<g ref="char:EOLhyphen"/>ditions of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>pane.</hi>
                  </note> namely, proportion conuerse, composed, deuided, and euerse: which were in continual quantitie, demonstrated in the 4. 17. 18. and 19. propositions of the fift booke. And first he demonstrateth conuerse proportion in this maner.</p>
               <p>Suppose that the number A be to the number B, as the number C is to the
<figure/>
number D.<note place="margin">
                     <hi>Two cases in this p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>positio<g ref="char:cmbAbbrStroke">̄</g>.</hi>
                  </note> Then I say, that conuersedly B is to A, as D is to C. For if A be lesse then, B, C also shall be lesse then D, and what part or partes A is of B, the selfe same par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or partes is C of D.<note place="margin">
                     <hi>The fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> case.</hi>
                  </note> Wherefore B is equemultiplex to A, as D is to C. Wherefore (by the 21. definition of this booke) as B is to A, so is D to C.</p>
               <p>
                  <note place="margin">
                     <hi>The second case.</hi>
                  </note> But if A be greater then B, C also is greater then D: and what part or partes B is of A, the selfe same part or partes is D of C. Wherefore (by the same definition) as B is to A, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> D to C: which was required to be proued.</p>
               <div type="section">
                  <head>Proportionalitie deuided, is thus demonstrated.</head>
                  <p>Suppose that the number AB be to the number B, as the number CD is to the number D. Then I say,<note place="margin">
                        <hi>Prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rtiona<g ref="char:EOLhyphen"/>lity deuided.</hi>
                     </note> that deuided also, as A is to B, so is C to D. For for that as AB is to B,
<figure/>
so is CD to D: ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore alternately (by the 14. of this booke) as AB is to CD, so is B to D. Wherefore (by the 11. of this booke) as AB is to CD, so is A to C. Wherefore as B is to D, so is A to C: and for that as A is to C, so is B to D, the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>efore alternately as A is to B, so is C to D.</p>
               </div>
               <div type="section">
                  <head>Proportionalitie composed, is thus demonstrated.</head>
                  <p>
                     <note place="margin">
                        <hi>Pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>portiona<g ref="char:EOLhyphen"/>li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>y composed.</hi>
                     </note>If A be vnto B, as C is to D: then shall AB be to B, as CD is to D. For al<g ref="char:EOLhyphen"/>ternately
<figure/>
A is to C, as B is to D. Wherefore (by the 13. of this booke) as AB, namely, all the antecedentes are to CD, namely, to all the consequentes, so is B to D, namely, one of the antecedentes to one of the consequentes. Wher<g ref="char:EOLhyphen"/>fore alte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nately as AB is to B, so is CD to D.</p>
               </div>
               <div type="section">
                  <head>Euerse proportionalitie, is thus proued.</head>
                  <p>
                     <note place="margin">
                        <hi>Euerse pro<g ref="char:EOLhyphen"/>portionality.</hi>
                     </note>Suppos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that AB be to B, as CD is to D: then shall AB be to A, as CD is to C. For alternately
<pb n="193" facs="tcp:7062:223"/>
AB is to CD, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> B is to D. Wheref<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> (by the 13. of this boo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>) A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> CD,
<figure/>
as A is to C. Wherefore alternately AB i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to A, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> CD i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to C: whi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ A proportion here added by <hi>Campane.</hi>
               </head>
               <p>
                  <q>If the proportion of the first to the second, be as the proportion of the third to the fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rth, and if the pro<g ref="char:EOLhyphen"/>porti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he fift to the second be as the prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rtion of the sixt to the fourth: then the proportion of the first and the fifth taken together, shall be to the second, as the proportion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>f the third and the sixt taken together to the fourth.</q>
               </p>
               <p>As if A be vnto B, as C is to D: and if also E be to B, as F is to D.<note place="margin">
                     <hi>This propos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLunhyphen"/>tion in discret quantity an<g ref="char:EOLhyphen"/>swereth to the</hi> 24. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g> of the fif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h boke in con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<g ref="char:EOLhyphen"/>nual quantity.</hi>
                  </note> Then shall A &amp; E taken together, be vnto B, as C and F taken together, are vnto <hi>D.</hi> For by conuerse proportionalitie, B is to <hi>E,</hi> as <hi>D</hi> is to F. Wherefore by proportion of equalitie, as
<figure/>
A is to E, so is C to F. Wherefore (by com<g ref="char:EOLhyphen"/>position) as A and E are to E, so are C and F to F. But (by supposition) as E is to B, so is F to D. Wherefore againe by proportion of equalitie, as A and E are to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, so are C and F to <hi>D</hi>: which was req<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ired to be proued.</p>
               <p>And after the same maner may you proue the conuerse of this Proposition. If B be to A, as D is to C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and if also B be vnto E, as <hi>D</hi> is to F: Then shall B be to AE, as D is to CF. For by conuerse proportio<g ref="char:EOLhyphen"/>nalitie,<note place="margin">
                     <hi>The conuers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the same pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>position.</hi>
                  </note> A is to B, as C is to D. Wherefore of equalitie, A is to E, as C is to F. Wherefore by composi<g ref="char:EOLhyphen"/>tion A and E are to E, as C and F are to F. Wherefore conuersedly, E is to A and E, as F is to C and F. But by supposition, B is to E, as <hi>D</hi> is to F. Wherefore agayne by Proportion of equalitie, B is to A and E, as D is to C and F:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>By this also it is manifest that if the proportion of numbers how many soeuer vnto the first, be as the proportion of as many other numbers vnto the second,<note place="margin">
                        <hi>A Corollary followi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>g th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>se proposi<g ref="char:EOLhyphen"/>tions ad<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ed by Campa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e.</hi>
                     </note> then shall the proportion of the numbers composed of all the numbers that were antecedentes to the first, be to the first, as the number composed of all the numbers that were antecedentes to the second is to the second. And also conuersedly if the proportion of the first to nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers how many soeuer, be as the proportion of the second to as many other numbers: then shall the proportion of the first to the number composed of all the numbers that were consequentes to it selfe, be as the proportion of the second to the number compo<g ref="char:EOLhyphen"/>sed of all the numbers that were consequen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es to it selfe.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 13. Theoreme. The 15. Proposition. If vnitie measure any number, and an other number do so many times mea<g ref="char:EOLhyphen"/>sure an other number: vnitie also shall alternately so many times measure the third number, as the second doth the fourth.</head>
               <p>SVppose that vnitie A do measure the number BC: and let an other nu<g ref="char:cmbAbbrStroke">̄</g>ber D so many times measure some other nu<g ref="char:cmbAbbrStroke">̄</g>ber, namely, EF. Then I say, that alternate<g ref="char:EOLhyphen"/>ly, vnitie A shall so many times measure the number D, as the number BC doth measure the number EF.<note place="margin">Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>str<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ctio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> For forasmuch as vnitie A doth so many times mea<g ref="char:EOLhyphen"/>sure BC, as D doth EF: therefore how many vnities there are in BC,
<figure/>
so many numbers are there in EF equall vnto D. Deuide (I say) BC in<g ref="char:EOLhyphen"/>to the vnities which are in it, that is, into BG, GH, and HC. And deuide likewise EF into the numbers equall vnto D, that is, into EK, KL, and LF. Now then the multitude of these BG, GH, and HC, is equall vnto the multitude of these EK, KL, LF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as these vnities BG, GH, and HC, are equall the one to the other, and these numbers EK, KL, &amp; LF, are also equall the one to the other, and the multitude of the vnities BG, GH, and HC, are equall vnto the multitude of the numbers EK, KL, &amp; LF: therefore as vnitie BG is to the num<g ref="char:EOLhyphen"/>ber EK, so is vnitie GH to the number KL, and also vnitie HC to the number LF. Wher<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:224"/>
(by the 12<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the seue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th) as one of the antecede<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s to one of the
<figure/>
consequentes, so are all the anteceden<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es to all the consequentes: Wherfore as vnitie BG is to the number EK, so is the number BC to the number EF. But vnitie BG is equall vnto vnitie A, and the number EK to the number D. VVherefore (by the <hi>7.</hi> common sentence) as vnitie A is to the number D, so is the number BC to the number EF. VVherefore vnitie A measureth the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber D, so many times, as BC measureth EF (by the 21 definition of this booke): which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 14. Theoreme. The 16. Proposition. If two numbers multiplying them selues the one into the other, produce any numbers: the numbers produced shall be equall the one into the other.</head>
               <p>SVppose that there be two numbers A and B: and let A multiplying B produce C, and let B multiplying A produce D. Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n I say, that the number C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equall vn<g ref="char:EOLhyphen"/>to the n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mber D.<note place="margin">Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note>Take any vnitie, namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> E. And forasmuch as A multiplying B produced C, therefore B measureth C by the vnities which are in A. And vni<g ref="char:EOLhyphen"/>tie E measureth the number A by those vnities which are in the number
<figure/>
A. VVhere<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore vnitie E so many times measureth A, as B measureth C. VVherefore alternately (by the 15. of the seuenth) vnitie E measureth the number B so many times as A measureth C. Againe, for that B mul<g ref="char:EOLhyphen"/>tiplying A produced D, therefore A measureth D by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vnities which are in B. And vnitie E measureth B by the vnities which are in B. VVherefore vnitie E so many times measureth the number B, as A mea<g ref="char:EOLhyphen"/>sureth D. But vnitie E so many times measureth the number B, as A measureth C. VVhere<g ref="char:EOLhyphen"/>fore A measureth either of these numbers C and D a like. VVherefore (by the 3. common s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ntence of this booke) C is equall vnto D: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The 15. Theoreme. The 17. Proposition. If one number multiply two numbers, and produce other numbers, the num<g ref="char:EOLhyphen"/>bers produced of them, shall be in the selfe same proportion, that the num<g ref="char:EOLhyphen"/>bers multiplied are.</head>
               <p>SVppose that the number A multiplieng two numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rs B and C, do produce the numbers D and E. Then I say that as B is to C, so is D to E. Take vnitie, name<g ref="char:EOLhyphen"/>ly, F. And <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as A multiplieng B produced D, therfore B measureth D by those vnities that are in A. And vnitie F measureth A by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> whih are in A.<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>emonstra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> Wherfore vnitie F so many times measureth the
<figure/>
number A, as B measureth D. VVherfore as vnitie I is to the number A. so is the number B to the number D (by the <hi>21</hi> de<g ref="char:EOLhyphen"/>finition of this booke) And by the same reason, as vnitie F is to the number A, so is the number C to the number E: wherefore also (by the <hi>7.</hi> common sentence of this booke) as B is to D, so is C to E. VVherfore alternately (by the <hi>15.</hi> of the seuenth) as B is to C, so is D to E. If therfore one number multiply two numbers<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and produce other numbers: the numbers produced of them, shall be in the selfe same propor<g ref="char:EOLhyphen"/>tion, that the numbers multiplied are: which was required to be proued.</p>
               <div type="section">
                  <pb n="194" facs="tcp:7062:224"/>
                  <head>Here Flu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>tes adde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ollary.</head>
                  <p>
                     <hi>If two number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> hauing one and the sam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> proporti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> with two other numbers do multiply th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e the other alternately,</hi>
                     <note place="margin">
                        <hi>A Corollary added by Fluss<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tes.</hi>
                     </note> 
                     <hi>and produce any numbers, the numbers produced of them shall be equall the one to the other.</hi>
                  </p>
                  <p>Suppose that there be two number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> and <hi>B,</hi> and also two other numbers <hi>C</hi> and <hi>D,</hi> hauing th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> same proportion that the numbers <hi>A</hi> and <hi>B</hi> haue: and let the numbers <hi>A</hi> and <hi>B</hi> multiply the number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>C</hi> &amp; <hi>D</hi> alternately, that is, let <hi>A</hi> multiplieng
<figure/>
                     <hi>D</hi> produce <hi>F,</hi> and let <hi>B</hi> multiplieng <hi>C</hi> produce <hi>E.</hi> Then I say that the numbers produced namely, <hi>E</hi> &amp; <hi>F</hi> are equall. Let <hi>A</hi> and <hi>B</hi> multiply the one the other in such sort, that let <hi>A</hi> multiplieng <hi>B</hi> pro<g ref="char:EOLhyphen"/>duce <hi>G,</hi> and let <hi>B,</hi> multiplieng <hi>A</hi> produce <hi>H,</hi> Now then the numbers <hi>G</hi> and <hi>H</hi> are equal by the 16. of this booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And forasmuch as <hi>A</hi> multipli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ng the two numbers <hi>B</hi> and <hi>D,</hi> produced the numbers <hi>G</hi> and <hi>F,</hi> therfore <hi>G</hi> is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, as <hi>B</hi> is to <hi>D</hi> by this proposition. So likewise <hi>B</hi> multiplieng the two numbers <hi>A</hi> and <hi>C</hi> produced the two numbers <hi>H</hi> and <hi>E.</hi> Wherfore by the same <hi>H</hi> is to <hi>E</hi> as <hi>A</hi> is to <hi>C.</hi> But alternately (by the 13. of this booke) <hi>A</hi> is to <hi>C</hi> as <hi>B</hi> is to <hi>D,</hi> but as <hi>A</hi> is to <hi>C</hi> so is <hi>H</hi> to <hi>E,</hi> and as <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is to <hi>D,</hi> so is <hi>G</hi> to <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> Wherfore by the seuenth common sentence, as <hi>H</hi> is to <hi>E</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> so is <hi>G</hi> to <hi>F.</hi> Wherfore a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ternately (by the 13. of this booke) <hi>H</hi> is to <hi>G</hi> as <hi>E</hi> is to <hi>F.</hi> But it is proued that <hi>G</hi> &amp; <hi>H</hi> are equall: Wherfore <hi>E</hi> and <hi>F</hi> (which haue the same proportion that <hi>A</hi> and <hi>B</hi> haue) are equall. If therefore there be two numbers, &amp;c. Which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 16. Theoreme. The 18. Proposition. If two numbers multiply any number, &amp; produce other numbers: the num<g ref="char:EOLhyphen"/>bers of them produced, shall be in the same proportion that the numbers multiplying are.</head>
               <p>
                  <hi>SVppose that two numbers A and B multiplieng the number C, doo produce the numbers D and E.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>Then I say that as A is to B, so is D to E. For forasmuch as A multiplieng C produced D, therfore C multipli<g ref="char:EOLhyphen"/>eng
<figure/>
A produceth also D (by the</hi> 16. <hi>of this booke.) And by the same reason C multiplieng B produceth E. Now then one number C multiplieng two numbers A and B, pro<g ref="char:EOLhyphen"/>duceth the numbers D and E. VVherfore by the</hi> 17. <hi>of the se<g ref="char:EOLhyphen"/>uenth, as A is to B, so is D to E: which was required to be de<g ref="char:EOLhyphen"/>monstrated.</hi>
               </p>
               <p>This Proposition, and the former touching two numbers, may be extended to num<g ref="char:EOLhyphen"/>bers how many soeuer.<note place="margin">
                     <hi>This proposi<g ref="char:EOLhyphen"/>tion and the former may be extended to numbers how many soeuer.</hi>
                  </note> So that if one number multiply numbers how many soeuer, and produce any numbers, the proportion of the numbers produced, and of the numbers multiplied, shall be one and the selfe same. Likewise if numbers how many soeuer mul<g ref="char:EOLhyphen"/>tiply one number, and produce any numbers, the proportion of the numbers produ<g ref="char:EOLhyphen"/>ced<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and of the numbers multiplieng shall be one and the selfe same: which thing by this and the former proposition repeted as often as is needefull, is not hard to proue.</p>
            </div>
            <div type="part">
               <head>¶The 17. Theoreme. The 19. Proposition. If there be foure numbers in proportion: the number produced of the first and the fourth, is equall to that number which is produced of the second and the third. And if the number which is produced of the first and the fourth be equall to that which is produced of the second &amp; the third: those foure numbers shall be in proportion.</head>
               <p>
                  <pb facs="tcp:7062:225"/>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion in num<g ref="char:EOLhyphen"/>bers demon<g ref="char:EOLhyphen"/>strateth that which the <hi>16.</hi> of the sixth booke demon<g ref="char:EOLhyphen"/>strateth in lines.</note>SVppose that there be foure numbers in proportion A, B, C, D, as A is to B, so let C be to D. And let A multiplieng D produce E: and let B multiplieng C pro<g ref="char:EOLhyphen"/>duce F. Then I say that the number E is equall vnto the number F. Let A mul<g ref="char:EOLhyphen"/>tiplieng C produce G. Now forasmuch as A multiplieng C, produceth G, &amp; mul<g ref="char:EOLhyphen"/>tiplieng D it produceth E: it followeth that the number A
<figure/>
multiplieng two numbers C and D, produceth G and E. VVherfore by the <hi>17</hi> of the seuenth, as C is to D, so is G to E. But as C is to D, so is A to B, wherfore as A is to B, so is G to E.<note place="margin">Construction.</note> Againe, forasmuch as A multiplieng C produced G, and B multiplieng C produced F:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore two numbers A and B, multiplieng one nu<g ref="char:cmbAbbrStroke">̄</g>ber C, do produce G &amp; F. VVher<g ref="char:EOLhyphen"/>fore by the <hi>18.</hi> of the seuenth, as A is to B, so is G to F. But as A is to B, so is G to E: wherfore as G is to E, so is G to F. VVherfore G hath to either of these E and F one &amp; the same proportion (But if one number haue to two numbers one and the same proportion, the said<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two numbers shall be equall). VVherfore E is equall vnto F.</p>
               <p>But now againe, suppose that E be equall vnto F. Then I say that as A is to B, so is C to D.<note place="margin">The second part of this proposition which is the conuerse of the first.</note> For the same order of construction remayning still, forasmuch as A multiplieng C &amp; D produced G and E, therfore by the <hi>17.</hi> of the seuenth, as C is to D so is G to E, but E is equall vnto F (But if two numbers be equall, one number shall haue vnto them on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the same proportion) wherfore as G is to E, so is G to F. But as G is to E, so is C to D. Wherefore as C is to D, so is G to F, but as G is to F, so is A to B by the <hi>18.</hi> of the seuenth, wherfore as A is to B so is C to D:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> which was required to be proued.</p>
               <p>
                  <note place="margin">An assumpt added by Campane.</note>
                  <hi>Here</hi> Campane <hi>addeth, that it is needeles to demonstrate, that if one number haue to two numbers one and the same proportion, the said two numbers shall be equall: or that if they be equal, one number hath to them one and the same proportion. For (saith he) if G haue vnto E and F one and the same proportion, the<g ref="char:cmbAbbrStroke">̄</g> either, what part or partes G is to E, the same part or parts is G also of F: or how multiplex G is to E, so multiplex is G to F (by the 21. definition) And therfore by the 2 and 3 common sentence, the said numbers shall be equall. And so conuersedly, if the two numbers E and F be equal, then shall the numbers E and F be either the selfe same parte or partes of the number G, or they shall be equemultiplices vnto it. And therfore by the same definition the number G shall haue to the numbers E and F one and the same proportion.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 18. Theoreme. The 20. Proposition. If there be three numbers in proportion, the number produced of the ex<g ref="char:EOLhyphen"/>tremes, is equall to the square made of the middle number. And if that nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber which is produced of the extremes, be equall to the square made of the middle number, those three numbers shall be in proportion.</head>
               <p>SVppose there be three numbers in proportion, A, B, C, as A is to B, so let B be to C.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion in num<g ref="char:EOLhyphen"/>bers demon<g ref="char:EOLhyphen"/>strateth that which the <hi>17.</hi> of the sixth demonstrateth in lines.</note> Then I say that the number produced of A and C is equall to the square num<g ref="char:EOLunhyphen"/>ber which is made of B. Put vnto B an equall number D.
<figure/>
Wherfore as A is to B, so is D to C. Wherfore that which is produced of A into C, is equal vnto that which is produced of B into D. But that which is produced of B into D is equal to that which is made of B (for B is equal vnto D) wherfore that which is produced of A into C is equal to that which is made of B.</p>
               <p>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note>But now suppose that that which is produced of A into C, be equall
<pb n="195" facs="tcp:7062:225"/>
to that which is made of B.<note place="margin">The second part which is the conuerse of the first.</note> Then I say that as A is to B, so is B to C. For (the same order of construction remayning) forasmuch as that which is produced of A into C, is equall to that which is made of B, but that which is made of B, is equal to that which is produced of B in<g ref="char:EOLhyphen"/>to D. (For B and D are by supposition equall): therfore that which is produced of A into C is equall to that which is produced of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> into D:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion</note> wherfore (by the second part of the former proposition) as A is to B, so is D to C: but D is equal to E. Wherefore as A is to B, so is the same B to C: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 19. Theoreme. The 21. Proposition. The left numbers in any proportion, measure any other nu<g ref="char:cmbAbbrStroke">̄</g>bers hauing the same proportion equally, the greater the greater, &amp; the lesse the lesse.</head>
               <p>SVppose that CD &amp; EF be the least numbers that haue one &amp; the same propor<g ref="char:EOLhyphen"/>tion with the numbers A and B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>lity.</note> Then I say, that the number CD so many times measureth the number A, as the number EF measureth the number B. For for<g ref="char:EOLhyphen"/>asmuch as by supposition CD is to EF, as A is to B, and CD and EF are also supposed to be lesse then A and B: therefore CD and EF are either part or partes of A and B (by the 4. of this booke, and by the 21. definition of the same). But they are not partes. For if it be possible, let CD be partes of A. VVherfore EF is the selfe same partes of B, that CD is of A. Wherefore how many partes of A there are in CD, so many partes are there of B in EF. Deuide CD into the partes of A, that is, into CG and GD. And likewise deuide EF into the partes of B, that is, into EH and HF. Now then the multitude
<figure/>
of these CG and GD, is equall vnto the multitude of these EH &amp; HF. And forasmuch as CG and GD are numbers equall the one to the other, and these numbers EH and HF are also equall the one to the other, and the multitude of these CG and GD, is equall to the multitude of these EH and HF: therefore as CG is to EH, so is GD to HF. Wherefore (by the 12. of the seuenth) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as CG is to EH, so is CD to EF. Wher<g ref="char:EOLhyphen"/>fore CG and DH <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re in the selfe same proportion that CD and EF are, being also lesse then CD and EF: which is impossible. For CD and EF are supposed to be the least that haue <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> and the same proportion with them. Wherefore CD is not partes of A: wherefore it is a part. Wherefore EF is of B <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he selfe same part, that CD is of A. Wherefore CD so many <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> measureth A as EF doth B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was req<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ired to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 20. Theoreme. The 22. Proposition. If there be three numbers, and other numbers equall vnto the<g ref="char:cmbAbbrStroke">̄</g> in multitude, which being compared two and two are in the selfe same proportion, and if also the proportion of them be perturbate, then of equalitie they shall be in one and the same proportion.</head>
               <p>SVppose that there be three numbers A, B, and C and
<figure/>
let the other numbers equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nto them in multitude be D, E, and F.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion in discret qua<g ref="char:cmbAbbrStroke">̄</g>titie ans<g ref="char:EOLhyphen"/>wereth to the <hi>23.</hi> proposi<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> of the fifth boke in conti<g ref="char:EOLhyphen"/>nual qua<g ref="char:cmbAbbrStroke">̄</g>titie.</note> And le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two and two compared toge<g ref="char:EOLhyphen"/>ther be in one and the same prop<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let the proportion of them be p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rturbate; so that <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> as B is to C, so let D be to E. <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> is to C, so is D to F. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or, for <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>
                  <pb facs="tcp:7062:226"/>
therefore that which is produced of A into F, is (by the 19. of
<figure/>
the seue<g ref="char:cmbAbbrStroke">̄</g>th) equall to that which is produced of B into E. Againe for that as B is to C, so is D to E, therefore that which is produ<g ref="char:EOLhyphen"/>ced of D into C, is equall to that which is produced of B into E. And it is proued, that that which is produced of A into F, is e<g ref="char:EOLhyphen"/>qual to that which is produced of B into E. Wherfore that which is produced of A into F, is equal to that which is produced of D into C. Wherefore (by the second part of the 19. of the seuenth) as A is to C, so is D to F: which was required to be proued.</p>
               <p>
                  <hi>The same may also be proued if in either order be more then three numbers: as it was proued in the 23. of the fift touching more magnitudes then three.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 21. Theoreme. The 23. Proposition. Numbers prime the one to the other: are y<hi rend="sup">e</hi> least of any numbers, that haue one and the same proportion with them.</head>
               <p>
                  <note place="margin">This and the eleuen propo<g ref="char:EOLhyphen"/>sitions follow<g ref="char:EOLhyphen"/>ing, declare the p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ssions and properties of<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> prime nu<g ref="char:cmbAbbrStroke">̄</g>bers.</note>SVppose that <hi>A</hi> and <hi>B</hi> be numbers prime the one to the other. Then I say that <hi>A</hi> and <hi>B</hi> are the least of any numbers that haue one and the same proportion with them. For if <hi>A</hi> and <hi>B</hi> be not the least of any numbers that haue one and the same propor<g ref="char:EOLhyphen"/>tion with them, then are there some numbers lesse then <hi>A</hi> and
<figure/>
                  <hi>B,</hi> being in the selfe same proportion that <hi>A</hi> and <hi>B</hi> are. Let the same be <hi>C</hi> and D. Now forasmuch as the least numbers in any proportion measure any other numbers hauing the same pro<g ref="char:EOLhyphen"/>portion equally, the greater the greater, and the lesse the lesse (by the <hi>21.</hi> of the seuenth) that is, the antecedent the antece<g ref="char:EOLhyphen"/>dent, and the consequent the consequent: therefore <hi>C</hi> so many times measureth <hi>A,</hi> as <hi>D</hi> measureth <hi>B.</hi> How many times <hi>C</hi> measureth <hi>A,</hi> so many vnities let there be in <hi>E.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>lity.</note> Wherefore <hi>D</hi> measureth <hi>B</hi> by those vnities which are in <hi>E.</hi> And forasmuch as <hi>C</hi> measureth <hi>A</hi> by those vnities which are in <hi>E,</hi> therefore <hi>E</hi> also measureth <hi>A</hi> by those vnities which are in <hi>C.</hi> And by the same reason <hi>E</hi> measureth <hi>B,</hi> by those vnities which are in <hi>D.</hi> Wherefore <hi>E</hi> measureth <hi>A</hi> and <hi>B</hi> being prime numbers the one to the other which (by the <hi>13.</hi> definition of the seuenth) is impossible. Wherefore there are no other numbers lesse then <hi>A</hi> and <hi>B,</hi> which are in the selfe same proportion that <hi>A</hi> and <hi>B</hi> are. Wherefore <hi>A</hi> and <hi>B</hi> are the least numbers that haue one and the same proportion with them: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 22. Theoreme. The 24. Proposition. The least numbers that haue one and the same proportion with them: are prin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the one to the other.</head>
               <p>SVppose that A and B be the least numbers that haue one and the same properti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> with them.<note place="margin">This is the co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>uerse of the former propo<g ref="char:EOLhyphen"/>sition.</note> Then I say that A and B be prime the one
<figure/>
to the other. For if A &amp; B be not prime the one to the other, then shall some one nu<g ref="char:cmbAbbrStroke">̄</g>ber measure A &amp; B. Let the same be C. And how oftentimes C measureth A, so many vnities let there be in D: and how ofte<g ref="char:cmbAbbrStroke">̄</g>times C measureth B, so many vni<g ref="char:EOLhyphen"/>ties let there be in E. And forasmuch as C measureth A by
<pb n="196" facs="tcp:7062:226"/>
those vnities that are in D, therefore C multiplying D produceth A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And by the same rea<g ref="char:EOLhyphen"/>son C multiplying E produceth B.<note place="margin">Demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Wherefore the number C multiplying two numbers D and E, produceth A &amp; B. Wherefore (by the <hi>17.</hi> of the seuenth) as D is to E, so is A to B. But the numbers D and E are lesse then A and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and are also in the selfe same proportion with the<g ref="char:cmbAbbrStroke">̄</g>, which is impossible. Wherefore no number measureth these numbers A and B. Wherefore A and B are prime the on to the other: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 23. Theoreme. The 25. Proposition. If two numbers be prime the on to the other: any number measuring one of them shalbe prime to the other number.</head>
               <p>SVppose that A and B be two prime numbers the on to the other. And let some num<g ref="char:EOLhyphen"/>ber, namely, C, measure A. Then I say that C and B are prime numbers the on to the other. For if C and B be not prime the one to the other, let some number measure C and B, and let the same be D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> And forasmuch as D measureth
<figure/>
C, and C measureth A, therfore D also measureth A (by the fift common sentence) and D measureth B. Wherefore D measu<g ref="char:EOLhyphen"/>reth A and B being numbers prime the on to the other, which is impossible (by the <hi>13.</hi> definition of the seuenth) wherefore no number measureth these numbers B and C. Wherefore B and C are numbers prime the on to the other. Which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 24. Theoreme. The 26. Proposition. If two numbers be prime to any one number, the number also produced of them shall be prime to the selfe same.</head>
               <p>SVppose that the two numbers <hi>A</hi> and <hi>B</hi> be prime to any one number, namely, to <hi>C,</hi> and let <hi>A</hi> multiplieng <hi>B</hi> produce <hi>D.</hi> Then I say that <hi>C</hi> and <hi>D</hi> are prime numbers the one to the other. For if <hi>C</hi> and <hi>D</hi> be not prime the one to the other then some number shall measure them. Let there be a number that measureth them, and let the same be <hi>E.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> And forasmuch as <hi>A</hi> and <hi>C</hi> are prime
<figure/>
numbers the one to the other, and the number <hi>E</hi> measureth <hi>C.</hi> Therfore <hi>E</hi> and <hi>A</hi> are (by the <hi>25.</hi> of the seuenth) prime the one to the other. And forasmuch as <hi>E</hi> measureth <hi>D.</hi> How many times <hi>E</hi> measureth <hi>D,</hi> so ma<g ref="char:EOLhyphen"/>ny vnities let there be in <hi>F.</hi> Wherfore <hi>F</hi> also measureth <hi>D</hi> by those vni<g ref="char:EOLhyphen"/>ties which are in <hi>E.</hi> Wherfore <hi>E</hi> multiplieng <hi>F</hi> produceth <hi>D.</hi> But <hi>A</hi> also multiplieng <hi>B</hi> produced <hi>D.</hi> Wherfore that which is produced of <hi>E</hi> into <hi>F</hi> is equall to that which is produced of <hi>A</hi> into <hi>B.</hi> But if that which is produced of the ex<g ref="char:EOLhyphen"/>tremes be equall to that which is produced of the meanes, then are those foure numbers in proportion (by the <hi>19.</hi> of the seuenth) wherefore <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <hi>E</hi> is to <hi>A,</hi> so is <hi>B</hi> to <hi>F.</hi> But <hi>A</hi> and <hi>E</hi> are prime the one to the other. Yea they are prime and the least in the same proportion (by the <hi>23</hi> of the seuenth.) But the least numbers in any proportion measure any other numbers hauing the same proportion equally, the greater the greater, and the lesse the lesse, that is, the antece<g ref="char:EOLhyphen"/>dent the antecedent, and the consequent the consequent by the <hi>21.</hi> of the seuenth wherfore <hi>E</hi> measureth <hi>B,</hi> and it also measureth <hi>C</hi>: wherfore <hi>E</hi> measureth <hi>C</hi> and <hi>B,</hi> which are by sup<g ref="char:EOLhyphen"/>position numbers prime the one to the other, which is impossible by the <hi>13.</hi> definition of the
<pb facs="tcp:7062:227"/>
seuenth. Wherfore no number measureth those numbers <hi>C</hi> and <hi>D.</hi> Wherfore <hi>C</hi> and <hi>D</hi> are prime the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 25. Theoreme. The 27. Proposition. If two numbers be prime the one to the other, that which is produced of the one into him selfe, is prime to the other.</head>
               <p>SVppose that there be two numbers prime the one to the other,
<figure/>
A and B, and let A multiplieng himselfe produce C. Then I say that B and C are prime the one to the other. For put vn<g ref="char:EOLhyphen"/>to A an equall number,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> namely, D. And forasmuch as A &amp; B are prime the one to the other, and A is equall vnto D: therefore D and B also are prime the one to the other. Wherfore either of these numbers D and A is prime to B. Wherfore that which is produced of D into A is prime vnto B by the former proposition. But that number which is produced of D into A is the number C. Wherefore C and B are prime numbers the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 26. Theoreme. The 28. Proposition. If two numbers be prime to two numbers, eche to either of both: the numbers produced of them shall be prime the one to the other.</head>
               <p>SVppose that there be two numbers A and B prime to two numbers C and D, either of both, to either of both: namely, let either of these A and B be prime to C, and also to D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And let A multiplieng B produce E, and let C
<figure/>
multiplieng D produce F. Then I say that E and F are prime numbers the one to the other. For forasmuch as either of these A &amp; B are prime vnto C, therfore that which is produced of A into B is prime vnto C by the <hi>26.</hi> of the seuenth. But that which is produced of A into B is the number E, therefore E and C are prime the one to the other. And by the same reason also E and D are prime the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore either of these num<g ref="char:EOLhyphen"/>bers C and D are prime vnto E: wherefore that also which is produced of C into D is prime vnto E by the same. But that which is produced of C into D, is the number F. Wherfore E and F are numbers prime the one to the other: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 27. Theoreme. The 29. Proposition If two numbers be prime the one to the other, and ech multiplying himselfe bring forth certaine numbers: the numbers of them produced shall be prime the one to the other. And if those numbers geuen at the beginning multi<g ref="char:EOLhyphen"/>plying the sayd numbers produced, produce any numbers: they also shall be prime the one to the other: and so shall it be continuing infinitely.</head>
               <p>
                  <pb n="197" facs="tcp:7062:227"/>SVppose that there be two numbers A and B prime the one to the other. And let A multiplying him selfe produce C: and multiplying C let it produce E. Likewise let B multiplying him selfe produce D, and multiplying D let it produce F. The<g ref="char:cmbAbbrStroke">̄</g> I say what C and D are numbers prime the one to the other. And likewise that E and F are numbers prime the one to the other. For forasmuch as A and B,<note place="margin">De<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>onstra<g ref="char:EOLhyphen"/>tion.</note> are prime the one to the other, and A multiplying him selfe produced C, therefore C and B are prime the one to the other (by the 27. of the seuenth). And by the same reason forasmuch as C and B
<figure/>
are prime the one to the other, and B multiplying him selfe produced D, therefore C and D are prime the one to the other. Againe forasmuch as A and B are prime the one to the other, and B multiplying him selfe produced D. Therefore (by the 27 of the seuenth) A and D are prime the one to the other. Now then forasmuch as two numbers A and C are prime to two numbers B and D, either of both to either of both: therefore (by the 28. of the seuenth) that which is produced of A into C is prime to that which is produced of B into D. But that which is produced of A into C is the number E, and that which is produced of B into D is the num<g ref="char:EOLhyphen"/>ber F. Wherefore E and F are numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rs prime the one to the other. And so alwayes if A &amp; B multiplying the numbers E and F do produce any numbers, the numbers produced, may by the former Proposition, be proued to be prime the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 28. Theoreme. The 30. Proposition. If two numbers be prime the one to the other: then both of them added to<g ref="char:EOLhyphen"/>gether, shall be prime to either of them. And if both of them added toge<g ref="char:EOLhyphen"/>ther be prime to any one of them, then also those numbers geuen at the be<g ref="char:EOLhyphen"/>ginning, are prime the one to the other.</head>
               <p>SVppose that these two numbers AB and BC being prime numbers be added to<g ref="char:EOLhyphen"/>gether. Then I say, that both these added together, namely, the number ABC, is prime to either of these AB, and BC. For if CA and AB be not prime the one to the other, some number then shall measure them.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part lea<g ref="char:EOLhyphen"/>ding to an absurditie.</note> Let some number mea<g ref="char:EOLhyphen"/>sure them, and let the same be D. Now then forasmuch as D
<figure/>
measureth the whole CA and the part taken away AB, it mea<g ref="char:EOLhyphen"/>sureth also the residue CB (by the 4. common sentence). And it measureth BA. Wherfore D measureth these numbers AB and BC, being prime the one to the other: which is impossible (by the 13. definition of the se<g ref="char:EOLhyphen"/>uenth). Wherefore no number measureth these numbers CA and AB. Wherefore CA and AB are prime the one to the other. And by the same reason also may it be proued, that CA and BC are prime the one to the other. Wherefore the number AC is to either of these num<g ref="char:EOLhyphen"/>bers AB and BC, prime.</p>
               <p>But now suppose that the numbers CA and AB be prime the one to the other. Then I say, that the numbers AB and BC are also prime the one to the other, For if, AB &amp; BC be not prime the one to the other: some one number measureth these numbers AB and BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <pb facs="tcp:7062:228"/>
Let some one number measure them, and let the same be D. And forasmuch as D measureth either of these numbers AB and BC, it shall also measure the whole CA (by the 6. common sentence).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part which is the con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se of the first, lean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ng also to an ab<g ref="char:EOLhyphen"/>surditi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> And it also measureth AB. Wherefore D measureth these numbers CA and AB being prime the one to the other: which is impossible (by the 13. definition of the seuenth). Wherefore no number measureth these numbers AB and BC. Wherefore AB and BC are prime the one to the other, which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 29. Theoreme. The 31. Proposition. Euery prime number is to euery number which it measureth not, prime.</head>
               <p>SVppose that there be geuen a prime number, namely, A, and let B be an other nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber, which it measureth not.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>sion leading to an absurditie.</note> Then I say, that the numbers B &amp; A are prime the one to the other. For if A and B be not prime the one to the other, then some number measureth them. Let there be a number that measureth them, and let the same be C. Now C is no vnitie. And forasmuch as C mea<g ref="char:EOLhyphen"/>sureth
<figure/>
B, but A measureth not B, therefore C is not one and the same number with A. And forasmuch as C measureth A &amp; B, it also measureth A being a prime number, and being not one and the same with it: which is impossible (by the 13. definition of the seuenth). Wherfore no number measureth these numbers A and B. And therefore A and B are prime the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 30. Theoreme. The 32. Proposition. If two numbers multiplying the one the other produce any number, and if also some prime number measure that which is produced of them: then shall it also measure one of those numbers which were put at the beginning.</head>
               <p>SVppose that two numbers A and B multiplying the one to the other do produce the nu<g ref="char:cmbAbbrStroke">̄</g>ber C: and let some prime number, namely, D measure C. The<g ref="char:cmbAbbrStroke">̄</g> I say that D mea<g ref="char:EOLhyphen"/>sureth one of these numbers either A or B. Suppose that it measure not A: now D is a prime number. Wherefore A and D are prime the one to the
<figure/>
other (by the proposition next going before). And how often D measureth C, so many vnities let there be in E.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>sion.</note> And forasmuch as D measureth C by those vnities which are in E, therfore D multiplying E produceth C: but A also multiplying B produced C wherfore that which is produced of D into E is equal to that which is produced of A into B. VVherfore (by the <hi>19.</hi> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the se<g ref="char:EOLhyphen"/>uenth) as A is to D, so is E to B: but D and A are prime numbers: and therefore the least numbers in that proportion: but the least in any proportion measure the numbers hauing the selfe same proportion with them equally, the greater, the greater, and the lesse, the lesse, that is the antecedent, the antecedent, and the consequent, the consequent (by the <hi>21.</hi> of the seue<g ref="char:cmbAbbrStroke">̄</g>th). wherefore the consequent D measureth the consequent B. In like sort may we proue that if D measure not B it measureth A. VVherfore D measureth one of these nu<g ref="char:cmbAbbrStroke">̄</g>bers A or B: which was required to to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ded by Campaue.</hi>
                     </note>Hereby it is manifest that if a number measure a number produced of two nu<g ref="char:cmbAbbrStroke">̄</g>bers multiplied the one into the other, or be commensurable to the same, it shall also either
<pb n="198" facs="tcp:7062:228"/>
measure one of the two number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> multiplied, or be comme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>surable with one of them.</p>
               </div>
            </div>
            <div type="part">
               <head>The 31. Theoreme. The 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Proposition. Euery composed number, is measured by some prime number.</head>
               <p>SVppose that A be a composed number. Then I say that A is measured by some prime number. For forasmuch as A is a composed number, some number must needes measure it (by the <hi>14.</hi> definition of the seuenth). Let there be a number that measureth it, and let the same be B. Now if B be a prime number then is that manifest which we seeke for:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> but if it be com<g ref="char:EOLhyphen"/>posed
<figure/>
number some number must needes measure it (by the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e same definition) Let there be a number that measureth it, and let the same be C. And for<g ref="char:EOLhyphen"/>asmuch as C measureth B, and B measureth A: therefore C also measureth A (by the 5. common sente<g ref="char:cmbAbbrStroke">̄</g>ce): and if C be a prime number then is that manifest which we sought for. But if it be a composed number some number shall mea<g ref="char:EOLhyphen"/>sure it: and the like consideration being had there shall at the length be found some prime number which measureth the number going before, which shall also measure A. For if there be not found any such prime number then shall infinite numbers decresing measure the sayd number A of which the one is lesse the<g ref="char:cmbAbbrStroke">̄</g> the other, which is impossible in numbers. VVherfore some prime number shall at the length be found which shall measure the number going before and which shall also measure the number A (by the <hi>5.</hi> common sentence). Euery co<g ref="char:cmbAbbrStroke">̄</g>posed num<g ref="char:EOLhyphen"/>ber therefore is measured by some prime number: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An other way.</head>
                  <p>Suppose that A be a composed number.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration.</note> Then I say that some prime number measureth it. For forasmuch as A is a composed number, some number shall measure it (by the <hi>14.</hi> de<g ref="char:EOLhyphen"/>finition of the seuenth) Let the least number that measureth it be B.
<figure/>
Then I say that B is a prime number. For if B be not a prime num<g ref="char:EOLhyphen"/>ber some number shall measure it. Let C measure it. Wherefore C is lesse then B. And forasmuch as C measureth B, and B measureth A, therfore C also measureth A being lesse then B, which by suppo<g ref="char:EOLhyphen"/>sition is the least number that measureth A, which is absurd. Wherfore B is not a composed number, but a prime number, which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>The 32. Theoreme. The 34. Proposition. Euery number is either a prime number, or els some prime number measu<g ref="char:EOLhyphen"/>reth it.</head>
               <p>SVppose that there be a number A.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that A is either a prime number, or els some prime number measureth it. For if A be a prime number
<figure/>
then is that had which is required. But if it be a composed num<g ref="char:EOLhyphen"/>ber, some prime number shall measure it (by the 33. of the seuenth). Euery number therfore is either a prime number, or els some prime number mea<g ref="char:EOLhyphen"/>sureth it: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:229"/>
               <head>¶The 3. Probleme. The 35. Proposition. How many numbers soeuer being geuen, to find out the least numbers that haue one and the same proportion with them.</head>
               <p>SVppose that there be a multitude of numbers geuen,
<figure/>
namely, <hi>A, B,</hi> and <hi>C.</hi> It is required to finde out the least numbers that haue one and the same proporti<g ref="char:EOLhyphen"/>on with these numbers <hi>A, B, C.</hi>
                  <note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> These numbers <hi>A, B, C,</hi> are either prime the one to the other, or not prime. If <hi>A, B, C,</hi> be prime the one to the other, then are they the least that haue one and the same proportion with them (by the 23. of the se<g ref="char:EOLhyphen"/>uenth).<note place="margin">The first case.</note>
               </p>
               <p>
                  <note place="margin">The second case.</note>But if they be not prime, take by the 3. of the seuenth, vnto <hi>A, B, C,</hi> the greatest common measure, which let be the number <hi>D.</hi> And how often <hi>D</hi> measureth euery one of these <hi>A, B, C,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so ma<g ref="char:EOLhyphen"/>ny vnities let there be in euery one of these numbers <hi>EFG.</hi> Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore these numbers <hi>E, F, G,</hi> do measure these numbers <hi>A, B, C,</hi> by those vnities which are in <hi>D.</hi> Wherfore these numbers <hi>E, F, G,</hi> measure these numbers <hi>A, B, C,</hi> equally. Wherfore <hi>E, F, G,</hi> are by the <hi>18.</hi> of the seuenth, in the selfe same proportion that <hi>A, B, C,</hi> are. Now then I say that they also are the least. For if <hi>E, F, G,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> be not the l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ast that haue one and the same proportion, with <hi>A, B, C,</hi> there shall the<g ref="char:cmbAbbrStroke">̄</g> be some numbers lesse then <hi>E, F, G,</hi> being in the selfe same proportion that <hi>A, B, C,</hi> are. Suppose that the same nu<g ref="char:cmbAbbrStroke">̄</g>bers be <hi>H, K, L,</hi> which shall measure the numbers <hi>A, B, C,</hi> equally. How many times H measureth <hi>A,</hi> so many vnities let there be in <hi>M.</hi> Wherfore either of these <hi>K</hi> and <hi>L</hi> measureth either of these <hi>B</hi> and <hi>C</hi> by those vnities which are in <hi>M</hi> (by the <hi>21.</hi> of the seuenth). And forasmuch as <hi>H</hi> measureth <hi>A</hi> by those vnities which are in <hi>M,</hi> therfore <hi>M</hi> also measureth <hi>A</hi> by those vnities which are in <hi>H.</hi> And by the same reason <hi>M</hi> measu<g ref="char:EOLhyphen"/>reth either of these <hi>B</hi> and <hi>C</hi> by those vnities which are in either of these <hi>K</hi> and <hi>L.</hi> Where<g ref="char:EOLhyphen"/>fore <hi>M</hi> measureth these numbers <hi>A, B, C.</hi> And forasmuch as <hi>H</hi> measureth <hi>A</hi> by those v<g ref="char:EOLhyphen"/>nities which are in <hi>M,</hi> therfore <hi>H</hi> multiplieng <hi>M,</hi> produceth <hi>A.</hi> And by the same reason <hi>E</hi> multiplieng <hi>D,</hi> produceth <hi>A.</hi> Wherfore that which is produced of <hi>E</hi> into <hi>D</hi> is equall to that which is produced of <hi>H</hi> into <hi>M.</hi> Wherfore (by the <hi>19.</hi> of the seuenth) as <hi>E</hi> is to <hi>H,</hi> so is <hi>M</hi> to <hi>D.</hi> But <hi>E</hi> is greater then <hi>H,</hi> wherfore <hi>M</hi> also is greater then <hi>D,</hi> and it measureth these num<g ref="char:EOLunhyphen"/>bers <hi>A, B, C,</hi> which is impossible. For <hi>D</hi> is supposed to be the greatest common measure vn<g ref="char:EOLhyphen"/>to <hi>A, B, C.</hi> Wherfore there shall be no other numbers lesse then <hi>E, F, G,</hi> and in the selfe same proportion with <hi>A, B, C.</hi> Wherfore <hi>E, F, G,</hi> are the lest numbers which haue one and the same proportion with <hi>A, B, C</hi>: which was required to be done.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>Hereby it is manifest that the greatest common measure to numbers how many soe<g ref="char:EOLhyphen"/>uer:<note place="margin">
                        <hi>A Corollary added by Campa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e.</hi>
                     </note> measureth the sayd numbers by the numbers in the least proportio<g ref="char:cmbAbbrStroke">̄</g> that the num<g ref="char:EOLhyphen"/>bers geuen are.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 4. Probleme. The 36. Proposition. Two numbers being geuen, to finde out the lest nu<g ref="char:cmbAbbrStroke">̄</g>ber which they measure.</head>
               <p>
                  <pb n="199" facs="tcp:7062:229"/>SVppose that the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>o numbers ge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be A and B. It is required to finde <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the lest number which they measure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> No<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A and B are ei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r prime <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he one to the other, or not.<note place="margin">Two cases in this propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> Suppose first that A and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be prime the one to the other: and let A multiplying B produce C: wherefore B, multiplying A produ<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> also C (by the 16. of the seuenth.)<note place="margin">The first case.</note> Wherefore A and B measure C. Now also I say, that C is the lest nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber which they measure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if it be not, those numbers A and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> measure some number lesse the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> C: let them measur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> some number lesse then C, and let the same be D: and how often A measureth D, so many vnities let there be in E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and how often <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> measureth D, so many v<g ref="char:EOLhyphen"/>nitie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> let there be in <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> Wherefore A multiplying E produce<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> D, and B multiplying F pro<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="6 letters">
                     <desc>••••••</desc>
                  </gap> also D. Wherefore that which is produced of A into <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,
<figure/>
is equall to that which is produced of B into <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>: wherefore (by the 19. of the seuenth) as A is to B, so is F to E. But A and B are prime : yea they are prime and also the lest in that propor<g ref="char:EOLhyphen"/>tion (by the 23. of the seuenth): but the lest numbers in any proportion measure those numbers that haue one and the same proportion with them equally: the greater the greater: and the lesse the lesse (by the 21 of the seuenth). Wherfore B measureth E, namely, the consequent, the consequent. And forasmuch as A multiplying B and E produced C and D: therefore (by the 17. of the seuenth) as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to E, so is C to D. But B measureth E. Wherefore C also measureth D, the greater, the lesse which is impossible. Wherefore if those numbers A and B be prime, they shall measure no number lesse then C. Wherefore C is the lest number which A and B measure.</p>
               <p>But now suppose that A and B be not prime the one to the other,<note place="margin">The second case<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> and take (by the 35. of the seuenth) the lest numbers that haue one and the same proportion with A and B, and let the same be F and E. Wherefore that which is produced of A into E, is equall to that which is produced of B into F (by the 19. of the seuenth). Let A multiplying E produce C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore B multiplying F produceth also C. Wherefore A and B measure C. Then I say, that C is the lest number that they measure. For if it be not, those numbers
<figure/>
A and B shall measure some number lesse then C: let them mea<g ref="char:EOLhyphen"/>sure a number lesse then C, and let the same be D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And how of<g ref="char:EOLhyphen"/>ten A measureth D, so many vnities let there be in G, And how often <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> measureth D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so many vnities let there be in H.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Now then A multiplying G produceth D. And B multiplying H pro<g ref="char:EOLhyphen"/>duceth also D. Wherefore that which is produced of A into G is equall to that which is produced of B into H. Wherefore (by the 19. of the seuenth) as A is to B, so is H to G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But as A is to B, so is F to E. Wherefore as F is to E, so is H to G: but the lest numbers in any proportion measure the numbers that haue the same proportion with them equally, the greater the grea<g ref="char:EOLhyphen"/>ter, &amp; the lesse the lesse (by the 21. of the seue<g ref="char:cmbAbbrStroke">̄</g>th). VVherfore E measureth G. And forasmuch as A multiplying G and E produced C and D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore (by the 17. of the fift) as E is to G, so is C to D. But E measureth G. VVherefore C also measureth D, the greater the lesse: which is impossible. Wherefore those numbers. A and B do not measure any number lesse then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Wherefore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is the lest number that is measured by A and B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> whi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h was required to be done.</p>
            </div>
            <div type="part">
               <head>The 33. Theoreme. The 37. Proposition. If two numbers measure any number, the least nu<g ref="char:cmbAbbrStroke">̄</g>ber also which they mea<g ref="char:EOLhyphen"/>sure, measureth the selfe same number.</head>
               <p>
                  <pb facs="tcp:7062:230"/>SVppose that there be two numbers geuen <hi>A</hi> and <hi>B</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let them measure the nu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>ber <hi>CD</hi>:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>.</note> and let the least number that they measure be <hi>E.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say that <hi>E</hi> also measureth the number <hi>CD.</hi> For if <hi>E</hi> do not mea<g ref="char:EOLhyphen"/>sure
<figure/>
                  <hi>CD,</hi> let <hi>E</hi> measuring <hi>CD,</hi> that is subtrahed out of <hi>CD</hi> as ofte<g ref="char:cmbAbbrStroke">̄</g> as you can, as for example, once, leaue a lesse then it selfe, namely, <hi>CF.</hi> And let the number subtrahed which <hi>E</hi> measu<g ref="char:EOLhyphen"/>reth be <hi>FD,</hi> and forasmuch as <hi>A</hi> and <hi>B</hi> measure <hi>E,</hi> and <hi>E</hi> measureth <hi>DF,</hi> therefore <hi>A</hi> and <hi>B</hi> also measure <hi>DF.</hi> And they measure the whole <hi>CD,</hi> wherefore by the <hi>4.</hi> common sentence of the seuenth they mea<g ref="char:EOLhyphen"/>sure also that which remayneth <hi>CF</hi> being lesse then <hi>E</hi>: which is impossible. Wherefore <hi>E</hi> of necessitie measureth <hi>CD,</hi> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 5. Probleme. The 38. Proposition. Three numbers being geuen, to finde out the least number which they measure.</head>
               <p>SVppose that there be three numbers geuen <hi>A, B, C.</hi> It is required to finde out the least number which they measure. Take (by the <hi>36.</hi> of the seuenth) the least number which <hi>A</hi> and <hi>B</hi> measure, and let the same be <hi>D.</hi>
                  <note place="margin">Two cases in this propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> Now then <hi>C</hi> either measureth <hi>D</hi> or els measureth it not.<note place="margin">The first case.</note> First let it measure it. And the numbers
<figure/>
also <hi>A</hi> and <hi>B</hi> measure <hi>D</hi>: wherefore <hi>A, B, C,</hi> measure <hi>D.</hi> Now then I say that <hi>D</hi> is the least number which they measure. For if not, let the numbers <hi>A, B, C,</hi> measure some number lesse then <hi>D,</hi> and let the same be <hi>E.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>g <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o an absur<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>e.</note> And forasmuch as <hi>A, B, C,</hi> measure <hi>E</hi> therefore also <hi>A</hi> and <hi>B</hi> measure <hi>E,</hi> wherefore (by the <hi>37.</hi> of the seuenth) the least number which those numbers <hi>A</hi> and <hi>B</hi> measure shall also measure <hi>E.</hi> But the least number which <hi>A</hi> and <hi>B</hi> measure is <hi>D.</hi> Wherfore D measureth E, the greater the lesse: which is impossible. Wherefore these numbers <hi>A, B, C,</hi> shall not measure any number lesse then <hi>D.</hi> Wherefore <hi>D</hi> is the least number that <hi>A, B, C,</hi> doo measure.</p>
               <p>But now suppose that <hi>C</hi> measure not <hi>D.</hi> And take (by the <hi>36.</hi> of the seuenth) the least number which those two numbers <hi>C</hi> and <hi>D</hi> do measure, and let the same be <hi>E.</hi>
                  <note place="margin">The second case.</note> And foras<g ref="char:EOLhyphen"/>much as <hi>A</hi> and <hi>B</hi> measure <hi>D,</hi> and <hi>D</hi> measureth <hi>E,</hi> therefore <hi>A</hi> and <hi>B</hi> also measure <hi>E,</hi> and <hi>C</hi> also measureth <hi>E,</hi> wherefore <hi>A, B, C,</hi> also measure <hi>E.</hi> I say moreouer that <hi>E</hi> is the least number which <hi>A, B, C</hi> measure. For if it be not, let there be some lesse number then <hi>E</hi> which they measure,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> and let the same be <hi>F.</hi> And foras<g ref="char:EOLhyphen"/>much
<figure/>
as <hi>A, B, C,</hi> measure <hi>F.</hi> Therefore <hi>A</hi> and <hi>B</hi> also measure <hi>F</hi>: wherefore the least number which these numbers <hi>A</hi> and <hi>B</hi> do measure doth also mea<g ref="char:EOLhyphen"/>sure <hi>F</hi> (by the <hi>37.</hi> of the seuenth). But the least num<g ref="char:EOLunhyphen"/>ber which <hi>A</hi> and <hi>B</hi> doo measure is <hi>D.</hi> Wherefore <hi>D</hi> measureth <hi>F.</hi> And C also measureth <hi>F.</hi> Wherfore <hi>D</hi> and <hi>C</hi> measure <hi>F.</hi> Wherefore the least number which <hi>C</hi> and <hi>D</hi> doo measure, shall also (by the selfe same) measure <hi>F.</hi> But the least <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mbe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which <hi>C</hi> &amp; <hi>D</hi> measure is <hi>E.</hi> Wherfore <hi>E</hi> measureth <hi>F,</hi> namely, the greater, the lesse, which is impossible. Wherefore these numbers <hi>A, B, C,</hi> do not measure any number lesse then <hi>E.</hi> Wherefore <hi>E</hi> is the least number which <hi>A, B, C,</hi> doo measure: which was required to be demonstrated.</p>
               <p>
                  <hi>In like maner also how many numbers soeuer being geuen, may be found out the
<pb n="200" facs="tcp:7062:230"/>
least number which they <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>easure. For if vnto the three n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>mbe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s A, B, C, be added a forth, then if the sayd forth number measure the number E, then is E the least number which the fower numbers geuen measure. But if it doo not measure E, the<g ref="char:cmbAbbrStroke">̄</g> by the 37. of this booke must you finde but the least number which E and the forth number measure. Which shall be the number sought for. And so likewise if there be fiue, sixe, or how ma<g ref="char:EOLhyphen"/>nysoeuer geuen.</hi>
               </p>
               <div type="section">
                  <head>Corollary.</head>
                  <p>Hereby it is manifest that the least commo<g ref="char:cmbAbbrStroke">̄</g> measure to numbers howmanysoeuer,<note place="margin">
                        <hi>A Corollary.</hi>
                     </note> measureth euery number which the sayd numbers how many soeuer measure.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 34. Theoreme. The 39. Proposition. If a number measure any number: the number measured shall haue a part after the denomination of the number measuring.</head>
               <p>SVppose that there be a number B, which let measure the number A. Then I say, that A hath a part taking his denomination of the number B. For how often B measureth A, so many vnities let there be in C. And let D be vnitie. And foras<g ref="char:EOLhyphen"/>much as B measureth A, by those vnities which are in C, and vnitie D measu<g ref="char:EOLhyphen"/>reth C by those vnities which are in C, therefore vnitie D, so
<figure/>
many times measureth the number C, as B doth measure A.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore alternately (by the 15. of the seuenth) vnitie D, so many times measureth B, as C doth measure A. Wherfore what part vnitie D is of the number B, the same part is C of A. But vnitie D is a part of B hauing his denomination of B. VVher<g ref="char:EOLhyphen"/>fore C also is a part of A hauing his denomination of B. VVherfore A hath C as a part taking his denomination of B: which was required to be proued.</p>
               <p>
                  <hi>The meaning of this Proposition is, that if three measure any number, that number hath a third part, and if foure measure any number the sayd number hath a fourth part. And so forth.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 35. Theoreme. The 40. Proposition. If a number haue any part: the number wherof the part taketh his deno<g ref="char:EOLhyphen"/>mination shall measure it.</head>
               <p>SVppose that the number A haue a part, namely, B: and let the part B haue his denomination of the number C. Then I say, that C measureth A. Let D be vni<g ref="char:EOLhyphen"/>tie. And forasmuch as B is a part of A, hauing his denomination of C:<note place="margin">The co<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>erse of the former proposition.</note> and D being vnitie is also a part of the number C, hauing his denominatio<g ref="char:cmbAbbrStroke">̄</g> of C: there<g ref="char:EOLhyphen"/>fore what part vnitie D is of the number C, the same part is also B of A:
<figure/>
wherefore vnitie D so many <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>es measureth the number C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> as B measu<g ref="char:EOLhyphen"/>r<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> A. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ref<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>ern<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ely (by <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e 15. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>f the s<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>nth) vnitie D so ma<g ref="char:EOLhyphen"/>ny <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mes m<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>reth the nu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> B, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> C meas<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>eth A. Wheref<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> C mea<g ref="char:EOLhyphen"/>sureth A: which was requi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>d to b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proued.</p>
               <p>
                  <hi>This Proposition is the conuerse of the former: and the meaning therof is, that e<g ref="char:EOLhyphen"/>uery number hauing a third part is measured of three, and hauing a fourth part is mea<g ref="char:EOLhyphen"/>sured of foure. And so forth.</hi>
               </p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:231"/>
               <head>¶The 6. Probleme. Th 41. Proposition. To finde out the least number, that containeth the partes geuen.</head>
               <p>SVppose that the partes geuen be A, B, C, namely, let A be an halfe part, B a third part, &amp; C a fourth part.<note place="margin">Construction.</note> Now it is required to finde out the least nu<g ref="char:cmbAbbrStroke">̄</g>ber which co<g ref="char:cmbAbbrStroke">̄</g>tai<g ref="char:EOLhyphen"/>neth the partes A, B, C. Let the said partes A, B, C, haue their denominations of the numbers D, E, F. And take (by the 38. of the seuenth) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he least number which the numbers D, E, F, measure, and let the same be G. And forasmuch as the
<figure/>
numbers D, E, F measure the number G, therfore the number G hath pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tes denominated of the nu<g ref="char:cmbAbbrStroke">̄</g>bers D, E, F (by the 39. of the seue<g ref="char:cmbAbbrStroke">̄</g>th). But the parts A, B, C, haue their denominatio<g ref="char:cmbAbbrStroke">̄</g> of the num<g ref="char:EOLunhyphen"/>bers D, E, F. Wherfore G hath those partes A, B, C. I say also that it is the least number which hath these partes.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ding to an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>bsu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ie.</note> For if G be not the least number which containeth those partes A, B, C, then let there be some number lesse then G which containeth the saide partes A, B, C. And suppose the same to be the number H. And forasmuch as H hath the said partes A, B, C, therfore the numbers that the partes A, B, C, take their denominations of, shall measure H (by the 40. of the seuenth) But the numbers whero<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the partes A, B, C, take their denominations of, are D, E, F. Wherfore the numbers D, E, F, m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>asu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> number H which is lesse then G, which is impossible. For G is supposed to be t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> number that the numbers D, E, F, do measure. Wherfore there is no number lesse then G, which containeth these partes A, B, C: which was required to be done.</p>
               <div type="section">
                  <head>Corrolary.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary ad<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ed by Campane.</hi>
                     </note>Hereby it is manifest that if there be taken the least number, that numbers how ma<g ref="char:EOLhyphen"/>ny soeuer do measure, the sayd number shall be the least which hath the partes denomi<g ref="char:EOLhyphen"/>nated of the sayd numbers how many soeuer.</p>
                  <p>
                     <hi>Campane</hi> after he hath taught to finde out the first least number that con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ayneth the partes geuen, teacheth also to finde out the second least number,<note place="margin">
                        <hi>How to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inde out the seconde least number and the third, and so <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>orth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>nitly.</hi>
                     </note> that is, which except the least of all is lesse then all other, and also the third least, and the fourth &amp;c. The se<g ref="char:EOLhyphen"/>cond is found out by doubling the number <hi>G.</hi> For the numbers which measure the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber <hi>G</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hall also measure the double therof (by the 5. commo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> sentence of the seuenth). But there cannot be geuen a number greater then the number <hi>G,</hi> &amp; lesse then the dou<g ref="char:EOLhyphen"/>ble therof, whom the partes geuen shall <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>asure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For forasmuch as the partes geuen do <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>easur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the whole, namely, which is lesse then the double, and they also measure the part taken away, namely, the number <hi>G,</hi> they should also measure the residue, namely, a number lesse then <hi>G,</hi> which is proued to be the lest number that they do measure, which is impossible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wherefore the second number which the said partes geuen do measure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> must, exceeding G, needes reache to the double of G, and the third to the treble, and the fourth to the quadruple, and so in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>initely, for those partes can neuer measure any number lesse then the number G.</p>
                  <p>By this Proposition also it is easie to find out the least number containing the partes geuen of partes. As if we would finde out the least number which contayneth one third <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> an hal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e part,<note place="margin">
                        <hi>How to si<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> out the least <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> a con<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ay<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>g <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>e pa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>s of parts.</hi>
                     </note> and one fourth part of a third part, reduce the said d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ers fracti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> into simple fraction (by the common <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> of reducing of fr<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>ions) namely, the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> of an hal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e into a <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> part of an whole<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he fourth of thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d into a twelfth part of an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> whole. And then by this Probleme search out the least number which contayneth a six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> part and a twelfth part, and so haue you done.</p>
               </div>
            </div>
            <trailer>The end of the seuenth booke of Euclides Elementes.</trailer>
         </div>
         <div n="8" type="book">
            <pb n="201" facs="tcp:7062:231"/>
            <head>¶ The eighthe booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>AFter that <hi>Euclide</hi> hath in the seuenth booke entreated of the proprieties of numbers in generall, and of certayne kindes thereof more specially, and of prime and composed numbers with others: now in this eight booke he prosecu<g ref="char:EOLhyphen"/>teth farther, and findeth out and demonstrateth the pro<g ref="char:EOLhyphen"/>perties and passions of certayne other kindes of numbers:<note place="margin">
                  <hi>The Argu<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap> of the eight books.</hi>
               </note> as of the least numbers in proportion, and how such may be found out infinitely in whatsoeuer proportion: which thing is both delectable, and to great vse. Also here is en<g ref="char:EOLhyphen"/>treated of playne numbers, and solide: and of theyr sides, and proportion of them. Likewise of the passions of num<g ref="char:EOLhyphen"/>bers square and cube, and of the natures and conditions of their sides, and of the meane proportionall numbers of playne, solide, square, and cube numbers, with many other thinges very requisite and necessary to be knowne.</p>
            <div type="part">
               <head>¶The first Theoreme. The first Proposition. If there be numbers in continuall proportion howmanysoeuer, and if their extremes be prime the one to the other: they are the least of all numbers that haue one and the same proportion with them.</head>
               <p>SVppose that the numbers in continuall proportion be <hi>A, B, C, D.</hi> And let their extreames namely, <hi>A</hi> and <hi>D</hi> be prime the one to the other. Then I say that the numbers <hi>A, B, C, D,</hi> are the least of all numbers that haue
<figure/>
one and the same pro<g ref="char:EOLhyphen"/>portion with the<g ref="char:cmbAbbrStroke">̄</g>. For if they be not, let <hi>E, F; G, H</hi> being lesse numbers then <hi>A, B, C, D,</hi> be in the selfe same proportion that <hi>ABCD</hi> are.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurd<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ie.</note> And forasmuch as the numbers <hi>A, B, C, D,</hi> are in the selfe same proportion that the numbers <hi>E, F, G, H,</hi> are, &amp; the mul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>itude of these numbers <hi>E, F, G, H,</hi> is equall to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ult<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ude of these <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mbers <hi>A, B, C, D,</hi> therefore of equalitie (by the <hi>14.</hi> of the seuenth) as <hi>A</hi> is to <hi>D,</hi> so is <hi>E</hi> to <hi>H.</hi> But <hi>A</hi> and <hi>D</hi> are prime the one to the other, yea they are prime and the least that ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the same proportion with them: But the least numbers in a<g ref="char:EOLhyphen"/>ny proportion measure the numbers that haue the same proportion with them equally, the antecedent the antecedent, and the conseque<g ref="char:cmbAbbrStroke">̄</g>t the cons<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>uen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (by the <hi>21.</hi> of the seuenth) wher<g ref="char:EOLunhyphen"/>fore <hi>A</hi> measureth <hi>E,</hi> the greater the lesse: which is impossible. Wherefore the numbers <hi>E, F, G, H,</hi> being lesse then <hi>A, B, C, D,</hi> are not in the same proportion that <hi>A, B, C, D,</hi> are, wher<g ref="char:EOLhyphen"/>fore <hi>A, B, C, D,</hi> are the least of all numbers which ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e one and the same proportion with
<pb facs="tcp:7062:232"/>
them<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 1. Probleme. The 2. Proposition. To finde out the least numbers in continuall proportion, as many as shall be required, in any proportion geuen.</head>
               <p>SVppose that the proportion geuen in the lest numbers be A to B. It is required to finde out the lest numbers in continuall proportion, as many as shall be requi<g ref="char:EOLhyphen"/>red, in the same proportion that A is to B.<note place="margin">Construction.</note> Let there be required foure. And let A multiplying him selfe produce C: and multiplying B let it produce D: and likewise let B multiplying him selfe produce E. And moreouer let A multiplying those num<g ref="char:EOLhyphen"/>bers C, D, E,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> produce F, G, H: and let B multiplying E produce K. And forasmuch as A multiplying him selfe produced C, and multiplying B produced D, now then the number A multiplying two numbers A and B produced C &amp; D. Wherefore (by the 17. of the seuenth) as A is to B, so is C to D. Againe, forasmuch as A multiplying B produced D, and B multi<g ref="char:EOLhyphen"/>plying him selfe produced E, therefore ech of those numbers A and B multiplying B, bring<g ref="char:EOLhyphen"/>eth forth these numbers D and E. VVherefore (by the 18. of the seuenth) as A is to B, so D to E. But as A is to B, so is C to D. Wherefore as C is to D, so is D to E. And forasmuch as A multiplying C and D produced F and G, therefore (by the 17. of the seuenth) as C is to D, so is F to G. But as C is to D, so is A to B. Wherefore as A is to B, so is F to G. Againe for<g ref="char:EOLhyphen"/>asmuch as A multiplying D and E produced G and H, therefore (by the 17. of the seuenth) as D is to E, so is G to H. But as D is to E, so is A to B. Wherefore as A is to B, so is G to H.
<figure/>
And forasmuch as those numbers A and B multiplying E produced H and K, therefore (by the <hi>18.</hi> of the seuenth) as A is to B, so is H to K. And it is proued, that as A is to B, so is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and G to H: wherefore as F is to G, so is G to H, and H to K. Wherefore these numbers C, D, E, and F, G, H, K, are proportionall in the same proportion, that A is to B. Now I say, that they are also the lest. For forasmuch as A and B are the lest of all numbers that haue the same proportion with them: but the lest numbers that haue one &amp; the same proportion with them are prime the one to the other (by the 24. of the seuenth): therefore A and B are prime the one to the other: and ech of these numbers A &amp; B multiplying him selfe produced these numbers C and E, and likewise multiplying ech of these numbers C and E they produced F and K. Wherefore (by the 29. of the seuenth) C, E, and F, K, are prime the one to the other. But if there be n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mbers in continuall proportion how many soeuer, and if their extremes be prime the one to the other, they are the lest of all nu<g ref="char:cmbAbbrStroke">̄</g>bers that haue the same proportion with them (by the first of the eight). Wherefore these numbers C, D, E, and F, G, H, K, are the lest
<pb n="202" facs="tcp:7062:232"/>
of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mber <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hat <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he same proportion with A and B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And forasmuch <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>9. of the seuenth) that alwaies happeneth touching the extre<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> namely, that A and B <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>lti<g ref="char:EOLhyphen"/>plying the numbers pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>duced <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> shall produce <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ther prime numbers, namely, the ex<g ref="char:EOLhyphen"/>tremes of fiue numbers in continuall proportion, therefore (by the first of this booke) all fiue are the lest of that proportion. And so infinitely: which was required to be done.</p>
               <div type="section">
                  <head>¶Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest, that if three numbers being in continuall proporti<g ref="char:EOLhyphen"/>on, be the lest of all numbers that haue the same proportion with them, their extremes are squares: and if there be foure their extremes are cubes.</hi> For the extremes of three are produced of the multiplying of the nu<g ref="char:cmbAbbrStroke">̄</g>bers A and B into them selues. And the extremes of foure are produced of the mul<g ref="char:EOLhyphen"/>tiplying of the rootes A and B into the squares C and E, whereby are made the cubes F and K.</p>
               </div>
            </div>
            <div type="part">
               <head>The 2. Theoreme. The 3. Proposition. If there be numbers in continuall proportion how many soeuer, and if they be the lest of all numbers that haue one and the same proportion with the<g ref="char:cmbAbbrStroke">̄</g>: their extremes shall be prime the one to the other.</head>
               <p>SVppose that the numbers in continuall proportion being the least of all numbers that haue the same proportion with them be A, B, C, D.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>uerse of the first.</note> Then I say that their extremes A and D are prime the one to the other. Take (by the <hi>2.</hi> of the eight, or by the 35. of the seuen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h) the two least numbers that are in the same proportion that A, B, C, D, are, and let the same be the numbers E, F. And after that take thre numbers
<figure/>
G, H, K, and so alwayes forward on more (by the former proposition) vntill the multitude <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>aken be equall to the multitude of the numbers geuen A, B, C, D. And let those numbers be L, M, N, O. Wherfore (by the <hi>29.</hi> of the seuenth) their extremes L, O,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> are prime the one to the other. For forasmuch as E and F are prime the one to the other, and eche of them mul<g ref="char:EOLhyphen"/>tiplieng
<pb facs="tcp:7062:233"/>
himselfe produced G and K, &amp; likewise ech of these, G &amp; K multiplieng himself pro<g ref="char:EOLhyphen"/>duced L &amp; O, therfore (by the <hi>29</hi> of the seuenth) G &amp; K are prime the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> other, &amp; so likewise are L and O prime the one to the other. And forasmuch as A, B, C, D, are the least of
<figure/>
all numbers that haue the same proportion with them, and likewise L, M, N, O, are the least of all numbers that are in the same proportion that A, B, C, D, are, and the multitude of these numbers A, B, C, D, is equall to the multitude of these L, M, N, O: therfore euery one of these A, B, C, D, is equall vnto euery one of these L, M, N, O. Wherefore A is equall vnto L, and D is equall vnto O. And forasmuch as L and O are prime the one to the other, and L is e<g ref="char:EOLhyphen"/>quall vnto A, and O is equall vnto D: therfore A and D are prime the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The 2. Probleme. The 4. Proposition. Proportions in the least numbers how many soeuer beyng geuen, to finde out the least numbers in continuall proportion in the said proportions geue<g ref="char:cmbAbbrStroke">̄</g>.</head>
               <p>SVppose that the proportions in the least numbers geuen, be A to B, C to D, and E to F. It is required to finde out the least numbers in continuall proportion, in the same proportion that A is to B, and that C is to D, and that E is to F. Take the least number whom B and C do measure, and let the same be G. And how often B measureth G, so many times let A measure H. And how ofte<g ref="char:cmbAbbrStroke">̄</g> C measureth G, so many times let D measure K. Now E either measureth K or measureth it not.<note place="margin">Two cases in this propositio<g ref="char:cmbAbbrStroke">̄</g>. The first case.</note> First let it measure it. And how often E measureth <hi>K,</hi> so many times let F measure L. And forasmuch as how often A measureth H, so many times doth B measure G: therfore by the <hi>17.</hi> of the seuenth, as A is to B, so is H to G. And by the same reaso<g ref="char:cmbAbbrStroke">̄</g> as C is to D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to K, and moreouer as E is to F, so is K to L. Wherfore these numbers H, G, K, L, are in continuall proportion, and in the same proportion that A is to B, and that C is to D, and moreouer that E is to F. I say also that they are the least in those proportions<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if H, G, K, L, be not the least num<g ref="char:EOLhyphen"/>bers in continuall proportion, and in the same proportions that A is to B, and C to D, and E to F, then are there some numbers less<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> then H, G, K, L, in the same proportions that A is to B and C to D, and E to F, let those numbers be N, X, M, O. And forasmuch as A is to
<pb n="203" facs="tcp:7062:233"/>
B, so is N to X<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and A and B are the lest, but the least measure
<figure/>
those numbers that haue one and the same proportion with them equally, the greater, the greater, and the lesse the lesse, that is, the antecedent, the antecedent, &amp; the consequent the consequent (by the 21. of the seuenth) therfore B measureth X. And by the same reason C also measureth X, wherfore B and C measure X. Wher<g ref="char:EOLhyphen"/>fore the least number whom B and C measure, shall also by the 37 of the seuenth measure X. But the least number whome B and C measure is G. Wherefore G measureth X, the greater, the lesse, which is impossible. Wherfore there shall not be any lesse numbers then H, G, K, L, in continuall proportion, and in the same pro<g ref="char:EOLhyphen"/>portions that A is to B, and C to D, and E to F.</p>
               <p>But now suppose that E measure not K.<note place="margin">The second case.</note> And by the 36. of the seuenth, take the least number whome E and K measure, and let the same be M. And how often K measureth M, so often let either of these G and H measure either of these N and X.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And how often E measureth M, so often let F measure O. And forasmuch as how often G measureth N, so often doth H measure X, therfore as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to G, so is X to N. But as H is to G, so is A to B Wherfore as A is to B, so is X to N. And by the same reason as C is to D, so is N to M. A<g ref="char:EOLhyphen"/>gaine, forasmuch as how often E measureth M, so often F measureth O, therfore as E is to F so is M to O. Wherfore X, N, M, O, are in continuall proportion, and in the same proporti<g ref="char:EOLhyphen"/>ons that A is to B, and C to D, and E to F. I say also that
<figure/>
they are the least in that proportion. For if X, N, M, O, be not the least in continuall proportion, and in the same pro<g ref="char:EOLhyphen"/>portions that A is to B, and C to D, and E to F, then shall there be some numbers lesse then X, N, M, O, in continuall proportion, and in the same proportions that A is to B, and C to D, and E to F. Let the same be the numbers P, R, S, T. And for that as P is to R, so is A to B, and A and B are the least, but the least numbers measure those numbers that haue one &amp; the same proportio<g ref="char:cmbAbbrStroke">̄</g> with them equally, the greater the greater, and the lesse the lesse, that is, the antece<g ref="char:EOLhyphen"/>dent the antecedent, and the consequent the consequent by the 21. of the seuenth, therfore B measureth R. And by the same reason also C measureth R. Wherefore B and C mea<g ref="char:EOLhyphen"/>sure R. Wherfore the least number wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>m B and C measure shall also measure R (by the <hi>37.</hi> of the seuenth). But the lest number whom B and C measure is G, wherfore G measureth R. And as G is to R, so is K to S. Wherfore K measureth S. And E also measureth S. Wherefore E and K measure S. Wherfore the least number whom E and K measure, shall (by the selfe same) measure S. But the least number whom E and K measures M. Wherefore M measureth S the greater the lesse, which is impossible. Wherfore there are no numbers lesse then X, N, M, O, in continuall proportion, &amp; in the same proportions that A is to B, and C to D, and E to F. Wherfore X, N, M, O are the least numbers in continuall proportion, and in the same proportions that A is to B, and C to D, and E to F: which was required to be done.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:234"/>
               <head>¶ The 3. Theoreme. The 5. Proposition. Playne or superficiall numbers are in that proportion the one to the other which is composed of the sides.</head>
               <p>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion in num<g ref="char:EOLhyphen"/>bers answe<g ref="char:EOLhyphen"/>reth to the of the sixth touching parel<g ref="char:EOLunhyphen"/>lelogrammes.</note>SVppose that <hi>A</hi> and <hi>B</hi> be playne or superficiall numbers, and let the sides of <hi>A</hi> be the numbers <hi>C</hi> and <hi>D,</hi> and let the sides of <hi>B,</hi> be the numbers E and <hi>F.</hi> Then I say that <hi>A</hi> is to <hi>B</hi> in that proportion that is composed of the sides. Take (by the fourth of the
<figure/>
eight) the least numbers in continuall proportion, and in the same proportions that <hi>C</hi> is to <hi>E,</hi> and <hi>D</hi> to <hi>F.</hi>
                  <note place="margin">Construction.</note> And let the same be the numbers <hi>G, H, K</hi>:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so that as <hi>C</hi> is to <hi>E,</hi> so let <hi>G</hi> be to <hi>H,</hi> &amp; as <hi>D</hi> is to <hi>F,</hi> so let <hi>H</hi> be to <hi>K.</hi> Wherefore those numbers <hi>G, H, K,</hi> haue the proporti<g ref="char:EOLhyphen"/>ons of the sides: but the pro<g ref="char:EOLhyphen"/>portion of <hi>G</hi> to <hi>K</hi> is con po<g ref="char:EOLhyphen"/>sed of that which <hi>G</hi> hath to <hi>H</hi> and of that which <hi>H</hi> hath to <hi>K</hi>: wherefore <hi>G</hi> is vnto <hi>K</hi> in that proportion which is composed of the sides. Now I say that as <hi>A</hi> is to <hi>B,</hi> so is <hi>G</hi> to <hi>K.</hi> For let <hi>D</hi> multiplying <hi>E</hi> produce <hi>L.</hi> And forasmuch as <hi>D</hi> multiplying <hi>C</hi> produced A, and mul<g ref="char:EOLhyphen"/>tiplying <hi>E</hi> produced <hi>L</hi>: therefore (by the <hi>17.</hi> of the seuenth) as <hi>C</hi> is to <hi>E,</hi> so is <hi>A</hi> to <hi>L.</hi> But as <hi>C</hi> is to <hi>E,</hi> so is <hi>G</hi> to <hi>H,</hi> wherefore as <hi>G</hi> is to <hi>H,</hi> so is <hi>A</hi> to <hi>L.</hi> Agayne forasmuch as <hi>E</hi> mul<g ref="char:EOLhyphen"/>tiplying <hi>D</hi> produced <hi>L,</hi> &amp; multiplying <hi>F</hi> produced <hi>B</hi>: therefore (by the <hi>17.</hi> of the seuenth) as <hi>D</hi> is to <hi>F,</hi> so is <hi>L</hi> to <hi>B.</hi> But as <hi>D</hi> is to <hi>F,</hi> so is <hi>H</hi> to <hi>K,</hi> wherefore as <hi>H</hi> is to <hi>K,</hi> so is <hi>L</hi> to <hi>B.</hi> And it is proued that as <hi>G</hi> is <hi>H,</hi> so is <hi>A</hi> to <hi>L.</hi> Wherefore of equalitie (by the <hi>14.</hi> of the seuenth) as <hi>G</hi> is to <hi>K,</hi> so is <hi>A</hi> to <hi>B.</hi> But <hi>G</hi> is vnto <hi>K</hi> in that proportion which is composed of the sides, wherefore <hi>A</hi> is vnto <hi>B</hi> in that proportion which is composed of the sides: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶An other demonstration of the same after <hi>Campane.</hi>
                  </head>
                  <p>Suppose that A and B be plaine numbers: and let the sides of A be the numbers C and D: and let the numbers E and F be the sides of the number B. And let D multiplying E produce the number G. The<g ref="char:cmbAbbrStroke">̄</g> I say that the proportio<g ref="char:cmbAbbrStroke">̄</g> of
<figure/>
A to B is co<g ref="char:cmbAbbrStroke">̄</g>posed of the propor<g ref="char:EOLunhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>s of C to E &amp; D to F that is,<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>ter Campane.</hi>
                     </note> of the sides of the superficial nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLunhyphen"/>ber A to the sides of the super<g ref="char:EOLhyphen"/>ficiall number B. For forasmuch as D multiplying E produced G, and multiplying C it produced A, therefore by (the 17. of the seuenth) A is to G as C is to E: agayne forasmuch as E multiplying D produced G and multiplying F it produ<g ref="char:EOLhyphen"/>ceth B, therefore by the same G is to B as D is to F. Wherefore the proportions of the sides namely, of C to E and of D to F are one and the same with the proportions of A to G and G to B. But (by the fifth definition of the sixth) the proportion of the extremes A to B is composed of the proportions of the meanes, namely, of A to G and G to B, which are proued to be one and the same with the proportions of the sides C to E, and D to F. Wherefore the proportion of the superficiall numbers A to B is co<g ref="char:cmbAbbrStroke">̄</g>posed of the proportions of the sides C to E, and D to F. Wherefore pla<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ne. &amp;c. which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <pb n="205" facs="tcp:7062:234"/>
               <head>¶The 4. Theoreme. The 6. Proposition. If there be numbers in continuall proportion how many soeuer, and if the first measure not the second, neither shall any one of the other measure any one of the other.</head>
               <p>SVppose that there be numbers how many soeuer in continuall proportion, name<g ref="char:EOLhyphen"/>ly, f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ue, A, B, C, D, E. And suppose that A measure not B. Then I say, that nei<g ref="char:EOLhyphen"/>ther shall any other of the numbers A, B, C, D, E, measure any one of the other. That A, B, C, D, E, do not in continuall order measure one the other, it is mani<g ref="char:EOLhyphen"/>fest: for A measureth not B. Now I say, that neither shall any other of them measure any o<g ref="char:EOLhyphen"/>ther of them. I say that A shall not measure C. For how many in multitude A, B, C, are, take so many of the lest numbers that haue one and the same proportion with A, B, C, (by the 35. of the seuenth) and let the same be F, G, H.<note place="margin">Demonstration.</note> And forasmuch as F, G, H are in the selfe same
<figure/>
proportion that A, B, C, are: and the multitude of these numbers A, B, C, is equall to the mul<g ref="char:EOLhyphen"/>titude of those numbers F, G, H, therefore of equalitie (by the 14. of the seuenth) as A is to C, so is F to H. And for that as A is to B, so is F to G, but A measureth not B, therefore nei<g ref="char:EOLhyphen"/>ther doth F measure G. Wherefore F is not vnitie. For if F were vnitie, it should measure any number. But F and H are prime the one to the other (by the 3. of the eight). Wherefore F measureth not H: &amp; as F is to H, so is A to C, wherefore neither doth A measure C. In like sort may we proue that neither shall any other of the numbers A, B, C, D, E, measure any other of the numbers A, B, C, D, E: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 5. Theoreme. The 7. Proposition. If there be numbers in continuall proportion how many soeuer, and if the first measure the last, it shall also measure the second.</head>
               <p>SVppose that there be a multitude of numbers in continuall proportion, namely, A, B, C, D. And let A the first measure D the last.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note>
                  <figure/>
Then I say, that A the first measureth B the second. For if A do not measure B, neither shall any o<g ref="char:EOLhyphen"/>ther measure any other (by the 7. of the eight): which (by supposition) is not true. For A is supposed to measure D. Now then A measuring D, shall also measure B: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:235"/>
               <head>¶ The 6. Theoreme. The 8. Proposition. If betwene two numbers there fall numbers in continuall proportion: how many numbers fall betwene them, so many also shall fall in continuall pro<g ref="char:EOLhyphen"/>portion betwene other numbers which haue the selfe same proportion.</head>
               <p>SVppose that betwene the two numbers A and B, do fall in continuall proporti<g ref="char:EOLhyphen"/>on the numbers C and D. And as A is to B, so let E be to F. Then I say, that how many numbers in continuall proportion do fall betwene A and B, so many numbers also in continuall proportion shall there fall betwene E and F. How many A, B, C, D, are in multitude, take (by the 35. of the seuenth) so many of the least num<g ref="char:EOLhyphen"/>bers that haue one and the same proportion with A, B, C, D, and let the same be G, H, K, L. Wherefore their extremes G and I. are prime the one to the other (by the 3. of the eight). And forasmuch as A and C, and D and B, are in the selfe same proportion that G &amp; H, and K and L are,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and the multitude of these numbers A, C, D, B, is equall to the multitude of these numbers G, H, K, L: therefore of equalitie (by the 14. of the seuenth) as A is to B, so is
<figure/>
G to L. But as A is to B, so is E to F. Wherefore as G is to L, so is E to F. But G and L are prime the one to the other: yea they are prime and the least. But the least numbers measure those numbers that haue the same proportion with them equally, the greater the greater, and the lesse the lesse (by the 21. of the seuenth) that is, the antecedent, the antecedent, &amp; the con<g ref="char:EOLhyphen"/>sequent, the consequent. Wherefore how many times G measureth E, so many times L mea<g ref="char:EOLhyphen"/>sureth F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> How often G measureth E, so often let H measure M, and K measure N Where<g ref="char:EOLhyphen"/>fore these numbers G, H, K, L, equally measure these numbers E, M, N, F. Wherefore (by the 18. of the seue<g ref="char:cmbAbbrStroke">̄</g>th) these nu<g ref="char:cmbAbbrStroke">̄</g>bers G, H, K, L, are in the selfe same proportion that E, M, N, F, are. But G, H, K, L, are in the selfe same proportion that A, C, D, B, are: wherefore those numbers A, C, D, B, are in the selfe same proportion that E, M, N, F, are<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But A, C, D, B, are in continuall proportion: wherefore also E, M, N, F, are in continuall proportion. Wherfore how many numbers in continuall proportion fall betwene A and B, so many also in continu<g ref="char:EOLhyphen"/>all proportion fall there betwene E and F: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary added by <hi>Flussates.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary added by Flussates.</hi>
                     </note>
                     <hi>Betwene two numbers whose proportion is superparticular, or superbipartient, there falleth no mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> proportio<g ref="char:EOLhyphen"/>nall.</hi> For the least numbers of that proportion differ the one from the other onely by vnitie or by two.
<pb n="206" facs="tcp:7062:235"/>
But if betwene the greater numbers of that proportion there should fall a meane proportionall then should there fall also a meane proportionall betwene the least numbers which haue the same proporti<g ref="char:EOLhyphen"/>on by this Proposition. But betwene numbers differing onely by vnitie or by two, there falleth no meane proportionall.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 7. Theoreme. The 9. Proposition. If two numbers be prime the one to the other, and if betwene them shall fall numbers in continuall proportion: how many numbers in continuall pro<g ref="char:EOLhyphen"/>portion fall betwene them, so many also shall fall in continuall proportion betwene either of those numbers and vnitie.</head>
               <p>SVppose that there be two numbers prime the one to the other A and B: and let there fall betwene them in continuall proportion these numbers C and D: and let E be vnitie. Then I say, that how many numbers in continuall proportion fall betwene A and B, so many also shall fall in continuall proportion betwene A and vnitie E:<note place="margin">Construction.</note> and likewise betwene B and vnitie E. Take (by the 35. of the seuenth) the two least numbers that are in the same proportion that A, C, D, B, are: and let the same be F and G: and then take three of the least nu<g ref="char:cmbAbbrStroke">̄</g>bers that are in the same proportion that A, C, D, B, are: and let the same be H, K, L: and so alwaies in order one more, vntill the multitude of them be equall to the multitude of these numbers A, C, D, B: and those being so taken let them be M, N, X, O. Now it is manifest, that F multiplying him selfe produced H, and multiplying N produced M.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And G multiplying him selfe produced L, and multiplying L produced O. And forasmuch as M, N, X, O, are (by supposition) the least of all numbers that haue the same proportion with G, F: and A, C, D, B, are (by the first of the eight) the least of all numbers that haue the same proportion with G, F: and the multitude of these numbers M, N, X, O, is equall to the multitude of these numbers A, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, D, B: therefore euery one of these numbers M, N, X, O, is equall to euery one of these numbers A, C, D, B. Wherefore M is equall vnto
<figure/>
A, and O is eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all vnto B. And forasmuch as F multiplying him selfe produced H: therfore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> measureth H by those vnities which are in F: and vnitie E measureth F by those vnities which are i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> F: wherfore (by the 15. of the seuenth) vnitie E, so many times measureth the number F, as F measureth H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore as vnitie E is to the number F, so is F to H. Againe forasmuch as F multiplying N produced M, therfore H measureth M by those vnities which are in F. And vnitie E measureth F by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se vnities which are in F: wherefore (by the self same) vnitie E so many times measureth F, as H measureth M. Wherefore as vnitie E is to the numbers F, so is H to M. But it is proued, that as vnitie E is to the number F, so is F to H: wherefore as vnitie E is to the number F, so is F to H, and H to M. But M is equall vnto A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore as vnitie E is to the number F, so is F to H, &amp; H to A. And by the same reason as vnitie E is to the number G, so is G to L and L to B. Wherefore how many num<g ref="char:EOLhyphen"/>bers fall in continuall proportion betwene A and B: so many numbers also in continuall pro<g ref="char:EOLhyphen"/>portion fall there betwene vnitie E and the number A, and likewise betwene vnitie E
<pb facs="tcp:7062:236"/>
and the number <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 8. Theoreme. The 10. Proposition. If betwene two numbers and vnitie fall numbers in continuall proportion: how many numbers in continuall proportion fal betwene either of them &amp; vnitie so many also shall there fall in continuall proportion betwene them.</head>
               <p>SVppose that betwene the two numbers <hi>A, B,</hi> and vnitie <hi>C</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> do fall these numbers in continuall proportion <hi>D, E,</hi> and <hi>F, G.</hi>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLhyphen"/>uerse of the former.</note> Then I say that how many numbers in con<g ref="char:EOLhyphen"/>tinuall proportion there are betwene either of these <hi>A, B,</hi> and vnitie <hi>C,</hi> so many
<figure/>
numbers also in continuall proportion shall there fall betwene <hi>A</hi> and <hi>B.</hi> Let <hi>D</hi> multiply<g ref="char:EOLhyphen"/>ing <hi>F</hi> produce <hi>H,</hi> and let <hi>D</hi> multiplying <hi>H</hi> produce <hi>K,</hi> and like wise let <hi>F</hi> multiplying <hi>H</hi> produce <hi>L.</hi>
                  <note place="margin">Construction.</note> And for that by supposition as vnitie <hi>C</hi> is to the number <hi>D,</hi> so is <hi>D</hi> to <hi>E,</hi> there<g ref="char:EOLhyphen"/>fore how many times vnitie <hi>C</hi> measureth the number <hi>D,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so many times doth <hi>D</hi> measure <hi>E.</hi> But vnitie <hi>C</hi> measureth <hi>D</hi> by those vnities which are in <hi>D</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore <hi>D</hi> measureth <hi>E</hi> by those vnities which are in <hi>D.</hi> Wherefore <hi>D</hi> multiplying himselfe produceth <hi>E.</hi> Againe for that as vnitie <hi>C</hi> is to the number <hi>D,</hi> so is <hi>E</hi> to <hi>A,</hi> therefore how many times vnitie <hi>C</hi> mea<g ref="char:EOLhyphen"/>sureth the number <hi>D,</hi> so many times <hi>E</hi> measureth <hi>A.</hi> But vnitie <hi>C</hi> measureth <hi>D,</hi> by those vnitie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which are in <hi>D,</hi> therefore <hi>E</hi> measureth <hi>A</hi> by those vnities which are in <hi>D.</hi> Where<g ref="char:EOLhyphen"/>fore <hi>D</hi> multilying <hi>E</hi> produced <hi>A.</hi> And by the same reason <hi>F</hi> multiplying himselfe produced <hi>G,</hi> and multiplying <hi>G</hi> produced <hi>B.</hi> And forasmuch as <hi>D</hi> multiplying himselfe produced <hi>E,</hi> and multiplying <hi>F</hi> produced <hi>H,</hi> therefore (by the <hi>17.</hi> of the seuenth) as <hi>D</hi> is to <hi>F,</hi> so is <hi>E</hi> to <hi>H.</hi> And by the same reason as <hi>D</hi> is to <hi>F,</hi> so is <hi>H</hi> to <hi>G.</hi> Wherefore as <hi>E</hi> is to <hi>H,</hi> so is <hi>H</hi> to <hi>G.</hi> Agayne forasmuch as <hi>D</hi> multiplying <hi>E</hi> produced <hi>A,</hi> and multiplying <hi>H</hi> produced <hi>K,</hi> there<g ref="char:EOLhyphen"/>fore (by the <hi>17.</hi> of the seue<g ref="char:cmbAbbrStroke">̄</g>th) as <hi>E</hi> is to <hi>H,</hi> so is <hi>A</hi> to <hi>K.</hi> But as <hi>E</hi> is to <hi>H,</hi> so is <hi>D</hi> to <hi>F,</hi> there<g ref="char:EOLhyphen"/>fore as <hi>D</hi> is to <hi>F,</hi> so is <hi>A</hi> to <hi>K.</hi> Againe forasmuch as <hi>D</hi> multiplying <hi>H</hi> produced <hi>K,</hi> and <hi>F</hi> multiplying <hi>H</hi> produced <hi>L,</hi> therefore (by the <hi>17.</hi> of the seuenth) as <hi>D</hi> is to <hi>F,</hi> so is <hi>K</hi> to <hi>L.</hi> But as <hi>D</hi> is to <hi>F,</hi> so is <hi>A</hi> to <hi>K,</hi> wherfore as <hi>A</hi> is to <hi>K,</hi> so is <hi>K</hi> to <hi>L.</hi> Againe forasmuch as <hi>F</hi> multiplying <hi>H</hi> produced <hi>L</hi> and multiplying <hi>G</hi> produced <hi>B,</hi> therefore (by the <hi>17.</hi> of the seuenth) as <hi>H</hi> is to <hi>G,</hi> so is <hi>L</hi> to <hi>B.</hi> But as <hi>H</hi> is to <hi>G</hi> so is <hi>D</hi> to <hi>F,</hi> wherefore as <hi>D</hi> is to <hi>F</hi> so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>L</hi> to <hi>B.</hi> And it is proued that as <hi>D</hi> is to <hi>F,</hi> so is <hi>A</hi> to <hi>K,</hi> and <hi>K</hi> to <hi>L,</hi> and <hi>L</hi> to <hi>B.</hi> Wherfore the numbers <hi>A, K, L, B,</hi> are continuall proportion. Wherefore how many numbers in conti<g ref="char:EOLhyphen"/>nuall proportion fall betwene either of these numbers <hi>A, B,</hi> &amp; vnitie <hi>C,</hi> so many also in con<g ref="char:EOLhyphen"/>tinuall proportion fall there betwene the numbers <hi>A</hi> and <hi>B</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 11. Proposition. Betwene two square numbers there is one meane proportional number. And
<pb n="207" facs="tcp:7062:236"/>
a square number to a square, is in double proportion of that which the side of the one is to the side of the other.</head>
               <p>SVppose that there be two square numbers A and B, and let the side of A be C, &amp; let the side of B be D. Then I say that betwene these square numbers A and B, there is one meane proportionall number, and also that A is vnto B in double proportion of that which C is to D. Let C multiplieng D produce E.<note place="margin">The first part of this propo<g ref="char:EOLhyphen"/>sition demon<g ref="char:EOLhyphen"/>strated.</note> And foras<g ref="char:EOLhyphen"/>much as A is a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, &amp; the side thereof is C,
<figure/>
therfore C multiplieng himselfe produced A. And by the same reason D multiplieng himselfe produced B. Now forasmuch as C multiplieng C produced A, and multiplieng D produced E, therfore (by the 17. of the seuenth) as C is to D, so is A to E. Againe forasmuch as C multiplieng D produced E, and D multiplieng himselfe produced B, therefore these two numbers C and D multiplieng one number, namely, D, produce E and B. Wherfore (by the 18. of the seuenth) as C is to D, so is E to B. But as C is to D, so is A to E. Wherefore as A is to E, so is E to B. Wherefore betwene these square numbers A and B, there is one meane proportionall number, namely, E.<note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note> Now also I say that A is vnto B in double proportion of that which C is to D. For forasmuch as there are three numbers in continuall proportion, A, E, B, therfore (by the <hi>10.</hi> definition of the fift) A is vnto B in double proportio<g ref="char:cmbAbbrStroke">̄</g> of that which A is to E. But as A is to E, so is C to D. Wherefore A is vnto B in double proportion of that which the side C is vnto the side D: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 10. Theoreme. The 12. Proposition. Betwene two cube numbers there are two meane proportionall numbers. And the one cube is to the other cube in treble proportion of that which the side of the one is to the side of the other.</head>
               <p>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>Vppose that there be two cube numbers A and B, and let the side of A be C, and let the side of B be D. Then I say that betwene those cube numbers A and B, there are two meane proportionall numbers, and that A is vnto B in treble proportion of that which C is to D. Let C multiplieng himselfe produce E, and multiplieng D let it produce F,<note place="margin">Construction.</note> and let D multiplieng himselfe produce G. And let C multiplieng F produce H, and let D
<figure/>
multiplieng F produce K.<note place="margin">The first part of this pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>po<g ref="char:EOLhyphen"/>sition de<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>strated.</note> And forasmuch as A is a cube number, and the side therof is C, &amp; C multiplieng himselfe produceth E, therfore C multiplieng E produceth A. And by the same
<pb facs="tcp:7062:237"/>
reason for that D multiplieng himselfe, produced G, therfore D multiplieng G produceth B. And forasmuch as C multiplieng C and D produced E and F: therfore by the <hi>17.</hi> of the fift, as C is to D, so is E to F. And by the same reason also, as C is to D, so is F to G. Againe foras<g ref="char:EOLhyphen"/>much as C multiplieng E and F produced A and H, therfore as E is to F, so is A to H. But as E is to F, so is C to D. Wherfore as C is to D: so is A to H. Againe forasmuch as eche of these
<figure/>
numbers C and D multiplieng F produced H and K, therfore (by the <hi>18.</hi> of the seuenth) as C is to D, so is H to K. Againe forasmuch as D multiplieng F and G produced K &amp; B: ther<g ref="char:EOLhyphen"/>fore (by the <hi>17.</hi> of the seuenth) as F is to G, so is K to B. But as F is to G, so is C to D, where<g ref="char:EOLhyphen"/>fore as C is to D, so is K to B. And it is proued that as C is to D, so is A to H, and H to K, and K to B: wherfore betwene these cube numbers A and B, there are two meane proportionall numbers, that is, H and K.</p>
               <p>Now also I say, that A is vnto B in treble proportion of that which C is to D.<note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note> For foras<g ref="char:EOLhyphen"/>much as there are foure numbers proportionall A, H, K, B, therfore (by the <hi>10.</hi> definition of the fift) A is vnto B in treble proportion of that which A is vnto H. But as A is vnto H, so is C to D, wherfore A is vnto B in treble proportion of that which C is to D: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 11. Theoreme. The 13. Proposition. If there be numbers in continuall proportion how many so euer, and ech multiplying himselfe produce certayne numbers, the numbers of them pro<g ref="char:EOLhyphen"/>duced shall be proportinall. And if those numbers geuen at the beginning multiplying the numbers produced, produce other numbers, they also shalbe proportionall: and so shall it be continuing infinitely.</head>
               <p>SVppose that there be a multitude of nu<g ref="char:cmbAbbrStroke">̄</g>bers in co<g ref="char:cmbAbbrStroke">̄</g>tinuall proportio<g ref="char:cmbAbbrStroke">̄</g>, namely, <hi>A, B, C,</hi> as <hi>A</hi> is to <hi>B,</hi> so let <hi>B</hi> be to <hi>C.</hi> And let <hi>A, B, C,</hi> multiplying ech himself bring forth the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>D, E, F,</hi> &amp; multiplying the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>D, E, F,</hi> let the<g ref="char:cmbAbbrStroke">̄</g> bring forth the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>bers <hi>G, H, K.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say that <hi>D, E, F,</hi> are in co<g ref="char:cmbAbbrStroke">̄</g>tinuall proportio<g ref="char:cmbAbbrStroke">̄</g>, and also that. <hi>G, H, K,</hi> are in co<g ref="char:cmbAbbrStroke">̄</g>tinuall proportio<g ref="char:cmbAbbrStroke">̄</g>.<note place="margin">Construction.</note> For it is manifest that the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>D, E, F,</hi> are square numbers, &amp; that the nu<g ref="char:cmbAbbrStroke">̄</g>bers <hi>G, H, K,</hi> are cube nu<g ref="char:cmbAbbrStroke">̄</g>bers. Let <hi>A</hi> multiplying <hi>B</hi> produce <hi>L.</hi> And let <hi>A</hi> &amp; <hi>B</hi> mul<g ref="char:EOLhyphen"/>tiplying <hi>L</hi> produce <hi>M</hi> and <hi>N.</hi> And againe let <hi>B</hi> multiplying <hi>C</hi> produce <hi>X</hi>: and let <hi>B</hi> and <hi>C</hi> multiplying <hi>X</hi> produce <hi>O</hi> and <hi>P.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now by the discourse of the proposition going before we
<pb n="208" facs="tcp:7062:237"/>
may proue that <hi>D, L, E,</hi> and also <hi>G, M,
<figure/>
N, H,</hi> are in continuall proportion and in the same proportion that <hi>A</hi> is to <hi>B</hi> and likewise that <hi>E, X, F,</hi> and also <hi>H, O, P, K,</hi> are in continuall proportion and in the same proportion that <hi>B</hi> is to <hi>C.</hi> But as <hi>A</hi> is to <hi>B,</hi> so is <hi>B</hi> to <hi>C.</hi> Wherefore <hi>D, L, E,</hi> are in one and the same proportion with <hi>E, X, F</hi> and moreouer <hi>G, M, N, H,</hi> are in one and the same proportion with <hi>H, O, P, K,</hi> and the multitude of these numbers <hi>D, L, E,</hi> is equall to the multitude of these numbers <hi>E, X, F,</hi> and likewise the multi<g ref="char:EOLhyphen"/>tude of these numbers <hi>G, M, N, H,</hi> is e<g ref="char:EOLhyphen"/>quall to the multitude of these numbers <hi>H, O, P, K,</hi> wherefore of equality (by the <hi>14.</hi> of the seuenth) as <hi>D</hi> is to <hi>E,</hi> so is <hi>E</hi> to <hi>F.</hi> And as <hi>G</hi> is to <hi>H,</hi> so is <hi>H</hi> to <hi>K</hi>: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 14. Proposition. If a square number measure a square number, the side also of the one shall measure the side of the other. And if the side of the one measure the side of the other, the square number also shall measure the square number.</head>
               <p>SVppose that there be two square numbers A and B, and let the sides of them be C and D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let A measure B. Wherefore C also shall measure D.<note place="margin">The first part of this propo<g ref="char:EOLhyphen"/>sition.</note> Let C mul<g ref="char:EOLhyphen"/>tiplying D produce E. Wherefore (by the 17. and 18. of the seuenth, and 13. of the eight) those numbers A, E, B, are in continuall proportion, and are in the same proportion that C is to D. And forasmuch as A, E, B, are in continuall proportion, and A measureth B; therefore (by the 7. of the eight) A measureth E. But as A is to E, so is C to D: wherefore C measureth D.</p>
               <figure/>
               <p>But now suppose that the side C do measure the side D.<note place="margin">The second part is the conuerse of the first.</note> Then I say, that the square number A also measureth the square number B. For the same order of construction remayning, we may in like sort proue, that the numbers A, E, B, are in continuall proportion, &amp; in the same proportion, that C is to D. And for that as C is to D, so is A to E, but C measureth D: ther<g ref="char:EOLhyphen"/>fore A measureth E: and A, E, B, are in continuall proportion: wherefore A measureth B. If therefore a square number measure a square number, the side also of the one shall measure the side o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the other. And if the side of the one measure the side of the other, the square num<g ref="char:EOLhyphen"/>ber also shall measure the square number: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 13. Theoreme. The 15. Proposition. If a cube number measure a cube number, the side also of the one shall mea<g ref="char:EOLhyphen"/>sure
<pb facs="tcp:7062:238"/>
the side of the other. And if the side of the one measure the side of the other, the cube number also shall measure the cube number.</head>
               <p>SVppose that the cube number A do measure the cube number B, and let the side of A be C, and the side of B be D. Then I say, that C measureth D. Let C mul<g ref="char:EOLhyphen"/>tiplying him selfe produce E, &amp; multiplying D let it produce F. And let D mul<g ref="char:EOLhyphen"/>tiplying him selfe produce G. And moreouer let C and D multiplying F produce H and K.<note place="margin">The first part of this proposition.</note> Now it is manifest (by the 17, and 18. of the seuenth, and 12. of the eight) that those numbers E, F, G, and also A, H, K, B, are in continuall proportion, &amp; in the same pro<g ref="char:EOLhyphen"/>portion that C is to D. And forasmuch as A, H, K, B, are in continuall proportion, and A measureth B, therefore (by the 7. of the eight) A also measureth H. But as A is to H, so is C to D. Wherefore C also measureth D.</p>
               <figure/>
               <p>
                  <note place="margin">The second part is the conuerse of the first.</note>But now suppose that the side C do measure the side D. Then I say, that the cube number A also measureth the cube number B. For the same order of construction being kept, in like sort may we proue, that A, H, K, B, are in continuall proportion, and in the same proportion that C is to D. And forasmuch as C measureth D, but as C is to D, so is A to H, therefore A measureth H: wherefore A also measureth B. If therefore a cube number measure a cube number, the side also of the one shall measure the side of the other. And if the side of the one measure the side of the other, the cube number also shall measure the cube number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 14. Theoreme. The 16. Proposition. If a square number measure not a square number, neither shall the side of the one measure the side of the other. And if the side of the one measure not the side of the other, neither shall the square number measure the square number.</head>
               <p>
                  <note place="margin">
                     <p>A negat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ue proportion.</p>
                     <p>The first part of this propo<g ref="char:EOLhyphen"/>sition.</p>
                  </note>SVppose that A and B be two square numbers, and let
<figure/>
the side of A be C: and let the side of B be D. And be it that A measureth not B. Then I say, that neither shall C measure D. For if C do measure D, then (by the 14. of the eight) A also measureth B. But A by supposition measureth not B: wherefore neither doth C measure D.</p>
               <p>
                  <note place="margin">The second part is the co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>uerse of the first.</note>But now againe suppose that the side C measure not the side D. Then I say, that neither shall the square number A measure the square number B. For if A do measure B, then shall
<pb n="209" facs="tcp:7062:238"/>
C (by the <hi>14.</hi> of the eight) measure D. But (by supposition) C measureth not D. Wherefore neither doth A measure B: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 15. Theoreme. The 17. Proposition. If a cube number measure not a cube number, neither shall the side of the one measure the side of the other. And if the side of the one measure not the side of the other, neither shall the cube nu<g ref="char:cmbAbbrStroke">̄</g>ber measure the cube number.</head>
               <p>SVppose that the cube number A do not measure the cube number B:<note place="margin">A negatiue proposition.</note> and let the side of A be C: and the side of B be D. Then I say, that C shall not measure D. For if C do measure D, then (by the <hi>15.</hi> of the eight) A also shall measure B.<note place="margin">The first part of this propo<g ref="char:EOLhyphen"/>sition.</note> But (by sup<g ref="char:EOLhyphen"/>position) A measureth not B: wherefore nei<g ref="char:EOLhyphen"/>ther
<figure/>
shall C measure D.</p>
               <p>But now suppose that the side C measure not the side D.<note place="margin">The second part is the co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>uerse of the first.</note> Then I say, that neither shall the cube number A measure the cube number B. For if A do measure B, then also (by the <hi>15.</hi> of the eight) shall C measure D. But (by suppo<g ref="char:EOLhyphen"/>sition) C measureth not D. Wherefore neither shall A measure B: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 16. Theoreme. The 18. Proposition. Betwene two like plaine or superficiall numbers there is one meane propor<g ref="char:EOLhyphen"/>tionall number. And the one like plaine number is to the other like plaine number in double proportion of that which the side of like proportion, is to the side of like proportion.</head>
               <p>SVppose that there be two like plaine or superficiall numbers A &amp; B. And let the sides of A be the nu<g ref="char:cmbAbbrStroke">̄</g>bers C, D: and the sides of B be the numbers E, F. And forasmuch as like plaine numbers are those which haue their sides proportionall (by the 22. defini<g ref="char:EOLhyphen"/>tion of the seuenth) therefore as C is to D, so is E to F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>st part of this proposition.</note> Then I say that betwene A and B there is one meane proportionall number, and that
<figure/>
A is vnto B in double proportio<g ref="char:cmbAbbrStroke">̄</g> of that which C is vnto E, or of that which D is vnto F, that is, of that which side of like proportion is to side of like proportion. For for that as C is to D, so is E to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, therefore alternately (by the <hi>13.</hi> of the seuenth) as C is to E, so is D to F. And forasmuch as A is a plaine or superfi<g ref="char:EOLhyphen"/>ciall number, and the sides thereof are C and D: therefore D multiplying C produced A. And by the same reason also E multiplying F produced B. Let D multiplying E produce G. And forasmuch as D multiplying C produced A, and multiplying E produced G, therefore (by the <hi>17.</hi> of the seuenth) as C is to E, so is A to G. But as C is to E, so is D to F, wherefore as D is to F, so is A to G. Againe forasmuch as E multiplying D produced G, and multiplying <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> produced B, therefore (by the <hi>17.</hi> of the seuenth) as D is to F, so is G to B. But it is proued
<pb facs="tcp:7062:239"/>
that as D is to F, so is A to G: wherfore as A is to G, so is G to B. Wherefore these numbers A, G, B, are in continuall prorortion. Wherefore betwene A and B there is one meane proportio<g ref="char:EOLhyphen"/>nall number.</p>
               <p>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part.</note>Now also I say that A is vnto B in dooble proportio<g ref="char:cmbAbbrStroke">̄</g> of that which side of like proporti<g ref="char:EOLhyphen"/>on is to side of like proportion, that is, of that which C is vnto E, or of that which D is vnto F. For forasmuch as A, G, B, are in continuall proportion, therefore (by the <hi>10.</hi> definition of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift) A is vnto B in double proportion of that which A is vnto G. But as A is to G, so is C to E, and D to F: wherefore A is vnto B in double proportion of that which C is to E, or D to F: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 17. Theoreme. Th 19. Proposition. Betwene two like solide numbers, there are two meane proportionall num<g ref="char:EOLhyphen"/>bers. And the one like solide number, is to the other like solide number in treble proportion of that which side of like proportion is to side of lyke proportion.</head>
               <p>SVppose that there be two like solide numbers A and B. And let the sides of the number A, be the numbers C, D, E. And let the sides of the number B, be the numbers F, G, H. And forasmuch as (by the 22. definition of the seuenth) lyke solide numbers haue their sides proportio<g ref="char:EOLhyphen"/>nall,
<figure/>
therfore as C is to D, so is F to G, and as D is to E, so is G to H. Then I say that betwene A and B, there are two meane pro<g ref="char:EOLhyphen"/>portionall numbers. And that A is vnto B in treble proportion of that which C is to F, or of that which D is to G, or moreouer of that which E is vnto H.</p>
               <p>For let C multiplieng D produce K. And let F multiplieng G, produce L.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part of this proposition.</note> And for<g ref="char:EOLhyphen"/>asmuch as C, D, are in the self same propor<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> that F, G, are, &amp; of C &amp; D is produced K, and of F and G is produced L, therefore K and L are like plaine numbers. And ther<g ref="char:EOLhyphen"/>fore betwene those numbers K and L, there is one meane proportionall number (by the <hi>18.</hi> of the se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth) Let the same be M. Wher<g ref="char:EOLunhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore M is produced of D and F, as it is ma<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>fest by the proposition goyng before. Wherfore as K is to M, so is M to L. And forasmuch as D multiplieng C produced K, and multiplieng F produced M: therfore (by the <hi>17.</hi> of the seuenth) as C is to F, so is K to M, but as K is to M, so is M to L. Wherfore these numbers K M, L, are in continuall proportion and in the same proportion that C is to D. And for that as C is to D, so is F to G, therfore alternately (by the <hi>13.</hi> of the seuenth) as C is to F, so is D to G. Againe, for that as D is to E, so is G to H, therfore alternately also as D is to G, so is E to H. Wherfore these numbers K, M, L, are in continuall proportion, and in the same propor<g ref="char:EOLhyphen"/>tion that C is to F, and that D is to G, and moreouer that E is to H. Now let E and H multi<g ref="char:EOLhyphen"/>plieng M produce N and X. And forasmuch as A is a solide number, and the sides thereof are C, D, E, therefore E multiplieng that which is produced of C and D, produceth A. But that which is produced of C and D is K. Wherfore E multiplieng K produceth A. And (by the same reason H multiplieng that which is produced of F and G, that is multiplieng L pro<g ref="char:EOLunhyphen"/>duceth
<pb n="210" facs="tcp:7062:239"/>
B. And forasmuch as E multiplieng K produced A and multiplieng M produced N, therfore (by the <hi>17.</hi> of the seue<g ref="char:cmbAbbrStroke">̄</g>th), as K is to M, so is A to N. But as K is to M, so is C to F, &amp; D to G. and moreouer E to H, therfore as C is to F, and D to G, and E to H, so is A to N. A<g ref="char:EOLhyphen"/>gayne, forasmuch as E multiplieng M produced N, and H multiplieng M, produced X, ther<g ref="char:EOLhyphen"/>fore (by the <hi>18.</hi> of the seuenth) as E is to H, so is N to X. But as E is to H, so is C to F, and D to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Wherfore as C is to F, and D to G, and E to H, so is A to N, and N to X. Againe foras<g ref="char:EOLhyphen"/>much as H multiplieng M, produced X, and multiplieng L produced B, therefore (by the <hi>17.</hi> of the seuenth) as M is to L, so is X to B. But as M is to L, so is C to F, and D to G, and E to H therfore as C is to F, and D to G, and E to H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is not onely X to B, but also A to N, and N to X. Wherfore these numbers A, N, X, B, are in continuall proportion, and that in the proporti<g ref="char:EOLhyphen"/>ons of the sides.<note place="margin">The second part.</note> I say moreouer that A is vnto B in treble proportion of that, which side of like proportion, is to side of like proportion, that is, of that which the number C hath to the num<g ref="char:EOLhyphen"/>ber F, or of that which D hath to G, or moreouer of that which E hath to H. For forasmuch as there are foure numbers in continual proportion, that is, A, N, X, B, therfore (by the <hi>10.</hi> de<g ref="char:EOLhyphen"/>finition of the fift) A is vnto B in treble proportion of that which A is vnto N. But as A is to N, so is it proued that C is to F, and D to G, and moreouer E to H. Wherefore A is vnto B in treble proportion of that which side of like proportion is vnto side of like proportion, that is of that which the number C is to the number F, and of that which D is to G, and moreouer of that which E is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o H: Which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 18. Theoreme. The 20. Proposition. If betwene two numbers there be one meane proportionall number: those numbers are like plaine numbers.</head>
               <p>SVppose that betwene the two numbers A and B there be one meane proportionall number, and let the same be C.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLhyphen"/>uerse of the <hi>18.</hi> proposition.</note> Then I say, that those numbers A and B are like plaine numbers. Take (by the 35. of the seuenth) two of the least numbers that haue one &amp; the same proportion with A, C, B: and let the same be the numbers D, E.<note place="margin">Construction.</note> Wherefore as D is to E, so is A to C, but as A is to C, so is C to B, wherefore as D is to E, so is C to B. Wherefore how many times D measureth A, so many times doth E measure C. How many times D measureth A, so many vnities let there be in F. Wherefore F multi<g ref="char:EOLhyphen"/>plying D produceth A, and multiplying E it
<figure/>
produceth C: wherefore A is a plaine number: and the sides therof are D and F (by the 17. de<g ref="char:EOLhyphen"/>finition of the seuenth). Againe forasmuch as D and E are the lest numbers that haue one &amp; the same proportion with C, B, therefore (by the <hi>21.</hi> of the seuenth) how many times D measu<g ref="char:EOLhyphen"/>reth C, so many times doth E measure B. How often E measureth B, so many vnities let there be in G. Wherefore E measureth B by those v<g ref="char:EOLhyphen"/>nities which are in G: wherefore G multiply<g ref="char:EOLhyphen"/>ing E produceth B: wherefore B is a plaine number (by the <hi>17.</hi> definition of the seuenth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And the sides thereof are E and G. Wherefore those two numbers A and B are two plaine numbers. I say moreouer that they are like. For forasmuch as F multiplying E produced C: and G multiplying E produced B: therefore (by the <hi>17.</hi> of the seuenth) as F is to G, so is C to B, but as C is to B, so is D to E, wherefore as D is to E, so is F to G. Wherefore A and B are like plaine numbers, for their sides are proportionall: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:240"/>
               <head>¶The 19. Theoreme. The 21. Proposition. If betwene two numbers, there be two meane proportionall numbers, those numbers are like solide numbers.</head>
               <p>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLhyphen"/>uerse of the <hi>19.</hi> proposition.</note>SVppose that betwene two numbers A and B, there be two meane proportionall numbers C, D. Then I say that A and B are like solide numbers. Take (by the 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the seuenth, or <hi>2.</hi> of the eight) three of the least numbers that haue one and the same proportion with A, C, D, B, and let the same be E, F, G. Wherefore (by the 3. of the eight) their extremes E, G are prime the one to the other. And forasmuch as betwene the numbers E and G there is one meane proportionall number:<note place="margin">Construction.</note> therfore (by the <hi>20</hi> of the eight) they are like plaine numbers. Suppose that the sides of E, be H and K. And let the sides of G, be L and M. Now it is manifest that these numbers E, F, G, are in continuall
<figure/>
proportion,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and in the same proportion that H is to L, and that K is to M. And forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> E, F, G are the least numbers that haue one and the same proportion with A, C, D, therefore of equalitie (by the <hi>14.</hi> of the seuenth) as E is to G, so is A to D. But E, G, are (by the <hi>3.</hi> of the eight) prime numbers, yea they are prime and the least, but the least numbers (by the <hi>21.</hi> of the seuenth) measure those numbers that haue one &amp; the same proportion with them equal<g ref="char:EOLhyphen"/>ly, the greater the greater, and the lesse the lesse, that is, the antecedent the antecedent, &amp; the consequent the consequent: therfore how many time<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> E measureth A, so many times G mea<g ref="char:EOLhyphen"/>sureth D. How many times E measureth A, so many vnities let there be in N. Wherefore N multiplieng E, produceth A. But E is produced of the numbers H, K. Wherfore N multipli<g ref="char:EOLhyphen"/>eng that which is produced of H, K, produceth A. Wherefore A is a solide number, and the sides therof are H, K, N. Agayne, forasmuch as E, F, G, are the least numbers that haue one and the same proportion with C, D, B, therefore how many times E measureth C, so many times G measureth B. How ofte<g ref="char:cmbAbbrStroke">̄</g>times G measureth B, so many vnities let there be in X. Wher<g ref="char:EOLhyphen"/>fore G measureth B by those vnities which are in X. Wherfore X multiplieng G produceth B. But G is produced of the numbers L, M. Wherefore X multiplieng that number which is pro<g ref="char:EOLhyphen"/>duced of L and M, produceth B. Wherfore B is a solide number, and the sides therof are L, M X. Wherfore A, B are solide numbers. I say moreouer that they are like solide numbers. For forasmuch as N and X multiplieng E produced A and C: therfore by the <hi>18.</hi> of the seuenth,
<pb n="211" facs="tcp:7062:240"/>
as N is to X, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A to C, that is E <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o F. But as E is to F, so is H to L, and K to M: therefore as H is to L, so is K to M, and N to X. And H, K, N, are the sides of A, and likewise L, M, X, a<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> sides of B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore A, B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are like solide numbers: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Theoreme. The 22. Proposition. If three numbers be in continuall proportion, and if the first be a square number, the third also shall be a square number.</head>
               <p>SVppose that there be three numbers in continuall proportion A, B, C, and let the first be a square number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that
<figure/>
the third is also a square number. For for<g ref="char:EOLhyphen"/>asmuch as betwene A and C there is one meane pro<g ref="char:EOLhyphen"/>portionall number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> namely B, therefore (by the <hi>20.</hi> of the eight) A and C are like playne numbers. But A is a square number. Wherefore C also is a square number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 21. Theoreme. The 23. Proposition. If foure numbers be in continuall proportion, and if the first be a cube nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber, the fourth also shall be a cube number.</head>
               <p>SVppose that there be foure numbers in <figure/>
continuall proportion A, B, C, D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And let A be a cube number. The<g ref="char:cmbAbbrStroke">̄</g> I say that D also is a cube number. For forasmuch as be<g ref="char:EOLhyphen"/>twene A and D there are two meane proportio<g ref="char:EOLhyphen"/>nall numbers B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> C. Therfore A, D are like solide numbers (by the 21. of this booke) But A is a cube number, wherfore D also is a cube number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 22. Theoreme. The 24. Proposition. If two numbers be in the same proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number is to a square number, and if the first be a square number, the second also shall be a square number.</head>
               <p>SVppose that two numbers A and B be in the same proportion, that the square number C is vnto the
<figure/>
squ<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> nu<g ref="char:cmbAbbrStroke">̄</g>ber D. And let A be a square nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that B also is a square number. For forasmuch as C and D are square numbers. Therfore G and D are like plaine numbers. Wherfore (by the <hi>18.</hi> of the eight) betwene C and D there is one meane proportionall number. But as C is to D, so is A to B. Wherfore betwene A and B there is one meane proportionall number (by the <hi>8.</hi> of the eight) But A is a square number. Wherfore (by the <hi>22.</hi> of the eight) B also is a square number which was <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>equired to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:241"/>
               <head>¶ The 23. Theoreme. The 25. Proposition. If two numbers be in the same proportion the one to the other, that a cube number is to a cube number, and if the first be a cube number, the second al<g ref="char:EOLhyphen"/>so shall be a cube number.</head>
               <p>SVppose that two numbers A and B be in the same proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, that the cube nu<g ref="char:cmbAbbrStroke">̄</g>ber C is vnto the cube number D. And let A be a cube number. Then I say that B also is a cube nu<g ref="char:cmbAbbrStroke">̄</g>ber. For forasmuch as C, D, are cube nu<g ref="char:cmbAbbrStroke">̄</g>bers, therfore C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> D are like solide numbers, wherfore (by the <hi>19.</hi> of the eight) betwene
<figure/>
C and D there are two proportionall numbers. But how many numbers fall in continual pro<g ref="char:EOLhyphen"/>portion betwene C and D, so many (by the <hi>8.</hi> of the eight) fal there betwene the numbers that haue the same proportion with them. Wherefore betwene A and B there are two meane pro<g ref="char:EOLhyphen"/>portionall numbers which let be E and F. And forasmuch as there are foure numbers in con<g ref="char:EOLhyphen"/>tinuall proportion, namely, A, E, F, B, and A is a cube number, therefore (by the 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the eight) B also is a cube number: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary added by Flussates.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary added by Flussates.</hi>
                     </note>Betwene a square number and a number that is not a square number, falleth not the proportion of one square number to an other. For if the first be a square number, the second also should be a square number which is contrary to the supposition. Likewis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> betwene a cube number, and a number that is no cube number falleth not the propor<g ref="char:EOLhyphen"/>tion of one cube number to an other. For if the first be a cube number, the second also should be a cube number, which is contrary to the supposition, &amp; therfore impossible.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 24. Theoreme. The 26. Proposition. Like playne numbers, are in the same proportion the one to the other, that a square number is to a square number.</head>
               <p>SVppose that A and B be like plaine numbers. Then I say that A is vnto B in the same proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number is to a square number. For forasmuch as A, B, are like plaine numbers,<note place="margin">Construction.</note> there<g ref="char:EOLhyphen"/>fore
<figure/>
betwene A and B there falleth one meane proportional nu<g ref="char:cmbAbbrStroke">̄</g>ber (by the <hi>18.</hi> of the eight). Let there fal such a number, and let the same be C. And (by the <hi>35.</hi> of the seuenth) take the three least num<g ref="char:EOLhyphen"/>bers that haue one and the same propor<g ref="char:EOLhyphen"/>tion with A, C, B, and let the same be
<pb n="212" facs="tcp:7062:241"/>
D, E, F: wherefore (by the corollary of the <hi>2</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the eight) their <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> that is D, F, are square numbers. And for that as D is to F, so is A to B, (by the <hi>14.</hi> of the seue<g ref="char:cmbAbbrStroke">̄</g>th): and D, F, are square numbers. Therfore A is vnto B in that proportion, that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber is vnto a square num<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The 25. Theoreme. The 27. Proposition. Like solide numbers are in the same proportion the one to the other, that a cube number is to a cube number.</head>
               <p>SVppose that a A and B be like solide numbers. Then I say that A is vnto B, in the same proportion, that a cube numbe is to to a cube number. For forasmuch as A, B, are like solide numbers.<note place="margin">Construction.</note> Therefore (by the 19. of the eight) betwene A and B there
<figure/>
fall two meane proportionall numbers. Let there fall two such numbers, and let the same be C and D. And take (by the <hi>35.</hi> of the seuenth) the least numbers that haue one and the same proportion with A, C, D, B,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and equall also with the<g ref="char:cmbAbbrStroke">̄</g> in multitude, and let the same be E, F, G, H. Wherfore (by the corollary of the 2. of the eight) their extreames, that is, EH, are cube numbers. But as E is to H, so is A to B. Wherefore A is vnto B in the same proportion, that a cube number is to a cube number: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶A Corollary added by <hi>Flussates.</hi>
                  </head>
                  <p>
                     <q>If two nnmbers be in the same proportion the one to the other that a square number is to a square number: those two numbers shall be like superficiall numbers.<note place="margin">
                           <hi>A Corollary added by Flussates.</hi>
                        </note> And if they be in the same propor<g ref="char:EOLhyphen"/>tion the one to the other that a cube number is to a cube number, they shall be like solide nu<g ref="char:cmbAbbrStroke">̄</g>bers.</q>
                  </p>
                  <p>First let the number A haue vnto the number B the same proportion, that the square number C hath to the square number D. Then I say, that A and B are like superficiall nu<g ref="char:cmbAbbrStroke">̄</g>bers. For forasmuch as betwene the square numbers C and D there falleth a meane proportionall (by the 11. of this booke) there shall
<figure/>
also betwene A and B (which haue one and the same proportion with C and D) fall a meane proportio<g ref="char:EOLhyphen"/>nall (by the 8. of this booke). Wherefore A and B are like superficiall nu<g ref="char:cmbAbbrStroke">̄</g>bers (by the 20. of this booke).</p>
                  <p>But if A be vnto B, as the cube number C, is to the cube number D. Then are A &amp; B like solide num<g ref="char:EOLhyphen"/>bers. For forasmuch as C and D are cube numbers, there falleth betwene them <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wo meane proportio<g ref="char:EOLhyphen"/>nall
<figure/>
numbers (by the 12. of this booke). And therefore (by the 8. of the same) betwene A and B (which are in the same proportion that C is to D) there falleth also two meane proportionall numbers. Wher<g ref="char:EOLhyphen"/>fore (by the 21. of this booke) A and B are like solide numbers.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:242"/>
                  <head>An other Corollary added also by <hi>Flussates.</hi>
                  </head>
                  <p>
                     <hi>If a number multiplying a square number, produce not a square number: the sayd number multiplying shall b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> no square number.</hi>
                     <note place="margin">
                        <hi>Another Co<g ref="char:EOLhyphen"/>rollary added by Flussates.</hi>
                     </note> For if it should be a square number, then should it and the number multiplied being like superficiall numbers (by reason they are square numbers) haue a meane proportionall (by the 18. of thys booke). And the number produced of the sayd meane should be equall to the number contayned vnder the extremes, which are square numbers (by the 20. of the seuenth). Wherefore the number produced of the extremes being equall to the square num<g ref="char:EOLhyphen"/>ber produced of the meane, should be a square number. But the sayd number by supposition, is no square number. Wherefore neither is the number multiplying the square number, a square number.</p>
                  <p>The first part of the first Corollary is the conuerse of the 26. Proposition of this booke, and hath some vse in the tenth booke. The second part of the same also is the conuerse of the 27. Proposi<g ref="char:EOLhyphen"/>tion of the same.</p>
               </div>
            </div>
            <trailer>The end of the eighth booke of Euclides Elementes.</trailer>
         </div>
         <div n="9" type="book">
            <pb n="213" facs="tcp:7062:242"/>
            <head>¶The ninth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>IN THIS NINTH BOOKE <hi>Euclide</hi> continueth his purpose touching numbers: partly prosecuting thynges more fully,<note place="margin">
                  <hi>The Argume<g ref="char:cmbAbbrStroke">̄</g>t of the ni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th booke.</hi>
               </note> which were before somewhat spoken of, as of square and cube numbers: and partly setting out the na<g ref="char:EOLhyphen"/>tures and proprieties of such kindes of number, as haue not yet bene entreated of: which yet are most necessary to be knowen. As are numbers euen, and odde: whose passions and conditions are in this booke largely taught, with their compositions, and subductions of the one from the other: with many other generall and speciall thinges to be noted, worthy the knowledge.</p>
            <div type="part">
               <head>¶The 1. Theoreme. The 1. Proposition. If two like plaine numbers multiplying the one the other produce any num<g ref="char:EOLhyphen"/>ber: the number of them produced shall be a square number.</head>
               <p>SVppose that A and B be two like plaine numbers. And let A multiplying B produce the number C. Then I say, that C is a square number. For let A multiplying him selfe produce D. Wherefore D is a square number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as A multi<g ref="char:EOLhyphen"/>plying him selfe produced D, and multiplying B produced C, therefore (by the 17. of the seuenth) as A is to B, so is D to C. And forasmuch as A, B, are like plaine numbers, therefore (by the 18. of the eight) betwene A and B there falleth a meane proportionall number. But if betwene two numbers fall num<g ref="char:EOLhyphen"/>bers in continuall proportion, how ma<g ref="char:EOLhyphen"/>ny
<figure/>
numbers fall betwene them, so many also (by the 8. of the eight) shall there fall betwene the numbers that haue the same proportion with them. Wherfore betwene C and D also there falleth a meane proportionall number. But D is a square number. Wherfore (by the 22. of the eight) C also is a square number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 2. Theoreme. The 2. Proposition. If two numbers multiplying the one the other produce a square number: those numbers are like plaine numbers.</head>
               <p>
                  <pb facs="tcp:7062:243" rendition="simple:additions"/>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLunhyphen"/>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rse o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e form<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</note>SVppose that two number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
A and B multiplying the one the other do produce C a square number. Then I say, that A and B are like plaine num<g ref="char:EOLhyphen"/>bers. For let A multiplying him selfe produce D. Wherefore D is a square number. And forasmuch as A multi<g ref="char:EOLhyphen"/>plying him selfe produced D,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and multiplying B produced C, therefore (by the 17. of the se<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth) as A is to B, so is D to C. And forasmuch as D is a square number, and so likewise is C, therefore D and C are like plaine numbers. Wherefore betwene D and C there is (by the 18. of the eight) one meane proportionall number. But as D is to C, so is A to B. Wherefore (by the 8. of the eight) betwene A and B there is one meane proportionall number. But if be<g ref="char:EOLhyphen"/>twene two numbers there be one meane proportionall number, those numbers are (by the 20. of the eight) like plaine numbers. Wherefore A and B are like plaine numbers: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <hi>H<gap reason="illegible" resp="#PDCC" extent="4 letters">
                           <desc>••••</desc>
                        </gap> it is manifest, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t two squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> numbers multiplyed the one into the, other do alwayes produce a squa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> num<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>r.</hi>
                     <note place="margin">
                        <hi>A Corollary a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ded by Campane.</hi>
                     </note> For they are like superficiall numbers, and therefore the num<g ref="char:EOLhyphen"/>ber produced of them, is (by the first of this booke) a square number. <hi>But a square num<g ref="char:EOLhyphen"/>ber mul<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>plye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> into a number not square, produceth a number not square.</hi> For if they should pro<g ref="char:EOLhyphen"/>duce a square number, they should be like superficiall numbers (by this Proposition). But they are not. Wherefore they produce a number not square. <hi>But if a square num<g ref="char:EOLhyphen"/>ber multiplyed into an other number produce a square number, that other number shall be a square number.</hi> For by this Proposition that other number is like vnto the square number which multiplyeth it, and therefore is a square number. <hi>But if a square number multiply<g ref="char:EOLhyphen"/>ed into an other number produce a number not square, neither shall that other number also be a square number.</hi> For if it should be a square number, then being multiplyed into the square number it should produce a square number, by the first part of this Corollary.</p>
               </div>
            </div>
            <div type="part">
               <head>The 3. Theoreme. The 3. Proposition. If a cube number multiplying himselfe produce a number, the number pro<g ref="char:EOLhyphen"/>duced shall be a cube number.</head>
               <p>SVppose that A being a cube number multiplieng himselfe, do produce the num<g ref="char:EOLhyphen"/>ber B. Then I say that B is a cube number. Take the side of A, and let the same be the number C, and let C multiplieng himselfe produce the number D. Now it is manifest that C multiplieng D produceth A (by the 20. definition of the seue<g ref="char:cmbAbbrStroke">̄</g>th)<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note>
                  <figure/>
And forasmuch as C multiplieng himselfe produced D, therfore C measureth D by those v<g ref="char:EOLhyphen"/>nities
<pb n="214" facs="tcp:7062:243"/>
which are in C. But vnitie also measureth C by those vnities which are in C. Wherfore as vnitie is to C, so is C to D. Againe forasmuch as C multiplieng D produceth A: therefore D measureth A by those vnities which are in C. But vnitie measureth C by those vnities which are in C: wherefore as vnitie is to C, so is D to A. But as vnitie is to C, so is C to D, wherfore as vnitie is to C, so is C to D &amp; D to A. Wherefore betwene vnitie &amp; A there are two meane proportionall numbers, namely, C, D. Againe forasmuch as A multiplieng him<g ref="char:EOLhyphen"/>selfe produced B, therefore A measureth B by those vnities which are in A. But vnitie also measureth A by those vnities which are in A. Wherfore as vnitie is to A, so is A to B. But be<g ref="char:EOLhyphen"/>twene A and vnitie, there are two meane proportionall numbers. Wherfore betwene A and B also there are two meane proportionall numbers by the 8. of the eight. But if betwene two numbers, there be two meane proportionall numbers, and if the first be a cube number, the fourth also shall be a cube number by the <hi>21.</hi> of the eight. But A is a cube number, wherefore B also is a cube number which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 4. Theoreme. The 4. Proposition. If a cube number multiplieng a cube number, produce any number, the number produced shall be a cube number.</head>
               <p>SVppose that the cube number A multiplieng the cube number B, do produce the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber C. Then I say that C is a cube number. For let A multiplieng himselfe produce D. Wherefore D is a cube number (by the proposition going before).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as A multiplieng
<figure/>
himselfe produced D, and mul<g ref="char:EOLhyphen"/>tiplieng B, it produced C: ther<g ref="char:EOLhyphen"/>fore (by the <hi>17.</hi> of the seuenth) as A is to B, so is D to C. And forasmuch as A and B are cube numbers, therfore A and B are like solide numbers. Wherfore betwene A and B (by the <hi>19.</hi> of the eight) there are two meane proportionall numbers. Wherefore also (by the 8. of the same) betwene D and C there are two meane proportionall numbers. But D is a cube number. Wher<g ref="char:EOLunhyphen"/>fore C also is a cube number (by the <hi>23.</hi> of the eight) which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 5. Theoreme. The 5. Proposition. If a cube number multiplying any number produce a cube nu<g ref="char:cmbAbbrStroke">̄</g>ber: the num<g ref="char:EOLhyphen"/>ber multiplyed is a cube number.</head>
               <p>SVppose that the cube number A, multiplying the number B, do produce a cube number, namely, C. Then I say, that B is a cube number. For let A multiplying him selfe produce D. Wherefore (by the 3. of the ninth) D is a cube nu<g ref="char:cmbAbbrStroke">̄</g>ber. And forasmuch as A multiplying him selfe produced D,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and multiplying B, it produ<g ref="char:EOLhyphen"/>ced C: therefore (by the <hi>17.</hi> of the seuenth) as A is to B, so is D to C. And forasmuch as D
<figure/>
and C are cube numbers, they are also like solide nu<g ref="char:cmbAbbrStroke">̄</g>bers. Wherefore (by the <hi>19.</hi> of the eight) betwene D and C there are two meane proportionall numbers. But as D is to C, so is A to B.
<pb facs="tcp:7062:244"/>
Wherefore (by the 8. of the eight) betwene A and B there are two meane proportionall num<g ref="char:EOLhyphen"/>bers. But A is a cube number. Wherefore B also is a cube number (by the 23. of the eight): which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <hi>Hereby it is manifest, that if a cube number multiply a number not cube, it shall produce a number not cube.</hi>
                     <note place="margin">
                        <hi>A Corollary added by Campane.</hi>
                     </note> For if it should produce a cube number, then the number multiplyed should also be a cube number (by this Proposition) which is contrary to the supposition. For it is supposed to be no cube number. <hi>And if a cube number multiplying a number produce a number not cube, the number multiplyed shall be no cube number.</hi> For if the number multi<g ref="char:EOLhyphen"/>plied should be a cube number, the number produced should also be a cube number (by the 4. of this booke): which is contrary to the supposition, and impossible.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 6. Theoreme. The 6. Proposition. If a number multiplieng himselfe produce a cube number: then is that num<g ref="char:EOLhyphen"/>ber also a cube number.</head>
               <p>SVppose that the number A multipli<g ref="char:EOLhyphen"/>eng
<figure/>
himself, do p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oduce B a cube nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber. Then I say that A also is a cube number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For let A multiplieng B produce C. And forasmuch as A multiplieng himselfe produced B, &amp; multiplieng B it produced C: therfore C is a cube number. And for that A multiplieng himselfe produced B, and multiplieng B it produced C, therfore (by the <hi>17.</hi> of the seuenth) as A is to B, so is B to C. And for that B and C are cube numbers, they are also like solide numbers. Wherfore (by the <hi>19.</hi> of the eight) betwene C and B there are two meane pro<g ref="char:EOLhyphen"/>portional numbers. But as B is to C, so is A to B: wherfore (by the 8. of the eight) betwene A and B there are two meane proportional nu<g ref="char:cmbAbbrStroke">̄</g>bers. But B is a cube number. Wherefore A also is a cube number by the <hi>23.</hi> of the eight: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 7. Theoreme. The 7. Proposition. If a composed number multiplieng any number, produce a number: the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber produced shall be a solide number.</head>
               <p>SVppose that the composed number A multiplieng the nu<g ref="char:cmbAbbrStroke">̄</g>ber B, do produce the num<g ref="char:EOLhyphen"/>ber C. Then I say that C is a solide number. For forasmuch as A is a composed nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber, therfore some number measureth it (by the <hi>14.</hi> definition). Let D measure it.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note>
                  <figure/>
And how o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ten D measureth A, so many vnities let there be in E. Wherefore E multiplieng D produceth A. And forasmuch as two numbers D and E, multiplieng themsel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es, produce A, which A againe multiplieng B produceth C: therfore C produced of three numbers mul<g ref="char:EOLhyphen"/>tiplieng the one the other, namely, D, E, and B is (by the <hi>18.</hi> definition of the seuenth) a so<g ref="char:EOLhyphen"/>lide number. And the sides therof are the numbers D, E, B. If therefore a composed number &amp;c, which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 8. Theoreme. The 8. Proposition. If from vnitie there be numbers in continuall proportion how many soeuer: the third number from vnitie is a square number, and so are all forwarde leauing one betwene. And the fourth number is a cube number, and so are all forward leauing two betwene. And the seuenth is both a cube number
<pb n="215" facs="tcp:7062:244"/>
and also a square number, and so are all forward leauing fiue betwene.</head>
               <p>SVppose that from vnitie there be these numbers in continuall proportion A, B, C, D, E, F. Then I say that the third number from vnitie, namely, B is a square number, and so are all forward leauing one betwene, namely, D and F.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the first part.</note> And that C the fourth number is a cube number, and so are all forwarde leauyng two be<g ref="char:EOLhyphen"/>twene. And that F the seuenth number is both a cube number and also a square number, and so are all forward leauing fiue betwene. For for that as vnitie is to A, so is A to B. Therefore how many times vnitie measureth A, so many times A measureth B. But vnitie measureth A by those vnities which are in A, wherefore A measureth B by those vnities which are in A. And forasmuch as A measureth B by those vnities which are in A. Therfore A multiplieng himselfe produceth B. Wherfore B is a square number. And forasmuch as these numbers B, C, D, are in continuall proportion, and B is a square number, therfore by the <hi>22.</hi> of the eight, D also is a square number. And by the same reason also F is a square number. And in like sort may we proue that leauing alwayes one betwene, all the rest forward are square num<g ref="char:EOLhyphen"/>bers.</p>
               <figure/>
               <p>Now also I say that the fourth number from vnitie, that is, C, is a cube number, and so are all forward leauing two betwene. For for that as vnitie is to the number A, so is B to C,<note place="margin">The second part demon<g ref="char:EOLhyphen"/>strated.</note> therefore how many times vnitie measureth the number A, so many times B measureth C. But vnitie measureth A by those vnities which are in A, wherfore B measureth C by those vnit<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s which are in A. Wherfore A multiplieng B produceth C. And forasmuch as A multi<g ref="char:EOLhyphen"/>pli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ng himselfe produced B, and multiplieng B it produced C, therefore C is a cube number. And forasmuch as C, D, E, F, are in continuall proportion. But G is a cube number, therefore (by the <hi>23.</hi> of the eight) F also is a cube number.</p>
               <p>And it is proued, that F being the seuenth number from vnitie is also a square number.<note place="margin">Demostratio<g ref="char:cmbAbbrStroke">̄</g> of the third part.</note> Wherfore F is both a cube number, and also a square number. In like sort may we proue, that lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ing alwaies fiue betwene, all the rest forwarde, are numbers both cube and also square: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 9. Proposition. If from<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vnitie be numbers in continuall proportion how many soeuer: and if th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> number which followeth next after vnitie be a square number, then all the rest following also be square numbers. And if that number which followeth next after vnitie be a cube number, then all the rest following shall be cube numbers.</head>
               <p>SVppose that from vnitie there be these numbers in continuall proportion A, B, C, D, E, F. And let A which followeth next vnto vnitie be a square number. Then I say, that all the rest following also are square numbers.<note place="margin">Demostra<g ref="char:EOLhyphen"/>tion of the first part of this proposition.</note> That the third number, namely, B, is a square number, &amp; so all forward leauing one betwene, it is plaine by the Proposition next going before. I say also that all the rest are square numbers. For, forasmuch as A, B, C, are in continuall proportion, and A is a square number, therfore (by the <hi>22.</hi> of the eight) C also is a square number. Againe forasmuch as B, C, D, are in con<g ref="char:EOLhyphen"/>tinuall
<pb facs="tcp:7062:245"/>
proportion, and B is a square num<g ref="char:EOLhyphen"/>ber,
<figure/>
therfore D also (by the <hi>22.</hi> of the eight) is a square number. In like sort may we proue, that all the rest are square numbers.</p>
               <p>
                  <note place="margin">The second p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rt demonstrated.</note>But now suppose that A be a cube num<g ref="char:EOLhyphen"/>ber. Then I say, that all the rest following are cube numbers. That the fourth from v<g ref="char:EOLhyphen"/>nitie, that is, C is a cube number, and so all forward leauing two betwene, it is plaine (by the Proposition going before). Now I say, that all the rest also are cube numbers. For, for that as vnitie is to A, so is A to B: therefore how many times vnitie measu<g ref="char:EOLhyphen"/>reth A, so many times A measureth B. But vnitie measureth A by those vnities which are in A. Wherefore A also measureth B by those vnities which are in A. Wherefore A multiplying him selfe produceth B. But A is a cube number. But if a cube number mutiply<g ref="char:EOLhyphen"/>ing him selfe produce any number, the number produced, is (by the 3. of the ninth) a cube number. Wherefore B is a cube number. And forasmuch as there are foure numbers in con<g ref="char:EOLhyphen"/>tinuall proportion A, B, C, D, and A is a cube number, therefore D also (by the <hi>23.</hi> of the eight) is a cube number. And by the same reason E also is a cube number, and in like sort are all the r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>st following: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 10. Theoreme. The 10. Proposition. If from vnitie be numbers in continuall proportion how many soeuer, and if that number which followeth next after vnitie be not a square num<g ref="char:EOLhyphen"/>ber, then is none of the rest following a square number, excepting the third from vnitie, and so all forward leauing one betwene. And if that number which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>olloweth next after vnitie be not a cube number, neither is any of the rest following a cube number, excepting the fourth from vnitie, and so all forward leauing two betwene.</head>
               <p>SVppose that from vnitie be these numbers in continuall proportion A, B, C, D, E, F. And let A which followeth next after vnitie be no square number. Then I say,<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the first part lea<g ref="char:EOLhyphen"/>uing to an absu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>itie.</note> that neither is any of the rest a square number, excepting the third from vnitie, &amp; so all forward leauing one betwene, namely, B, D, F, which are square numbers (by the 8. of this booke). For if it be possible, let C be a square number. But B also is a square number. Wher<g ref="char:EOLhyphen"/>fore B is vnto C in that proportion that a square number is to a square number. But as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is
<figure/>
                  <pb n="216" facs="tcp:7062:245"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o C, so is A to B. Wherefore A is vnto B in th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> proportion that a square number is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> square number. But B is a square number. Wherefore A also is a square number (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>4. of the eigth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which is contrary to the supposition. Wherefore C is not a square number. And by the same reason none of all the other is a square number, excepting the third from vnitie, and so all forward leauing one betwene.</p>
               <p>But now suppose that A be not a cube number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Then I say,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>cond p<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> lea<g ref="char:EOLhyphen"/>ding al<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o to an absurditie.</note> that none of all the rest is a cube number, excepting the fourth from vnitie, &amp; so all forward leauing two betwene, name<g ref="char:EOLhyphen"/>ly, C, and F, which (by the 8. of this booke) are cube numbers. For if i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> possible, l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> D be a cube number. But C also is a cube number (by the 8. of the ninth). For it is the fourth from vnitie. But as C is to D, so is B to C. Wherefore B is vnto C, in that proportion tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a cube number is to a cube number. But C is a cube number. Wherefore B also is a cube number (by the 25. of the eight). And as vnitie is to A, so is A to B. But vnitie measureth A by those vnities which are in A. Wherefore A measureth B by those vnities which are in A. Wherfore A multiplying him<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>selfe produceth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> B a cube number. But if a number multiplying him selfe produce a cube number, then is that number also a cube number (by the 6. of the ninth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore A is a cube number: which is contrary to the supposition. Wherefore D is not a cube number. In like sort may we proue, that neither is any of the rest a cube number, excep<g ref="char:EOLhyphen"/>ting the fourth from vnitie, and so all forward leauing two betwene: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 11. Theoreme. The 11. Proposition. If from vnitie be numbers in continuall proportion how many soeuer, the lesse measureth the greater by some one of them which are before in the said proportionall numbers.</head>
               <p>SVppose that from vnitie A be these numbers in continuall proportion B, C, D, E. Then I say that of these numbers B, C, D, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> E being the lesse, measureth E the grea<g ref="char:EOLunhyphen"/>ter by one of these numbers C or D. For for that as vnitie A is vnto the number B, so is D to E,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore how many times vnitie A measureth
<figure/>
the number B, so many times D measureth E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore alternately (by the <hi>15.</hi> of the seuenth) how many times vni<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A measureth the number D, so many times <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> mea<g ref="char:EOLhyphen"/>sureth E. But vnitie A measureth D by those vnities which are in D. Wherefore B also measureth E by those vnities which are in D. Wherefore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lesse, measu<g ref="char:EOLhyphen"/>reth. E the greater by some one of the numbers which went before E in the proportionall numbers. And so likewise may we proue that E measureth D by some one of the numbers <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, C, D, namely, by C. And so of the rest. If therfore from vnitie &amp;c. Which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 12. Proposition. If from vnitie be numbers in continuall proportion how many soeuer, how many prime numbers measure the least<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so many also shal measure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the num<g ref="char:EOLhyphen"/>ber which followeth next after vnitie.</head>
               <p>
                  <pb facs="tcp:7062:246"/>SVppose that from vnitie be these numbers in continuall proportion A, B, C, D. Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> I say that how many prime nu<g ref="char:cmbAbbrStroke">̄</g>bers measure D, so many also do measure A. Suppose that some prime number namely, E, do measure D. The<g ref="char:cmbAbbrStroke">̄</g> I say that E also measureth A, which is next vnto vnitie. For if E do not measure A, and E is a prime number, but eue<g ref="char:EOLhyphen"/>ry number is to euery number which it measureth not a prime number (by the <hi>31.</hi> of the se<g ref="char:EOLhyphen"/>uenth). Wherefore A and E are prime numbers the one to the other. And forasmuch as E measureth D, let it measure D by the number F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Wherefore E multiplieng F produceth D. Againe forasmuch as A measureth D by those vnities which are in C, therefore A multipli<g ref="char:EOLhyphen"/>eng C produceth D. But E also multiplieng F produced D, wherfore that which is produced of the numbers A, C is equall to that which is produced of the numbers E, F. Wherfore as A is to E, so is F to C. But A, E, are prime numbers, yea they are prime and the least. But the lest numbers measure the numbers that haue one and the same proportion with them equally by the <hi>21.</hi> of the seuenth, namely, the antecedent the antecedent, and the consequent the conse<g ref="char:EOLhyphen"/>quent. Wherfore E measureth C. Let it measure it by G. Wherefore E multiplieng G produ<g ref="char:EOLhyphen"/>ceth C. But A also multiplieng B produceth C. Wherfore that which is produced of the num<g ref="char:EOLhyphen"/>bers
<figure/>
A, B, is equall to that which is produced of the numbers E, G. Wherfore as A is to E, so is G to B. But A, E are prime numbers, yea they are prime and the least. But the least num<g ref="char:EOLhyphen"/>bers (by the <hi>21.</hi> of the seuenth) measure the numbers that haue one and the same proportion with the<g ref="char:cmbAbbrStroke">̄</g> equally, namely, the antece<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s the antecede<g ref="char:cmbAbbrStroke">̄</g>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, &amp; the co<g ref="char:cmbAbbrStroke">̄</g>seque<g ref="char:cmbAbbrStroke">̄</g>t the conseqe<g ref="char:cmbAbbrStroke">̄</g>t. Wherfore E measureth B. Let it measure it by H. Wherefore B multiplieng H produceth B. But A also multiplieng himselfe produceth B, wherfore that which is produced of the numbers E, H, is equall to that which is produced of the number A. Wherfore as E is to A, so is A to H. But AE are prime nu<g ref="char:cmbAbbrStroke">̄</g>bers, yea they are prime &amp; the least, but the least numbers (by the <hi>21.</hi> of the se<g ref="char:EOLhyphen"/>uenth) measure the numbers that haue one and the same proportion with the<g ref="char:cmbAbbrStroke">̄</g> equally, name<g ref="char:EOLhyphen"/>ly, the antecede<g ref="char:cmbAbbrStroke">̄</g>t the antecedent, and the co<g ref="char:cmbAbbrStroke">̄</g>sequent the consequent. Wherfore E measureth A and it also doth not measure it by <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>pp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>sition, which is impossible. Wherfore A and E are not prime the one to the other, wherfore they are composed. But all composed numbers are measu<g ref="char:EOLhyphen"/>red of some prime number, wherfore A and E are measured by some prime number. And for<g ref="char:EOLunhyphen"/>asmuch as E is supposed to be a prime number. But a prime number is not (by the definition) measured by any other number but of himselfe. Wherfore E measureth A and E, wherfore B measureth A, and it also measureth D. Wherfore E measureth these numbers A and D. And in like sort may we proue that how many prime numbers measure D, so many also shall mea<g ref="char:EOLhyphen"/>sure A: which was required to be proued.</p>
               <div type="section">
                  <head>An other more briefe demonstration after <hi>Flussates.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>ter Flussates.</hi>
                     </note>Suppose that from vnitie be nu<g ref="char:cmbAbbrStroke">̄</g>bers in co<g ref="char:cmbAbbrStroke">̄</g>
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>nuall proportion how many so euer, namely, A, B, C, D. And let some prime nu<g ref="char:cmbAbbrStroke">̄</g>ber, namely, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> measure the last nu<g ref="char:cmbAbbrStroke">̄</g>ber which is D. The<g ref="char:cmbAbbrStroke">̄</g> I say that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> same E mea<g ref="char:EOLhyphen"/>sureth A which is the next number vnto vnitie. For if E doo not measure A, then are they prime the one to the other by the 31. of the seuenth. And forasmuch as A, B, C, D, are proportionall from vnitie,
<pb n="217" facs="tcp:7062:246"/>
therefore A multiplying himselfe produceth B. Wherfore B and E ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> prim<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.
<figure/>
numbers (by the 27. of the seuenth). And forasmuch as A multiplying B produceth C, therefore C is to E also a prime number by the 26. of the se<g ref="char:EOLhyphen"/>uenth. And likewise infinitely A multiplying C produceth D: wherefore D and E are prime numbers the one to the other (b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> same <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>6. of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>uenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wherefore E measureth not D as it was supposed, which is absurd, wherefore the prime number E measureth A, whiche is nexte vnto vnities which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 13. Theoreme. The 13. Proposition. If from vnitie be numbers in continuall proportion how many soeuer, and if that which followeth next after vnitie be a prime number: then shall no other number measure the greatest number, but those onely which are be<g ref="char:EOLhyphen"/>fore in the sayd proportionall numbers.</head>
               <p>SVppose that from vnitie be these numbers in continuall proportion A, B, C, D, and let that which followeth next after vnitie, that is, A, be a prime number. Then I say, that no other number besides these numbers A, B, C, measureth the greatest number of them which is D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if it be possible, let E measure D. And let E be none of these numbers A, B, C, D. Now it is manifest that E is not a prime number. For if E be a prime number, &amp; do also measure D, it shall likewise measure A being a prime number and not being one and the same with A, by the former Proposition: which is im<g ref="char:EOLhyphen"/>possible. Wherefore E is not a prime number. Wherefore it is a composed number. But euery composed number (by the 33. of the seuenth) is measured by some prime number.</p>
               <figure/>
               <p>Now I say, that no other prime nu<g ref="char:cmbAbbrStroke">̄</g>ber besides A shall measure E. For if any other prime nu<g ref="char:cmbAbbrStroke">̄</g>ber do measure E, &amp; E measureth D, therfore that number also shall measure D (by the 5. common sentence of the seuenth). Wherfore it shal also measure A (by the proposition next going before) being a prime number and not being one and the same with A: which is im<g ref="char:EOLhyphen"/>possible. Wherefore euery the prime number A measureth E which measureth the greatest number D.</p>
               <p>And forasmuch as E measureth D, let it measure it by F. Now I say, that F is none of these numbers A, B, C. For if F be one and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> with any of these numbers A, B, C, and is measureth D by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, therefore one of these numbers A, B, C, measureth D by E. But one of these numbers A, B, C, measureth D by some one of these numbers A, B, C, therefore E is one and the same with one of these numbers A, B, C, which is cont<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ry <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> supposition. Wher<g ref="char:EOLhyphen"/>fore F<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> one and the same with any of these <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>bers A, B, C.</p>
               <p>In like sort may we pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e, that onely the prime number A measureth F, pro<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>g first th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t
<pb facs="tcp:7062:247"/>
F is not a prime number. For if F be a prime number, and it measureth D, therefore it also measureth A being a prime number, and not being one and the same with A, by the former Proposition: which is impossible. Wherefore F is not a prime number: wherefore it is a com<g ref="char:EOLhyphen"/>posed number, and therefore some prime number shall measure it. Now I say, that no other prime number besides A shall measure it. For if any other prime number do measure F, and F measureth D, therefore that number shall measure D (by the 5. common sentence of the se<g ref="char:EOLhyphen"/>uenth). Wherefore it shall also measure A (by the former Proposition) being a prime num<g ref="char:EOLhyphen"/>ber and not being one and the same with A: which is impossible. Wherefore onely the prime number A measureth F. And forasmuch as E measureth D by F, therefore E multiplying F produceth D, But A also multiplying C produceth D, therefore that which is produced
<figure/>
if A into C, is equall to that which is produced of E into F. Wherfore proportionally as A is to E, so is F to C. But A measureth E. Wherefore F measureth C. Let F measure C by G. And in like sort may we proue, that G is not one and the same with any of these numbers A, B, C, and that G is a composed number, and also that onely the prime number A measureth it. And forasmuch as F measureth C by G, therefore G multiplying F produced C. But A al<g ref="char:EOLhyphen"/>so multiplying B produced C. Wherefore that which is produced of A into B, is equall to that which is produced of F into G. Wherefore proportionally as A is to F, so is G to B. But A measureth F. Wherefore G also measureth B. Let G measure B by H. Now in like sort as be<g ref="char:EOLhyphen"/>fore may we proue, that H is not one and the same with A, and that H is a composed number, and measured onely of the prime number A. And forasmuch as G measureth B by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se vni<g ref="char:EOLhyphen"/>ties which are in H, therfore G multiplying H produced B. But A multiplying him selfe pro<g ref="char:EOLhyphen"/>duced B. Wherfore that which is produced of H into G, is equall to the square number which is produced of A. Wherefore as H is to A, so is A to G. But A measureth G. Wherfore H mea<g ref="char:EOLhyphen"/>sureth A being a prime number and not being one and the same with it: which is absurde. Wherfore no other number besides these numbers A, B, C, measureth the greatest number D: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration of the same after Campane.</head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>ter Campane.</hi>
                     </note>Suppose that E not being one and the same with the numbers A, B, C, D doo measure the num<g ref="char:EOLhyphen"/>ber D. And let it measure it by the number P. And forasmuch as A being a prime number measureth the number D, which is produced of E into F: therefore by the 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the fourth, A measureth either <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> or <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Let it measure E. Now forasmuch as D is produced of <hi>A</hi> into G, and also of E into F: therefore by the second part of the 19. of the seuenth, <hi>A</hi> is to E, as F is to C. But <hi>A</hi> measureth E: wherefore F measu<g ref="char:EOLhyphen"/>reth C. Let it measure it by G. Wherefore by the 32. of the seuenth <hi>A</hi> shall measure either F or G. Let it measure F. Wherefore as before by the second part of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the seuenth G shall measu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e B. Let it measure it by H. Now then as before it followeth by the 32. of the seuenth that <hi>A</hi> shall measure either G or H: suppose that it measure G. Wherefore by the second part of the 20. of the seuenth H shall mea<g ref="char:EOLhyphen"/>sure <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> (fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of <hi>A</hi> into himselfe is produced B, and of H into G also it produced B) I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore H be not
<pb n="218" facs="tcp:7062:247"/>
equall vnto <hi>A, A</hi> shall be no prime number. Which is contrary to the supposition. <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>t if it be equall vn<g ref="char:EOLhyphen"/>to <hi>A,</hi> then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uery one of these numbers <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, F, E, shall be some one of the numbers <hi>A, B,</hi> C, D, by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 11.
<figure/>
proposition of the ninth repeted a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>en as neede requireth. Wherefore E is not a number diuerse from them, but is one and the same with some one of them: which is contrary to the supposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion; wher<g ref="char:EOLunhyphen"/>fore that is manifest which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 14. Theoreme. The 14. Proposition. If there be geuen the least number, whom certayne prime numbers geuen, do measure: no other prime number shall measure that nu<g ref="char:cmbAbbrStroke">̄</g>ber, besides those prime numbers geuen.</head>
               <p>SVppose that the least number whom these prime numbers B, C, D, do measure, be A. Then I say that no other prime number besides B, C, D, measureth A. For if it be possible, let E being a prime number measure A, and let E be none of these numbers B, C, D.<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stration leading to an absurditie.</note> And forasmuch as E measu<g ref="char:EOLhyphen"/>reth
<figure/>
A, let it measure it by F. Wherfore E multiplieng I produceth A: And these prime numbers B, C, D, measure A. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut if two numbers multiplieng the one the other produce any number. And if some prime number measure that which is produced, it shall also measure one of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se numbers which were put at the beginning (by the 32. of the seuenth) Wherfore those numbers B, C, D, measure one of these numbers E or F. But they measure not E, for E is a prime number, and is not one and the same with any one of these numbers B, C, D. Wherfore they measure F being lesse then A which is impossible. For A is supposed to be the least whom B, C, D, measure. Wherefore no prime number besides B, C, D, measureth A: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A proposition added by Campane.</head>
                  <p>
                     <hi>If there be numbers how many soeuer in continuall proportion being the least in that proportion: a number measuring one of them, shall be a number not prime to one of the two least numbers in that proportion.</hi>
                  </p>
                  <p>Suppose that there be numbers in continuall proportion how many soe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>r namely <hi>A,</hi> B, C, <hi>D,</hi> E which let be the least that haue the same proportion with them:<note place="margin">
                        <hi>A propositio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> added by Campane.</hi>
                     </note> and let the two least numbers in that proportion be F and G. And let some number as H measure some one of the numbers <hi>A, B,</hi> C, D, E, namely, C. Then I say that H is a number not prime either to F or G. Take (by the 2. of the eight) the
<pb facs="tcp:7062:248"/>
three least numbers in
<figure/>
the proportion of <hi>A</hi> to <hi>B</hi>: which let be P, Q, R. And afterward fower (by the same) which let be K, L, M, N: &amp; so for<g ref="char:EOLhyphen"/>ward till you come to the multitude of the numbers geuen <hi>A, B, C, D, E.</hi> Now it is manifest (by the demonstration of the second of the eight) that F multiplyed by P, Q, and R produceth K, L, M: and that F multiplyed by K, L, M, N produc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th <hi>A, B,</hi> C, D. And forasmuch as H measureth C: therfore H is either to F or to M not prime (by the corollary of the 32. of the seuenth added by Ca<g ref="char:cmbAbbrStroke">̄</g>pane) If it be not prime vnto F: the<g ref="char:cmbAbbrStroke">̄</g> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that manifest which was required to be proued. But if H be not prime vnto M. The<g ref="char:cmbAbbrStroke">̄</g> shal it not be prime either to F or to R (by the same corollary). If agayne it be not prime vnto F, then is that proued which was required. But if it be not prime vnto R, than (by the same corollary) shall it be a number not prime vnto G (which produceth R by the 2. of the eight)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> but G is one of the two least numbers F or G which are in the proportion of the numbers geuen at the beginning <hi>A,</hi> B, C, D, E. If therefore there be num<g ref="char:EOLhyphen"/>bers how many soeuer. &amp;c. which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 15. Theoreme. The 15. Proposition. If three numbers in continuall proportion be the least of all numbers that haue one and the same proportion with them: euery two of them added to<g ref="char:EOLhyphen"/>gether shall be prime to the third.</head>
               <p>
                  <hi>SVppose that there be three numbers in continuall proportion A, B, C, being the left of all numbers that haue one and the same proportion with them. Then I say, that euery two of these numbers A, B, C, added together,</hi>
                  <note place="margin">
                     <hi>Construc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion.</hi>
                  </note> 
                  <hi>are prime to the third: name<g ref="char:EOLhyphen"/>ly, that A, B, is prime to C, and B, C, to A, and A, C, to B. Take (by the 35. of the seuenth) two of the numbers that haue one and the same proportion with A, B, C, &amp; let the same be the numbers DE, and EF.</hi>
               </p>
               <p>
                  <hi>Now it is manifest (by the sayd 35. Proposition) that DE multiplying him selfe produced A,</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>and multiplying EF produced B, and moreouer EF multiplying him selfe produced C.</hi>
               </p>
               <p>
                  <hi>And forasmuch as DE and EF are the least in that proportion, they are also prime the one to the other (by the 24. of the seuenth). But
<figure/>
i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> two numbers be prime the one to the other, then both of the<g ref="char:cmbAbbrStroke">̄</g> added together, shall be prime to either of them (by the 30. of the seuenth). Wherefore the whole number DF is prime to either of these nu<g ref="char:cmbAbbrStroke">̄</g>bers DE &amp; EF. But DE also is prime vnto EF. Wherfore DF, &amp; DE are prime vnto EF. Wherfore that which is produced of DF into DE, is (by the 26. of the se<g ref="char:EOLunhyphen"/>ue<g ref="char:cmbAbbrStroke">̄</g>th) prime vnto EF. But if two nu<g ref="char:cmbAbbrStroke">̄</g>bers be prime the one to the other, that which is produced of the one of the<g ref="char:cmbAbbrStroke">̄</g> into himselfe, is prime to the other (by the 27. of the seue<g ref="char:cmbAbbrStroke">̄</g>th). Wherfore that which is produced of DF into DE, is prime to that which is produced of EF into himsel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e. But that which is produced of FD into DE, is the square nu<g ref="char:cmbAbbrStroke">̄</g>ber which is produced of DE into himsel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e together with that which is produced of DE into EF (by the 3. of the second). Wherfore the square nu<g ref="char:cmbAbbrStroke">̄</g>ber which is produced of DE together with that which is produced of DE into EF, is prime to that which is produced of EF into himself. But that which is produ<g ref="char:EOLunhyphen"/>ced of DE into him selfe, is the number A, &amp; that which is produced of DE into EF, is the number B: and that which is produced of EF into himselfe, is the number C. Wherefore the numbers A, B, added together are prime vnto C.</hi>
               </p>
               <p>
                  <pb n="219" facs="tcp:7062:248"/>
                  <hi>By the like demonstration also may we
<figure/>
proue, that the numbers B, C, are prime vnto the number A.</hi>
               </p>
               <p>
                  <hi>Now also I say, that the numbers A, C, are prime vnto the number B.</hi>
               </p>
               <p>
                  <hi>For forasmuch as DF is prime to either of
<figure/>
these DE and EF: therefore that which is produced of DF into him self, is prime to that which is produced of DE into EF. But that which is produced of DF into him selfe, is e<g ref="char:EOLhyphen"/>quall to the square numbers which are produced of DE and EF together with that number which is produced of DE into EF, twise (by the 4. of the second). Wherefore the square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers which are produced of DE and EF together with that which is produced of DE into EF twise are prime to that which is produced of DE into EF. And by diuision also (by the 30. of the seuenth) the square numbers produced of DE and EF, together with that which is produced of DE into EF once are prime to that which is produced of DE into EF. A<g ref="char:EOLhyphen"/>gaine (by the same 30. of the seuenth) the square nu<g ref="char:cmbAbbrStroke">̄</g>bers produced of DE and EF, are prime to that which is produced of DE into EF. But that which is produced of DE into him selfe is A, and that which is produced of EF into him selfe is C, and that which is produced of DE into EF, is B. Wherefore the numbers A, C, added together are prime vnto the num<g ref="char:EOLhyphen"/>ber B: which was required to be demonstrated.</hi>
               </p>
               <p>This latter part of the demonstration, which proueth that the numbers A, &amp; C are prime vnto B, is somewhat obscurely put of <hi>Theon.</hi> And therefore I will here make it playner.</p>
               <p>Forasmuch as either of the numbers DE, and EF is prime to the whole DF:<note place="margin">
                     <hi>Demonstrati<g ref="char:EOLhyphen"/>on to proue that the num<g ref="char:EOLhyphen"/>bers A and C are prime to B.</hi>
                  </note> (as hath before b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne proued) therefore that which is produced of DE into EF (which is the number B) is prime vnto DF, by the 26. of the seuenth. Wherefore by the 27. of the same that which is produced of DF into himselfe (which is the number composed of A and C and of the double of B by the 4. of the second) shall be prime vnto B. Wherefore it followeth that the number composed of A and C is prime vnto B. For if a number composed of two numbers, be prime to one of the said two numbers, as here the number com<g ref="char:EOLunhyphen"/>posed of A and C taken as one number and of the double of B, is prime vnto the double of B: then the two numbers whereof the number is composed, namely, the number composed of A and C, and the double of B shall be prime the one to the other (by the 30 of the seuenth). And therefore the number composed of A and C shall be prime to B taken once. For if any number should measure the two num<g ref="char:EOLhyphen"/>bers, namely the number composed of A and C, and the number B, it should also measure the number composed of A and C, and the double of B (by the 5. common sentence of the seuenth): which is not possible, for that they are proued to be prime numbers.</p>
               <p>Here haue I added an other demonstration of the former Proposition after <hi>Cam<g ref="char:EOLhyphen"/>pane,</hi> which proueth that in nu<g ref="char:cmbAbbrStroke">̄</g>bers how many soeuer, which is there proued onely tou<g ref="char:EOLhyphen"/>ching three numbers: and the demonstration seement somwhat more perspicous then <hi>Theons</hi> demonstration. And thus he putteth the proposition.</p>
               <p>
                  <hi>If numbers how many soeuer being in continuall proportion be the least that haue one &amp; the same proportion with them: euery one of them shalbe to the number composed of the rest prime.</hi>
               </p>
               <p>Suppose that there be numbers in continuall proportion how many soeuer,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurdi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ie.</hi>
                  </note> and the least in their proportion: namely, <hi>A,</hi> B, C, D. Then I say that euery one of them, as for example <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> D, is prime to the number composed of the rest, namely, of <hi>A, B,</hi> C. For if it be not, let some number, namely E mea<g ref="char:EOLunhyphen"/>sure D, and the number composed of <hi>A, B,</hi> C. Take the two least numbers in the same proportion that <hi>A, B,</hi> C, D are (by the 35. of the seuenth) which let
<figure/>
be F, G. And forasmuch as E measureth one of these number <hi>A,</hi> B, C, D, the same E shalbe a number not prime either to F or to G (by the proposition before added by Campane after the 14. proposition) wher<g ref="char:EOLhyphen"/>fore some number shall measure E and one of these number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> For G: which let be H. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd forasmuch as H measureth But shall also measure D, which num<g ref="char:EOLhyphen"/>ber D the number E also measureth (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. com<g ref="char:EOLhyphen"/>mon sentence of the seuenth). Moreouer forasmuch is H (by supposition) measureth one of these num<g ref="char:EOLhyphen"/>bers
<pb facs="tcp:7062:249"/>
F or G, the same H shall measure all the meanes betwen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A and D by the same co<g ref="char:cmbAbbrStroke">̄</g>mon sente<g ref="char:cmbAbbrStroke">̄</g>ce. For e<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>er of these numbers F or G produceth all the meanes by the next numbers in continuall proporti<g ref="char:EOLhyphen"/>on and in the same proportion with them (as by L, I, K) by the second of the eight. Agayne forasmuch as H measureth E, which (by supposition) measureth the whole A, B, C: the same H shall also measure the whole, A, B, C (by the foresayd common sentence) and it measureth the part taken away, namely, the meanes B, C (as it hath bene proued) wherefore it also measureth the residue A (by the 4. common sentence o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the seuenth) wherefore H measureth the extreames D and A, which are prime the one to the other (by the 3. of the eight) which were absurd. Wherefore D is a number prime to the number composed of the rest, namely, of A, B, C.</p>
               <p>Secondly I say that this is so in euery one of them: namely that C is a prime number to the num<g ref="char:EOLhyphen"/>ber composed of A, B, D. For if not, then as before let E measure C, and the number composed of A, B, <hi>D</hi>: which <hi>E</hi> shalbe a number not prime either to F or to G (by the former proposition added by Cam<g ref="char:EOLhyphen"/>pane) wherefore let H measure them. And forasmuch as H measureth <hi>E,</hi> it shall also measure the whole A, B, C, D whom E measureth. And forasmuch as H measureth one of these numbers F or G, it shall measure one of the extreames A or D: which are produced of F or G (by the second of the eight) if they be multipl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ed into the meanes L or K. And moreouer the same H shall measure the meames, BC (by the 5. common sentence of the seuenth) when as by supposition it measureth either F or G. which mea<g ref="char:EOLhyphen"/>sure B, C (by the second of the eight). But the same H measureth the whole A, B, C, D as we haue pro<g ref="char:EOLhyphen"/>ued, for that it measureth E. Wherefore it shall also measure the residue, namely, the number composed of the extreames A and D (by the 4. common sentence of the seuenth). And it measureth one of these A or D (for it measureth one of these F or G which produce A and D) wherefore the same H shall measure one of these A or D and also the other of them (by the former common sentence) which num<g ref="char:EOLunhyphen"/>bers A and D are by the 3. of the eight prime the one to the other. Which were absurd. This may also be proued in euery one of these numbers A, B, C, D. Wherefore no number shall measure one of these numbers A, B, C, D and the numbers composed of the rest. Wherefore they are prime the one to the o<g ref="char:EOLhyphen"/>ther: If therefore numbers how many soeuer. &amp;c: which was required to be proued.</p>
               <p>Here as I promised, I haue added <hi>Campanes</hi> demonstrations of those Propositions in numbers, which <hi>Eucl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>de</hi> in the second booke demonstrated in lines. And that in thys place so much the rather, for that <hi>Theon</hi> as we see in the demonstration of the 15. Pro<g ref="char:EOLhyphen"/>position seemeth to alledge the 3. &amp; 4. Proposition of the second boke: which although they concerne lines onely, yet as we there declared and proued, are they true also in numbers.</p>
               <div type="section">
                  <head>¶ The first Proposition added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <q>That number which is produced of the multiplication of one number into numbers how many so<g ref="char:EOLhyphen"/>euer: is equall to that number which is produced of the multiplication of the same number into the number composed of them.</q>
                  </p>
                  <p>This proueth that in numbers which the first of the second proued touching lines. Suppo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> that the number A being multiplyed into the number B, and into the number C,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tiou.</hi>
                     </note> and into the number D, doo produce the numbers E, F and G. Then I say that the number produced of A multiplyed into the number composed of B, C, and D is equall to the number composed of E, F, and G. For by the con<g ref="char:EOLhyphen"/>uerse of the definition of a number multiplyed, what part vnitie is of A, the selfe same part is B of E, and C of F, and also D of G. Wherefore
<figure/>
by the 5. of the seuenth what part vnitie is of A, the selfe same part is the num<g ref="char:EOLhyphen"/>ber composed of B, C, and D, of the num<g ref="char:EOLunhyphen"/>ber composed of E, F, and G. Wherfore by the definition that which is produ<g ref="char:EOLhyphen"/>ced of A into the number composed of B, C, D, is equall to the number compo<g ref="char:EOLhyphen"/>sed of E, F, G: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The second Proposition.</head>
                  <p>
                     <q>That number which is produced of the multiplication of numbers how many soeuer into one nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber: is equall to that number which is produced of the multiplication of the number composed of them into the same number.</q>
                  </p>
                  <p>
                     <pb n="220" facs="tcp:7062:249"/>This is the conuerse of the former As if the <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>bers <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and G and D multiplyed into the number <hi>A</hi> doo produce the numbers E and F and G.<note place="margin">
                        <hi>This proposi<g ref="char:EOLhyphen"/>tion is the co<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>uerse of the former.</hi>
                     </note>
                     <figure/>
Then the number composed of B, C, D. mul<g ref="char:EOLhyphen"/>tiplyed into the number <hi>A</hi> shall produce the number composed of the numbers E, F, G. Which thing is easly proued by the 16. of the seuenth and by the former proposition.</p>
               </div>
               <div type="section">
                  <head>¶ The third Proposition.</head>
                  <p>
                     <q>That number which is produced of the multiplication of numbers how many soeuer into other numbers how many soeuer, is equall to that number which is produced of the multiplication of the number composed of those first numbers, into the number composed of these latter numbers.</q>
                  </p>
                  <p>As if the numbers A, B, C doo multiply the numbers D, E, F, ech one eche other, and if the num<g ref="char:EOLhyphen"/>bers produced be added together. Then I say that the
<figure/>
number composed of the numbers produced is equall to the number produced of the number composed of the numbers A, B,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> C into the number composed of the num<g ref="char:EOLhyphen"/>bers D, E, F. For by the former propositio<g ref="char:cmbAbbrStroke">̄</g> that which is produced of the number composed of A, B, C into D is equall to that which is produced of euery one of the sayd numbers into D: and by the same reason that which is produced of the number composed of A, B, C into E, is equal to that which is pro<g ref="char:EOLhyphen"/>duced of euery one of the sayd numbers into E: and so likewise that which is produced of the number composed of A, B, C into F is equall to that which is produced of euery one of the sayd numbers into F. But by the first of these propositions th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> which is produced of the number composed of these num<g ref="char:EOLunhyphen"/>bers A, B, C into euery one of these numbers D, E, F is equall to that which is produced of the number composed into the number composed: wherefore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The fourth Proposition.</head>
                  <p>
                     <q>If a number be deu<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>d into partes how many soeuer: that nu<g ref="char:cmbAbbrStroke">̄</g>ber which is produced of the whole into him selfe, is equall to that number which is produced of the same number into all his partes.</q>
                  </p>
                  <p>This pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eth in numbers that which the second of the second proued in lines. As if the number A, be deuided into the numbers B and C, and D.<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the 2. of the second.</hi>
                     </note> Then I
<figure/>
say, that that which is produced of <hi>A</hi> into himselfe, is e<g ref="char:EOLhyphen"/>quall to that which is produced of <hi>A</hi> into all the sayd num<g ref="char:EOLunhyphen"/>bers B, C, and D. For putting the number <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall to the number A, it is manifest by the first of these propositions that that which is produced of E into A,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> is equall to that which is produced of E into all the partes of A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> But by the common sentence that which is produced of E into A is equal to that which is produced of A into himselfe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and that which is produced of E into the partes of A is equall to that which is pro<g ref="char:EOLhyphen"/>duced of A into the selfe same partes. Wherefore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The fift Proposition.</head>
                  <p>
                     <q>If a number be deuided into two partes that which is produced of the whole in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o one of the partes, as equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o that which is produced of the selfe same par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> into him selfe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and into the other part.</q>
                  </p>
                  <p>This proueth in numbers that which in the 3. of the second was proued in lines.<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the 3. of the thirds.</hi>
                     </note> For let the num<g ref="char:EOLhyphen"/>ber A be deuided into the numbers B and C. Then I say that that which is
<figure/>
produced of A into C, is equall to that which is produced of C into him<g ref="char:EOLhyphen"/>selfe and into B. For by the 16. of the seuenth, that which is produced of A into C is equall to that which is produced of C into A. Now then put the number D equall to the number C.<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> Wherefore that which is produced of A into C is equall to that which is produced of D into A. But by the first of these propositions that which is produced of D into A is equall to that which is produced of D into B and of D into C. Wher<g ref="char:EOLhyphen"/>fore forasmu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> that which is produced of D into A and into B and into C is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>qu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll to that which is produced of C into A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and into B, and into himselfe, by reason of the equalitie of C and D: that is ma<g ref="char:EOLhyphen"/>nifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:250"/>
                  <head>¶The sixt Proposition.</head>
                  <p>
                     <q>If a number be deuided into two partes: that which is produced of the multiplication of the whole <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nto him selfe, is equall to that which is produced of the multiplication of either of the partes into him selfe, and of the one into the other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>wise.</q>
                  </p>
                  <p>This proueth in numbers that which the fourth of the second proued touching lines. As if the number A be deuided into the numbers B and C. Then I say that
<figure/>
that which is produced of A into himself is equal to that which is produced of B into himselfe,<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>th to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> 4. <hi>of the second.</hi>
                     </note> and of C into himselfe, and of B into C twise. For by the 4 of these propositions, that which is produced of A into himselfe, is equall to that which is produced of A into B, and into C. But that which is produced of A into B, is equall to that which is produced of B into himselfe and into C (by the former proposition). And by the same that which is produced of A into C is equall to that which is produced of C into himselfe and into B.<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> And forasmuch as that which is produced of C into B is equall to that which is produced of B into C by the 16. of the seuenth, it is manifest that that is true which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The seuenth Proposition.</head>
                  <p>
                     <q>If a number be deuided into two equall partes, &amp; into two vnequall partes: that which is produ<g ref="char:EOLhyphen"/>ced of the greater of the vnequall partes into the lesse, together with the square nu<g ref="char:cmbAbbrStroke">̄</g>ber of the num<g ref="char:EOLhyphen"/>ber set betwene, is equall to the square number produced of the halfe of the whole.</q>
                  </p>
                  <p>This proueth in numbers that which the 5. of the second proued in lines. As if the number AB be deuided into two equall numbers, which let be AC, and CB: and also in two vnequal numbers namely,<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the</hi> 5. <hi>of the second.</hi>
                     </note> AD and DB, of which let AD be the greater, and
<figure/>
DB the lesse. Then I say that that which is produced of the whole AD into DB together with the square number of CD, is equall to the square number of CB. For by the former propositio<g ref="char:cmbAbbrStroke">̄</g> the square of CB is equall to the square of CD and to the square of DB, and to that which is produced of BD into CD twise. But that which is produced of <hi>B</hi>D into himselfe and into CD is equall to that which is produced of BD into C<hi>B</hi> by the first of these propositions,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> and therefore vnto that which is produced of <hi>B</hi>D into AC. Wherefore that which is produced of BD into himselfe and into CD twise is equall so that which is produced of BD into AD. Wherefore by the same the square of C<hi>B</hi> exceedeth that which is produced of BD into AD by the square of CD: wher<g ref="char:EOLhyphen"/>fore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The 8. Proposition.</head>
                  <p>
                     <q>If a number be deuided into two equall part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s, and if vnto it be added an other number: that which is produced of the multiplication of the whole being composed into the number added, to<g ref="char:EOLhyphen"/>gether with the square of the halfe, is equall to the square of the number composed of the halfe and the number added.</q>
                  </p>
                  <p>This proueth in numbers that which the 6. of the second proued touching lines. For suppose that the number <hi>AB</hi> be deuided into equall numbers, which let be <hi>AC</hi> and <hi>CB:</hi> and vnto it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> adde the number BD.<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the</hi> 6. <hi>of the second.</hi>
                     </note> Then I say, that that which is produced of the whole <hi>AD</hi> into <hi>DA</hi> together with the square of BC, is equall to the square of <hi>CD.</hi> For by the 6. of these
<figure/>
propositio<g ref="char:cmbAbbrStroke">̄</g>s the square of <hi>CD</hi> is equal to the square of <hi>DB,</hi> &amp; to the square of <hi>BC,</hi> and to that which is produced of <hi>DB</hi> into <hi>BC</hi> twise. But by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of these propositi<g ref="char:EOLhyphen"/>ons,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> that which is produced of <hi>BD</hi> into himselfe and into <hi>BC</hi> twise is equall to that which is produ<g ref="char:EOLhyphen"/>ced of <hi>BD</hi> into <hi>DA</hi> (for <hi>AC</hi> and <hi>CB</hi> are equall) wherefore the square of <hi>CD</hi> exceedeth that which is produced of <hi>BD</hi> into <hi>DA</hi> by the square of <hi>CB.</hi> Wherefore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The 9. Proposition.</head>
                  <p>
                     <q>If a number be deuided into two partes: that which is produced of the whole number into him selfe together with that which is produced of one of the partes into him self, is equall to that which is produced of the whole into the sayd part twise together with that which is produced of the other part into him selfe.</q>
                  </p>
                  <p>
                     <pb n="221" facs="tcp:7062:250"/>This proueth in numbers that which the 7. of the second proued in lines. For let the number <hi>A</hi> be deuided into the numbers <hi>B</hi> and <hi>D.</hi>
                     <note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the</hi> 7. <hi>of the second.</hi>
                     </note> Then I say that the square of <hi>A</hi> together with the square of <hi>D</hi> is equall to that which is produced of <hi>A</hi> into <hi>D</hi> twise together with the
<figure/>
square of <hi>B.</hi> For it is manifest by the 6. of these propositions that the square of <hi>A</hi> is equall to the squares of <hi>B</hi> and <hi>D</hi> together with that which is produced of <hi>B</hi> into <hi>D</hi> twise. Wherefore the square of <hi>A</hi> together with the square of <hi>D,</hi> is equall to two squares of <hi>D</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> and to that which is produced of <hi>D</hi> into <hi>B</hi> twise together with the square of <hi>B.</hi> But by the first of these propositions two squares of <hi>D,</hi> and that which is produced of <hi>D</hi> into <hi>B</hi> twise is e<g ref="char:EOLhyphen"/>quall to that which is produced of <hi>D</hi> into <hi>A</hi> twise. Wherfore that which is produced of <hi>D</hi> into <hi>A</hi> twise together with the square of <hi>B,</hi> is equall to the square of <hi>A</hi> together with the square of <hi>D</hi>: wherfore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶The 10. proposition.</head>
                  <p>
                     <q>If a number be deuided into two partes, and vnto it be added a number equall to one of the parts: the square of the whole number composed, is equall to the quadruple of that which is produced of the first number into the number added, together with the square of the other part.</q>
                  </p>
                  <p>This proueth in numbers, that which the 8. of the second proued in lines. Suppose that the num<g ref="char:EOLhyphen"/>ber <hi>AB</hi> be deuided into the numbers <hi>AC</hi> and <hi>CB,</hi> vnto which
<figure/>
adde the number <hi>BD,</hi> which let be equall to the number <hi>CB.</hi>
                     <note place="margin">
                        <hi>This answe<g ref="char:EOLunhyphen"/>reth to the</hi> 8. <hi>of the second.</hi>
                     </note> Then I say that the square of the whole number composed, namely, <hi>AD,</hi> is equall to that which is pro<g ref="char:EOLhyphen"/>duced of <hi>AB</hi> into <hi>BD</hi> fower times together with the square of <hi>AC.</hi> For by the 6. of these propositions the square of <hi>AD,</hi> is equall to the square <hi>AB</hi> and to the square of <hi>BD</hi> together with that which is pro<g ref="char:EOLhyphen"/>duced of <hi>AB</hi> into <hi>BD</hi> twise.<note place="margin">
                        <hi>Demonstrati<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> And forasmuch as the square of <hi>BD</hi> is equal to the square <hi>CB</hi>: therfore the square of <hi>AD</hi> shall be equall to the square of <hi>AB</hi> and to the square of <hi>CB</hi> together with that which is produced of <hi>AB</hi> into <hi>BD</hi> twise. But by the former proposition the square of <hi>AB</hi> together with the square of <hi>CB,</hi> is equall to the square of <hi>AC</hi> together with that which is produced of <hi>AB</hi> into <hi>BC</hi> twise wherfore the square of <hi>AD</hi> is equal to that which is produced of <hi>AB</hi> into <hi>BD</hi> twise, and to that which is produced of <hi>AB</hi> into <hi>BC</hi> twise together with the square of <hi>AC.</hi> And for that that which is produced of <hi>AB</hi> into <hi>BC</hi> is equall to that which is produced of <hi>AB</hi> into <hi>BD,</hi> therefore is that manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The 11. proposition.</head>
                  <p>
                     <q>If a number be deuided into two equall partes, and into two vnequall partes: the squares of the two vnequall partes taken together, are double to the square of the halfe, and to the square of the excesse of the greater part aboue the lesse, the sayd two squares being added together.</q>
                  </p>
                  <p>This proueth in numbers that which the 9. of the second proued in lines. For suppose that the number <hi>AB</hi> be deuided into two equall partes: which let be <hi>AC</hi> and <hi>CB,</hi> and into two vnequall parts, which let be <hi>AD</hi> and D<hi>B.</hi>
                     <note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> 9. <hi>of the second.</hi>
                     </note> Then I say that the squares of the two numbers <hi>AD</hi> &amp; <hi>DB,</hi> taken together, are double to the two squares of the two numbers <hi>AC</hi> and <hi>CD,</hi> taken together. For by the 6. of these propositions the square of <hi>AD</hi> is equall to the squares of
<figure/>
                     <hi>AC</hi> and <hi>CD,</hi> and to that which is produced of <hi>AC</hi> into <hi>CD</hi> twise. And forasmuch as the square of <hi>AC</hi> is equal to the square of <hi>CB,</hi> the square of <hi>AD</hi> shal be equall to the square of <hi>BC</hi> &amp; to the square of <hi>CD</hi> together with that which is produced of <hi>BC</hi> into <hi>CD</hi> twise.<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> Wherefore the square of <hi>AD</hi> together with the square of <hi>BD,</hi> is equall to the square of <hi>BC,</hi> and to the square of <hi>CD</hi> and to that which is produced of <hi>BC</hi> into <hi>CD</hi> twise together with the square of <hi>BD.</hi> But that which is produced of <hi>BC</hi> into <hi>CD</hi> twise together with the square of <hi>BD,</hi> is equall to the square of <hi>BC</hi> and to the square of <hi>CD</hi> by the <hi>9.</hi> of these propositions. Wherfore the squares of the two numbers <hi>AD</hi> and <hi>DB</hi> are equall to the squares of the two numbers <hi>BC</hi> and <hi>CD,</hi> taken twise. And therefore the squares of the two numbers <hi>AD</hi> and <hi>DB</hi> are double to the squares of the two numbers <hi>BC</hi> and <hi>CD,</hi> that is <hi>AC</hi> and <hi>CD</hi> (for the numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rs <hi>AC</hi> and <hi>BC</hi> are by supposition equall) wherfore that is manifest which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The 12. proposition.</head>
                  <p>
                     <q>If a number be deuided into two equall partes, and vnto it be added an other number: the square of the whole number composed together with the square of the number added, is double to the square of the halfe, together with the square of the number composed of the halfe and the num<g ref="char:EOLhyphen"/>ber added.</q>
                  </p>
                  <p>
                     <pb facs="tcp:7062:251"/>This proueth in numbers that which the 10. of the second proued in lines. Suppose that the num<g ref="char:EOLhyphen"/>ber <hi>AB</hi> be deuided into two equall numbers <hi>AC</hi> and <hi>CB</hi>:<note place="margin">
                        <hi>This answe<g ref="char:EOLhyphen"/>reth to the</hi> 10. <hi>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the second.</hi>
                     </note> and vnto it adde the number <hi>BD.</hi> Then I say that the square of <hi>AD</hi> together with the square of <hi>BD,</hi> is
<figure/>
double to the square of <hi>AC</hi> together with the square of <hi>CD.</hi> For forasmuch as the number <hi>CD</hi> is deuided into two partes, and vnto it is added the number <hi>AC</hi> which is equall to one of the partes (namely, to <hi>CB</hi>) therefore by the 10. of thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> propositions,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> the square of <hi>AD</hi> is equall to that which is produced of <hi>CD</hi> into <hi>CA</hi> foure times to<g ref="char:EOLhyphen"/>gether with the square of <hi>BD.</hi> And forasmuch as <hi>AC</hi> is equall to <hi>CB,</hi> therefore the square of <hi>AD,</hi> is equall to t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>t which is produced of <hi>DC</hi> into <hi>CB</hi> fower times together with the square of <hi>BD.</hi> Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r<g ref="char:EOLhyphen"/>fore the square of <hi>AD</hi> together with the square of <hi>DB,</hi> is equall to that which is produced of <hi>DC</hi> into <hi>CB</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ower times together with two squares of <hi>BD.</hi> But by the 9. of these propositions, that which is produced of <hi>DC</hi> into <hi>CB</hi> fower times together with two squares of <hi>BD</hi> is double to the square of <hi>CD</hi> together with the square of <hi>CB</hi> (for the square of <hi>CD</hi> together with the square of <hi>CB</hi> is equal to that which is produced of <hi>DC</hi> into <hi>CB</hi> twise together with one square of <hi>CB</hi>). Wherefore forasmuch as the square of <hi>CB</hi> is equall to the square of <hi>AC,</hi> that is mani<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>est: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>The 13. proposition.</head>
                  <p>
                     <q>It is impossible to deuide a number in such sorte: that that which is contayned vnder the whole and one o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the partes, shall be equall to the square of the other part.</q>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>A negatiue propositi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n.</hi>
                     </note>That which the 11. of the second taught to be doone in lines is here proued to be impossible to be doone in numbers. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or suppose that there be a number whatsoeuer namely <hi>A</hi>B. Then I say, that it is impossible to deuide it in such sort as is required in the proposition. For so should it be deuided ac<g ref="char:EOLhyphen"/>cording to a proportion hauing a meane and two ex<g ref="char:EOLhyphen"/>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eames.
<figure/>
But i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be possible,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion lea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ing to an impossibi<g ref="char:EOLhyphen"/>litie.</hi>
                     </note> let the number AB be so deuided in C. And as AB is to BC, so let BC be to C<hi>A.</hi> Wherefore AC shall be lesse then CB. Now then take away from CB a number equall to AC which let be CD. And forasmuch as the proportion of the whole AB to the whole BC, is as the proportion of the part taken away from AB, namely, BC to the part taken away from BC, namely, CD: therefore the proportion of the residue of AB, namely, AC, to the residue of BC, namely, to BD, is as the proportion of the whole AB to the whole BC (by the 11. of the seuenth). Wherefore BC is to CD, as CD is to DB. Wherfore CD is greater then DB. Wherefore sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rahing DE out of CD, so that let DE be equall to DB: the proportion of BC to CD is as the proportio<g ref="char:cmbAbbrStroke">̄</g> of CD to DE. Wherfore the residue of CB, namely, DB, shal be to the residue of CD, namely, to CE, as the proportion of the whole <hi>B</hi>C to the whole CD. Wherfore CE may be subtra<g ref="char:EOLhyphen"/>hed out of ED: wherfore there shalbe no end of this subraction: which is impossible.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 16. Theoreme. The 16. Proposition. If two numbers be prime the one to the other, the second shall not be to any other number, as the first is to the second.</head>
               <p>SVppose that these two numbers A and B be prime the one to the other. Then I say that B is not to any other nu<g ref="char:cmbAbbrStroke">̄</g>ber as A is to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if it be possible, as A is to B, so let B be to C. Now A and B are prime numbers, yea they are prime and the lest by the 23. of the seuenth. But (by the 21. of the seuenth) the least measure the numbers that haue one and the same proportion with them equally, the
<figure/>
antecedent the antecedent, and the consequent the consequent. Where<g ref="char:EOLhyphen"/>fore the antecedent A, measureth the antecedent B, and it measureth also it selfe. Wherfore A measureth these numbers A &amp; B being prime the one to the other, which is impossible. Wherfore as A is to B, so is not B to C: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 17. Theoreme. The 17. Proposition. If there be numbers in continuall proportion how many soeuer, and if theyr
<pb n="222" facs="tcp:7062:251"/>
extremes be prime the one to the other, the lesse shall not be to any other number, as the first is to the second.</head>
               <p>SVppose that there be these numbers in continuall proportion A, B, C, D, and let their extremes A and D be prime the one to the other. Then I say that D is not to any other number as A is to B. For if it be possible, as A is to B, so let D be to E. Wherfore alternately by the <hi>13.</hi> of the seuenth, as A is to D, so is B to E. But A and D are prime,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an abjurditie.</note> yea they are prime and
<figure/>
the least. But the least numbers (by the 21. of the seuenth) measure the numbers that haue one and the same proportio<g ref="char:cmbAbbrStroke">̄</g> with them equal<g ref="char:EOLhyphen"/>ly, the antecedent the antecedent, and the con<g ref="char:EOLunhyphen"/>sequent. Wherefore the antecedent A measu<g ref="char:EOLhyphen"/>reth the antecedent B: but as A is to B, so is B to C. Wherfore B also measureth C. Wherfore A also measureth C (by the 5. commo<g ref="char:cmbAbbrStroke">̄</g> sentence of the seuenth) and forasmuch as B is to C, so is C to D, but B measureth C. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore C mea<g ref="char:EOLhyphen"/>sureth D. But A measureth C. Wherfore A also measureth D by the same common sentence, and it also measureth it selfe. Wherefore A measureth these numbers A and D being prime the one to the other, which is impossible. Wherfore D is not to any other number as A is to B: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 18. Theoreme. The 18. Proposition. Two numbers being geuen, to searche out if it be possible a third number in proportion with them.</head>
               <p>SVppose that the two numbers geuen be A and B. It is required to searche out if it be possible a third number proportionall with them. Now A, B are either prime the one to the other or not prime.<note place="margin">Three cases in this proposi<g ref="char:EOLhyphen"/>tion.</note> If they be prime, then (by the 16. of the ninth) it is mani<g ref="char:EOLhyphen"/>nifest that it is impossible to finde out a third number proportional with them. But now sup<g ref="char:EOLhyphen"/>pose that AB be not prime the one to the other. And let B multiplieng himselfe produce C. Now A either measureth C, or measureth it not. First,<note place="margin">The first case.</note> let it measure it and that by D. Wher<g ref="char:EOLhyphen"/>fore A multiplieng D produceth
<figure/>
C.<note place="margin">The second case.</note> But B also multiplieng himself produced C. Wherfore that which is produced of A into D, is equall to that which is produced of B in<g ref="char:EOLhyphen"/>to himselfe. Wherefore (by the se<g ref="char:EOLhyphen"/>cond part of the <hi>19.</hi> of the seue<g ref="char:cmbAbbrStroke">̄</g>th) as A is to B, so is B to D. Wherfore vnto these numbers A, B is found out a third number in proportion, namely, D.</p>
               <p>But now suppose that A do not measure C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">The third case.</note> Then I say that it is impossible to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inde out a third nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber in proportion with these num<g ref="char:EOLhyphen"/>bers A, B. For if it be possible, let there be found out such a number, and let the same be D.
<pb facs="tcp:7062:252"/>
Wherfore that which is produced of A into D, is equall to that which is produced of B into himselfe, but that which is produced of B into himselfe is C. Wherfore that which is produced of A into D is equall vnto C. Wherfore A multiplieng D produced C. Wherefore A measu<g ref="char:EOLhyphen"/>reth G by D. But it is supposed also not to measure it, which is impossible. Wherefore it is not possible to finde out a third number in proportion with A &amp; B, whensoeuer A measureth not C: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 19. Theoreme. The 19. Proposition. Three numbers beyng geuen, to search out if it be possible the fourth num<g ref="char:EOLhyphen"/>ber proportionall with them.</head>
               <p>
                  <note place="margin">Diuert cases <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n this proposi<g ref="char:EOLhyphen"/>tion.</note>SVppose that the three numbers geuen be A, B, C. It is required to search out if it be possible a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ourth number proportionall with them. Now A, B, C, are either in continuall proportion, and their extremes A, C are prime the one to the other or they are not in continuall proportion, and their extremes are yet prime the one to the other: or they are in continuall proportion, and their extremes are not prime the one to the other: or they are neither in continuall proportion, nor their extremes are prime the one to the other.</p>
               <p>
                  <note place="margin">The first case.</note>If A, B, C, be in continuall proportion, and their extremes be prime the one to the other, it is manifest (by the <hi>17.</hi> of the ninth) that it is impossible to finde out a fourth number pro<g ref="char:EOLhyphen"/>portionall with them.</p>
               <p>But now suppose that A, B, C, be not in continuall
<figure/>
proportion,<note place="margin">The second case.</note> and yet let their extremes be prime the one to the other. Then I say that so also it is impossible to finde out a fourth number proportional with the<g ref="char:cmbAbbrStroke">̄</g>. For if it be possible, let there be found such a number, and let the same be D. So that as A is to B, so let C be to D, and as B is to C, so let D bet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> E. And for that as A is to B, so is C to D, and as B is C, so is D to E, therfore of equallitie (by the <hi>14.</hi> of the se<g ref="char:EOLhyphen"/>uenth) as A is to C, so is C to E. But A and C are prime
<figure/>
the one to the other, yea they are prime, and the least: but the lest measure the numbers that haue one &amp; the same proportion with them equally, the antecedent, the ante<g ref="char:EOLhyphen"/>cedent, and the consequent the consequent (by the <hi>21.</hi> of the seuenth). Wherfore A measureth C, namely, the an<g ref="char:EOLhyphen"/>tecedent the antecedent, and it also measureth it selfe. Wherfore A measureth these numbers A and C being prime the one to the other, which is impossible. Wherfore it is not possible to finde out a fourth number proportionall with these numbers A, B, C.</p>
               <p>But now againe suppose that A, B, C, be in
<figure/>
co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tinuall proportion,<note place="margin">The third case.</note> and let A and C not be prime the one to the other. Then I say that it is pos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ible to finde out a fourth number pro<g ref="char:EOLhyphen"/>portionall with them. For let B multiplieng C produce D. Now A either measureth D, or measureth it not. First let it measure it, and that by E. Wherfore A multiplieng E produced D. But B also multiplieng C produced D. Wherfore that which is produced of AE is equal to that which is produced of BC: wherfore in what proportio<g ref="char:cmbAbbrStroke">̄</g> A is to B, in y<hi rend="sup">e</hi> same is C to E. Wher<g ref="char:EOLunhyphen"/>fore there is found out a fourth number, namely, E, proportionall with these nu<g ref="char:cmbAbbrStroke">̄</g>bers A, B, C.</p>
               <p>
                  <pb n="232" facs="tcp:7062:252"/>But now suppose that A do not measure D. Then I say that it is not possible to finde out a fourth number proportionall with these numbers A, B, C. For if it be possible, let there be found such a number, and let the same be E. Wherfore that which is produced of A into E is equall to that which is produced of B into C. But that which is produced of B into C is D. Wherfore that which is produced of A into E is equall vnto D. Wherefore A multiplieng E produced D, wherfore A measureth D, but it also measureth it not, which is impossible. Wher<g ref="char:EOLunhyphen"/>fore it is impossible to finde out a fourth number proportionall, with these numbers A, B, C, whensoeuer A measureth not D.</p>
               <p>But now suppose
<figure/>
that A, B,<note place="margin">The fourth case.</note> C be nei<g ref="char:EOLhyphen"/>ther in continuall proportio<g ref="char:cmbAbbrStroke">̄</g>, neither al<g ref="char:EOLunhyphen"/>so their extremes be prime the one to the other. And let B mul<g ref="char:EOLunhyphen"/>tiplieng C produce D. And in like sorte may we proue that if A do measure D, it is possible to finde out a fourth number proportionall with them. But if it do not measure D, the<g ref="char:cmbAbbrStroke">̄</g> is it vn<g ref="char:EOLhyphen"/>possible: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Theoreme. The 20. Proposition. Prime numbers being geuen how many soeuer, there may be geuen more prime numbers.</head>
               <p>SVppose that the prime numbers geuen be A, B, C.<note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> Then I say, that there are yet more prime numbers besides A, B, C. Take (by the 38. of the seuenth) the lest number whom these numbers A, B, C do measure, and let the same be DE. And vnto DE adde vnitie DF. Now EF is either a prime number or not. First let it be a prime number,<note place="margin">The first case.</note> then are there found
<figure/>
these prime numbers A, B, C, and EF more in multi<g ref="char:EOLhyphen"/>tude then the prime numbers <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst geuen A, B, C.</p>
               <p>But now suppose that EF be not prime.<note place="margin">The second case.</note> Wherefore some prime number measureth it (by the 24. of the se<g ref="char:EOLhyphen"/>uenth). Let a prime number measure it, namely, G. Then I say, that G is none of these numbers A, B, C. For if G be one and the same with any of these A, B, C. But A, B, C, measure the nu<g ref="char:cmbAbbrStroke">̄</g>ber DE: wher<g ref="char:EOLhyphen"/>fore G also measureth DE: and it also measureth the whole EF. Wherefore G being a num<g ref="char:EOLhyphen"/>ber shall measure the residue DF being vnitie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which is impossible. Wherefore G is not one and the same with any of these prime numbers A, B, C: and it is also supposed to be a prime number. Wherefore there are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ound these prime numbers A, B, C, G, being more in multitude then the prime numbers geuen A, B, C: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>By thys Proposition it is manifest, that the multitude of prime numbers is infinite.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 21. Theoreme. The 21. Proposition. If euen nu<g ref="char:cmbAbbrStroke">̄</g>bers how many soeuer be added together: the whole shall be eue<g ref="char:cmbAbbrStroke">̄</g>.</head>
               <p>
                  <pb facs="tcp:7062:253"/>SVppose that these euen numbers AB, BC; CD, and DE, be added together. Then I say, that the whole number, namely, AE, is an euen number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For foras<g ref="char:EOLhyphen"/>much as euery one of these numbers AB, BC, CD, and DE, is an euen num<g ref="char:EOLhyphen"/>ber, therefore euery one of them hath an halfe. Wherefore the whole AE also hath an halfe. But an euen
<figure/>
number (by the definition) is that which may be deuided in<g ref="char:EOLhyphen"/>to two equall partes. Where<g ref="char:EOLhyphen"/>fore AE is an euen number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 22. Theoreme. The 22. Proposition. If odde numbers how many soeuer be added together, &amp; if their multitude be euen, the whole also shall be euen.</head>
               <p>SVppose that these odde numbers AB, BC, CD, and DE, being euen in multitude, be added together. Then I say, that the whole AE is an euen number. For foras<g ref="char:EOLhyphen"/>much as euery one of these numbers AB, BC, CD, and DE, is an odde number, is ye take away vnitie from e<g ref="char:EOLhyphen"/>uery
<figure/>
one of them,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> that which remayneth o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> euery one of the<g ref="char:cmbAbbrStroke">̄</g> is an euen number. Wherefore they all added together, are (by the 21. of the ninth) an euen number: and the multitude of the vnities taken away is euen. Wherefore the whole AE is an euen number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 23. Theoreme. The 23. Proposition. If odde numbers how many soeuer be added together, and if the multitude of them be odde, the whole also shall be odde.</head>
               <p>SVppose that these odde numbers, AB, BC, and CD being odde in multitude be ad<g ref="char:EOLhyphen"/>ded together. Then I say that the whole AD is an odde number. Take away from CD, vnitie DE, wherefore that which remaineth CE is an euen number. But AC also (by the 22. of the
<figure/>
ninth) is an euen num<g ref="char:EOLhyphen"/>ber.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion</note> Wherfore the whole AE is an euen number. But DE which is vnitie being added to the euen number AE, maketh the whole AD a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> odde number: which was required to be proued<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </p>
            </div>
            <div type="part">
               <head>¶ The 24. Theoreme. The 24. Proposition. If from an euen number be take<g ref="char:cmbAbbrStroke">̄</g> away an euen number, that which remai<g ref="char:EOLhyphen"/>neth shall be an euen number.</head>
               <p>SVppose that AB be an euen number, and from
<figure/>
it take away an euen number CB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that that which remayneth, namely, AC is an euen number. For forasmuch as AB is an euen
<pb n="224" facs="tcp:7062:253"/>
euen number, it hath an halfe, and by the same reason also BC hath an halfe. Wherfore the residue CA hath an halfe. Wherfore AC is an euen number: which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 25. Theoreme. The 25. Proposition. If from an euen number be taken away an odde number, that which remai<g ref="char:EOLhyphen"/>neth shall be an odde number.</head>
               <p>SVppose that AB be an euen number, and
<figure/>
take away from it BC an odde number. Then I say that the residue CA is an odde number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Take away from BC vnitie CD. Wherfore DB is an euen number. And AB also is an euen number, wherefore the residue AD is an euen num<g ref="char:EOLhyphen"/>ber (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ormer proposition) But CD which is vnitie, being taken away from the euen nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber AD maketh the residue AC an odde number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 26. Theoreme. The 26. Proposition. If from an odde number be taken away an odde number, that which re<g ref="char:EOLhyphen"/>mayneth shall be an euen number.</head>
               <p>SVppose that AB be an odde number, and from it
<figure/>
take away an odde number BC. The<g ref="char:cmbAbbrStroke">̄</g> I say that the residue CA is an euen number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as AB is an odde number, take away from it vnitie BD. Wherfore the residue AD is euen. And by the same reason CD is an euen number: wherfore the residue CA is an euen number (by the 24. of this booke)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 27. Theoreme. The 27. Proposition. If from an odde number be taken a way an euen number, the residue shall be an odde number.</head>
               <p>SVppose that AB be an odde number, and from it
<figure/>
take away an euen number BC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that the residue CA is an odde number. Take away fro<g ref="char:cmbAbbrStroke">̄</g> AB vnitie AD. Wherfore the residue DB is an eue<g ref="char:cmbAbbrStroke">̄</g> number, &amp; BC is (by supposition) euen. Wherfore the residue CD is an euen number. Wherefore DA which is vnitie, beyng added vnto CD which is an euen number maketh the whole AC an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dde number: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 28. Theoreme. The 28. Proposition. If an odde number multiplieng an euen number produce any number, the number produced shall be an euen number.</head>
               <p>
                  <pb facs="tcp:7062:254"/>SVppose that A being an odde number multiplieng B
<figure/>
being an euen number, do produce the number C. Then I say that C is an euen number.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as A multiplieng B produced C, therfore C is composed of so many numbers equall vnto B as there be in vnities in A. But B is an euen nu<g ref="char:cmbAbbrStroke">̄</g>ber: wherfore C is composed of so many euen numbers, as there are vnities in A. But if eue<g ref="char:cmbAbbrStroke">̄</g> numbers how ma<g ref="char:EOLhyphen"/>ny soeu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r be added together, the whole (by the 21. of the ninth) is an euen number: wherfore C is an euen number: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 29. Theoreme. The 29. Proposition. I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> an odde number multiplying an odde number produce any number, the number produced shalbe an odde number</head>
               <p>SVppose that A being an odde number multiplying B being also an odde number, doo produce the number C. Then I say that C is an odde number. For forasmuch as A multiplying B produced C, therefore C is composed of so many numbers equall vnto B as there be vnities in A.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> But either of these num<g ref="char:EOLunhyphen"/>bers
<figure/>
A and B is an odde number. Wherefore C is com<g ref="char:EOLhyphen"/>posed of odde numbers, whose multitude also is odde. Wherfore (by the 23. of the ninth) C is an odde nu<g ref="char:cmbAbbrStroke">̄</g>ber: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A proposition added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>A proposition added by Campa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e.</hi>
                     </note>
                     <hi>If an odde number measure an euen number, it shall measure it by an euen number.</hi>
                  </p>
                  <p>For if it should measure it by an odde number, then of an odde number multiplyed into an odde number should be produced an odde number, which by the former proposition is impossible.</p>
               </div>
               <div type="section">
                  <head>An other proposition added by him.</head>
                  <p>
                     <note place="margin">
                        <hi>An other ad<g ref="char:EOLhyphen"/>ded by him.</hi>
                     </note>
                     <hi>If an odde number measure an odde number, it shall measure it by an odde number.</hi>
                  </p>
                  <p>For if it should measure it by an euen number, then of an odde number multiplyed into an euen number should be produced an odde number which by the 28. of this booke is impossible.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 30. Theoreme. The 30. Proposition. If an odde number measure an euen number, it shall also measure the halfe thereof.</head>
               <p>SVppose that A being an odde number doo measure B being an euen number. Th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> I say that it shall measure the halfe thereof. For forasmuch as A measureth B let i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> measure it by C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> The<g ref="char:cmbAbbrStroke">̄</g> I say that C is an euen number. For if not then, if it be possible le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be odde. And forasmuch as A measureth B by C: ther<g ref="char:EOLhyphen"/>fore
<figure/>
A multiplying C produceth B. Wherfore B is composed of odde numbers whose multitude also is odde. Wherfore B is an odde number (by the <hi>29.</hi> of this booke) which is ab<g ref="char:EOLhyphen"/>surd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for it is supposed to be euen: wherefore C is an euen num<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er. Wherefore A measureth B by an euen number: and C measureth B by A. But either
<pb n="225" facs="tcp:7062:254"/>
of these numbers C and B hath an halfe part wherfore as C is to B, so is the halfe to the halfe. But C measureth B by A. Wherefore the halfe of C measureth the halfe of B by A: wherfore A multiplying the halfe of C produceth the halfe of B. Wherfore A measureth the halfe of B: and it measureth it by the halfe of C. Wherefore A measureth the halfe of the number B: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 31. Theoreme. The 31. Proposition. If an odde number be prime to any number, it shal also be prime to the dou<g ref="char:EOLhyphen"/>ble thereof.</head>
               <p>SVppose that A being an odde number be prime vnto the number B: and let the dou<g ref="char:EOLhyphen"/>ble of B be C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say, that A is prime vnto C. For if A and C be not prime the one to the other, some one number measureth them both. Let there be such a num<g ref="char:EOLhyphen"/>ber which measureth them both, and let the same be D.
<figure/>
But A is an odde number. Wherefore D also is an odde number. (For if D which measureth A should be an euen number, then should A also be an euen number (by the 21. of this booke): which is co<g ref="char:cmbAbbrStroke">̄</g>trary to the suppositi<g ref="char:EOLhyphen"/>on. For A is supposed to be an odde nu<g ref="char:cmbAbbrStroke">̄</g>ber: &amp; therefore D also is an odde number). And forasmuch as D being an odde number measureth C, but C is an eue<g ref="char:cmbAbbrStroke">̄</g> number (for that it hath an halfe, namely, B). Wherfore (by the Proposition next going before) D measureth the halfe of C. But the halfe of C is B. Wherefore D measureth B: and it also measureth A. Wherefore D measureth A and B being prime the one to the o<g ref="char:EOLhyphen"/>ther: which is absurde. Wherefore no number measureth the numbers A &amp; C. VVherfore A is a prime number vnto C. VVherefore these numbers A and C are prime the one to the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 32. Theoreme. The 32. Proposition. Euery nu<g ref="char:cmbAbbrStroke">̄</g>ber produced by the doubling of two vpward, is euenly euen onely.</head>
               <p>SVppose that A be the number two: and from A vpward double numbers how many soeuer; as B, C, D. Then I say, that B, C, D, are numbers euenly euen onely. That e<g ref="char:EOLhyphen"/>uery one of them is euenly euen, it is manifest:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> for euery one of them is produced by the doubling of two. I say also, that euery one of them is euenly
<figure/>
euen onely. Take vnitie E. And forasmuch as from vnitie are certaine numbers in continuall proportion, &amp; A which follow<g ref="char:EOLhyphen"/>eth next after vnitie is a prime number, therefore (by the 13. of the third) no number measureth D being the greatest number of these numbers A, B, C, D, besides the selfe same numbers in proportion. But euery one of these numbers A, B, C, is euenly euen. VVherefore D is euenly euen onely. In like sort may we proue, that euery one of these numbers A, B, C, is euenly euen onely: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 33. Theoreme. The 33. Proposition. A number whose halfe part is odde, is euenly odde onely.</head>
               <p>
                  <pb facs="tcp:7062:255"/>SVppose that A be a number whose halfe part is odde. Then I say that A is euenly od onely. That it is euenly odde it is manifest: for his halfe being odde measureth him by an eue<g ref="char:cmbAbbrStroke">̄</g> number, namely, by <hi>2.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> (by the defini<g ref="char:EOLhyphen"/>tion).
<figure/>
I say also that it is euenly odde onely. For if A be euen<g ref="char:EOLhyphen"/>ly euen, his halfe also is euen. For (by the definition) an euen number measureth him by an euen number. Wherefore that euen number which measureth him by an euen number shall also measure the halfe thereof being an odde number by the <hi>4.</hi> common sentence of the seuenth which is absurd. Wherfore A is a number euenly odde onely: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration to proue the same.</head>
                  <p>Suppose that the number A haue to his halfe an od nu<g ref="char:cmbAbbrStroke">̄</g>ber, namely, B. The<g ref="char:cmbAbbrStroke">̄</g> I say that A is eue<g ref="char:cmbAbbrStroke">̄</g>ly od onely. That it is euenly odde needeth no profe: forasmuch as the number 2. an euen number measu<g ref="char:EOLhyphen"/>reth it by the halfe thereof which is an odde number.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration.</hi>
                     </note> Let C be the number 2. by which <hi>B</hi> measureth A
<figure/>
(for that A is supposed to be double vnto <hi>B</hi>). And let an euen number, namely, D measure A (which is possible for that A is an euen number by the definition) by F. And forasmuch as that which is produced of C into B is equall to that which is produced of D into F, therefore by the 19. of the seuenth, as C is to D, so is <hi>B</hi> to F. But C the number two measureth D being an euen number: wherfore F also measureth B which is the halfe of A. Wherfore F is an odde number. For if F were an euen number then should it in the <hi>B</hi> whome it measureth an odde number also by the 21. of this booke, which is contrary to the supposition. And in like maner may we proue that all the eue<g ref="char:cmbAbbrStroke">̄</g> nu<g ref="char:cmbAbbrStroke">̄</g>bers which measure the number A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> do measure it by odde numbers. Wherefore A is a number euenly odde onely: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 34. Theoreme. The 34. Proposition. If a number be neither doubled from two, nor hath to his half part an odde number, it shall be a number both euenly euen, and euenly odde.</head>
               <p>
                  <hi>SVppose that the nu<g ref="char:cmbAbbrStroke">̄</g>ber A be a nu<g ref="char:cmbAbbrStroke">̄</g>ber neither doubled fro<g ref="char:cmbAbbrStroke">̄</g> the nu<g ref="char:cmbAbbrStroke">̄</g>ber two, neither also let it haue to his halfe part an odde nu<g ref="char:cmbAbbrStroke">̄</g>ber. Then I say that A is a nu<g ref="char:cmbAbbrStroke">̄</g>ber both euenly euen, and euenly odde.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>That A is euenly euen it is manifest, for the halfe therof is not odde, and is measured by the number 2. which is an euen number. Now I say that it is euenly odde also. For if we deuide A into two equall partes, and so conti<g ref="char:EOLhyphen"/>nuing still, we shall at the length light vpon a certaine
<figure/>
odde number which shall measure A by an euen num<g ref="char:EOLhyphen"/>ber. For if we should not light vpon such an odde nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber, which measureth A by an euen number, we should at the length come vnto the number two, and so should A be one of those numbers which are doubled from two vpward, which is contrary to the supposition. Wherfore A is euenly odde. And it is proued that it is euenly eue<g ref="char:cmbAbbrStroke">̄</g>: wherfore A is a number both euenly euen and euenly odde: whiche was required to be de<g ref="char:EOLhyphen"/>monstrated.</hi>
               </p>
               <p>This proposition and the two former manifestly declare that which we noted vp<g ref="char:EOLunhyphen"/>pon the tenth definition of the seuenth booke namely, that <hi>Campane</hi> and <hi>Flussates</hi> and diuers other interpreters of <hi>Euclide</hi> (onely <hi>Theon</hi> except) did not rightly vnderstand the 8. and 9. definitions of the same booke concerning a number euenly euen, and a num<g ref="char:EOLhyphen"/>ber euenly odde. For in the one definition they adde vnto <hi>Euclides</hi> wordes extant in
<pb n="226" facs="tcp:7062:255"/>
the Greeke this word <hi>onely</hi> (as we there noted) and in the other this word <hi>all.</hi> So that af<g ref="char:EOLhyphen"/>ter their definitions a number can not be euenly euen vnlesse it be measured onely by euen numbers: likewise a number can not be euenly odde vnlesse all the euen numbers which doo measure it, doo measure it by an odde number. The contrary whereof in this proposition we manifestly see. For here <hi>Euclide</hi> proueth that one number may be both euenly euen and euenly odde. And in the two former propositions he proued that some numbers are euenly euen onely, and some euenly odde onely: which word onely had bene in vaine of him added, if no number euenly euen could be measured by an odde number, or if all the numbers that measure a number euenly odde must needes measure it by an odde number. Although <hi>Campane</hi> and <hi>Flussates</hi> to auoyde this absurdi<g ref="char:EOLhyphen"/>ty haue wreasted the 32. proposition of this booke fro<g ref="char:cmbAbbrStroke">̄</g> the true sence of the Greeke and as it is interpreted of <hi>Theon.</hi> So also hath <hi>Flussates</hi> wreasted the 33. proposition. For wheras <hi>Euclide</hi> sayth, <hi>Euery nu<g ref="char:cmbAbbrStroke">̄</g>ber produced by the doubling of two vpward, is eue<g ref="char:cmbAbbrStroke">̄</g>ly eue<g ref="char:cmbAbbrStroke">̄</g> only:</hi> they say, <hi>onely the numbers produced by the doubling of two, are euenly euen.</hi> Likewise whereas <hi>Euclide</hi> saith, <hi>A number whose hafle part is odde, is euenly odde onely, Flussates</hi> sayth, <hi>onely a number whose halfe part is od, Is eue<g ref="char:cmbAbbrStroke">̄</g>ly od.</hi> Which their interpretatio<g ref="char:cmbAbbrStroke">̄</g> is not true, neither can be applyed to the propositions as they are exta<g ref="char:cmbAbbrStroke">̄</g>t in the Greeke. In dede the sayd 32. and 33. propositi<g ref="char:EOLhyphen"/>ons as they put the<g ref="char:cmbAbbrStroke">̄</g> are true touching those numbers which are euenly euen onely, or eue<g ref="char:cmbAbbrStroke">̄</g>ly od onely. For no number is euenly euen onely, but those onely which are doubled from two vpward. Likewise no numbers are euenly odde onely, but those onely whose halfe is an odde number. But this letteth not, but that a number may be euenly euen although it be not doubled from two vpward &amp; also that a number may be eue<g ref="char:cmbAbbrStroke">̄</g>ly odde although it haue not to his halfe an odde number. As in this 34. propositio<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>Euclide</hi> hath plainly proued. Which thing could by no meanes be true, if the foresayd 32. &amp; 33. pro<g ref="char:EOLhyphen"/>positons of this booke should haue that sence and meaning wherein they take it.</p>
            </div>
            <div type="part">
               <head>¶ The 35. Theoreme. The 35. Proposition. If there be numbers in continuall proportion how many soeuer, and if from the second and last be taken away numbers equall vnto the first, as the ex<g ref="char:EOLhyphen"/>cesse of the second is to the first, so is the excesse of the last to all the nu<g ref="char:cmbAbbrStroke">̄</g>bers going before the last.</head>
               <p>SVppose that these numbers A, BC, D, and EF, be in continuall proportion be<g ref="char:EOLhyphen"/>ginning at A the least. And from BC, which is the second, take away CG equall vnto the first, namely, to A, and likewise from EF the last take away FH e<g ref="char:EOLhyphen"/>quall also vnto the first, namely, to A. Then I say, that as the excesse BG is to A the first, so is HE the excesse, to all the numbers D, BC, and A, which go before the last number, namely, EF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Forasmuch as EF is the greater (for the second is supposed greater then the first) put the number FL equall to the number D, and likewise the number FK equall to the number BC. And forasmuch as FK is equall vnto CB, of which FH is equall vnto GC, therefore the residue HK is equall vnto the residue GB. And for that as the whole F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is to the whole FL, so is the part taken away FL, to the part taken away FK, therefore the residue LE is to
<figure/>
the residue KL, as the whole <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E is to the whole FL (by the <hi>11.</hi> of the seuenth). So likewise for that FL is to FK, as FK is to FH, KL shall be to HK, as the whole FL is to the whole FK (by the same Proposition). But as FE is to FL, and as FL is to FK, and FK to FH, so were FE to D, and D to BC, and BC <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> A. Wherefore as LE is to KL, and as KL is to HK, so is D to BC. Wherefore alter<g ref="char:EOLhyphen"/>nately (by the <hi>23.</hi> of the seuenth) as LE is to D, so is KL to be BC, and as KL is to BC, so is HK to A. Wherefore also as one of the antecedentes is to one of the consequentes; so are
<pb facs="tcp:7062:256"/>
all the antecedentes to all the consequentes. Wherefore as KH is to A, so are HK, KL, and LE, to D, BC, and A (by the <hi>12.</hi> of the seuenth). But it is proued, that KH is e<g ref="char:EOLhyphen"/>quall vnto BG. Wherefore as BG, which is the excesse of the second, is to A, so is EH the excesse of the last vnto the numbers going before D, BC, and A. Wherefore as the excesse of the second is vnto the first, so is the excesse of the last to all the numbers going before the last: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 36. Theoreme. The 36. Proposition. If from vnitie be taken numbers how many soeuer in double proportion continually, vntill the whole added together be a prime number, and if the whole multiplying the last produce any number, that which is produced is a perfecte number.</head>
               <p>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion teach<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th how to finde out a perfect number.</note>SVppose that from vnitie be taken these numbers A, B, C, D, in double proportion continually, so that all those numbers A, B, C, D, &amp; vnitie added together, make a prime number: and let E be the number composed of all those numbers A, B, C, D, &amp; vnitie added together: and let E multiplying D, which is the last num<g ref="char:EOLhyphen"/>ber, produce the number FG. Then I say, that FG is a perfect number.</p>
               <p>
                  <figure/>
               </p>
               <p>
                  <note place="margin">Construction.</note>How many in multitude A, B, C, D, are, so many in continuall double proportion take be<g ref="char:EOLhyphen"/>ginning at E, which let be the numbers E, HK, L, and M. VVherefore of equalitie (by the 13. of the seuenth) as A is to D, so is E to M. VVherefore that which is produced of E into D, is equall to that which is produced of A into M. But that which is produced of E in<g ref="char:EOLhyphen"/>to D, is the number FG. VVherefore that which is produced of A into M, is equall vnto FG. VVherefore A multiplying M produceth FG. VVherefore M measureth FG by those vnities which are in A. But A is the number two. VVherefore FG is double to M. And the num<g ref="char:EOLhyphen"/>bers M, L, HK, and E, are also in continuall double proportion. VVherefore all the num<g ref="char:EOLhyphen"/>bers E, HK, L, M, and FG, are continually proportionall in double proportion. Take from the second number KH, and from the last FG a number equall vnto the first, namely, to E: and let those numbers taken be HN, &amp; FX.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> VVherefore (by the Proposition going before) as the excesse of the second number is to the first number, so is the excesse of the last to all the
<pb n="227" facs="tcp:7062:256"/>
numbers going before it. VVherefore as NK is to E, so is XG to these numbers M, L, KH, and E. But NK is equall vnto E (for it is the halfe of HK, which is supposed to be double to E). VVherefore XG is equall vnto these numbers M, L, HK, and E. But XF is equall vnto E, and E is equall vnto these numbers A, B, C, D, and vnto vnitie. Wherfore the whole number FG is equall vnto these numbers E, HK, L, M,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> and also vnto these num<g ref="char:EOLhyphen"/>bers A, B, C, D, and vnto vnitie. Moreouer I say, that vnitie and all the numbers A, B, C, D, E, HK, L, and M, do measure the number FG. That vnitie measureth it, it needeth no proufe. And forasmuch as FG is produced of D into E, therefore D and E do measure it. And forasmuch as the double from vnitie, namely, the nu<g ref="char:cmbAbbrStroke">̄</g>bers A, B, C, do measure the num<g ref="char:EOLhyphen"/>ber D (by the 13. of this booke) therefore they shall also measure the number FG (whom D measureth) by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. common sentence. By the same reason forasmuch as the nu<g ref="char:cmbAbbrStroke">̄</g>bers E, HK,
<figure/>
L, and M, are vnto FG, as vnitie and the numbers A, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, C, are vnto D (namely, in subdu<g ref="char:EOLhyphen"/>ple proportion) and vnitie and the numbers A, B, C, do <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> D, therefore also the num<g ref="char:EOLhyphen"/>bers E, HK, L, and M, shall measure the number FG: No<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> I say also, that no other num<g ref="char:EOLhyphen"/>ber measureth FG besides these numbers A, B, C, D, E, HK, L, M, and vnitie. For if it be possible, let O measure FG. And let O not be any of these numbers A, B, C, D, E, HK, L, and M. And how often O measureth FG, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> vnities let there be in P. Wherefore O mul<g ref="char:EOLhyphen"/>tiplying P produceth FG. But E also multiplying D produced FG. Wherefore (by the 19. of the seuenth) as E is to O, so is P to D. Wherefore alternately (by the 9. of the seuenth) as E is to P, so is O to D. And forasmuch as from vnitie are these numbers in continuall pro<g ref="char:EOLhyphen"/>portion A, B, C, D, and the number A which is next after vnitie is a prime number, therfore (by the <hi>13.</hi> of the ninth) no other number measureth D besides the numbers A, B, C. And it is supposed that O is not one and the same with any of these nu<g ref="char:cmbAbbrStroke">̄</g>bers A, B, C. Wherefore O mea<g ref="char:EOLhyphen"/>sureth not D. But as O is to D, so is E to P. Wherefore neither doth E measure P. And E is a prime number. But (by the 31. of the seuenth) euery prime number, is to euery number that it measureth not, a prime number. Wherefore E and P are prime the one to the other: yea they are prime and the least. But (by the <hi>21.</hi> of the seuenth) the least measure the numbers that haue one and the same proportion with them equally, the antecedent the antecedent, and the consequent the consequent. And as E is to P, so is O to D. Wherefore how many times E measureth O, so many times P measureth D. But no other number measureth D besides the numbers A, B, C (by the <hi>13.</hi> of this booke). Wherefore P is one and the same with one of these numbers A, B, C. Suppose that P be one and the same with B, &amp; how many B, C, D, are
<pb facs="tcp:7062:257"/>
in multitude, so many take from E vpward, namely, E, HK, and L. But E, HK, and L, are in the same proportion that B, C, D, are. VVherefore of equalitie, as B is to D, so is E to L. VVherefore that which is produced of B into L, is equall to that which is produced of D into E. But that which is produced of D into E, is equall to that which is produced of P into O. VVherefore that which is produced of P into O, is equall to that which is produced of B into L. VVherefore as P is to B, so is L to O: and P is one &amp; the same with B: wherefore L also is one and the same with O: which is impossible. For O is supposed not to be one and the same with any of the numbers geuen. VVherefore no num<g ref="char:EOLhyphen"/>ber measureth FG besides these numbers A, B, C, D, E, HK, L, M, and vnitie. And it is proued, that FG is equall vnto these num<g ref="char:EOLhyphen"/>bers A, B, C, D, E, HK, L, M, and vnitie, which are the partes therof (by the 39. of the seuenth). But a perfect nu<g ref="char:cmbAbbrStroke">̄</g>ber (by the definition) is that which is equall vnto all his partes. VVherfore FG is a perfect number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <trailer>The end of the ninth booke of Euclides Elementes.</trailer>
         </div>
         <div n="10" type="book">
            <pb n="228" facs="tcp:7062:257"/>
            <head>¶The tenth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>IN THIS TENTH BOOKE doth <hi>Euclide</hi> entreat of lines and other magnitudes rationall &amp; irrationall,<note place="margin">
                  <hi>The Argume<g ref="char:cmbAbbrStroke">̄</g>t of the tenth booke.</hi>
               </note> but chiefly of irrationall magnitudes, commensura<g ref="char:EOLhyphen"/>ble and incommensurable: of which hitherto, in al his former 9. bokes he hath made no mention at all. And herein differeth number from magnitude,<note place="margin">
                  <hi>Difference betwene num<g ref="char:EOLunhyphen"/>ber and mag<g ref="char:EOLhyphen"/>nitude.</hi>
               </note> or Arith<g ref="char:EOLhyphen"/>metike from Geometry: for that although in Arith<g ref="char:EOLhyphen"/>metike, certayne numbers be called prime numbers in consideration of themselues, or in respect of an other, and so are called incommensurable, for that no one number measureth them, but onely vnitie. Yet in dede and to speake absolutely and truely, there are no two numbers incommensurable, but haue one common measure which measureth the<g ref="char:cmbAbbrStroke">̄</g> both, if none other, yet haue they vnitie, which is a common part and measure to all numbers, and all numbers are made of vnities, as of their partes.<note place="margin">
                  <hi>A line is not made of points as number is made of vni<g ref="char:EOLhyphen"/>ties.</hi>
               </note> As hath before bene shewed in the declaration of the definitio<g ref="char:cmbAbbrStroke">̄</g>s of the seuenth booke. But in mag<g ref="char:EOLhyphen"/>nitude it is farre otherwise, for although many lines, plaine figures, and bodies, are commensurable, and may haue one measure to measure them, yet all haue not so, nor can haue. For that a line is not made of pointes, as number is made of vnities, and therfore cannot a point be a common part of all lines, and measure them, as v<g ref="char:EOLhyphen"/>nitie is a common part of all numbers and measureth them. Vnitie taken certayne tymes maketh any number. For there are not in any number infinite vnities: but a point taken certayne tymes, yea as often as ye list, neuer maketh any line, for that in euery line there are infinite pointes. Wherfore lines, figures, and bodies in Geo<g ref="char:EOLhyphen"/>metry, are oftentymes incommensurable and irrationall. Now which are rationall, and which irrationall, which commensurable, and which incommensurable, how many and how sundry sortes and kindes there are of them, what are their natures, passions, and properties, doth Euclide most manifestly shew in this booke, and de<g ref="char:EOLhyphen"/>monstrate them most exactly.</p>
            <p>This tenth booke hath euer hitherto of all men, and is yet thought &amp; accomp<g ref="char:EOLhyphen"/>ted, to be the hardest booke to vnderstand of all the bookes of Euclide.<note place="margin">
                  <hi>This booke the hardest to vn<g ref="char:EOLhyphen"/>derstand of all the bookes of Euclide.</hi>
               </note> Which co<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>mon receiued opinion, hath caused many to shrinke, and hath (as it were) deterred them from the handeling and treatie thereof. There haue bene in deede in times past, and are presently in these our dayes, many which haue delt, and haue taken great and good diligence in commenting, amending, and restoryng of the sixe first bookes of Euclide, and there haue stayed themselues and gone no farther, beyng deterred and made afrayde (as it seemeth, by the opinion of the hardnes of this booke) to passe forth to the bookes following. Truth it is that this booke hath in it somewhat an other &amp; strau<g ref="char:cmbAbbrStroke">̄</g>ger maner of matter entreated of,<note place="margin">
                  <hi>In this booke is entreated of a straunger maner of mat<g ref="char:EOLhyphen"/>ter then in the former.</hi>
               </note> the<g ref="char:cmbAbbrStroke">̄</g> the other bokes before had, and the demonstrations also thereof, &amp; the order seeme likewise at the first somewhat straunge and vnaccustomed, which thinges may seeme also to cause the obscuritie therof, and to feare away many from the reading and diligent study of the same, so much that many of the well learned haue much complayned of the darkenes and difficultie thereof, and haue thought it a very hard thing, and in ma<g ref="char:EOLhyphen"/>ner
<pb facs="tcp:7062:258"/>
impossible to attayne to the right and full vnderstanding of this booke,<note place="margin">
                  <hi>Many euen of the well lear<g ref="char:EOLhyphen"/>ned haue thought that this booke can not well be vnderstanded without Al<g ref="char:EOLhyphen"/>gebra.</hi>
               </note> with<g ref="char:EOLhyphen"/>out the ayde and helpe of some other knowledge and learnyng, and chiefly with<g ref="char:EOLhyphen"/>out the knowledge of that more secret and subtill part of Arithmetike, commonly called Algebra, which vndoubtedly first well had and knowne, would geue great light therunto: yet certainly may this booke very well be entred into, and fully vn<g ref="char:EOLhyphen"/>derstand without any straunge helpe or succour, onely by diligent obseruation of the order, and course of Euclides writinges. So that he which diligently hath peru<g ref="char:EOLhyphen"/>sed and fully vnderstandeth the 9. bookes goyng before, and marketh also earnest<g ref="char:EOLhyphen"/>ly the principles and definitions of this <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>enth booke, he shal well perceiue that Eu<g ref="char:EOLhyphen"/>clide is of himselfe a sufficient teacher and instructer,<note place="margin">
                  <hi>The nine for<g ref="char:EOLhyphen"/>mer bookes &amp; the principles of this <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ooke well vnder<g ref="char:EOLhyphen"/>stoode, this booke will not be hard to vnderstand.</hi>
               </note> and needeth not the helpe of any other, and shall soone see that this tenth booke is not of such hardnes and ob<g ref="char:EOLhyphen"/>scuritie, as it hath bene hetherto thought. Yea, I doubt not, but that by the trauell and industry taken in this translation, and by addicions and emendations gotten of others, there shall appeare in it no hardnes at all, but shall be as easie as the rest of his bookes are.</p>
            <div type="part">
               <head>Definitions.</head>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rst definition.</hi>
                     </note>1 Magnitudes commensurable are suchwhich one and the selfe same mea<g ref="char:EOLhyphen"/>sure doth measure.</head>
                  <p>First he sheweth what magnitudes are commensurable one to an other. To the better and more cleare vnderstanding of this definition, note that that measure whereby any magnitude is measured, is lesse then the magnitude which it measureth, or at least e<g ref="char:EOLhyphen"/>quall vnto it. For the greater can by no meanes measure the lesse. Farther it behoueth, that that measure if it be equall to that which is measured, taken once make the mag<g ref="char:EOLhyphen"/>nitude which is measured: if it be lesse, then oftentimes taken and repeted, it must pre<g ref="char:EOLhyphen"/>cisely render and make the magnitude which it measureth. Which thing in numbers is easely sene, for that (as was before said) all numbers are commensurable one to an o<g ref="char:EOLhyphen"/>ther. And although Euclide in this definition comprehendeth purposedly, onely mag<g ref="char:EOLhyphen"/>nitudes which are continuall quantities, as are lines, superficieces, and bodies, yet vn<g ref="char:EOLhyphen"/>doubtedly the explication of this and such like places, is aptly to be sought of numbers as well rationall as irrationall. For that all quantities commensurable haue that pro<g ref="char:EOLhyphen"/>portion the one to the other, which number hath to numbers. In numbers therfore, 9 and 12 are commensurable, because there is one common measure which measureth them both, namely, the number 3. First it measureth 12, for it is lesse then 12. and be<g ref="char:EOLhyphen"/>ing taken certaine times, namely, 4 times, it maketh exactly 12: 3 times 4 is 12, it also measureth 9, for it is lesse then 9. and also taken certaine times, namely, 3 times, it ma<g ref="char:EOLhyphen"/>keth precisely 9: 3 times 3 is 9. Likewise is it in magnitudes, if one magnitude measure two other magnitudes, those two magnitudes so measured, are said to be commensura<g ref="char:EOLhyphen"/>ble. As for example, if the line C being dou<g ref="char:EOLhyphen"/>bled,
<figure/>
do make the line B, and the same lyne C tripled, do make the line A, then are the two lines A and B, lines or magnitudes com<g ref="char:EOLunhyphen"/>mensurable. For that one measure, namely, the line C measureth the<g ref="char:cmbAbbrStroke">̄</g> both. First, the line C is lesse the<g ref="char:cmbAbbrStroke">̄</g> the line A, and alsolesse the<g ref="char:cmbAbbrStroke">̄</g> the line B, also the line C taken or repeted certaine times, namely, 3 times, maketh precisely the line A, and the same line C taken also certain times, namely, two times, maketh pre<g ref="char:EOLhyphen"/>cisely the line B. So that the line C is a common measure to them both, and doth mea<g ref="char:EOLhyphen"/>sure them both. And therfore are the two lines A and B lines commensurable. And so i<g ref="char:EOLhyphen"/>magine ye of magnitudes of other kyndes, as of superficiall figures, and also of bodies.</p>
               </div>
               <div type="section">
                  <pb n="229" facs="tcp:7062:258"/>
                  <head>2 Incommensurable magnitudes are such, which no one common measure doth measure.<note place="margin">
                        <hi>The second definition.</hi>
                     </note>
                  </head>
                  <p>This diffinition neadeth no explanation at all,<note place="margin">
                        <hi>Contraryes made manifest by the compa<g ref="char:EOLhyphen"/>ring of the one to the other.</hi>
                     </note>
                     <figure/>
it is easely vnderstanded by the diffinition going before of lines commensurable. For contraries are made manifest by comparing of the one to the o<g ref="char:EOLhyphen"/>ther: as if the line C, or any other line oftentimes iterated, doo not render precisely the line A, nor the line B, the<g ref="char:cmbAbbrStroke">̄</g> are the lines A and B incommensurable. Also if the line C, or any other line certayne times repeted, doo exactly render the line A, and doo not measure the line B: or if it measure the line B, and measureth not also the line A, the lines A and B, are yet lines inco<g ref="char:cmbAbbrStroke">̄</g>mensurable: &amp; so of other mag<g ref="char:EOLhyphen"/>nitudes as of superficieces, and bodyes.</p>
               </div>
               <div type="section">
                  <head>3 Right lines commensurable in power are such, whose squares one and the selfe same superficies, <hi>area,</hi> or <hi>plat</hi> doth measure.<note place="margin">
                        <hi>The thirde definition.</hi>
                     </note>
                  </head>
                  <p>To the declaration of this diffinition we must first call to minde what is vnder<g ref="char:EOLhyphen"/>standed &amp; ment by the power of a line: which as we haue before in the former bookes noted is nothing ells but the square thereof, or any other plaine figure equall to the square therof. And so great power &amp; habilitie <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s a line said to haue, as is the quantitie of the square, which it is able to describe, or a figure superficial equal to the square therof.<note place="margin">
                        <hi>What the power of a line is.</hi>
                     </note> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> This i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> also to be noted that of lines, some are commensurable in length, the one to the other, and some are commensurable the one to the other in power. Of lines com<g ref="char:EOLunhyphen"/>mensurable in length the one to the other, was geuen an example in the declaration of the first diffinitio<g ref="char:cmbAbbrStroke">̄</g>, namely, the lines A and B, which were commensurable in length, one and the selfe measure, namely, the line C measured the length of either of them. Of the other kinde is geuen this diffinition here set: for the opening of which take this exam<g ref="char:EOLhyphen"/>ple. Let there be a certaine line, namely, the
<figure/>
line BC, and let the square of that line be the square BCDE. Suppose also an other line, namely, the line FH, &amp; let the square thereof be the square FHIK, and let a certayne superficies, namely, the superfi<g ref="char:EOLhyphen"/>cies A, measure the square BCDE, taken 16. times: which is the number of the litle areas, squares, plats, or superficieces co<g ref="char:cmbAbbrStroke">̄</g>tai<g ref="char:EOLhyphen"/>ned and described within the sayd squares ech of which is equall to the superficie: A. Agayne let the same superficies A mea<g ref="char:EOLhyphen"/>sure the square FHIK 9. times taken, ac<g ref="char:EOLhyphen"/>cording to the number of the field<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s or su<g ref="char:EOLhyphen"/>perficieces contayned and described in the same. Ye see the<g ref="char:cmbAbbrStroke">̄</g> that one and the selfe same superficies, namely, the superficies A, is a common measure to both these squares, and by certayne repeticions thereof, measureth them both. Wherefore the two lines BC and FH, which are the sides or lines producing these squares, and whose powers these squares are, are by this diffinition lines commensurable in power.</p>
               </div>
               <div type="section">
                  <head>4 Lines incommensurable are such, whose squares no one plat or superfi<g ref="char:EOLhyphen"/>cies doth measure.<note place="margin">
                        <hi>The fourth definition.</hi>
                     </note>
                  </head>
                  <p>
                     <pb facs="tcp:7062:259"/>This diffinition is easy to be vnderstanded by that which was sayd in the diffini<g ref="char:EOLhyphen"/>tion last set before this, and neadeth no farther declaration. And thereof take this ex<g ref="char:EOLhyphen"/>ample. If neither the superficies A,
<figure/>
nor any other superficies doo mea<g ref="char:EOLhyphen"/>sure the two squares B<hi>C</hi>DE, and FHIK: or if it measure the one, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rely BCDE, and not the other FHIK, or if it measure the square FHIK, and not the square BCDE: the two lines BC and FH, are in power incommensurable, and ther<g ref="char:EOLhyphen"/>fore also incomme<g ref="char:cmbAbbrStroke">̄</g>surable in length. For whatsoeuer lines are incomme<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>surable in power, the same are also incommensurable in length as shall afterward in the 9. proposition of this booke be proued. And therfore such lines are here defined to be absolutely incommensurable. These thinges thus standing it may easely appeare, that if a line be assigned and layd before vs, there may be innumerable other lines commensurable vnto it, and other incommensurable vnto it: of commensurable lines some are commensurable in length and power, and some in power onely.</p>
               </div>
               <div type="section">
                  <head>5 And that right line so set forth is called a rationall line.</head>
                  <p>
                     <note place="margin">
                        <hi>The fifth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>efi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>tion.</hi>
                     </note>These principles, diffinitions and groundes of this tenth booke ought well to be paysed, they are somewhat more strange and more hard, then are the diffinitions and principles of the other bookes of <hi>Euclide</hi> going before, and therefore at the first sight or reading are not straight way conceiued, but ought often to be repeted,<note place="margin">
                        <hi>The principles of this booke ought well to be pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>sed, for that they are more straun<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e then the prin<g ref="char:EOLunhyphen"/>ciples of the former bokes.</hi>
                     </note> and by vse to be confirmed. For the propositions following, bring vnto them much light, and facili<g ref="char:EOLhyphen"/>tie of vnderstanding. First there is a line supposed, and layd before vs, which may be a<g ref="char:EOLhyphen"/>ny line whatsoeuer, of what length, or shortnes ye will: this line thus first supposed i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> imagined to haue such diuisions and so many partes as ye list to conceiue in minde, as 3. 4. 5. and so forth, which may be applied to any kinde of measure, as it shall happen, as to inches, feete, pases, and such other. Vnto this line saith <hi>Euclide</hi> may be co<g ref="char:cmbAbbrStroke">̄</g>pared innu<g ref="char:EOLhyphen"/>merable lines, of which some shalbe commensurable, and other some incommensura<g ref="char:EOLhyphen"/>ble: of commensurable lines, some are commensurable both in length and in power, other some are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in pow<g ref="char:EOLunhyphen"/>er
<figure/>
onely. As if any part of the line pro<g ref="char:EOLunhyphen"/>posed which let be the line AB, doo mesure also the line EF and againe if any one superficies do measure the square of the line AB, which let be the square ABCD: and also doo measure the square of the line EF, which let be the square EFGH: the<g ref="char:cmbAbbrStroke">̄</g> is the line EF to the supposed line &amp; first set, name<g ref="char:EOLunhyphen"/>ly, to the line AB, a line commensura<g ref="char:EOLhyphen"/>ble both in length and power, as ye may see in the first example here set.</p>
                  <p>And if it so be that one &amp; the selfe same superficies do measure both the squares of these two lines AB and EF, namely the squares ABCD and EFGH: and no one line do measure the lines AB and EF: then is the line
<pb n="230" facs="tcp:7062:259"/>
EF (compared to the supposed <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>d
<figure/>
first line AB) commensurable in pow<g ref="char:EOLhyphen"/>er onely. As in this example ye may easely perceiue. For the triangle or superficies ACD, twise taken, mea<g ref="char:EOLhyphen"/>sureth the square ABCD, namely, the square of the line AB: and the selfe same triangle ACD taken foure times measureth the other square, namely, the square of the line EF. But no one measure or line can be as<g ref="char:EOLhyphen"/>signed to measure both the lines AB and EF, because the sides of a square and the diameter are incommensura<g ref="char:EOLhyphen"/>ble in length the one to the other, as afterward shalbe shewed. Wherefore they are in length incommensurable, &amp; commensurable in power onely, that is by rea<g ref="char:EOLhyphen"/>son of their squares, which are commensurable the one to the other.</p>
                  <p>Agayne if it so be that no one line may
<figure/>
be found to be a co<g ref="char:cmbAbbrStroke">̄</g>mon measure, to measure both the first line, namely, AB, and also the line EF: nor yet any one superficies to mea<g ref="char:EOLhyphen"/>sure the square or powers of these two lines, then is the line EF, to the first line set and supposed, incommensurable both in length and in power. As is supposed to be in this example.</p>
                  <p>Thus may ye see, how to the supposed line first set may be compared infinite lines,<note place="margin">
                        <hi>Vnto the supposed line first set may be compared infi<g ref="char:EOLhyphen"/>nite lines.</hi>
                     </note> some commensurable both in length &amp; pow<g ref="char:EOLhyphen"/>er, and some commensurable in power onely, and incommensurable in length, and some incommensurable both in power &amp; in length. And this first line so set, whereunto, and to whose squares the other lines and their squares are compared, is called a rationall line, commonly of the most part of writers.<note place="margin">
                        <hi>Why some mislike that the line first set should be called a ratio<g ref="char:EOLhyphen"/>nal line.</hi>
                     </note> But some there are which mislike that it should be called a rationall line, &amp; that not without iust cause. In the Greeke copy it is called <gap reason="foreign">
                        <desc>〈 in non-Latin alphabet 〉</desc>
                     </gap>, <hi>rete,</hi> which signifieth a thing that may be spoke<g ref="char:cmbAbbrStroke">̄</g>, &amp; expressed by word, a thing certayne, graunted and appoynted. Wherefore <hi>Flussates,</hi> a man which bestowed great trauell and diligence in restoring of these elementes of <hi>Euclide,</hi> leauing this word ratio<g ref="char:EOLhyphen"/>nall, calleth this line supposed and first set, a line certaine, because the partes thereof in<g ref="char:EOLhyphen"/>to which it is deuided are certaine,<note place="margin">
                        <hi>Flussates cal<g ref="char:EOLhyphen"/>leth this line, a line certaine.</hi>
                     </note> and known, and may be expressed by voyce, and also be coumpted by number: other lines, being to this line incommensurable, whose parts are not distinctly known, but are vncertayne, nor can be expressed by name nor assignd by number, which are of other men called irrationall, he calleth vncertaine and surd lines. <hi>Petrus Montaureus</hi> although he doth not very wel like of the name, yet he altereth it not, but vseth it in al his booke. Likewise wil we doo here, for that the word hath bene and is so vniuersally receiued. And therefore will we vse the same name, and call it a ra<g ref="char:EOLhyphen"/>tionall line. For it is not so great a matter what names we geue to thinges, so that we fully vnderstand the thinges which the names signifie.</p>
                  <p>This rationall line thus here defined, is the ground and foundation of all the propo<g ref="char:EOLhyphen"/>sitions almost of this whole tenth booke.<note place="margin">
                        <hi>This rational line the grou<g ref="char:cmbAbbrStroke">̄</g>d in a maner of all the proposi<g ref="char:EOLhyphen"/>tions in this tenth booke.</hi>
                     </note> And chiefly from the tenth proposition for<g ref="char:EOLhyphen"/>wardes. So that vnlesse ye first place this rationall line, and haue a speciall and continu<g ref="char:EOLhyphen"/>all regard vnto it before ye begin any demonstration, ye shall not easely vnderstand it. For it is as it were the touch and triall of all other lines, by which it is known whether any of them be rationall or not.<note place="margin">
                        <hi>Note.</hi>
                     </note> And this may be called the first rationall line,<note place="margin">
                        <hi>The line Rati<g ref="char:EOLhyphen"/>onall of pur<g ref="char:EOLhyphen"/>pose.</hi>
                     </note> the line rationall of purpose or a rationall line set in the first place, and so made distinct and se<g ref="char:EOLhyphen"/>uered from other rationall lines, of which shall be spoken afterwarde. And this must ye well commit to memory.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:260"/>
                  <head>
                     <note place="margin">
                        <hi>The sixth de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inition.</hi>
                     </note>6 Lines which are commensurable to this line, whether in length and power, or in power onely, are also called rationall.</head>
                  <p>This definition needeth no declaration at all, but is easily perceiued, if the first de<g ref="char:EOLhyphen"/>finition be remembred, which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>heweth what magnitudes are commensurable, and the third, which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>heweth what lines are commensurable in power. Here not<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, how aptly &amp; naturally, Euclide in this place vseth these wordes commensurable either in length and power, or in power onely. Because that all lines which are commensurable in length, are also commensurable in power<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> when he speaketh of lines commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth, he euer addeth and in power, but when he speaketh of lines commensurable in power, he addeth this worde Onely, and addeth not this worde in length, as he in the other added this worde in power. For not all lines which are commensurable in power, are straight way commensurable also in length. Of this definition, take this example. Let the first line rationall of purpose,
<figure/>
which is supposed and laide forth, whose partes are certaine &amp; known, and may be expressed, named, and nu<g ref="char:cmbAbbrStroke">̄</g>bred be <hi>AB,</hi> the quadrate wher<g ref="char:EOLhyphen"/>of let be <hi>ABCD</hi>: then suppose a<g ref="char:EOLhyphen"/>gaine an other lyne, namely, the line <hi>EF,</hi> which let be commensurable both in length and in power to the first rationall line, that is, (as before was taught) let one line measure the length of eche line, and also l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t one super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies measure the two squares of the said two lines, as here in the example is supposed and also ap<g ref="char:EOLhyphen"/>peareth to the eie, then is the line <hi>E F</hi> also a rationall line.</p>
                  <p>Moreouer if the lyne <hi>EF</hi> be commensurable in power onely to the rationall line <hi>AB</hi> first set and supposed, so that no one line do measure the two lines <hi>AB</hi> and <hi>EF</hi>: As in example y<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> see to be (for that the line <hi>EF,</hi> is made equall to the line AD, which is the diameter of the square <hi>ABCD,</hi> of
<figure/>
which square the line <hi>AB</hi> is a side, it is certayne that the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ide of a square is inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the diameter of the same square: if there be yet founde any one su<g ref="char:EOLhyphen"/>perficies, which measureth the two squares <hi>ABCD,</hi> and <hi>EFGH</hi> (as here doth the triangle <hi>ABD,</hi> or the triangle <hi>ACD</hi> noted in the square <hi>ABCD,</hi> or any of the foure triangles noted in the square <hi>EFGH,</hi> as appeareth somwhat more manifestly in the second example, in the declaration of the last definition going before) the line <hi>EF</hi> is also a rational line. Note that these lines which here are called rationall lines, are not rational lines of pur<g ref="char:EOLhyphen"/>pose, or by supposition, as was the first rationall line, but are rationall onely by reason of relation and comparison which they haue vnto it, because they are commensurable vn<g ref="char:EOLhyphen"/>to it either in length and power, or in power onely. Farther here is to be noted, that these wordes length, and power, and power onely, are ioyned onely with these worde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> commensurable or incommensurable, and are neuer ioyned with these woordes ratio<g ref="char:EOLhyphen"/>nall or irrationall. So that no lines can be called rational in length, or in power, nor like wise can they be called irrationall in length, or in power. Wherin vndoubtedly Campa<g ref="char:EOLhyphen"/>nus was deceiued,<note place="margin">
                        <hi>Camp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>nus <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ath caused much o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>scuri<g ref="char:EOLhyphen"/>tie in this tenth booke.</hi>
                     </note> who vsing those wordes &amp; speaches indifferently, caused &amp; brought in great obscuritie to the propositions and demonstrations of this boke, which he shall
<pb n="231" facs="tcp:7062:260"/>
easily see which marketh with diligence the demonstrations of <hi>Campanus</hi> in this booke.</p>
               </div>
               <div type="section">
                  <head>7 Lines which are incommensurable to the rationall line, are called ir<g ref="char:EOLhyphen"/>rationall.<note place="margin">
                        <hi>The seuenth definition.</hi>
                     </note>
                  </head>
                  <p>By lines incommensurable to the rationall line supposed in this place, he vnder<g ref="char:EOLhyphen"/>standeth such as be incommensurable vnto it both in length and in power. For there are no lines incommensurable in power onely: for it cannot be that any lines should so be incommen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>urable in power onely, that they be not also incommensurable in length. What so euer lines be incomme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>surable in power, the same be also incommensurable in length. Neither can <hi>Euclide</hi> here in this place meane lines incommensurable in length onely, for in the diffinition before, he called them rationall lines, n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ither may they be placed amongst irrationall lines. Wherfore it remayneth that in this diffintion he speaketh onely of those lines which are incommensurable to the rationall line first geuen and supposed, both in length, and in power. Which by all meanes are incommen<g ref="char:EOLunhyphen"/>surable to the rationall line, &amp; therfore most aptly are they called irrationall lines. This diffinition is easy to be vnderstanded by that which hath bene sayd before. Yet for the more plainenes see this example. Let the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rst rationall line supposed, be the line AB, whose square or quadrate, let be ABCD. And
<figure/>
let there be geuen an other line EF which l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t be to the rationall line incommensurable in length and power, so that let no one line mea<g ref="char:EOLhyphen"/>sure the length of the two lines, AB and EF: and let the square of the line EF be EFGH. Now if also there be no one superficies which measureth the two squares ABCD, and EFGH, as is supposed to be in this example, the<g ref="char:cmbAbbrStroke">̄</g> is the line EF an irrationall line, which word irrational (As before did this word rational) misliketh many learned in this knowledge of Geometry. <hi>Flussates,</hi>
                     <note place="margin">
                        <hi>Flussates in steede of this word irratio<g ref="char:EOLhyphen"/>nall vseth this word vncer<g ref="char:EOLhyphen"/>tayne.</hi>
                     </note> as he left the word ratio<g ref="char:EOLhyphen"/>nall, and in steade thereof vsed this word cer<g ref="char:EOLhyphen"/>taine, so here he leaueth the word irrationall, and vseth in place thereof this word <hi>vncertaine,</hi> and euer nameth these lines vncertaine lines. <hi>Petrus Montaureus</hi> also misliking the word irrationall, would rather haue them to be called surd lines, yet because this word irrationall hath euer by custome and long vse, so generally bene receiued<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> he vseth continually the same. In Greeke such lines are called <gap reason="foreign">
                        <desc>〈 in non-Latin alphabet 〉</desc>
                     </gap>, <hi>alogoi,</hi> which signifieth nameles, vnspeakeable, vncertayne, in determinate,<note place="margin">
                        <hi>Why they are called irratio<g ref="char:EOLhyphen"/>nall lines.</hi>
                     </note> and with out proportion: not that these irrationall lines haue no proportion at all, ei<g ref="char:EOLhyphen"/>ther to the first rationall line, or betwene them selues: but are so named, for that theyr proportions to the rationall line cannot be expressed in number. That is vndoubtedly very vntrue, which many write, that their proportions are vnknowne both to vs and to nature. Is it not thinke you a thing very absurd to say that there is any thing in nature, and produced by nature, to be hidde from nature, and not to be knowne of nature? it can not be sayd that their proportions are vtterly hidde and vnknowne to vs (much lesse vnto nature) although we cannot geue them their names, and distinctly expresse them by numbers: otherwise should <hi>Euclide</hi> haue taken all this trauell and wonderfull diligence bestowed in this booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in vaine and to no vse<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in which he doth nothing ell<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> but teach the proprieties and passions of these irrationall lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and sheweth the pro<g ref="char:EOLhyphen"/>portions which they haue the one to the other.</p>
                  <p>Here is also to be noted, which thing also <hi>Tartalea</hi> hath before diligently noted<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that <hi>Campanus</hi> and many other writers of Geometry<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> ouer much <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>ed and were decei<g ref="char:EOLhyphen"/>ued in that they wrote and taught, that all these lines whose squares were not s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gnified and mought be expressed by a square number (although they migh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by any other number: as by 11. 12. 14. and such others not square numbers) are irrationall lines. Which is manifestly repugnant to the groundes and principles of <hi>Euclide,</hi> who wil, that all lines which are commensurable to the rationall line, whether it be in length and
<pb facs="tcp:7062:261"/>
power, or in power onely, should be rationall. Vndoubtedly this hath bene one of the chiefest and greatest causes of the wonderfull confusion and darkenes of this booke,<note place="margin">
                        <hi>The cause of the obscurity and confused<g ref="char:EOLhyphen"/>nes in this booke.</hi>
                     </note> which so hath tossed, and tormoyled the wittes of all both writers and readers, masters and scholers, and so ouerwhelmed them, that they could not with out infinite trauell and sweate, attayne to the truth and perfect vnderstanding thereof.</p>
               </div>
               <div type="section">
                  <head>
                     <note place="margin">
                        <hi>The eighth definition.</hi>
                     </note>8 The square which is described of the rationall right line supposed, is rationall.</head>
                  <p>Vntill this diffinition hath <hi>Euclide</hi> set forth the nature and proprietie of the first kinde of magnitude, namely, of lines how they are rationall or irrationall, now he b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>ginneth to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hew how the second kinde of magnitudes, namely superficies are one to the other rationall or irrationall. This diffinition is very playne.
<figure/>
Suppose the line AB, to be the rationall line, hauing his parts and diuisions certaynely knowne: the square of which line let be the square ABCD. Now because it is the square of the rationall line AB, it is also called rationall: and as the line AB, is the first rationall line, vnto which other lines compa<g ref="char:EOLhyphen"/>red are coumpted rationall, or irrationall so is the quadrat or square thereof, the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst rationall superficies vnto which all other squares or figures compared, are coumpted and na<g ref="char:EOLhyphen"/>med rationall or irrationall.</p>
               </div>
               <div type="section">
                  <head>9 Such which are commensurable vnto it, are rationall.</head>
                  <p>
                     <note place="margin">
                        <hi>The ninth definit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>on.</hi>
                     </note>In this di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>inition, where it is sayd, such as are commensurable to the square of the rationall line, are not vnderstand onely other squares or quadrates, but all other kindes of rectiline figures playne plats &amp; superficieses. What so euer so that if any such figure be co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto
<figure/>
that rationall square<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> it is also rationall. As sup<g ref="char:EOLunhyphen"/>pose that the square of the rationall line, which is also rationall, be ABCD: suppose <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> so some other square as the square EFGH, to be commensurable to the same: the<g ref="char:cmbAbbrStroke">̄</g> is the square EFGH also ra<g ref="char:EOLhyphen"/>tional. So also if the rectiline figure KLMN, which is a figure on the one side longer, be commensurable vnto the sayd square as is supposed in this example<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> it is also a rational superficies and so of all other superficieses.</p>
               </div>
               <div type="section">
                  <head>10 Such which are incommensurable vnto it, are irrationall.</head>
                  <p>
                     <note place="margin">
                        <hi>The tenth de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inition.</hi>
                     </note>Where it is sayd in this diffini<g ref="char:EOLunhyphen"/>tion
<figure/>
such which are incommensu<g ref="char:EOLhyphen"/>rable, it is generally to be taken: as was this word co<g ref="char:cmbAbbrStroke">̄</g>mensurable in the diffinitio<g ref="char:cmbAbbrStroke">̄</g> before. For al super<g ref="char:EOLhyphen"/>ficieses, whether they be squares or figures on the one side longer, or otherwise what maner of right lined figure so euer it be, if they be incommensurable vnto the ratio<g ref="char:EOLhyphen"/>nall square supposed, the<g ref="char:cmbAbbrStroke">̄</g> are they
<pb n="232" facs="tcp:7062:261"/>
irrationall. As let th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> square ABCD be the square of the supposed rationall line which square therefore is also rationall: suppose also also an other square, namely the square E, suppose also any other figure, as for example sake a figure of one side longer, which let be <hi>F</hi>: Now if the square E and the figure F, be both incommensurable to the ratio<g ref="char:EOLhyphen"/>nall square ABCD, then is <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> of these figures E &amp; F irrationall. And so of other.</p>
               </div>
               <div type="section">
                  <head>11 And these lines whose poweres they are, are irrationall. If they be squares, then are their sides irrationall. If they be not squares,<note place="margin">
                        <hi>The eleuenth de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inition.</hi>
                     </note> but some other rectiline figures, then shall the lines, whose squares are equall to these rectiline figures, be irrationall.</head>
                  <p>Suppose that the rationall
<figure/>
square be ABCD. Suppose also an other square, name<g ref="char:EOLunhyphen"/>ly the square E, which let be inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable to the ra<g ref="char:EOLunhyphen"/>tionall square, &amp; therefore is it irrationall: and let the side or line which produ<g ref="char:EOLhyphen"/>ceth this square be the line FG: then shall the line FG by this diffinition be an ir<g ref="char:EOLhyphen"/>rationall line: because it is the side of an irrationall square. Let also the figure H being a figure on the one side longer (which may be any other rectiline figure rectangled or not rectangled, triangle, pentagone, trapezite, or what so euer ells) be incommensurable to the rationall square ABCD, then because the figure H is not a square, it hath no side or roote to produce it yet may there be a square made equall vnto it: for that all such figures may be reduced into triangles, and so into squares, by the 14. of the second. Suppose that the square Q be equall to the irrationall figure H. The side of which figure Q let be the line KL: then shall the line KL be also an irrational line, because the power or square thereof, is equal to the irrationall figure H: and thus conceiue of others the like.</p>
                  <p>These irrationall lines and figures are the chiefest matter and subiect, which is en<g ref="char:EOLhyphen"/>treated of in all this tenth booke: the knowledge, of which is deepe, and secret, and pertaineth to the highest and most worthy part of Geometrie, wherein standeth the pith and mary of the hole science: the knowlede hereof bringeth light to all the bookes following, with out which they are hard and cannot be at all vnderstoode. And for the more plainenes, ye shall note, that of irrationall lines there be di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ers sortes and kindes. But they, whose names are set in a table here following, and are in number 13. are the <milestone type="tcpmilestone" unit="unspecified" n="1"/> chiefe, and in this te<g ref="char:cmbAbbrStroke">̄</g>th boke sufficiently for Euclides principall purpose, discoursed on.<milestone type="tcpmilestone" unit="unspecified" n="2"/>
                  </p>
                  <list>
                     <item>A mediall line.<milestone type="tcpmilestone" unit="unspecified" n="3"/>
                     </item>
                     <item>A binomiall line.<milestone type="tcpmilestone" unit="unspecified" n="4"/>
                     </item>
                     <item>A first bimediall line.<milestone type="tcpmilestone" unit="unspecified" n="5"/>
                     </item>
                     <item>A second bimediall line.<milestone type="tcpmilestone" unit="unspecified" n="6"/>
                     </item>
                     <item>A greater line.<milestone type="tcpmilestone" unit="unspecified" n="7"/>
                     </item>
                     <item>A line containing in power a rationall superficies and a mediall superficies.<milestone type="tcpmilestone" unit="unspecified" n="8"/>
                     </item>
                     <item>A line containing in power two mediall superficieces.<milestone type="tcpmilestone" unit="unspecified" n="9"/>
                     </item>
                     <item>A residuall line.<milestone type="tcpmilestone" unit="unspecified" n="10"/>
                     </item>
                     <item>A first mediall residuall line.<milestone type="tcpmilestone" unit="unspecified" n="11"/>
                     </item>
                     <item>A second mediall residuall line.<milestone type="tcpmilestone" unit="unspecified" n="12"/>
                     </item>
                     <item>A lesse line.<milestone type="tcpmilestone" unit="unspecified" n="13"/>
                     </item>
                     <item>A line making with a rationall superficies the whole superficies mediall.</item>
                     <item>A line making with a mediall superficies the whole superficies mediall.</item>
                     <item>Of all which kindes the diffinitions together with there declarations shalbe set here after in their due places.</item>
                  </list>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:262"/>
               <head>¶ The 1. Theoreme. The 1. Proposition. Two vnequall magnitudes being geuen, if from the greater be taken away more then the halfe, and from the residue be againe taken away more then the halfe, and so be done still continually, there shall at length be left a cer<g ref="char:EOLhyphen"/>taine magnitude lesser then the lesse of the magnitudes first geuen.</head>
               <p>SVppose that there be two vnequall magnitudes AB, and C, of which let AB be the greater. Then I say, that if from AB, be taken away more then the halfe,<note place="margin">Construction.</note> and from the residue be taken a<g ref="char:EOLhyphen"/>gaine more then the halfe, and so still continually, there shall at the length be left a certaine magnitude, lesser then the lesse mag<g ref="char:EOLhyphen"/>nitude geue<g ref="char:cmbAbbrStroke">̄</g>, namely, then C. For forasmuch as C is the lesse mag<g ref="char:EOLhyphen"/>nitude, therefore C may be so multiplyed, that at the length it will be greater then the magnitude AB (by the 5. definition of the fift booke).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Let it be so multiplyed, and let the multiplex of C grea<g ref="char:EOLhyphen"/>ter then AB, be DE. And deuide DE into the partes equall
<figure/>
vnto C, which let be DF, FG, and GE. And from the mag<g ref="char:EOLhyphen"/>nitudes AB take away more then the halfe, which let be BH: and againe from AH, take away more then the halfe, which let be HK. And so do continually vntill the diuisions which are in the magnitude AB, be equall in multitude vnto the diui<g ref="char:EOLhyphen"/>sions which are in the magnitude DE. So that let the diuisions AK, KH, and HB, be equall in multitude vnto the diuisions DF, FG, and GE. And forasmuch as the magnitude DE is greater then the magnitude AB, and from DE is taken away lesse then the halfe, that is, EG (which detraction or taking a<g ref="char:EOLhyphen"/>way is vnderstand to be done by the former diuision of the mag<g ref="char:EOLhyphen"/>nitude DE into the partes equall vnto C: for as a magnitude is by multiplication increased, so is it by diuision diminished) and from AB is taken away more then the halfe, that is, BH: therefore the residue GD is greater then the residue HA (which thing is most true and most easie to conceaue, if we reme<g ref="char:cmbAbbrStroke">̄</g>ber this principle, that the residue of a greater magnitude, after the ta<g ref="char:EOLhyphen"/>king away of the halfe or lesse then the halfe, is euer greater then the residue of a lesse magni<g ref="char:EOLhyphen"/>tude, after the taking away of more then the halfe). And forasmuch as the magnitude GD is greater then the magnitude HA, and from GD is taken away the halfe, that is, GF: and from AH is taken away more then the halfe, that is, HK: therefore the residue DF is greater then the residue AK (by the foresayd principle). But the magnitude DF is equall vnto the magnitude C (by supposition). Wherefore also the magnitude C is greater then the magnitude AK. Wherefore the magnitude AK is lesse then the magnitude C. Wherefore of the magnitude AB is left a magnitude AK lesse then the lesse magnitude geuen, namely, then C which was required to be proued. In like sort also may it be proued if the halfes be taken away.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note>Of this Proposition it followeth, that any magnitude being geuen how litle soeuer it be, there may be geuen a magnitude lesse then it: so that it is impossible that any mag<g ref="char:EOLhyphen"/>nitude
<pb n="233" facs="tcp:7062:262"/>
should be geuen then which can be geuen no losse.</p>
                  <div type="subsection">
                     <head>¶An other demonstration of the same.</head>
                     <p>Suppose that the two vnequall magnitudes geuen be AB and C. And let C be the lesse. And forasmuch as C is the lesse, ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>fore C may so be multiplyed, that it shall at the length be greater then A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.<note place="margin">Construction.</note> Let it be so multiplyed, and let the multiplex of C exceding AB be the magnitude FM. And deuide FM into his partes equall vnto C, that is, into the magni<g ref="char:EOLhyphen"/>tudes MH, HG, and GF. And from AB take away more then the halfe, which let be the magnitude BE: and likewise from EA take away againe more then the halfe, name<g ref="char:EOLhyphen"/>ly, the magnitude ED. And thus do continually vntill the diuisions which are in the magnitude FM, be equall in multitude to the diuisions which are in the magnitude AB: and let those diuisions be the magnitudes BE, ED, and DA. And how multiplex the magnitude FM is to the magnitude C, so multiplex let the magnitude KX be to the magnitude DA.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And deuide the magnitude KX into the magnitudes equall to the magnitude DA: which let be KL, LN, and NX. Now then the diuisions which are in the magnitude KX, are equall vnto the diuisi<g ref="char:EOLhyphen"/>ons
<figure/>
which are in the magnitude MF. And forasmuch as BE is greater then the halfe of AB, therefore BE is greater then the residue EA Wherefore BE is much more greater then DA. But DA is equall vnto XN. Wherefore BE is greater then XN. Againe forasmuch as DE is greater then the halfe of EA, therefore DE is greater then the residue DA: but DA is equall vnto LN: wherefore DE is greater then LN. Wher<g ref="char:EOLhyphen"/>fore the whole magnitude DB is greater the<g ref="char:cmbAbbrStroke">̄</g> the whole magnitude XL. But DA is equall vnto LK. Where<g ref="char:EOLhyphen"/>fore the whole magnitude AB is greater then the whole magnitude KX. And the magnitude MF is greater then the magnitude BA: wherefore MF is much greater the<g ref="char:cmbAbbrStroke">̄</g> KX. And forasmuch as those mag<g ref="char:EOLhyphen"/>nitudes XN, NL, &amp; LK, are equall the one to the o<g ref="char:EOLhyphen"/>ther, &amp; likewise these magnitudes MH, HG, and GF, are equal the one to the other, &amp; the multitude of those magnitudes which are in MF, is equall to the multi<g ref="char:EOLhyphen"/>tude of those magnitudes which are in KX: therfore as KL is to FG, so is LN to GH, and NX to HM. Wherefore (by the 12. of the fift) as one of the antecedentes, namely, KL, is to one of the consequences, namely, to FG, so are all the antecedentes, namely, the whole KX to all the consequentes, namely, to the whole FM. But FM is greater then KX. Wherefore FG is greater then LK. But FG is equall vnto C: and KL vnto DA (by supposition). Wherefore the magnitude C is greater then the magnitude AD: which was required to be proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 2. Theoreme. The 2. Proposition. Two vnequall magnitudes being geuen, if the lesse be continually taken from the greater, &amp; that which remayneth measureth at no time the mag<g ref="char:EOLhyphen"/>nitude going before: then are the magnitudes geuen incommensurable.</head>
               <p>
                  <pb facs="tcp:7062:263"/>
                  <note place="margin">This proposi<g ref="char:EOLunhyphen"/>tion teacheth that inconti<g ref="char:EOLhyphen"/>nuall quanti<g ref="char:EOLhyphen"/>tie which the first of the se<g ref="char:EOLhyphen"/>uenth taught in discrete quantity.</note>SVppose that there be two vnequall magnitudes AB, and
<figure/>
CD, and let AB be the lesse: and taking away continu<g ref="char:EOLhyphen"/>ally by a certaine alternate detraction the lesse from the greater, let not the residue measure the magnitude going before. Then I say, that those two magnitudes AB and CD, are in<g ref="char:EOLhyphen"/>commensurable. For if they be commensurable, then (by the first de<g ref="char:EOLhyphen"/>finition of the tenth) some one magnitude shall measure them both: Let there be such a magnitude, if it be possible, and let the same be E. And let AB measuring DF,<note place="margin">Construction.</note> leaue a lesse then it selfe, namely, CF (<hi>that is, from the greater magnitude CD, take away a certayne part as DF, which let be equall to AB, or if it be not equall vnto it, yet let it be such, that that lesse magnitude AB being more then once repeated may make the mag<g ref="char:EOLhyphen"/>nitude DF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For this is the meaning of this</hi> let AB measuring DF. &amp;c. <hi>And this detraction made, of the lesse I say out of the greater, let there be left of the greater a certaine portion CF, lesse then the magnitude AB. And this is the meaning of that which in the Theoreme was said,</hi> And that which remaineth mea<g ref="char:EOLunhyphen"/>sureth at no time the magnitude going before). Likewise let CF measuring BG leaue a lesse then it selfe, namely, AG: and do this continually as often as neede requireth, vntill there be found such a magnitude that is lesse then E: which must needes at the length happen (by the Proposition going before)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Let there be found such a magnitude lesse then E, which let be AG. And forasmuch as the magnitude E measureth the magnitude AB, but AB measureth DF: therefore E measu<g ref="char:EOLhyphen"/>reth the magnitude DF (by this common sentence, If a magnitude measure an other magnitude it shall also measure euery magnitude whom it measureth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an ab<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>urditie.</note> And it measureth also the whole CD (for it is supposed to be a common measure to the magnitudes AB and CD). Wherefore also it measureth the residue CF (by this common sentence, If a magnitude measure an whole and a part taken away, it shall also measure the residue). And forasmuch as E measureth CF, but CF mea<g ref="char:EOLhyphen"/>sureth BG, therefore E also measureth BG (by the first foresaid common sentence): and it measureth the whole AB. Wherefore it shall also measure the residue AG (by the other fore<g ref="char:EOLhyphen"/>sayd common sentence) namely, the greater magnitude shall measure the lesse: which is im<g ref="char:EOLhyphen"/>possible. Wherefore no magnitude measureth these magnitudes AB and CD. Wherefore the magnitudes AB and CD are incommensurable. If therefore two vnequall magnitudes being geuen, the lesse be continually taken from the greater, and that which remayneth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> mea<g ref="char:EOLhyphen"/>sureth at no time the magnitude going before, then are the magnitudes geuen incommensu<g ref="char:EOLhyphen"/>rable: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶A Corollary added by Montaureus.</head>
                  <p>By this proposition it is manifest, that if the two vnequall magnitudes geuen be not incommensurable, but commensurable, then the lesse being continually subtrahed out of the greater, the residue shall of necessitie measure that which went before.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 1. Probleme. The 3. Proposition. Two magnitudes commensurable being geuen, to finde out, their greatest common measure.</head>
               <p>SVppose that the two commensurable magnitudes geuen be AB &amp; CD, of which let AB be the lesse. It is required to finde out the greatest common measure of the magnitudes AB and CD.<note place="margin">Two cases in this propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> Now AB either measureth CD, or not. If it measure it, and seing it also measureth it selfe,<note place="margin">The first case.</note> wherfore AB is a common measure vnto
<pb n="234" facs="tcp:7062:263"/>
the magnitudes AB and CD.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion teacheth, that in conti<g ref="char:EOLhyphen"/>nual quanti<g ref="char:EOLhyphen"/>ty, which the <hi>2.</hi> of the s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ith taught in numbers.</note> And it is manifest that it is the grea<g ref="char:EOLhyphen"/>test
<figure/>
common measure to them. For no magnitude greater then AB can measure AB. But now suppose that AB do not measure CD. And taking continually the lesse from the greater that which remai<g ref="char:EOLhyphen"/>neth shall at length measure that which goeth before (by the corolla<g ref="char:EOLhyphen"/>ry before added), for that AB and CD are co<g ref="char:cmbAbbrStroke">̄</g>mensurable. Now then let AB measuring ED which is a part of the magnitude CD, leaue a magnitude lesse then it selfe, namely, EC.<note place="margin">The second case.</note> And let EC measuring the magnitude FB, which is a part of the magnitude AB, leaue a lesse the<g ref="char:cmbAbbrStroke">̄</g> it selfe, namely, FA, and let FA precisely measure the mag<g ref="char:EOLhyphen"/>nitude CE (<hi>And this is the meaning of this,</hi> That which remayneth shall at the length measure that which goeth before, <hi>when there is nothing left after the measuring made</hi>) And forasmuch as AF measureth the magnitude CE, but CE measureth FB: wherfore AF also measureth FB, &amp; it measureth it selfe: wherfore AF measureth also the whole AB. But AB measureth DE, wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore AF also measureth DE. And AF also measureth CE, wherfore it measureth the whole magnitude CD. Wherfore the magnitude AF measureth both the magnitudes AB and CD. Wherfore AF is a common measure vnto AB and CD. I say also that it is the greatest co<g ref="char:cmbAbbrStroke">̄</g>mon measure vnto them. For if not, then is there some magnitude greater the<g ref="char:cmbAbbrStroke">̄</g> the magnitude AF which measureth both the magnitudes AB and CD. Let there be such a one if it be possible, and let the same be the mag<g ref="char:EOLunhyphen"/>nitude G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> And forasmuch as G measureth AB, and AB measureth ED: therefore G also measureth ED, and by supposition, it measureth the whole CD, wherfore G measureth also the residue CE. But CE measureth FB, wherfore G also measureth FB. And by supposition it measureth the whole AB. Wherfore it measureth the residue AF, namely, the greater mag<g ref="char:EOLunhyphen"/>nitude the lesse, which is impossible. Wherefore no magnitude greater then AF, measureth these magnitudes AB and CD. Wherefore AF is the greatest common measure vnto the magnitudes AB and CD. Wherfore vnto two commensurable magnitudes geuen, namely, AB and CD is found out their greatest common measure, namely, the magnitude AF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be done.</p>
               <div type="section">
                  <head>¶Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest that if a magnitude measure two magnitudes,</hi>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note> 
                     <hi>it shall also measure their greatest common measure.</hi> For if it measure the wholes and the partes taken away, it shal also measure the residues, of which one is the greatest common measure, as we may see by the latter ende of the former demonstration.</p>
                  <div type="subsection">
                     <head>
                        <hi>Monta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>reus</hi> reduceth this Probleme into a Theoreme after this maner.</head>
                     <p>Two vnequall and commensurable magnitudes being geuen, if the lesse do measure the greater,<note place="margin">This Pro<g ref="char:EOLhyphen"/>bleme reduced to a Theo<g ref="char:EOLhyphen"/>reme.</note> then is it the greatest common measure to them both. But if not, then the lesse being continually by a mutuall detraction (as before hath bene taught) taken out of the greater, whensoeuer the residue precisely measureth that which w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nt before leauing nothing, the said residue shall be the greatest common measure to both the magnitudes geuen.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 2. Probleme. The 4. Proposition. Three magnitudes commensurable beyng geuen, to finde out their greatest common measure.</head>
               <p>
                  <pb facs="tcp:7062:264"/>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion teacheth, that in conti<g ref="char:EOLhyphen"/>nual quanti<g ref="char:EOLhyphen"/>ty, which the <hi>3.</hi> of the second taught in numbers.</note>SVppose that the three commensurable magnitudes geuen be A, B, C. It is required of these three magnitudes to finde out the greatest common measure. Take (by the former proposition) the greatest common measure of the two magnitudes A, B and let the same be D. Now then this magnitude D either measureth the third magnitude, which is C, or not. First let it measure C. And forasmuch as
<figure/>
D measureth C, and it measureth also the magni<g ref="char:EOLhyphen"/>tudes A, B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore D measureth the three magni<g ref="char:EOLhyphen"/>tudes A, B, C. Wherfore D is a common measure vn<g ref="char:EOLhyphen"/>to the magnitudes A, B, C. And it is manifest, that it is the greatest common measure.<note place="margin">Construction.</note> For no magnitude greater then D can measure the magnitudes A, B, C.<note place="margin">Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</note> For if it be possible let the magnitude E being grea<g ref="char:EOLhyphen"/>ter then the magnitude D, measure the magnitude A, B, C.<note place="margin">The first case.</note> And forasmuch as E measureth the magni<g ref="char:EOLhyphen"/>tudes A, B, C, it measureth the two first magnitudes AB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Wherfore it shall also (by the Corollary of the former proposition) measure the greatest co<g ref="char:cmbAbbrStroke">̄</g>mon mea<g ref="char:EOLhyphen"/>sure of the magnitudes A, B which is D, namely, the greater shall measure the lesse, which is impossible.</p>
               <p>
                  <note place="margin">The second case.</note>But now let D not measure the magnitude C. First I say that the magnitudes C, D, are commensu<g ref="char:EOLhyphen"/>rable. For forasmuch as the magnitudes A, B, C are commensurable, some magnitude shall measure them, which shall also measure the magnitudes AB taken a part.<note place="margin">A Le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ma neces<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ary to be pr<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>d be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all to the demo<g ref="char:cmbAbbrStroke">̄</g>
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ration.</note> Wherfore it shall also (by the corollary going before) measure the greatest common "measure of A, B, that is, D: and (by supposition) it measureth the magnitude
<figure/>
C. Wherfore that said magnitude shall measure the magnitudes C and D.<note place="margin">Construction.</note> Wherfore the magni<g ref="char:EOLhyphen"/>tudes C, D are comme<g ref="char:cmbAbbrStroke">̄</g>surable. Take by the third of this tenth, their greatest common measure, &amp; let the same be the magnitude E. And foras<g ref="char:EOLhyphen"/>much as E measureth D, but D measureth the magnitudes A, B. Wherfore E also measureth the magnitudes A, B, and it also measureth the magnitude C. Wherfore E is a common measure to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he three magnitudes A, B, C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> I say also that it is the greatest commo<g ref="char:cmbAbbrStroke">̄</g> measure. For if it be possi<g ref="char:EOLhyphen"/>ble, let there be a magnitude, namely, F, grea<g ref="char:EOLhyphen"/>ter then the magnitude E. And let F measure the three magnitudes A, B, C. A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d forasmuch as F measureth the magnitudes A, B, C, it also measureth the two first magnitudes A, B. Wher<g ref="char:EOLunhyphen"/>fore (by the corollary going before) it shall also measure the greatest common measure of the magnitudes A, B. But the greatest common measure of the magnitudes A, B, is D. Wherfore F measureth D, and it also measureth C. Wherfore F measureth the magnitudes CD. Wher<g ref="char:EOLhyphen"/>fore F shall also measure the greatest common measure of the magnitudes C, D. But the grea<g ref="char:EOLhyphen"/>test common measure of the magnitudes C, D, is E. Wherfore F measureth E, namely, the greater, the lesse, which is impossible. Wherfore no magnitude greater then E measureth the magnitudes A, B, C. Wherfore E is the greatest common measure of the magnitudes ABC. If D do not measure the magnitude C. But if D do measure C, then is D the greatest commo<g ref="char:cmbAbbrStroke">̄</g> measure. Wherfore three magnitudes commensurable being geuen, there is found their grea<g ref="char:EOLhyphen"/>test common measure: which was required to be done.</p>
               <div type="section">
                  <pb n="235" facs="tcp:7062:264"/>
                  <head>¶ Corollary.</head>
                  <p>Hereby it is manifest, that if a magnitude measure three magnitudes,<note place="margin">A Corollary.</note> it shall also measure their greatest common measure. In like sort also in magnitudes comme<g ref="char:cmbAbbrStroke">̄</g>surable how many soeuer being geue<g ref="char:cmbAbbrStroke">̄</g>, may be found out their greatest common measure, and the corollary will euer be true.</p>
                  <div type="subsection">
                     <head>This Probleme also <hi>Montaureus</hi> reduceth into a Theoreme after this maner.<note place="margin">This Pro<g ref="char:EOLhyphen"/>bleme reduced to a Theo<g ref="char:EOLhyphen"/>reme.</note>
                     </head>
                     <p>Three magnitude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> being commensurable, if the greatest common measure to two of them, do measure the third it shall be the greatest common measure to all the three magnitudes geuen. But if it do not measure it, the greatest common measure of the third and of the greatest common measure of the two first, is the greatest common mea<g ref="char:EOLhyphen"/>sure of all the three magnitudes.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 3. Theoreme. The 5. Proposition. Magnitudes commensurable, haue such proportion the one to the other, as number hath to number.</head>
               <p>SVppose that A and B be magnitudes co<g ref="char:cmbAbbrStroke">̄</g>mensurable. Then I say that A hath vnto B such proportion as nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to number. For forasmuch as A and B are co<g ref="char:cmbAbbrStroke">̄</g>mensu<g ref="char:EOLhyphen"/>rable, therfore some magnitude measureth them, let there be a magnitude that mea<g ref="char:EOLunhyphen"/>sureth them, and let the same be C.<note place="margin">Construction.</note> And how often C measureth A, so many vnities let there be in the number D. And how often C measureth B, so many vnities let there be in the num<g ref="char:EOLhyphen"/>ber E, and let F be vnitie.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the magnitude C measureth the magnitude A by those vnities which are in the number D, and
<figure/>
vnitie F measureth the number D by those vnities which are in the number D: therefore how many times vnitie measureth the number D, so many times doth the magnitude C measure the magnitude A: wherfore as the magnitude C is to the magnitude A, so is vnitie F to the number D. Wherefore contrary wise (by the corollary of the fourth proposition of the fift booke) as the magnitude A is to the magnitude C, so is the number D to vnitie F. Againe forasmuch as the magnitude C measureth the magnitude B by those vnities which are in the number E, and vnitie F measureth the number E by those vnities which are in the number E: therfore how many times vni<g ref="char:EOLhyphen"/>tie F measureth the number E, so many tymes doth the magnitude C measure the magnitude B. Wherfore as the magnitude C is to the magni<g ref="char:EOLhyphen"/>tude B, so is vnitie F to the number E, and it is proued that as the magnitude A is to the magnitude C, so is the number D to vnitie F. Wherfore of equalitie (by the 22. of the fifte) as the magnitude A is to the magnitude B, so is the number D to the number E. Wherefore the commensurable magnitudes A and B haue that proportion the one to the other that the number D hath to the number E. Magnitudes therfore commensurable, haue such pro<g ref="char:EOLhyphen"/>portion the one to the other, as number hath to number: which was required to be proued.</p>
               <p>
                  <pb facs="tcp:7062:265"/>
                  <hi>Magnitudes are sayd to haue such proportion the one to the other, as nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to number,</hi>
                  <note place="margin">How magni<g ref="char:EOLhyphen"/>tudes are sayd to be in pro<g ref="char:EOLhyphen"/>portion the on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the o<g ref="char:EOLhyphen"/>ther, as num<g ref="char:EOLhyphen"/>ber is to num<g ref="char:EOLhyphen"/>ber.</note> 
                  <hi>when as what soeuer proportion is betwene those magnitudes, the same is found betwene some certaine numbers: as if a magnitude be vnto a magnitude either equall, as the number 2. is to the number 2, or double, as the number 4. to the number 2, or triple, as 6. to 2, or in any other multiplex proportion. And so also touching the other kindes o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportion, either superparticular, or superpartient.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 4. Theoreme. The 6. Proposition. I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two magnitudes haue such proportion the one to the other, as number hath to number: those magnitudes are commensurable.</head>
               <p>SVppose that these two magnitudes A and B haue that proportion the one to the other,<note place="margin">This pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>osi<g ref="char:EOLhyphen"/>tion is the conuerse of form<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r.</note> that the number D hath to the number E. Then I say, that the magni<g ref="char:EOLhyphen"/>tudes A, B, are commensurable. How many vnities there are in the number D, into so many equall partes deuide the magnitude A (by the 9. of the sixt) and let the magnitude C be equall to one o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the partes therof. And how many vnities there are in the number E, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so many magnitudes equall vnto the magnitude C let the magnitude F be co<g ref="char:cmbAbbrStroke">̄</g>posed. And let G be vnitie.<note place="margin">Con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>truction.</note> Now forasmuch as how many vnities there are in the nu<g ref="char:cmbAbbrStroke">̄</g>ber D, so many magnitudes also are there in the
<figure/>
magnit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de A equall vnto the magnitude C:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore what part vnite G is of the nu<g ref="char:cmbAbbrStroke">̄</g>ber D, the same part is the magnitude C of the mag<g ref="char:EOLhyphen"/>nitude A. Wherefore as the magnitude C is to the magnitude A, so is vnitie G to the num<g ref="char:EOLhyphen"/>ber D. But vnitie G measureth the number D. Wherefore the magnitude C also measu<g ref="char:EOLhyphen"/>reth the magnitude A. And for that as the magnitude C is to the magnitude A, so is v<g ref="char:EOLhyphen"/>nitie G to the number D: therefore contrary<g ref="char:EOLhyphen"/>wise (by the Corolla<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y of the fourth of the fift) as the magnitude A is to the magnitude C, so is the number D to vnitie G. Againe foras<g ref="char:EOLhyphen"/>much as how many vnities there are in the number E, so many magnitudes also are there in the magnitude F equall vnto the magni<g ref="char:EOLhyphen"/>tude C: therefore as the magnitude C is to the magnitude F, so is vnitie G to the number E. And it is proued that as the magnitude A is to the magnitude C, so is the number D to vni<g ref="char:EOLhyphen"/>tie G. Wherefore of equalitie (by the 22. of the fift) as the magnitude A is to the magnitude F, so is the number D to the number E. But as the number D is to the number E, so is the magnitude A to the magnitude B. Wherefore (by the 11. of the fift) as the magnitude A is to the magnitude B, so is the same magnitude A to the magnitude F: wherefore A hath vn<g ref="char:EOLhyphen"/>to either of these magnitudes B and F one and the same proportion. Wherefore (by the 9. of the fift) the magnitude B is equall vnto the magnitude F. But the magnitude C measureth the magnitude F: wherefore it also measureth the magnitude B: and it likewise measureth the magnitude A. Wherefore the magnitude C measureth the magnitudes A and B. Wher<g ref="char:EOLhyphen"/>fore the magnitudes A &amp; B are commensurable. If therefore two magnitudes haue such pro<g ref="char:EOLhyphen"/>portion the one to the other, as number hath to number, those magnitudes are co<g ref="char:cmbAbbrStroke">̄</g>mensurable: which was required to be proued.</p>
               <div type="section">
                  <pb n="236" facs="tcp:7062:265"/>
                  <head>Corollary.</head>
                  <p>Hereby it is manifest, that if there be two nu<g ref="char:cmbAbbrStroke">̄</g>bers,<note place="margin">A Corollary.</note> as D and
<figure/>
E, and a right line, as A, it is possible to geue an other line, vnto which the line A shall haue the same proportion, that the number D hath to the number E. For deuide the line A into so many equall partes as there are vnities in the number D (by the 9. of the sixt). And take an other line, as F, which let be composed of so many partes equall to the partes of the line A, as there be vnities in the number E. Wherefore the line A shall be to the line F, as the number D is to the number E. And by this meanes you may vnto any line geue<g ref="char:cmbAbbrStroke">̄</g> geue an other line com<g ref="char:EOLhyphen"/>mensurable in length. For if two lines be in proportion the one to the other, as number is to number, they shall also be commen<g ref="char:EOLhyphen"/>surable in length, by this 6. Theoreme.</p>
               </div>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>Two numbers being geuen, and also a right line: as the one number is to the other, so to make the square of y<hi rend="sup">e</hi> line geuen to be to the square of an other line.</p>
                  <p>Suppose that the numbers geuen be D and E: and let the right line geuen be A. It is required, as the number D is to the number E, so to make the square of the line A to be to the square of an other line. As the number D is to the number E, so let the line A be to the line F (by the former Corollary).<note place="margin">Construction.</note> And take
<figure/>
betwene those two lines A and F the meane proportionall (by the 13. of the sixt) which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>et be the line B:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now for that as the number D is to the number E, so is the line A to the line F: and as the line A is to the line F, so is the square of the line A to the square of the line B (by the second Corollary of the 20. of the sixt). Wherefore as the number D is to the number E, so is the square of the line A to the square of the line B: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>¶ An other demonstration of the 6. Proposition.</head>
                  <p>Suppose that these two magnitudes geuen A and B,
<figure/>
haue that proportion the one to the other, that the num<g ref="char:EOLhyphen"/>ber C hath to the nu<g ref="char:cmbAbbrStroke">̄</g>ber D. The<g ref="char:cmbAbbrStroke">̄</g> I say, that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se magni<g ref="char:EOLhyphen"/>tudes are comme<g ref="char:cmbAbbrStroke">̄</g>surable.<note place="margin">Construction.</note> How many vnities there are in the number C, into so many equall partes let the magni<g ref="char:EOLhyphen"/>tude A be deuided, &amp; let the magnitude E be equall vn<g ref="char:EOLhyphen"/>to one of those partes.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore as vnitie is to the num<g ref="char:EOLunhyphen"/>ber C, so is the magnitude E to the magnitude A. And as the number C is to the number D, so is the magnitude A to the magnitude B. Wherefore of equalitie (by the 22. of the fift) as vnitie is to the number D, so is the magni<g ref="char:EOLhyphen"/>tude E to the magnitude B. But vnitie measureth the
<pb facs="tcp:7062:266"/>
number D. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore the magnitude E measureth the magnitude B. And it also measu<g ref="char:EOLhyphen"/>reth the magnitude A (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or that vnitie measureth the number C). Wherefore the magnitude E measureth either of these magnitudes A and B. Wherefore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he magnitu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es A and B are commensurable, and the magnitude E is their common measure.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 5. Theoreme. The 7. Proposition. Magnitudes incommensurable, haue not that proportion the one to the o<g ref="char:EOLhyphen"/>ther, that number hath to number.</head>
               <p>SVppose that the magnitudes A and B be incommensurable. Then I say, that A hath not to B,<note place="margin">Demon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tra<g ref="char:EOLhyphen"/>tion leading to an ab<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>urdi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ie.</note> that proportion that number hath to number. For if A haue vn<g ref="char:EOLhyphen"/>to B that proportion that number hath to number, then is A comensurable vn<g ref="char:EOLhyphen"/>to B (by the 6. of this tenth). But (by supposition) it is not. Wherefore A hath not vnto B that proportion that number hath to
<figure/>
number. Magnitudes incommensurable therfore haue not that proportion the one to the other that number hath to number: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 6. Theoreme. The 8. Proposition. If two magnitudes haue not that proportion the one to the other that num<g ref="char:EOLhyphen"/>ber hath to number, those magnitudes are incommensurable.</head>
               <p>SVppose that these two magnitudes A and B, haue not that proportion the one to the other that number hath to number. Then I say, that A and B are magni<g ref="char:EOLhyphen"/>tudes incommensurable.<note place="margin">This is the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> de<g ref="char:EOLhyphen"/>mons<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ation.</note> For if A and B be commensurable, then shall A haue vnto B, that proportion that number hath to number (by the 5. of this tenth). But (by supposition) it hath not that proportion that
<figure/>
number hath to number. Wherefore A and B are incommensurable magnitudes. If therfore two mag<g ref="char:EOLhyphen"/>nitudes haue not that proportion the one to the o<g ref="char:EOLhyphen"/>ther that number hath to nu<g ref="char:cmbAbbrStroke">̄</g>ber, those magnitudes are incommensurable: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 7. Theoreme. The 9. Proposition. <milestone type="tcpmilestone" unit="unspecified" n="1"/> Squares described of right lines commensurable in length, haue that pro<g ref="char:EOLhyphen"/>portion the one to the other, that a square number hath to a square number. <milestone type="tcpmilestone" unit="unspecified" n="2"/> And squares which haue that proportion the one to the other that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, shall also haue their sides co<g ref="char:cmbAbbrStroke">̄</g>mensurable in <milestone type="tcpmilestone" unit="unspecified" n="3"/> length. But squares described of right lines incommensurable in length, haue not that proportion the one to the other, that a square number hath to <milestone type="tcpmilestone" unit="unspecified" n="4"/> a square number. And squares which haue not that proportion the one to the other that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square number, haue not their sides commensurable in length.</head>
               <p>
                  <pb n="237" facs="tcp:7062:266"/>SVppose that A and B be lines
<figure/>
co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length.<note place="margin">The first part demonstrat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d.</note> Then I say that the square of the line A hath vnto the square of the line B, that proportion that a square num<g ref="char:EOLhyphen"/>ber hath to a square number. For foras<g ref="char:EOLunhyphen"/>much as the lines A and B are commen<g ref="char:EOLunhyphen"/>surable in length: therefore the line A hath vnto the line B that proportion that number hath to number (by the <hi>5.</hi> of this tenth). Let the line A haue vnto the line B that proportion, that the number C hath to the number D, Now for that as the line A is to the line B, so is the number C to the number D: but the square of the line A is vnto the square of the line B in double proportio<g ref="char:cmbAbbrStroke">̄</g> of that which the line A is vnto the line B (for like rectiline figures (by the first corollary of the <hi>20.</hi> of the sixt) are in double proportion of that which the sides of like proportion are) and likewise the square number produced of C is to the square number produced of D, in double proportion of that which the number C is to the number D (for by the <hi>11.</hi> of the eight betwene two square numbers there is one meane proportionall number &amp; a square number is to a square number in double proportion of that which side is vnto side). Wherefore as the square of the line A is to the square of the line B, so is the square number produced of the number C, to the square number produced of the number D.</p>
               <div type="section">
                  <head>An other demonstration to proue the same.</head>
                  <p>Forasmuch as the lines A and B are commensurable, therefore (by the <hi>5</hi> of this tenth) A hath vnto B the same proportion that number hath to number.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the first part.</note> Let them haue that proportio<g ref="char:cmbAbbrStroke">̄</g> that the number C hath to the number D. And let the number C multiplying himselfe pro<g ref="char:EOLhyphen"/>duce the number E, and multiplying the number D, let it produce the number F: and let the number D multiplying himselfe produce the number G. And forasmuch as the number C multiplying himselfe produced the number E, and multiplying the number D it produced the number F: therefore (by the <hi>17.</hi> of the seuenth) as the number C is to the number D, that is, the line A to the
<figure/>
line B so is the num<g ref="char:EOLhyphen"/>ber E to the nu<g ref="char:cmbAbbrStroke">̄</g>ber F. But as the line A is to the line B, so is the square of the line A to the parallelograme contained vnder the lines A and B (by the first of the sixt) Wher<g ref="char:EOLunhyphen"/>fore as the square of the line A is to that which is contayned vnder the lines A
<pb facs="tcp:7062:267"/>
and B, so is the number E to the number F. Agayne for as muche as the number C multi<g ref="char:EOLhyphen"/>plying the number D produced the number F, &amp; the number D multiplying himselfe pro<g ref="char:EOLhyphen"/>duced the number G: therefore (by the <hi>17.</hi> of the seuenth) as the number C is to the num<g ref="char:EOLhyphen"/>ber D, that is, as the line A is to the line B, so is the number F to the number G. But as the line A is to the line B, so is parallellograme contained vnder the lines A and B to the square of the line B (by the first of the sixt). Wherefore as that which is contained vn<g ref="char:EOLhyphen"/>der the lines A and B is to the square of the line B, so is the number F to the number G. But as the square of the line A is to that which is contayned vnder the lines A and B, so is the number E to the number F. Wherefore of equalitie (by the <hi>22.</hi> of the fifte) as the square of the line A is to the square of the line B, so is the number E to the number G. But either of these numbers E and G is a square number. For E is produced of the number C multiplyed into him selfe, and G is produced of the number D multiplyed into him selfe. Wherefore the square of the line A hath vnto the square of the line B that proportion that a square number hath to a square number: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>An other demonstration of the same first part after <hi>Montaureus.</hi>
                  </head>
                  <p>Suppose that there be two lines commensurable in length A and B. Then I say that the squares described of those lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s shalbe in proportion the one to the other as a square number is to a square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber.<note place="margin">
                        <hi>An oth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>r de<g ref="char:EOLhyphen"/>mon<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the same first part after Montaureus.</hi>
                     </note> For forasmuch as the lines A and B are comme<g ref="char:cmbAbbrStroke">̄</g>surable in length, they shalbe in proportion the one to the other as number is to number (by the 5. of this booke). Let A be to B in duple proportion, which is in such proportion as number is to number, namely, as 4. is to 2, and 6. to 3, and so of many other. And (by the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cond o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the eight) take three least numbers in continuall proportion, and in duple propor<g ref="char:EOLunhyphen"/>tion, &amp; let the same be the numbers 4.2.1: whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore by
<figure/>
the corrollary of the second of the eight, the numbers 4. and 1. shalbe square numbers. (For as 4. is a square number produced of 2. multiplied into him selfe, so is 1. also a square number, for it is produced of vnitie multi<g ref="char:EOLhyphen"/>plied into him selfe.) I say moreouer that those are the square numbers, whose proportion the squares of the lines A and B haue the one to the other. For as the number 4. is to the number 2. so is the line A to the line B (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or either proportion is double by supposition): but as the line A is to the line B, so is the square of the line A to the parallelograme contained vnder the lines A and B (by the first of the sixt). Wherefore as the number 4. is to the number 2: so shall the square of the line A be to the parallelo<g ref="char:EOLhyphen"/>grame contained vnder the lines A and B. Likewise as the number 2. is to the number 1. so is the line A to the line B (For either proportion is duple by supposition): but as the line A is to the line B, so is the parallelograme contained vnder the lines A and B to the square of the line B (by the selfe same first of the sixt). Wherefore as the number 2. is to 1. so is the parallelograme contained vnder the lines A and B to the square of the line B: wherefore of equallitie (by the 22. of the fifth) as the square of the line A is to the square of the line B, so is the number 4. to 1. which are proued to be square numbers.</p>
                  <p>
                     <note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the se<g ref="char:EOLhyphen"/>conde part which is the co<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>erse of the former.</hi>
                     </note>
                     <hi>But now suppose that the square of
<figure/>
the line A, be vnto the square of the line B, as the square nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of the number C is to the square number produced of the number D. Then I say that the lines A &amp; B are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length. For for that as the square of the line A is to the square of the line B, so is the square number produced of the number C to the square number pro<g ref="char:EOLhyphen"/>duced of the number D: but the propor<g ref="char:EOLunhyphen"/>tion of the square of the line A, is vnto the square of the line B double to that proportion which the line A hath vnto the line B (by the corollary of the</hi> 20. <hi>of
<pb n="238" facs="tcp:7062:267"/>
the sixt). And the proportion of the square number which is produced of the number C to the square number produced of the number D is (by the</hi> 11. <hi>of the eight) double to that pro<g ref="char:EOLhyphen"/>portion which the number C hath vnto the number D. Wherefore as the line A is to the line B, so is the number G to the number D. Wherefore the line A hath vnto the line B the same proportion that the number C hath to the number D. Wherefore (by the 6. of this booke) the lines A and B are commensurable in length: which was required to be proued.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>An other demonstration to proue the same.</head>
                  <p>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>Vppose againe that the square of the line A haue vnto the square of the line B,<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of the second part.</note> the same proportion, that the square number E hath to the square number G. Then I say that the lines A and B are commensurable in length. For suppose that the side of the square number E
<figure/>
be the nu<g ref="char:cmbAbbrStroke">̄</g>ber C, &amp; let the side of the square number G be the nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber D: and let the number C multiply<g ref="char:EOLhyphen"/>ing the number D produce the number F. Wherefore these numbers E, F, G, are in continuall propor<g ref="char:EOLhyphen"/>tion, aud in the same proportion that the number C is to the number D (by the <hi>17.</hi> and <hi>18.</hi> of the seuenth).<note place="margin">This Assump<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> followeth as a Corollary of the <hi>25</hi>: but (so as it might al<g ref="char:EOLhyphen"/>so be here in Methode, pla<g ref="char:EOLhyphen"/>ced) you shall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>inde it after the <hi>53.</hi> of this booke, abso<g ref="char:EOLhyphen"/>lutely demon<g ref="char:EOLhyphen"/>strated: for there it ser<g ref="char:EOLhyphen"/>ueth to the <hi>54.</hi> his demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the meane proportionall betwene the squares of the lines A and B is that which is contained vnder the lines A and B. (Which though it might briefely be proued: yet we take it as now). And likewise the meane proportionall betwene the numbers E and G is the number F (by the <hi>20.</hi> of the seuenth): therefore as the square of the line A is to that which is contained vnder the lines A and B, so is the number E to the num<g ref="char:EOLhyphen"/>ber F, and as that which is contained vnder the lines A and B is to the square of the line B, so is the number F to the number G. But as the square of the line A is to that which is con<g ref="char:EOLhyphen"/>tained vnder the lines A and B, so is the line A to the line B (by the first of the sixt). Where<g ref="char:EOLhyphen"/>fore the lines A and B are in the same proportion that the number E is to the number F, that is, that the number C is to the number D. Wherefore the lines A and B are commensu<g ref="char:EOLhyphen"/>rable in length (by the 6. of this tenth) which was required to be proued.</p>
                  <p>But now suppose that the lines A and B be incom<g ref="char:EOLhyphen"/>mensurable
<figure/>
in length. Then I say that the square of the line A hath not vnto the square of the line B that proportion that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber.<note place="margin">Demo<g ref="char:cmbAbbrStroke">̄</g>stratio<g ref="char:cmbAbbrStroke">̄</g> of the third part.</note> For if the square of the line A haue vnto the square of the line B, the same proportion that a square number hath to a square number, the<g ref="char:cmbAbbrStroke">̄</g> shall the lines A and B be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length (by the second part of this proportion). But by supposition they are not. Wherefore the square of the line A, hath not vnto the square of the lyne B that proportion that a square number hath to a square number: which was requi<g ref="char:EOLhyphen"/>red to be proued.<note place="margin">Demo<g ref="char:cmbAbbrStroke">̄</g>stratio<g ref="char:cmbAbbrStroke">̄</g> of the fourth part which is the co<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>uerse of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</note>
                  </p>
                  <p>Againe suppose that the square of the line A haue not vnto the square of the line B, the
<pb facs="tcp:7062:268"/>
same proportion that a square number hath to a square
<figure/>
number. Then I say that the lines A and B are incom<g ref="char:EOLhyphen"/>mensurable in length. For if the lines A and B be com<g ref="char:EOLhyphen"/>mensurable in length, then the square of the line A should haue vnto the square of the line B, the same proportion that a square number hath to a square number, by the first part of this proposition, but by supposition it hath not, wherfore the lines A and B are not commensurable in length.<note place="margin">Conclusion of the whole pro<g ref="char:EOLhyphen"/>position.</note> Wher<g ref="char:EOLhyphen"/>fore they are incomensurable in length. Wherfore squares made of right lines commensura- in length, haue that proportion the one to the other, that a square number hath to a square number. And squares which haue that proportion the one to the other, that a square number hath to a squa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e number, shall also haue the sides commensurable in length. But squares de<g ref="char:EOLhyphen"/>scribed of right lines incommensurable in length, haue not that proportion the one to the o<g ref="char:EOLhyphen"/>ther that a square number hath to a square number. And squares which haue not that pro<g ref="char:EOLhyphen"/>po<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tion the one to the other, that a s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uare number hath to a square number, haue not also their sides comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nsurable in length: which was all that was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ Corrollary.</head>
                  <p>
                     <q>Hereby it is manifest, that right lines co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length, are also euer commensurable in power. But right lines commensurable in power, are not alwayes commensurable in length.<note place="margin">
                           <hi>A Cor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>llary.</hi>
                        </note> And right lines inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth are not alwayes incommensurable in power. But right lines incommensura<g ref="char:EOLhyphen"/>ble in power, are euer also incommensurable in length.</q>
                  </p>
                  <p>
                     <hi>For forasmu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h as squares made of right lines commensurable in length, haue that pro<g ref="char:EOLhyphen"/>portion the one to the other,</hi>
                     <note place="margin">
                        <hi>Pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e of the first part of the Corollary.</hi>
                     </note> 
                     <hi>that a square number hath to a square number (by the first part of this proposition), but magnitudes which haue that proportion the one to the other, that number simply hath to number, are (by the sixt of the tenth) commensurable. Wherfore right lines commensurable in length, are commensurable not onely in length, but also in power.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Profe of the second part.</hi>
                     </note>
                     <hi>Againe forasmuch as there are certaine squares which haue not that proportion the one to the other that a square number hath to a square number, but yet haue that proportion the one to the other which number simply hath to number: their sides in dede are in power com<g ref="char:EOLhyphen"/>mensurable, for that they describe squares which haue that proportion which number sim<g ref="char:EOLhyphen"/>ply hath to number, which squares are therfore commensurable (by the 6. of this booke): but the said sides are incommensurable in length by the latter part of this proposition. Wher<g ref="char:EOLunhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>fore it is t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ue that lines commensurable in power, are not straight way commensurable in length also.</hi>
                     <note place="margin">
                        <hi>Profe of the third p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rt.</hi>
                     </note>
                  </p>
                  <p>
                     <hi>And by the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e same reason is proued also that third part of the corollary, that lines incommensurable in length, are not alwayes incommensurable in power. For they may be in<g ref="char:EOLhyphen"/>comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nsurable in length, but yet commensurable in power. As in those squares which are in proportion the one to the other, as number is to number, but not as a square number is to a square number.</hi>
                  </p>
                  <p>
                     <note place="margin">
                        <hi>Pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the fourth part.</hi>
                     </note>
                     <hi>But right lines incommensurable in power, are alwayes also incommensurable in length. For i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> they be commensurable in length, they shal also be commensurable in power by the first part of this Corollary. But they are supposed to be incommensurable in length, which is ab<g ref="char:EOLhyphen"/>surde. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ore right lines incommensurable in power, are euer incommensurable in length<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                  </p>
                  <p>
                     <pb n="239" facs="tcp:7062:268"/>For the better vnderstanding of this proposition and the other following, I haue here added certayne annotacions taken out of <hi>Montaureus.</hi>
                     <note place="margin">
                        <hi>Certayne an<g ref="char:EOLhyphen"/>notations <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ut of Montau<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>us.</hi>
                     </note> And first as touching the signi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ication o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wordes and termes herein vsed, wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ch ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> such, that vnlesse they be well marked and peysed, the matter will be obscure and hard, and in a maner inexplicable.</p>
                  <p>First, this ye must note, that lines to be commensurable in length, and lines to be in proportion the one to the other, as number is to number is all one. So that whatsoeuer lines are commensurable in length, are also in proportion the one to the other, as num<g ref="char:EOLhyphen"/>ber is to number. And conuersedly what so euer lynes are in proportion the one to the other, as number is to number, are also comme<g ref="char:cmbAbbrStroke">̄</g>surable in length, as it is manifest by the 5 and 6 of this booke. Likewise lines to be incommensurable in length, and not to be in proportion the one to the other, as number is to number is all one, as it is manifest by the 7. and 8. of this booke. Wherfore that which is sayd in this Theoreme, ought to be vnderstand of lines commensurable in length, and incommensurable in length.</p>
                  <p>This moreouer is to be noted, that it is not all one, numbers to be square numbers, and to be in proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, as a square number is to a square number. For although square numbers be in proportion the one to the other, as a square num<g ref="char:EOLhyphen"/>ber is to a square number, yet are not all those numbers which are in proportion the one to the other, as a square number is to a square number, square numbers. For they may be like superficiall numbers, and yet not square numbers, which yet are in propor<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, as a square number is to a square number. Although all square numbers are like superficiall numbers. For betwene two square numbers there <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>alleth one meane proportionall number (by the 11. of the eight). But if betwene two num<g ref="char:EOLhyphen"/>bers, there fall one meane proportionall number, those two numbers are like superfici<g ref="char:EOLhyphen"/>all numbers (by the 20. of the eight). So also if two numbers be in proportion the one to the other, as a square number is to a square number, they shall be like superficiall nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers by the first corollary added after the last proposition of the eight booke.</p>
                  <p>And now to know whether two superficiall numbers geuen, be like superficiall num<g ref="char:EOLhyphen"/>bers or no,<note place="margin">
                        <hi>Rules to know whether two superficiall numbers be like or no.</hi>
                     </note> it is thus found out. First if betwene the two numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rs geuen, there fall no meane proportionall, then are not these two numbers like superficiall numbers (by the 18. of the eight. But if there do fall betwene them a meane proportionall, then are they like super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iciall numbers (by the 20. of the eight) Moreouer two like superficiall num<g ref="char:EOLhyphen"/>bers multiplied the one into the other, do produce a square number (by the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the ninth). Wherfore if they do not produce a square number, then are they not like super<g ref="char:EOLhyphen"/>ficiall numbers. And if the one being multiplied into the other, they produce a square number, then are they like superficiall (by the 2. of the ninth). Moreouer if the said two superficial numbers be in superperticular, or superbipartient proportion, then are they not like superficiall numbers. For if they should be like, then should there be a meane proportionall betwene them (by the 20. of the eight). But that is contrary to the Co<g ref="char:EOLhyphen"/>rollary of the 20. of the eight.</p>
                  <p>And the easilier to conceiue the demonstrations following, take this example of that which we haue sayd.</p>
                  <p>Suppose that there be a line, namely, C, which imagine to be foure fote long: and let there be an other line D, which let be three foote long. And (by the 13. of the sixt) take the meane propor<g ref="char:EOLhyphen"/>tionall betwene the lines A, D, which let be the line B. Wherefore the square of the line B shall be equall to the rectangle parallelogramme contayned vnder the line C and D (by the 17. of the sixt). Which square shall contayne 12. foote, &amp; so much also shall the parallelogramme described of the lines C &amp; D containe. Take also two other lines E and F, of which let E be 3. foote long, and let F be a foote long. And let the meane proportionall betwene the lines E and F, be the line A. Now then the square of the line A shall containe 3. foote, as also doth the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me described of the lines E, F. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the square of the line B, which co<g ref="char:cmbAbbrStroke">̄</g>taineth 12. foote, is to the square of the line A, which contayneth 3. foote, in that proportion that a square number is to a square number. For as the number 12. is to the number 3, so is the square of the line B, which containeth 12. foote, to the square of the line A, which contayneth 3 foote. But the numbers 12. and 3. are like superficiall numbers, for the sides of 12. which are 2. and 6, are proportionall with the sides of 3. which are 1. and 3. Wherefore the square of the line B, which contayneth 12. foote, shall be vnto the square of the line A, which contayneth 3. foote, in that proportion that a like superficiall number is to a like superfici<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll number. But like superficiall numbers are in proportion the one to the other, as a square number i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb facs="tcp:7062:269"/>
to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber,
<figure/>
which square num<g ref="char:EOLhyphen"/>bers are 4. and 1. (by the 26. of the eight). Wherefore the square of the line B, which con<g ref="char:EOLhyphen"/>tayneth 12. foote, is to the square of the line A, which co<g ref="char:cmbAbbrStroke">̄</g>tai<g ref="char:EOLhyphen"/>neth 3. foote, in that proportion that a square number is to a square number, namely, that the number 4. is to the number 1: which proportion is qua<g ref="char:EOLhyphen"/>druple. For the greater square whi<g ref="char:EOLhyphen"/>che is 12, contay<g ref="char:EOLhyphen"/>neth the lesse square which is 3, foure times. Wherefore the side of the square 12, which is the line B, is double to the side of the square 3, which is the line A. Wher<g ref="char:EOLhyphen"/>fore the line B is to the line A, in that proportion that number is to num<g ref="char:EOLhyphen"/>ber. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e (by the 5. of this booke) the lines B &amp; A are commensurable in length. Which is a supposition necessary to conclude the first part of this Theoreme, namely, that the squares of such lines are in proportion the one to the other, that a square number is to a square number.</p>
                  <p>So also the nu<g ref="char:cmbAbbrStroke">̄</g>ber which denominateth the greater terme of the proportion of the line B to the line A, which is 2, if it be multiplyed into it selfe, it maketh a square number, namely, 4. Likewise the num<g ref="char:EOLhyphen"/>ber which denominateth the lesse terme, namely, 1. if it be multiplyed into it selfe, it maketh no more but 1. Which vnitie is also in power a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore the square of the line B, is to the square of the line A, in that proportion that a square number is to a square number, namely, that 4. is to 1. By this you see (which thing was before noted) that it is not all one, numbers to be square numbers, and to be in proportion the one to the other, as a square number is to a square number. For it is manifest, that the numbers 12. and 3. are not square numbers, when yet the squares expressed by those num<g ref="char:EOLhyphen"/>bers are in that proportion. But the side of the square 12. although it can not of it selfe be expressed by number distinctly, to say that the side thereof is so many foote long, which feete square taken, make the whole square 12: yet b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ing referred or compared to an other thyng, namely, to the side of the square 3, which side also of it selfe can not be expressed by number, it is vnto the sayde side of the square 3, in double proportion. For the one square being quadruple to the other square (as is the square of the line B, which contayneth 12. foote, to the square of the line A, which contay<g ref="char:EOLhyphen"/>neth 3. foote) hath his side double to the side of the other square, by this generall Corollary of the 10. of the sixt, <hi>like rectiline figures are in double proportion the one to the other that their sides of like proportion are.</hi> Now if a man will say, that the side of the square 12. may be measured, for that hys proportion which it hath to the side of the square 3, is measured by 2 (forasmuch as it is dupla propor<g ref="char:EOLhyphen"/>tion: this is to be considered, that in so saying, you say not, that that magnitude can of it selfe be measu<g ref="char:EOLhyphen"/>red, but the proportion therof. For, that magnitude, namely, the side of the square 12, should by it selfe be measured, when without any respect of the proportion of it to an other thing, we may say that the side of the square, which contayneth 12. foote, is so many foote long, the number of which foote multi<g ref="char:EOLhyphen"/>plyed into it selfe should make that number 12. But this is not possible, for that 12. is not a square num<g ref="char:EOLhyphen"/>ber. Wherefore thus you may say: In asmuch as that square 12. is considered by it self, without hauing any respect of the proportion of it to any other thing, but onely as it is 11. foote, it hath no side which of it selfe can be expressed by number. But if it be compared to any other thing, namely, to the square of 3. foote, then may you say that the side of the square 12. is 2, and the side of the square 3. is 1. But thys is the denomination of that propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tion which is called duple, which proportion can not be or considered
<pb n="240" facs="tcp:7062:269"/>
in fewer termes then two, when as it is a relation of one thing to an other thing: wherefore 2. is not the number of such feete, of which there are 12. in the square. Agayne, if the number 2. shoulde be the side of the square 12, so that that side should be 2, then of the multiplication of 2. into it selfe, shoulde not be made that square 12, but an other square which should be 4. foote: as of the number 2. multiply<g ref="char:EOLhyphen"/>ed into him selfe is produced the square number 4. Neither also if any other number. measure the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the square 12, and the sayd number be multiplyed into him selfe shall it euer make the number 12. When yet all numbers denominating the side of any square number, if they be multiplyed into them selues, they make the number which denominateth the square, whose sides they denominate. As 2. multiplyed into him selfe maketh 4 : 3. maketh 9 : 4. maketh 16: and so likewise of all others. Where<g ref="char:EOLhyphen"/>fore it is not all one, magnitudes to be in proportion the one to the other, as number is to number, and euery one of them to be measured b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> him selfe without any respect had of the proportion. As here the side of the square 12. can of it selfe by no meanes be measured, but being compared to any other magni<g ref="char:EOLhyphen"/>tude, namely, to the side of the square 3, the proportion thereof is expressed by number. So also the side of the square 3, and of all other square figures, whose areas yet can not be expressed by square numbers. And that which we here say, is manifest euen by the wordes of <hi>Euclide</hi> in the 5.6. 7. and 8. Theoremes of this booke. Where he sayth not, that magnitudes commensurable and incommensurable are of the<g ref="char:cmbAbbrStroke">̄</g> selues or of their owne nature expressed by numbers,<note place="margin">
                        <hi>Note.</hi>
                     </note> but that either they haue or haue not that propor<g ref="char:EOLhyphen"/>tion which number hath to number. Which thing not being well considered, it should seme hath cau<g ref="char:EOLhyphen"/>sed many to erre as hereafter shall be made manifest. And in deede they which haue demonstrated this Theoreme, may seme to some rather to haue demo<g ref="char:cmbAbbrStroke">̄</g>strated it particularly &amp; not vniuersally. And doubt<g ref="char:EOLhyphen"/>les I iudge there are some which vnderstand their sayinges otherwise then they ment: when as they thinke, that they suppose certayne lines not onely commensurable in length, as they are supposed to be in the Propositio<g ref="char:cmbAbbrStroke">̄</g>, but also such, that ech of
<figure/>
them apa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> may be expressed by some cer<g ref="char:EOLhyphen"/>tayne number. Wherfore for want of right vndersta<g ref="char:cmbAbbrStroke">̄</g>ding, this mought they say of their demonstrations: that wheras they thought that they had concluded that generally, which is in this theoreme of <hi>Euclide, Squares described of lines co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length, are in proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other that a square nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber is to a square number:</hi> they conclude par<g ref="char:EOLhyphen"/>ticularly, thys onely: <hi>Squares described of lines which may by them selues be expressed by some certaine number, are in proportion. &amp;c.</hi> which yet is otherwise, and their demo<g ref="char:cmbAbbrStroke">̄</g>stra<g ref="char:EOLhyphen"/>tions are right &amp; agreable with the Theoreme. Onely the picture of the figures which the Greeke boke hath, may seeme to bring some doubt. For the squares are so described with certayn litle areas, that the number of them may be denominated by a square number: whereby it mought seeme that the lines A &amp; B which describe the squares, ought to
<figure/>
be such that they may be expressed by some certaine nu<g ref="char:cmbAbbrStroke">̄</g>ber. As the line A to be 5. foote, and the line B 3. foote. As the two former fi<g ref="char:EOLunhyphen"/>gures here set declare. Which thyng yet <hi>Eu<g ref="char:EOLhyphen"/>clide</hi> supposeth not, but only requireth that they be commensurable in length, as in the former example of the two squares, the whole area of one of which is 12, and the whole area of the other is 3. For although their sides ca<g ref="char:cmbAbbrStroke">̄</g> not by them selues be expres<g ref="char:EOLhyphen"/>sed by some certayne number, yet are they commensurable in length. Moreouer thys describing of the squares of the lines A and B, deuided by certaine litle areas, may cause this error, that a man should thinke that it is all o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e two numbers to be square numbers, and to be in proportion the one to the other as a square number is to a square number. For the number of the areas in the square of the line A is a square number, namely, 25. produced of the roote 5, which is the length of the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Likewise the number of the areas of the square of the line B, is a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, namely, 9, which is pro<g ref="char:EOLhyphen"/>duced of the roote 3, which is the length of the line B. But we haue before declared that it is not all one, numbers to be called square numbers, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d to be in proportion the one to the other, as a square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber is to a square number. Wherefore as touching those areas contayned in the greater square, which is of the line A, and which are in number 25, they do expresse that square number 25, which is produced of the number 5. multiplyed into him selfe, which number 5. is the greater extreme of the proportion betwene 5. and 3, which is the proportion of the lines A and B. And this proportion, namely, of 5. to 3. causeth that the lines A and B are commensurable in length (by the 6. of this booke). The same may be sayd also of the areas of the lesse square. Neither is it of necessitie that you vnderstand those area<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to be squares, as either fe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>te square or pases square which make the whole square, although in deede they
<pb facs="tcp:7062:270"/>
may be such, so that the sides of those squares be so many foote long, as 5. foote or 3. foote. Howbeit thys is of necessitie that the numbers which expresse the number of the feete square or pases square, co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tayned in the squares, be either both of the
<figure/>
square numbers, as in these square figures of the lines A, B, or that both of them be like supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ficiall numbers, as in the former squares which were 12. and 3: of which nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>bers it is manifest by that which hath be<g ref="char:EOLhyphen"/>fore bene said, that they are like superficiall numbers, and therefore haue that propor<g ref="char:EOLhyphen"/>tion the one to the other, that a square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber hath to a square number. And therfore you may describe the squares of the lines A, B, without any distinction of such litle areas, so that the squares may be voyde and emptie, and contayned onely of foure right lines, as in the figure here put.</p>
               </div>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>Forasmuch as in the eight booke in the 26. proposition it was proued, that like playne numbers haue that proportion the one to the other, that a square number hath to a square number: and likewise in the <hi>24.</hi> of the same booke it was proued, that if two numbers haue that proportion the one to the other,<note place="margin">This assumpt is the conuerse of the <hi>26.</hi> of the eight.</note> that a square number hath to a square number, those numbers are like plaine numbers. Hereby it is manifest, that vnlike plaine numbers, that is, whose sides are not proportionall, haue not that proportion the one to the other, that a square number hath to a square number. For if they haue, then should they be like plaine numbers, which is contrary to the supposition. Wherfore vnlike plaine numbers haue not that propor<g ref="char:EOLhyphen"/>tion the one to the other, that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. And therfore squares which haue that proportion the one to the other, that vnlike plaine numbers haue, shall haue their sides incommensurable in length (by the last part of the former proposition) for that those squares haue not that proportion the one to the other that a square number hath to a square number.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 8. Theoreme. The 10. Proposition. If foure magnitudes be proportionall, and if the first be commensurable vnto the second, the third also shal be commensurable vnto the fourth. And if the first be incommensurable vnto the second, the third shall also be in<g ref="char:EOLhyphen"/>commensurable vnto the fourth.</head>
               <p>SVppose that these foure magnitudes A, B, C, D, be proportionall. As A is to B, so let C be to D, and let A be commensurable vnto B. Then I say that C is also commensurable vnto D.<note place="margin">Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>io<g ref="char:cmbAbbrStroke">̄</g> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the first part.</note> For forasmuch as A is commensurable vnto B, it hath (by the fift of the tenth) that proportion that number hath to number. But as A is to B, so is C to D. Wherfore C also hath vnto D that pro<g ref="char:EOLunhyphen"/>portion
<figure/>
that number hath to number. Wherfore C is commen<g ref="char:EOLhyphen"/>surable vnto D (by the 6. of the tenth). But now suppose that the magnitude A be incommensurable vnto the magnitude B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> Then I say that the magnitude C also is incommensurable vnto the magnitude D. For forasmuch as A is incommensura<g ref="char:EOLhyphen"/>ble vnto B, therfore (by the 7. of this booke) A hath not vnto B such proportion as number hath to number. But as A is to B, so is C to D. Wherefore C hath not vnto D such proportion as number hath to number. Wherfore (by the 8. of the tenth) C is incommensurable vnto D. If therefore there be foure magni<g ref="char:EOLhyphen"/>tudes proportionall, and if the first be commensurable vnto the
<pb n="241" facs="tcp:7062:270"/>
second, the third also shall be commensurable vnto the fourth. And if the first be incommen<g ref="char:EOLhyphen"/>surable vnto the second, the third shall also be incommensurable vnto the fourth: which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary added by Montaureus.</head>
                  <p>
                     <hi>If there be foure lines proportionall, and if the two first, or the two last be commensurable in power onely, the other two also shall be commensurable in power onely.</hi>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note> This is proued by the 22. of the sixt, and by this tenth proposition. And this Corollary Euclide vseth in the 27. and 28. propositions of this booke, and in other propositions also.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 3. Probleme. The 11. Proposition. Vnto a right line first set and geuen (which is called a rationall line) to finde out two right lines incommensurable, the one in length onely, and the other in length and also in power.</head>
               <p>SVppose that the right line first set and geuen, which is called a rationall line of purpose be A. It is required vnto the said line A, to finde out two right lines in<g ref="char:EOLhyphen"/>commensurable, the one in length onely, the other both in length and in power.<note place="margin">To finde out the first line incommensu<g ref="char:EOLhyphen"/>rable in length onely to the line geuen.</note> Take (by that which was added after the 9. proposition of this booke) two num<g ref="char:EOLhyphen"/>bers B and C, not hauing that proportion the one to the other, that a square number hath to a square number, that is, let them not be like plaine numbers (for like plaine numbers by the <hi>26.</hi> of the eight haue that proportion the one to the other that a square number hath to a square number). And as the number B is to the number C, so let the square of the line A be vnto the square of an other line, namely, of D (how to do this was taught in the assumpt put before the 6. proposition of this booke.) Wherfore the square of the line A, is vnto the square of the line D commensurable (by the sixt of the tenth.)
<figure/>
And forasmuch as the number B hath not vnto the num<g ref="char:EOLunhyphen"/>ber C, that proportion that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, therfore the square of the line A hath not vnto the square of y<hi rend="sup">e</hi> line D, that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore by the 9. of the tenth, the line A is vnto the line D incommensurable in length onely. And so is found out the first line, namely, D incom<g ref="char:EOLhyphen"/>mensurable in length onely to the line geuen A.<note place="margin">To finde out the second line incommensu<g ref="char:EOLhyphen"/>rable both in length and in power to the line geuen.</note> Agayne take (by the 13. of the sixt) the meane proportionall be<g ref="char:EOLhyphen"/>twene the lines A and D, and let the same be E. Wherfore as the line A is to the line D, so is the square of the line A to the square of the line E (by the Corollary of the 20. of the sixt). But the line A is vnto the line D incommen<g ref="char:EOLhyphen"/>surable in length. Wherfore also the square of the line A is vnto the square of the line E incommensurable by the second part of the former proposition. Now forasmuch as the square of the line A is inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable to the square of the line E, it followeth (by the definition of incommensurable lynes) that the line A is incommensurable in power to the line E. Wherfore vnto the right line ge<g ref="char:EOLhyphen"/>uen, and first set, A, which is a rationall line, and which is supposed to haue such diuisions and so many partes as ye list to conceyue in minde, as in this example 11, whereunto, as was de<g ref="char:EOLhyphen"/>clared in the 5. definition of this booke, may be compared infinite other lines, either commen<g ref="char:EOLhyphen"/>surable
<pb facs="tcp:7062:271"/>
or incommensurable, is found out the line D incommensurable in length onely. Wher<g ref="char:EOLunhyphen"/>fore the line D is rationall (by the sixt definitio<g ref="char:cmbAbbrStroke">̄</g> of this booke) for that it is incommensura<g ref="char:EOLhyphen"/>ble in length onely to the line A, which is the first line set, and is by suppositio<g ref="char:cmbAbbrStroke">̄</g> rational. There is also found out the line E, which is vnto the same line A incommensurable, not onely in length but also in power, which line E compared to the rationall line A, is by the definition irrationall. For Euclide alwayes calleth those lines irrationall, which are incommensurable both in length and in power to the line first set, and by supposition rationall.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 12. Proposition. Magnitudes commensurable to one and the selfe same magnitude: are also commensurable the one to the other.</head>
               <p>SVppose that either of these magnitudes A and B, be commensurable vnto the mag<g ref="char:EOLhyphen"/>nitude C: Then I say that the magnitude A is commensurable vnto the magnitude B.<note place="margin">Construction.</note> For <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as the magnitude A is commensurable vnto the magnitude C, therefore (by the <hi>5.</hi> of the tenth) A hath vnto C such
<figure/>
proportion as number hath to number. Let A haue vnto C that proportion that the number D hath to the number E. Againe forasmuch as B is commen<g ref="char:EOLunhyphen"/>su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>able vnto C, therefore (by the selfe same) C hath vnto B that proportion that number hath to num<g ref="char:EOLhyphen"/>ber. Let C haue vnto B that proportion that the number F hath vnto the number G. Now then take the least numbers in continuall proportion and in these proportions geuen, namely, that the number D hath to the number E, and that the number F hath to the number G (by the <hi>4.</hi> of the eight): which let be the numbers <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, K, L.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> So that as the number D is to the number E, so let the number H be to the number K, and as the nu<g ref="char:cmbAbbrStroke">̄</g>ber F is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber G, so let the nu<g ref="char:cmbAbbrStroke">̄</g>ber K be to the nu<g ref="char:cmbAbbrStroke">̄</g>ber L. Now for that as A is to C, so is D to E, but as D is to E so is H to K, therfore as A is to C, so is H to K. Againe for that as C is to B, so is F to G, but as F is to G so is K to L: therefore as C is to B, so is K to L. But it is now proued that as A is to C, so is H to K. Wherefore of equalitie (by the <hi>22.</hi> of the fift) as A is to B, so is the number H to the number L. Wherefore A hath vnto B such propor<g ref="char:EOLhyphen"/>tion as number hath to number. Wherefore (by the sixt of the tenth) the magnitude A is com<g ref="char:EOLunhyphen"/>mensurable vnto the magnitude B. Magnitudes therefore commensurable to one and the selfe same magnitude,<note place="margin">T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>is is wi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h Zambert an A<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>mpt, but v<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ly impro<g ref="char:EOLhyphen"/>perly: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ssate<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> ma<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eth i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a Co<g ref="char:EOLhyphen"/>rollary, but the Gree<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e and Montaureus ma<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e it a pro<g ref="char:EOLhyphen"/>position: but e<g ref="char:EOLhyphen"/>uery way an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nfallible truth <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>.</note> are also commensurable the one to the other: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>
                     <q>If there be two magnitudes compared to one and the selfe same magnitude, and if the one of them be commensurable vnto it, and the other incommen<g ref="char:EOLhyphen"/>surable:
those magnitudes are incommensurable the one to the other.</q>
                  </p>
                  <p>
                     <pb n="242" facs="tcp:7062:271"/>SVppose that there be two magnitudes, namely,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> A and B and let C be a certayne other magnitude. And let A <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e commensurable
<figure/>
vnto C, and let B be comme<g ref="char:cmbAbbrStroke">̄</g>surable vnto the selfe same C. Then I say that the magnitude A is in<g ref="char:EOLhyphen"/>commensurable vnto B. For if A be commensurable vnto B, forasmuch as A is also comme<g ref="char:cmbAbbrStroke">̄</g>surable vnto C therefore (by the <hi>12.</hi> of the tenth) B is also comme<g ref="char:cmbAbbrStroke">̄</g>sura<g ref="char:EOLhyphen"/>ble vnto C: which is contrary to the supposition.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 10. Theoreme. The 13. Proposition. If there be two magnitudes commensurable, and if the one of them be in<g ref="char:EOLhyphen"/>commensurable to any other magnitude: the other also shall be incommen<g ref="char:EOLhyphen"/>surable vnto the same.</head>
               <p>SVppose that these two magnitudes A, B be commensurable the one to the other, and let the one of them, namely A, be incommensurable vnto an other magnitude,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurdit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</note> namely, vnto C. Then I say that the other magnitude also, namely B, is incommen<g ref="char:EOLhyphen"/>surable vnto C. For if B be commensurable vnto C,
<figure/>
then forasmuch as A is commensurable vnto B, therefore (by the <hi>12.</hi> of the tenth) the magnitude A also is commensurable vnto the magnitude C. But it is supposed to be incommensurable vnto it, which is impossible. Wherefore the magnitudes B and C are not commensurable. Wherefore they are incommensurable. If therefore there be two magni<g ref="char:EOLhyphen"/>tudes commensurable, and if the one of them be incommensurable to any other magnitude, the other also shalbe incommensurable vnto the same: which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary added by Montaureus.</head>
                  <p>Magnitudes commensurable to magnitudes incomme<g ref="char:cmbAbbrStroke">̄</g>surable, are also incommen<g ref="char:EOLhyphen"/>surable the one to the other.<note place="margin">
                        <hi>A Corollary.</hi>
                     </note>
                  </p>
                  <p>Suppose that the magnitudes A and B be incommensurable the one to the other, and let the mag<g ref="char:EOLhyphen"/>nitud<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> C be co<g ref="char:cmbAbbrStroke">̄</g>mensurable to A, and let the magnitude D be co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>mensurable
<figure/>
vnto B. Then I say that the magnitu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>s C and D are incommensurable the one to the other. For A and C are commen<g ref="char:EOLhyphen"/>surable, of which the magnitude A is incommensurable vnto B, wherefore by this 13. proposition the magnitudes C and B are also incommensurable: but the magnitude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> B and D are co<g ref="char:cmbAbbrStroke">̄</g>mensurable wherefore by the same, or by the former assumpt, the magnitudes C and D are incommensurable the one to the other. This corolla<g ref="char:EOLhyphen"/>ry, <hi>Theon</hi> vseth often times as in the 22. 26. and 36 propositions of this booke, and in other propositions also.</p>
               </div>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>Two vnequall right lines being geuen, to fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>de out how much the greater is in power more then the lesse.</p>
                  <p>Suppos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that the two vnequall right lines geuen, be AB and C, of which let AB be the
<pb facs="tcp:7062:272"/>
greater. It is required to finde out how much more in power the line AB is then the line C. Describe vpon the line AB a semicircle ADB.<note place="margin">Construction</note> And vnto it from the point A apply (by the first of the fourth) a right line AD, equall vnto the line
<figure/>
C, and draw a right line from D to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now it is mani<g ref="char:EOLhyphen"/>fest that the angle ADB is a right angle (by the 31. of the third): and that the line AB is in power more then the line AD, that is, then the line C, by the line DB, by the 47. of the first.</p>
                  <p>
                     <note place="margin">A Corollary.</note>
                     <hi>And like in sorte,</hi> two right lines being geuen, by this meanes may be founde out a right lyne which contayneth them both in power. Suppose that the two right lines geuen be AD and DB. It is required to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inde out a right lyne that contayneth them both in power. Let the lines AB and DB be so put, that they compre<g ref="char:EOLhyphen"/>hend a right angle ADB, and draw a right line from A to B. Now agayne it is manifest (by the 47. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst) that the line AB contayneth in power the lines AD and DB.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 11. Theoreme. The 14. Proposition. If there be sower right lines proportionall, and if the first be in power more then the second by the square of a right line commensurable in length vnto the first, the third also shalbe in power more then the fourth, by the square of a right line commensurable vnto the third. And if the first be in po<g ref="char:EOLhyphen"/>wer more then the second by the square of a right line incommensu<g ref="char:EOLhyphen"/>rable in length vnto the first, the third also shall be in power more then the fourth by the square of a right line incommensurable in length to the third.</head>
               <p>SVppose that these foure right lines A, B, C, D, be proportionall. As A is to B, so let C be to D. And let A be in power more then B, by the square of the line E. And likewise let C be in power more then D, by the square of the line F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Then I say that if A be commensurable in length vnto the line E, C also shall be commen<g ref="char:EOLhyphen"/>surable in length vnto the line F. And if A be incommen<g ref="char:EOLhyphen"/>surable
<figure/>
in length to the line E, C also shall be incommen<g ref="char:EOLhyphen"/>surable in length to the line F. For for that as A is to B, so is C to D, therefore as the square of the line A is to the square of the line B, so is the square of the line C to the square of the line D (by the 22. of the sixt). But by suppo<g ref="char:EOLhyphen"/>sition vnto the square of the line A are equall, the squares o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lines E and B, and vnto the square of the line C are equall the squares of the of the lines D and F: Where<g ref="char:EOLhyphen"/>fore as the squares of the lines E and B (which are equall to the square of the line A) are to the square of the line B, so are the squares of the lines D and F (which are equall to the square of the line C) to the square of the line D (by the seuenth of the fift). Wherfore (by the 17. of the fift) as the square of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the square of the line B, so is the square of the line F to the square of the line D. Wherfore also as the line E is to the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, so is the line F to the line D (by the second parte of the 22. of the sixt) wherefore contrari<g ref="char:EOLhyphen"/>wise (by the Corollary of the fourth of the fift) as B is to E so is D to F. But (by supposition)
<pb n="233" facs="tcp:7062:272"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s A is to B, so is C to D, Wherfore of equallitie (by the <hi>22.</hi> of the fift) a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, so is C is F. If therfore A be commensurable in length vnto E, C also shall be comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nsurable in le<g ref="char:cmbAbbrStroke">̄</g>gth vnto F: and if it ba incommensurable in length vnto E, C also shalbe incommensurablel in length vnto F, by the 10. of this booke. If therfore there be foure right tlines proportionall, and if the first be in power more then the secondby the square of a right line commensurable in length vnto the first, the third also shall be in power more then the fourth, by the square of a right line commensurable in length vnto the third and if the first be in power more the<g ref="char:cmbAbbrStroke">̄</g> the second, by the square of a right line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the first, the third also shall be in power more the the fourth, by the square of a right line incommensurable in length to the third: which was required to be proued.</p>
               <p>
                  <hi>Note that the line A may be proued to be in proportion to the line E, as the line C is to the line F, by an other way, namely, by conuersion of proportion (of some<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as we haue before noted,</hi>
                  <note place="margin">An other way to proue that the lines A, E, C, F, are pro<g ref="char:EOLhyphen"/>portionall.</note> 
                  <hi>called inuerse proportion) by the 19. of the fift. For, forasmuch as the foure lines A, B, C, D, are proportionall: ther<g ref="char:EOLhyphen"/>fore (by the 22. of the sixt) their squares also are proportionall. And forasmuch as the antecedent, name<g ref="char:EOLhyphen"/>ly, the square of the line A excedeth the consequent, namely, the square of the line B, by the square of the line E: and the other antecedent, namely, the square of the line C, excedeth the other consequent, namely, the square of the line D, by the square of the line F, therefore as the square of the line A is to the excesse, namely, to the square of the line E, so is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he square of the line C to the excesse; namely, to the square of the line F. Wherefore (by the second part of the 22. of the sixt) as the line A is to the line E, so is the line C to the line F.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 15. Proposition. If two magnitudes commensurable be composed, the whole magnitude com<g ref="char:EOLhyphen"/>posed also shall be commensurable to either of the two partes. And if the whole magnitude composed be commensurable to any one of the two partes, those two partes shall also be commensurable.</head>
               <p>LEt these two commensurable magnitudes AB and BC, be composed or added toge<g ref="char:EOLhyphen"/>ther. Then I say, that the whole magnitude AC is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to either of these partes AB and BC: For forasmuch as AB and BC are commensurable, therfore (by the first definition of the tenth) some one magnitude measureth them both.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</note> Let there be a magnitude that measureth them, and
<figure/>
let the same be D. Now forasmuch as D measureth AB and BC, it shall also measure the whole magni<g ref="char:EOLhyphen"/>tude composed AC, by this common sentence, what soeuer magnitude measureth two other magnitudes, shall also measure the magnitude composed of them. But the same D measureth AB and BC (by supposition). Wherefore D measureth AB, BC, and AC. Wherefore AC is commensu<g ref="char:EOLhyphen"/>rable to either of these magnitudes AB and BC.</p>
               <p>But now suppose that the whole composed magnitude AC be commensurable to any one of these two magnitudes AB or BC, let it be commensurable I say vnto AB.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t which is the conuerse of the first.</note> Then I say, that the two magnitudes AB and BC are commensurable. For forasmuch as AB and AC are commonsurable, some one magnitude measureth them (by the first definition of the tenth). Let some magnitude measure them, and let the same be D. Now forasmuch as D measureth AB and AC, it also measureth the residue BC, by this common sentence, what soeuer mea<g ref="char:EOLhyphen"/>sureth the whole and the part taken away, shall also measure the residu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. But the same D mea<g ref="char:EOLhyphen"/>sureth the magnitude AB (by supposition). Wherefore D measureth either of these magni<g ref="char:EOLhyphen"/>tudes AB and BC. Wherefore the magnitudes AB and BC are commensurable. If ther<g ref="char:EOLhyphen"/>fore two magnitudes commensurable be composed, the whole magnitude composed also shall be commensurable to either of the two partes. And if the whole magnitude composed be com<g ref="char:EOLhyphen"/>mensurable
<pb facs="tcp:7062:273"/>
to any one of the two partes, those two partes shall also be commensurable: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Montaureus.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note>
                     <hi>If an whole magnitude be commensurable to one of the two magnitudes which make the whole magnitude, it shall also be commensurable to the other of the two magnitudes.</hi> For if the whole mag<g ref="char:EOLhyphen"/>nitude AC be commensurable vnto the magnitude BC, then by the 2. part of thys 15 Proposition: the magnitudes AB and BC are commensurable. Wherefore (by the first part of the same) the magnitude AC shall be commensurable to either of these magnitudes AB and BC. This Corollary <hi>Theon</hi> vseth in the demonstration of the 17. Proposition and also of other Propositions. Howbeit <hi>Euclide</hi> left it out, for that it seemed easie as in a maner do all other Corollaryes.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 13. Theoreme. The 16. Proposition. If two magnitudes incommensurable be composed, the whole magnitude also shall be incommensurable vnto either of the two partes co<g ref="char:cmbAbbrStroke">̄</g>ponentes. And if the whole be incommensurable to one of the partes componentes, those first magnitudes also shall be incommensurable.</head>
               <p>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first part by an argument leadindg to an absurditie.</note>LEt these two incommensurable magnitudes AB &amp; BC, be composed, or added together. Then I say, that the whole magnitude AC, is incommensurable to either of these magnitudes AB and BC. For if AC and AB be not incom<g ref="char:EOLhyphen"/>mensurable, then some one magnitude measureth them (by the first definition of the tenth). Let there be such a magnitude, if it be possible, and let the same be D. Now forasmuch as D measureth CA and AB,
<figure/>
it also measureth the residue BC, &amp; it likewise measu<g ref="char:EOLhyphen"/>reth AB. Wherefore D measureth AB and BC. Wher<g ref="char:EOLhyphen"/>fore (by the first definitio<g ref="char:cmbAbbrStroke">̄</g> of the tenth) the magnitudes AB and BC are commensurable. But it is supposed that they are incommensurable: which is impossible. Wherefore no magnitude doth measure the magnitudes AB and AC. Where<g ref="char:EOLhyphen"/>fore the magnitudes CA and AB are incommensurable. In like sort also may we proue, that the magnitudes AC and CB are incommensurable.</p>
               <p>
                  <note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t leading also to an impossibi<g ref="char:EOLhyphen"/>litie. And this second part is the conuerse of the first.</note>But now suppose that the magnitude AC be incommensurable to one of these magnitudes AB or BC, and first let it be incommensurable vnto AB. Then I say, that the magnitudes AB and BC are incommensurable. For if they be commensurable some one magnitude mea<g ref="char:EOLhyphen"/>sureth them. Let some one magnitude measure them, &amp; let the same be D. Now forasmuch as D measureth AB and BC, it also measureth the whole magnitude AC. And it measu<g ref="char:EOLhyphen"/>reth AB: Wherefore D measureth these magnitudes CA and AB. Wherefore CA &amp; AB. are commensurable. And they are supposed to be inco<g ref="char:cmbAbbrStroke">̄</g>mensurable<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which is impossible. Wher<g ref="char:EOLhyphen"/>fore no magnitude measureth AB and BC. Wherefore the magnitudes AB and BC are in<g ref="char:EOLhyphen"/>commensurable. And in like sort may they be proued to be incommensurable, if the magni<g ref="char:EOLhyphen"/>tude AC be supposed to be incommensurable vnto BC. If therefore there be two magnitudes incommensurable composed, the whole also shall be incommensurable vnto either of the two partes component, and if the whole be incommensurable to one of the partes component, those first magnitudes shall be incommensurable: which was required to be proued.</p>
               <div type="section">
                  <head>¶A Corollary added by <hi>Montaureus.</hi>
                  </head>
                  <p>
                     <hi>If an whole magnitude bee incommensurable to one of the two magnitudes which make the whole magnitude, it shall also be incommensurable to the other of the two magnitudes.</hi> For if the
<pb n="244" facs="tcp:7062:273"/>
whole magnitude AC be inco<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto the magnitude BC, then by the 2 part of this 16. Theo<g ref="char:EOLhyphen"/>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>me, the magnitudes AB and BC shall be incommensurable. Wherefore by the first part of the same Theoreme, the magnitude AC shall be incommensurable to either of these magnitudes AB and BC. This Corollary <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> vseth in the demonstration of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>3. Theoreme, &amp; also of other Propositions.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>If vpon a right line be applied a parallelogramme wanting in figure by a square: the parallelogramme so applied, is equall to that parallelogramme which is contayned vnder the segmentes of the right line, which segmentes are made by reason of that application.</q>
                  </p>
                  <p>Suppose that vpon a right line AB be applied a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AG, wanting in forme by the square GB. Then I say, that AG is equall vnto that which is contayned vnder AD and DB, which thing is of it self manifest. For
<figure/>
forasmuch as GB is a square, therefore the line DG is equall vnto the line DB: and the pa<g ref="char:EOLhyphen"/>rallelogramme AG is that which is contayned vnder the lines AD and DG, that is, vn<g ref="char:EOLhyphen"/>der the lines AD and DB. If therefore vp<g ref="char:EOLhyphen"/>on a right line be applied a parallelogramme wanting in figure by a square: the parallelo<g ref="char:EOLhyphen"/>gra<g ref="char:cmbAbbrStroke">̄</g>me applied is equall to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me which is contayned vnder the segmentes of the right line, which are made by reason of that application. which was required to be demonstrated.</p>
                  <p>
                     <hi>This Assumpt I before added as a Corollary out of</hi> Flussates <hi>after the 28. Pro<g ref="char:EOLhyphen"/>position of the sixt booke.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 14. Theoreme. The 17. Proposition. If there be two right lines vnequall, and if vpon the greater be applied a parallelogramme equall vnto the fourth part of the square of the lesse line, and wanting in figure by a square, if also the parallelogramme thus appli<g ref="char:EOLhyphen"/>ed deuide the line where vpon it is applied into partes commensurable in length: then shall the greater line be in power more then the lesse, by the square of a line commensurable in length vnto the greater. And if the grea<g ref="char:EOLhyphen"/>ter be in power more then the lesse by the square of a right line commensu<g ref="char:EOLhyphen"/>rable in length vnto the greater, and if also vpon the greater be applied a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me equall vnto the fourth part of the square of the lesse line, and wanting in figure by a square: then shall it deuide the greater line in<g ref="char:EOLhyphen"/>to partes commensurable.</head>
               <p>SVppose that these two right lines A and BC, be vnequall: of which let BC be the greater. And vpon the line BC let there be applied (by the 28. of the sixt) a parallelogramme equall vnto the fourth part of the square of the line A being the lesse (that is, equall vnto the square described vpon halfe of
<pb facs="tcp:7062:274"/>
the line A) and wanting in figure by a square. And let the same parallelogramme be th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> which is contained vnder the lines BD and DC. And (by supposition) let the lines BD and DC be commensurable in length. The<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> I say, that the line BC, is in power more then the line A, by the square of a line commensurable in
<figure/>
length vnto the sayd line BC. Deuide (by the 10. of the first) the line BC into two equall partes in the point E.<note place="margin">Construction.</note> And (by the third of the first) vnto the line DE p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> an equall line EF. Wherefore the resi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ue DC is equall vnto the residue BF. And forasmuch as the right line BC is deuided into two equall partes in the point E,<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the first part.</note> and into two vnequall partes in the point D, therefore (by the <hi>5.</hi> of the second) the rectangle figure comprehended vnder the lines BD and DC together with the square of the line ED, is equall to the square of the line EC. And in the same proportion are they eche being taken fower times by the <hi>15.</hi> of the fifth. Wherfore that which is contained vnder the lines BD and DC taken fower times together with the square of the line ED taken also fower times, is equall to the square of the line EC taken fower times. But vnto that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines BD &amp; DC foure times is equal the square of the line A by supposition: for the parallelograme contained vnder the lines BD and DC once is supposed to be equall to the fourth part of the square of the line A. And vnto the square of the line DE taken fower times is equall the square of the line DF, for the line DF is double to the line DE. And vnto the square of the line EC fower times ta<g ref="char:EOLhyphen"/>ken, is equall the square of the line BC, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or the line BC is also double to the line CE. Where<g ref="char:EOLhyphen"/>fore the squares o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lines A and DF are equall vnto the square of the line BC. Wherefore the square of the line BC, is greater then the square of the line A, by the square of the line. DF. Wherefore the greater line BC is in power more then the lesse line A, by the square of the line DF. Now resteth to proue that the line BC is commensurable in length vnto the line DF. Forasmuch as by suposition the line BD is commensurable in length vnto DC, there<g ref="char:EOLhyphen"/>fore (by the <hi>15.</hi> of the tenth) the whole line BC is commensurable in length vnto the line DC: but the line DC equall to the line BF. Wherefore the whole line BC is commensurable in length vnto the lines BF &amp; CD. Let the two lines BF and CD be imagined to be so compo<g ref="char:EOLhyphen"/>sed that they make one line. Now forasmuch as the whole line BC is commensurable in length to the two lines BF and CD taken as one line: therefore the lines BF and CD taken as one line are commensurable in length to the line FD (by the 2. part of the 15. of the tenth). Wher<g ref="char:EOLunhyphen"/>fore also the whole line BC is commensurable in length to the line FD by the first part of the same: this may also be proued by the corollary put after the 16. proposition of this booke. Wherefore the line BC is in power more then the line A by the square of a line commmensu<g ref="char:EOLhyphen"/>rable in length vnto the line BC.</p>
               <p>But now suppose that the line BC be in power more then the line A, by the square of a line commensurable in length vnto the line BC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part which is the conuerse of the first.</note> And vpon the line BC let there be applied a rectangle parallelograme equall vnto the fourth part of the square of the line A, and wan<g ref="char:EOLhyphen"/>ting in figure by a square, and let the sayd parallelograme be that which is contained vnder the lines BD and DC. Then must we proue that the line BD is vnto the line DC commen<g ref="char:EOLhyphen"/>surable in length. The same constructions and suppositions, that were before, remayning, we may in like sort proue that the line BC is in power more then the line A, by the square of the line FD. But by suppositio<g ref="char:cmbAbbrStroke">̄</g> the line BC is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line A by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto it in length. Wherfore the line BC is vnto the line FD co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length. Wherefore the line composed of the two lines BF and DC is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line FD (by the second part of the <hi>15.</hi> of the tenth). Wherefore (by the 12. of the tenth or by the first part of the <hi>15.</hi> of the tenth) the line BC is commensurable in length to the line composed of BF and DC. But the whole line conposed BF and DC is commensura<g ref="char:EOLhyphen"/>ble in length vnto DC. For BF (as before hath bene proued) is equall to DC. Wherefore the
<pb n="245" facs="tcp:7062:274"/>
line BC is commensurable in length vnto the line DC (by the <hi>12.</hi> of the tenth). Wh<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>fore also the line BD is commensurable in length vnto the line DC (by the second part of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 15. of the te<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th). If therfore there be two right lines vnequall, and if vpon the greater be appli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d a parallelograme equall vnto the fourth part of the square of the lesse and wanting in figure by a square, if also the parallelograme thus applied deuide the line whereupon it is applied into partes commensurable in length: then shall the greater line be in power more then the lesse by the square of a line commensurable in length vnto the greater. And if the greater be in power more then the lesse by the square of a line comme<g ref="char:cmbAbbrStroke">̄</g>surable in length vnto the greater, and if also vpon the greater be applied a parallelograme equall vnto the fourth part of the square made of the lesse and wanting in figure by a square: then shall it deuide the greater line into partes commensurable in length: which was required to be proued.</p>
               <p>Campan<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>after this proposition reacheth how we may redily apply vpon the line BC a parallelograme equall to the fourth part of the square of halfe of the line A, and wanting in figure by a square after this maner.</hi>
               </p>
               <p>
                  <hi>Deuide the line BC into two lines in such sort that halfe of the line A shalbe the meane propor<g ref="char:EOLhyphen"/>tionall betwene those two lines, which is possible, when as the line BC is supposed to be greater then the line A, and may thus be done.</hi>
                  <note place="margin">How to de<g ref="char:EOLhyphen"/>uide the line BC redely in such sort as i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> required in the propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> 
                  <hi>Deuide the line BC into two equal partes in the point E and describe vpon the line BC a semicircle BHC. And vnto the line BC, and from the point C erect a perp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>dicular line CK and put the line CK equall to halfe of the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <figure/>
And by the point K draw vnto the line EC a parallel line KH cutting the semicircle in the point H, (which it must needes cut, fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asmuch as the line BC is greater then the line A). And fro<g ref="char:cmbAbbrStroke">̄</g> the point H draw vnto the line BC a per<g ref="char:EOLunhyphen"/>pendicular li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e HD: which line HD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> forasmuch as by the 34: of the first it is equall vnto the line KC, shall also be e<g ref="char:EOLhyphen"/>quall to halfe of the line A: draw the lines BH and HC. Now then by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>. of the third the angle BHC is a right a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gle. Wherefore by the corollary of the eight of the sixt booke the line HD is the meane proportionall betwene the lines BD and DC. Wherefore the halfe of the line A which is equall vnto the line HD is the meane proportio<g ref="char:EOLhyphen"/>nall betwene the lines BD and DC. Wherefore that which is contained vnder the lines BD and DC is equall to the fourth part of the square of the line A. And so if vpon the line BD be described a rect<g ref="char:EOLhyphen"/>angle parallelograme hauing his other side equall to the line DC, there shalbe applied vpon the line BC a rectangle parallelograme equall vnto the square of halfe of the line A, and wanting in figure by a square: which was required to be done.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 15. Theoreme. The 18. Proposition. If there be two right lines vnequall, and if vpon the greater be applied a parallelograme equall vnto the fourth part of the square of lesse, and wan<g ref="char:EOLhyphen"/>ting in figure by a square, if also the parallelograme thus applied deuide the line whereupon it is applied into partes incommensurable in length: the greater line shalbe in power more then the lesse line by the square of a line incommensurable in length vnto the greater line. And if the grea<g ref="char:EOLhyphen"/>ter line be in power more then the lesse line, by the square of a line incomme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable in length vnto the greater, and if also vpon the greater be applied a parallelograme equall vnto the fourth part of the square of the lesse and wanting in figure by a square: then shall it deuide the greater line into partes incommensurable in length.</head>
               <p>
                  <pb facs="tcp:7062:275"/>SVppose that these two right lines A and BC be vnequall the one to the other, of which let BC be the greater. And vpon the same BC apply a parallelograme equall vnto the fourth part of the square of the line A, and wanting in figure by a square: which how to doo was before taught in the end of the
<figure/>
former proposition. And let the sayd parallelograme be that which is contained vnder the lines BD and DC. And let BD be incommensurable in length vn<g ref="char:EOLunhyphen"/>to DC. Then I say that the line BC is in power more then the line A, by the square of a right line incommensurable in length vnto the line BC. First let the same order of construction, and demonstration be obserued in this which was in the former proposition. And we may in like sort proue that the line BC is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line A by the square of the line DF. Now then must we proue that the lines BC and DF are incommensurable in length.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> of the first part.</note> Forasmuch as by supposition the line BD is incommensu<g ref="char:EOLhyphen"/>rable in le<g ref="char:cmbAbbrStroke">̄</g>gth vnto the line DC: therefore (by the <hi>16.</hi> of the tenth) the line BC is incommen<g ref="char:EOLhyphen"/>surable in length vnto the line CD. But DC is commensurable to these two lines BF and DC added together. For BF is equall vnto DC. Wherefore (by the <hi>13.</hi> of the tenth) BC is incommensurable vnto these two lines BF and DC composed. Wherefore by the second part of the <hi>16.</hi> of the tenth the line composed of the lines BF and DC taken as one line in incom<g ref="char:EOLhyphen"/>mensurable in length vnto the line FD. Wherefore by the first part of the same <hi>16.</hi> propositi<g ref="char:EOLhyphen"/>on the line BC is incommensurable in length vnto the line FD. Wherefore the line BC is in power more then the line A by the square of a line incommensurable in length vnto the line BC.</p>
               <p>But now suppose that the line BC be in power more then the line A by the square of a line incommensurable in length vnto BC.<note place="margin">Demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ti<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond part which is the conuerse of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e former.</note> And vpon the line BC let there be applied a parallelo<g ref="char:EOLhyphen"/>gramme equall vnto the fourth part of the square of the line A, and wanting in figure by a square, and let the said parallelogramme be that which is contained vnder the lines BD &amp; DC. Then must we proue that the line BD is vnto the line DC incommensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth. The same order of construction and demonstration being kept, we may in like sort proue that the line BC is in power more then the line A by the square of the line FD. But now (by sup<g ref="char:EOLhyphen"/>position) the line BC is in power more then the line A by the square of a line incommensu<g ref="char:EOLhyphen"/>rable in length vnto BC. Wherefore the line BC is vnto the line FD incommensurable in length. Wherfore the line composed of BF and DC taken as one line, shall be incommensu<g ref="char:EOLhyphen"/>rable in length to the line FD (by the second part of the 16. of the tenth): wherefore also by the first part of the same, the line BC shall be incommensurable in length to the line compo<g ref="char:EOLhyphen"/>sed of the lines BF and DC. But the line composed of the lines BF and DC is commensu<g ref="char:EOLhyphen"/>rable in length to the line DC (for that BF (as before hath bene proued) is equall to DC). Wherfore the line BC is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line DC (by the 13. of the tenth). Wherefore by the second part of the <hi>16.</hi> of the tenth, the line BD is incomme<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth vnto the line DC. If therfore there be two right lines vnequall, and if vpon the greater be applied a parallelogramme equall vnto the fourth part of the square of the lesse line, &amp; wan<g ref="char:EOLhyphen"/>ting in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure by a square, if also the parallelogramme thus applied deuide the line wherupon it is applied into partes incommensurable in length: the greater line shall be in power more then the lesse line by the square of a line incommensurable in length vnto the greater. And if the greater line be in power more then the lesse, by the square of a line incommensu<g ref="char:EOLhyphen"/>rable in length vnto the greater, and if also vpon the greater be applied a parallelogramme equall vnto the fourth part of the square of the lesse line, and wanting in figure by a square, then shall it deuide the greater line into partes incommensurable in length: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
               <p>
                  <hi>This Proposition may also be demonstrated by the former proposition, namely, the first part of this
<pb n="246" facs="tcp:7062:275"/>
by the second part of the former, and the second part of this by the first part of the former, by an argu<g ref="char:EOLhyphen"/>ment leading to an absurditie.</hi>
                  <note place="margin">An other de<g ref="char:EOLhyphen"/>monstration<g ref="char:EOLunhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y an argume<g ref="char:cmbAbbrStroke">̄</g>t leading to an absurditie.</note> 
                  <hi>For as touching the first part of this proposition, the line BC contayning in power more then the line A by the square of the line FD, if the line BG be not incommensurable vn<g ref="char:EOLunhyphen"/>to the line FD, then is it commensurable vnto it. Wherfore (by the second part of the 17. proposition) the lines BD and DC also are commensurable, which is impossible, for they are supposed to be incom<g ref="char:EOLhyphen"/>mensurable. So likewise as touching the second parte of the same, the line BC contayning in power more then the line A by the square of the line FD, if the line DB be not incommensurable to the lyne DC, then is it commensurable vnto it: wherfore (by the first part of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>, proposition) the lines BC and FD are also commensurable, which were absu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>de. For the lines BC and FD are supposed to be in<g ref="char:EOLhyphen"/>commensu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>able: which was required to be proued.</hi>
               </p>
               <div type="section">
                  <head>¶An assumpt.</head>
                  <p>Forasmuch as it hath bene proued that lines commensurable in length,<note place="margin">An Assumpt.</note> are alwayes also commensurable in power, but lines commensurable in power are not alwayes commensurable in length, but may be in length both commensurable and also incommensurable: it is mani<g ref="char:EOLhyphen"/>fest, that if vnto the line propounded, which is called rationall of purpose, a certayne line be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length, it ought to be called rationall and co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto it, not only in length, but also in power: for lines commensurable in length are also alwayes commensura<g ref="char:EOLhyphen"/>ble in power. But if vnto the line propounded which is called rationall of purpose, a certayne line be commensurable in power, then if it be also commensurable vnto it in length, it is cal<g ref="char:EOLhyphen"/>led rationall and commensurable vnto it both in length and in power. But againe if vnto the said line geuen which is called rationall, a certayne line be commensurable in power, and in<g ref="char:EOLhyphen"/>commensurable in length, that also is called rationall, commensurable in power onely.</p>
               </div>
               <div type="section">
                  <head>An annotacion of Proclus.</head>
                  <p>He calleth those lines rationall, which are vnto the rationall line first set co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length &amp; in power, or in power only. And there are also other right lines, which are vnto the rationall line first set, incommensurable in length, and are vnto it commensurable in power only, and therfore they are called rationall, &amp; comme<g ref="char:cmbAbbrStroke">̄</g>surable the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for which cause they are rationall. But euen these lines may be commensurable the one to the other, either in length, and therefore in power, or els in power onely. Now if they be commensu<g ref="char:EOLhyphen"/>rable in length, then are those lines called rationall, commensurable in length, but yet so that they be vnderstand to be in power commensurable: but if they be commensurable the one to the other in power onely, they also are called rationall commensurable in power onely.</p>
               </div>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>And that two lines or more being rationall and commensurable in length to the rational line first set, are also commensurable the one to the other in length, hereby it is manifest:<note place="margin">
                        <hi>A Corollary added by Montaureu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> for forasmuch as they are rationall and commensurable in length to the rationall line first set, but those magnitudes whiche are commensurable to one and the selfe same magnitude, are also commensurable the one to the other (by the 12. of the tenth) wherfore the rationall lines, commensurable in length to the rational lyne first set, are also commensurable in length the one to the
<figure/>
other. And as touching those which are rationall com<g ref="char:EOLunhyphen"/>mensurable in power onely to the rationall line first set, they also must needes be at the least commensura<g ref="char:EOLhyphen"/>ble in power the one to the other. For forasmuch as their squares are rationall they shall bee commensu<g ref="char:EOLhyphen"/>rable to the square of the rationall line first set. Wher<g ref="char:EOLhyphen"/>fore by the 12. of this booke, they are also commensu<g ref="char:EOLhyphen"/>rable the one to the other. Wherefore their lines are at the least commensurable in power the one to the other. And it is possible also that they may be comme<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth the one to the other. For suppose that A be a rationall li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e first set, and let the line B be vnto the same rationall line A commensurable in power onely, that is, incommensurable in length vnto it. Let there be also an other line C commen<g ref="char:EOLhyphen"/>surable
<pb facs="tcp:7062:276"/>
in length to the lyne B (which is possible by the principles of this booke.) Now by the 13. of the tenth, it is manifest that the line C is incommensurable in length vnto the line A. But the square of the line A is co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable to the square of the line B by supposition, and the square of the line C is also commensurable to the square of the line B by supposition. Wherefore by the 12. of this booke, the square of the line C is commensurable to the square of the line A. Wherfore by the definition, the line C shall be rationall commensurable in power onely to the line A, as also is the line B. Wherefore there are geuen two rationall lines commensurable in power onely to the rationall line first set, and comme<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>surable in length the one to the other.</p>
                  <p>Here is to be noted, which thing also we before noted in the definitions, that <hi>Cam<g ref="char:EOLhyphen"/>pane</hi> and others which followed him, brought in these phrases of speaches, to call some lynes rationall in power onely,<note place="margin">
                        <hi>Cause.</hi>
                     </note> and other some rationall in length and in power, which we cannot finde that Euclide euer vsed. For these wordes in length and in power are ne<g ref="char:EOLhyphen"/>uer referred to rationalitie or irrationalitie, but alwayes to the commensurabilitie or incommensurablitie of lines. Which peruerting of wordes (as was there declared) hath much increased the difficulty and obscurenes of this booke.<note place="margin">
                        <hi>Cause of in<g ref="char:EOLhyphen"/>creasing the difficulty of this booke.</hi>
                     </note> And now I thinke it good agayne to put you in minde, that in these propositions which follow, we must euer haue before our eyes the rationall line first set,<note place="margin">
                        <hi>Note.</hi>
                     </note> vnto which other lines compared are either rationall or irrationall, according to their commensurability or incommensurabilitie.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 16. Theoreme. The 19. Proposition. A rectangle figure comprehended vnder right lines commensurable in lengthe, being rationall according to one of the foresaide wayes: is ra<g ref="char:EOLhyphen"/>tionall.</head>
               <p>
                  <hi>SVppose that this rectangle figure AC be comprehended vnder these right lines AB and BC being commensurable in length, and rationall according to one of the foresaid wayes. Then I say that the superficies AC is rationall, describe (by the 46. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the first) vpon the line AB a square AD.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>
                  <figure/>
                  <hi>Wherfore that square AD is rationall by the definition.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>And forasmuch as the line AB is commensurable in length vnto the line BC, and the line AB is equall vnto the lyne BD, therefore the lyne BD is commensurable in length vnto the line BC. And as the line BD is to the line BC, so is the square DA to the superficies AC (by the first of the sixt): but it is proued that the line BD is commensurable vnto the line BC, wherfore (by the</hi> 10. <hi>of the tenth) the square DA is commensurable vnto the rectangle superficies AC. But the square DA is rationall, wherfore the rectangle su<g ref="char:EOLhyphen"/>perficies AC also is rationall by the definition. A rectangle figure therfore comprehended vnder right lines commensurable in length, beyng rationall accordyng to one of the foresayd wayes is rationall: which was required to be proued.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Diuers ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>es in this propo<g ref="char:EOLhyphen"/>sition.</hi>
                  </note>Where as in the former demonstration the square was described vpon the lesse line, we may also demonstrate the Proposition, if we describe the square vpon the greater line, and that after thys maner. Suppose that the rectangle superficies BC be contayned of these vnequall lines AB and AC, which let be rationall commensurable the one to the other in length. And let the line AC be the greater.<note place="margin">
                     <hi>The second case.</hi>
                  </note> And
<pb n="247" facs="tcp:7062:276"/>
vpon the line AC describe the square DC. Then
<figure/>
I say, that the parallelogramme BC is rationall.<note place="margin">
                     <hi>The first kind of rationall lines com<g ref="char:EOLhyphen"/>mensurable in length.</hi>
                  </note> For the line AC is commensurable in length vnto the line AB by supposition, and the line DA is e<g ref="char:EOLhyphen"/>quall to the line AC. Wherefore the line DA is commensurable in length to the line AB. But what proportion the line DA hath to the line AB, the same hath the square DC to the para<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>le<g ref="char:EOLhyphen"/>logramme C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (by the first of the sixt). Wherefore (by the 10. of this booke) the square DC is com<g ref="char:EOLhyphen"/>mensurable to the parallelogramme CB. But it is manifest, that the square DC is rationall, for that it is the square of a rationall line, namely, AC. Wherefore (by the definition) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me also CB is rationall.</p>
               <p>Moreouer, forasmuch as those two former demonstrations seeme to speake of that parallelogra<g ref="char:cmbAbbrStroke">̄</g>me which is made of two lines, of which any one may be the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e first set, which is called the first rationall line, from which (we sayd) ought to be taken the measures of the other lines compared vnto it, and the other is comme<g ref="char:cmbAbbrStroke">̄</g>surable in length to the same first rationall line,<note place="margin">
                     <p>
                        <hi>This particle in the proposi<g ref="char:EOLhyphen"/>tion (accor<g ref="char:EOLhyphen"/>ding to any of the foresayde wayes) was not in vayne put.</hi>
                     </p>
                     <p>
                        <hi>The second kinde of rati<g ref="char:EOLhyphen"/>onall lines co<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>mensurable in leng<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>h.</hi>
                     </p>
                  </note> which is the first kinde of rationall lines co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length: I thinke it good here to set an other case of the other kinde of rationall lines, of lines I say rationall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length compared to an other rationall line first set, to declare the generall truth of this Theoreme, and that we might see that this particle <hi>according to any of the foresayd wayes</hi> was not here in vaine put. Now then suppose first a rationall line AB. Let there be also a paral<g ref="char:EOLhyphen"/>lelogra<g ref="char:cmbAbbrStroke">̄</g>me CD contayned vnder the lines
<figure/>
CE and ED, which lines let be rationall, that is commensurable in length to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst rationall line propounded AB. How<g ref="char:EOLhyphen"/>beit, let those two lines CE and ED be diuers and vnequall lines vnto the first ra<g ref="char:EOLhyphen"/>tionall line AB. Then I say, that the pa<g ref="char:EOLhyphen"/>rallelogramme CD is rationall.<note place="margin">
                     <hi>The third case.</hi>
                  </note> Describe the square of the line DE, which let be DF. First it is manifest (by the 12. of this booke) that the lines CE &amp; ED, are com<g ref="char:EOLhyphen"/>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth the one to the other. For either of them is supposed to be com<g ref="char:EOLhyphen"/>mensurable in length vnto the line AB. But the line ED is equall to the line EF. Wherefore the line CE is commensurable in length to the line BF. But <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the line CE is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> F, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o is the paralle<g ref="char:EOLhyphen"/>logramme CD to the square DF (by the first of the sixt). Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>refore (by the 10. of this booke) the pa<g ref="char:EOLhyphen"/>rallelogramme CD shall be commensurable to the square DF. But the square DF is commensurable to the square of the line AB which is the first rationall line propounded. Wherfore (by the 12. of this booke) the parallelogramme CD is commensurable to the square of the line AB. But the square of the line AB is rationall (by the definition). Wherfore by the definition also of rationall figures, the pa<g ref="char:EOLhyphen"/>rallelogramme CD shall be rationall.</p>
               <p>Now resteth an other ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e of the thirde kinde of rationall lines commensurable in length the one to the other, which are to the rationall line AB first set commensurable in power onely, and yet are therfore rationall lines. And let the lines CE and ED be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length the one to the other.<note place="margin">
                     <hi>The third kinde of ratio<g ref="char:EOLhyphen"/>nall lines com<g ref="char:EOLhyphen"/>mensurable in length.</hi>
                  </note> Now then let the selfe same construction remaine that was in the former: so that let the lines CE and ED be rationall commensurable in power onely vnto the line AB. But let them be commensurable in length the one to the other. Then I say, that in this case also the parallelogramme CD is rationall. First it may be proued as before, that the parallelogramme CD is commensurable to the square DF. Wher<g ref="char:EOLhyphen"/>fore (by the 12. of this booke) the parallelogramme CD shall be commensurable to the square of the line AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But the square of the line AB is rationall.<note place="margin">
                     <hi>The fourth case.</hi>
                  </note> Wherefore (by the definition) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me CD shall be also rationall. This case is well to be noted. For it serueth to the demonstration and vn<g ref="char:EOLhyphen"/>derstanding of the 25. Proposition of this booke.</p>
            </div>
            <div type="part">
               <head>¶The 17. Theoreme. The 20. Proposition. If vpon a rationall line be applied a rationall rectangle parallelogramme: the other side that maketh the breadth thereof shall be a rationall line and commensurable in length vnto that line wherupon the rationall parallelo<g ref="char:EOLhyphen"/>gramme is applied.</head>
               <p>
                  <pb facs="tcp:7062:277"/>SVppose that this rationall rectangle parallelogramme AC, be applied vpon the line AB, which let be rationall according to any one of the foresaid wayes (whether it be the first rationall line set,<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLunhyphen"/>uerse of the former propo<g ref="char:EOLhyphen"/>sition.</note> or any other line commensurable to the rationall line first set, and that in length and in power, or in power onely: for one of these three wayes, as was declared in the Assumpt put before the 19. Proposition of this booke, is a line called rationall) and making in breadth the line BC. Then I say, that the line BC is rationall and commensurable in length vnto the line BA.<note place="margin">Construction.</note> Desrcribe (by the 46. of the first) vpon the line BA a square AD. Wher<g ref="char:EOLhyphen"/>fore
<figure/>
(by the 9. definitio<g ref="char:cmbAbbrStroke">̄</g> of the tenth) the square AD is rationall. But the parallelogramme AC also is rationall (by supposition).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Where<g ref="char:EOLhyphen"/>fore (by the conuersion of the definition of ra<g ref="char:EOLhyphen"/>tionall figures, or by the 12. of this booke) the square DA is commensurable vnto the pa<g ref="char:EOLhyphen"/>rallelogramme AC. But as the square DA is to the parallelogramme AC, so is the line DB to the line BC (by the first of the sixt). Wher<g ref="char:EOLhyphen"/>fore (by the 10. of the tenth) the line DB is commensurable vnto the line BC. But the line DB is equall vnto the line BA. Wherefore the line AB is co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto the line BC. But the line AB is rationall. Wherefore the line BC also is rationall and commensurable in length vnto the line BA. If therefore vpon a rationall line be applied a rationall rectangle parallelogramme, the other side that maketh the breadth therof shall be a rationall line com<g ref="char:EOLhyphen"/>mensurable in length vnto that line whereupon the rationall parallelogramme is applied: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>A line contayning in power an irrationall superficies, is irrationall.</q>
                  </p>
                  <p>
                     <note place="margin">An Assumpt.</note>Suppose that the line AB co<g ref="char:cmbAbbrStroke">̄</g>taine in power an irrationall superficies, that is, let the square described vpon the line AB, be equall vnto an irrationall superficies. Then I say, that the
<figure/>
line AB is irrationall. For if the line AB be rationall, the<g ref="char:cmbAbbrStroke">̄</g> shall the square of the line AB be also rationall. For so was it put in the definitions. But (by supposition) it is not. Wherefore the line AB is irrationall. A line therefore contayning in power an irrationall superficies, is ir<g ref="char:EOLhyphen"/>rationall.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 18. Theoreme. The 21. Proposition. A rectangle figure comprehended vnder two rationall right lines com<g ref="char:EOLhyphen"/>mensurable in power onely, is irrationall. And the line which in power contayneth that rectangle figure is irrationall, &amp; is called a mediall line.</head>
               <p>
                  <hi>SVppose that this rectangle figure AC be comprehended vnder these rationall right lines AB and BC commensurable in power onely. Then I say, that the su<g ref="char:EOLhyphen"/>perficies AC is irrationall: and the line which contayneth it in power is irratio<g ref="char:EOLhyphen"/>nall, and is called a mediall line.</hi>
                  <note place="margin">
                     <hi>Const<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uction.</hi>
                  </note> 
                  <hi>Describe (by the 46. of the first) vpon the line AB a square AD.</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>Wherefore the square AD is rationall. And forasmuch as the line AB is vnto the line BC incommensurable in length, for they are supposed to be commensurable in power onely, and the line AB is equall vnto the line BD, therefore also the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> BD is
<pb n="248" facs="tcp:7062:277"/>
vnto the line BC incommensurable in length. And <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> is to the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> C, so <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the square AD to the parallelogramme AC (by the first of the fiu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>). Wherefore (by the 10. of the tenth) the square DA is vnto the parallelogramme AC incommensurable. But the square DA is rationall. Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme
<figure/>
AC is irrationall. Where<g ref="char:EOLhyphen"/>fore also the line that contayneth the super<g ref="char:EOLhyphen"/>ficies AC in power, that is whose square is equall vnto the parallelogramme AC, is (by the Assumpt going before) irrationall. And it is called a mediall line, for that the square which is made of it, is equall to that which is contayned vnder the lines AB and BC, and therefore it is (by the second part of the 17. of the sixt) a meane proportionall line betwene the lines AB and BC. A rectangle fi<g ref="char:EOLhyphen"/>gure therefore comprehended vnder rationall right lines which are commensurable in power onely, is irrationall. And the line which in power contayneth that rectangle figure is irra<g ref="char:EOLhyphen"/>tionall, and is called a mediall line.</hi>
               </p>
               <p>At this Proposition doth <hi>Euclide</hi> first entreate of the generation and production of irrationall lines. And here he searcheth out the first kinde of them, which he calleth a mediall line. And the definition therof is fully gathered and taken out of this 21. Pro<g ref="char:EOLhyphen"/>position, which is this. A mediall line is an irrationall line whose square is equall to a rectangled figure contayned of two rationall lines commensurable in power onely.<note place="margin">
                     <hi>Diffinition of a mediall line.</hi>
                  </note> It is called a mediall line, as <hi>Theon</hi> rightly sayth, for two causes, first for that the power or square which it produceth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is equall to a mediall superficies or parallelogramme. For as that line which produceth a rationall square, is called a rationall line, and that line which produceth an irrationall square, or a square equall to an irrationall figure gene<g ref="char:EOLhyphen"/>rally is called an irrationall line: so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line which produceth a mediall square, or a square equall to a mediall superficies, called by speciall name a mediall line. Secondly it is called a mediall line, because it is a meane proportionall betwene the two lines co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mensurable in power onely which comprehend the mediall superficies.</p>
               <div type="section">
                  <head>¶A Corollary added by <hi>Flussates.</hi>
                  </head>
                  <p>
                     <hi>A rectangle parallelogramme contayned vnder a rationall line and an <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rrationall line, is irrationall.</hi>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note> For if the line AB be rationall, and if the line CB be irrationall, they shall be incommensurable. But as the line BD (which is equall to the line BA) is to the line BC, so is the square AD to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AC. Wherefore the parallelogramme AC shall be incommensurable to the square AD which is ra<g ref="char:EOLhyphen"/>tionall (for that the line AB wherupon it is described is supposed to be rationall). Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme AC which is contayned vnder the rationall line AB, and the irrationall line BC, is irrationall.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>If there be two right lines, as the first is to the second, so is the square which is described vpon the first to the parallelograme which is contained vnder the two right lines.</q>
                  </p>
                  <p>Suppose that there be two right lines AB and BC.<note place="margin">This assumpt is nothing els but a part of the first pro<g ref="char:EOLhyphen"/>position of the sixt booke.</note> Then I say that as the line AB is to the line BC, so is the square of the line AB, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> that which is contained vnder the lines AB and BC. Describe (by the <hi>46.</hi> of the first) vpon the line AB a square AD. And make perfect the parallelograme AC. Now for that as the line AB is to the line BC (for the line AB, is equall to the line BD); so is the square AD to the parallelograme CA by the first of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <pb facs="tcp:7062:278"/>
and AD is the square which is made of the
<figure/>
line AB, and AC is that which is contained vnder the lines BD and BC, that is, vnder the lines AB &amp; BC: therfore as the line AB is to the line BC, so is the square described vp<g ref="char:EOLunhyphen"/>pon the the line AB to the rectangle figure contained vnder the lines AB &amp; BC. And conuersedly as the parallelograme which is contained vnder the lines AB and BC is to the square of the line AB, so is the line CB to the line BA.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 19. Theoreme. The 22. Proposition. If vpon a rationall line be applied the square of a mediall line: the other side that maketh the breadth thereof shalbe rationall, and incommensura<g ref="char:EOLhyphen"/>ble in length to the line wherupon the parallelograme is applied.</head>
               <p>SVppose that A be a mediall line, and let BC be a line rationall, and vpon the line BC describe a rectangle parallelograme equall vnto the square of the line A, and let the same be BD making in breadth the line CD. Then I say that the line CD is rationall and inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line CB.<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>
                  </note> For forasmuch as A is a mediall line, it containeth in power (by the <hi>21.</hi> of the tenth) a rectangle parallelograme comprehen<g ref="char:EOLhyphen"/>ded vnder rationall right lines commensurable in power onely. Suppose that is containe in power the parallelograme GF: and by supposition it also containeth in power the parallelo<g ref="char:EOLhyphen"/>grame BD. Wherefore the parallelo<g ref="char:EOLhyphen"/>grame
<figure/>
BD is equall vnto the parallelo<g ref="char:EOLhyphen"/>grame GF: and it is also equiangle vn<g ref="char:EOLhyphen"/>to it, for that they are ech recta<g ref="char:cmbAbbrStroke">̄</g>gle. But in parallelogrames equall and equiangle the sides which containe the equall an<g ref="char:EOLhyphen"/>gles, are reciprocall (by the 14. of the sixt): Wherfore what proportio<g ref="char:cmbAbbrStroke">̄</g> the line BC hath to the line EG, the same hath the line EF to the line CD. Therefore (by the <hi>22.</hi> of the sixt) as the square of the line BC is to the square of the line EG, so is the square of the line EF to the square of the line CD. But the square of the line BC is commensurable vnto the square of the line EG (by supposition). For either of them is rationall. Wherefore (by the the 10. of the tenth) the square of the line EF is commensurable vnto the square of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CD. But the square of the line EF is rationall. Wherefore the square of the line CD is likewise ratio<g ref="char:EOLhyphen"/>nall. Wherefore the line CD is rational. And forasmuch as the line EF is inco<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mensurable in length vnto the line EG (for they are supposed to be commensurable in power onely). But as the line EF is to the line EG, so (by the assumpt going before) is the square of the line EF to the parallelograme which is contained vnder the lines EF and EG. Wherefore (by the <hi>10.</hi> of the tenth) the square of the line EF is incommensurable vnto the parallelograme which is contained vnder the lines FE and EG. But vnto the square of the line EF the square of the line CD is commensurable, for it is proued that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ither of them is
<pb n="249" facs="tcp:7062:278"/>
a rationall lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And that which is contained vnder the lines DC and CB is commensurable vnto that which is contained vnder the lines FE and EG. For they are both equall to the square of the line A. Wherefore (by the 13. of the tenth) the square of the line CD is incom<g ref="char:EOLhyphen"/>mensurable to that which is contained vnder the lines DC and CB. But as the square of the line CD is to that which is contained vnder the lines DC and CB, so (by the assumpt going before) is the line DC to the line CB. Wherefore the line DC is incommensurable in length vnto the line CB. Wherefore the line CD is rationall and incommensurable in length vnto the line CB. If therefore vpon a rationall line be applied the square of a mediall line, the o<g ref="char:EOLhyphen"/>ther side that maketh the breadth thereof shalbe rationall, and incommensurable in length to the line whereupon the parallelogramme is applied: which was required to be proued.</p>
               <p>
                  <hi>A square is sayd to be applied vpon a line, when it, or a parallelograme equall vnto it, is applied vpon the sayd line.</hi>
                  <note place="margin">How a square is sayde to be applied vppon a line.</note> 
                  <hi>If vpon a rationall line geuen we will apply a rectangle parallelograme equall to the square of a mediall line geuen, and so of any line geuen, we must, by the 11. of the sixt, finde out the third line proportionall with the rationall line and the mediall line geuen: so yet that the rationall line be the first, and the mediall line geuen, (which containeth in power the square to be applied) be the second. For then the supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ficies contained vnder the first and the third, shalbe equall to the square of the midle line, by the 17. of the sixt.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 20. Theoreme. The 23. Proposition. A right line commensurable to a mediall line, is also a mediall line.</head>
               <p>SVppose that A be a mediall line. And vnto the line A let the line B be commen<g ref="char:EOLhyphen"/>surable, either in length, &amp; in power, or in power only.<note place="margin">Construction.</note> Then I say that B also is a mediall line. Let there be put a rationall line CD. And vpon the line CD, apply a rectangle parallelograme CE, equall vnto the square of the line A, and ma<g ref="char:EOLhyphen"/>king in breadth the line ED. Wherefore (by the proposition going before) the line ED is ra<g ref="char:EOLhyphen"/>tionall and incommensurable in length vnto the line CD.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And againe vpon the line CD apply a rec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>angle parallelograme CF equall vnto the square of the line B, and making in breadth the line DF. And forasmuch as
<figure/>
the line A is commensurable vnto the line B, therefore the square of the line A is commensurable to the square of the line B. But the parallelograme EC is equall to the square of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A, and the parallelo<g ref="char:EOLhyphen"/>grame CF is equall to the square of the line B: wherefore the parallelograme EC is co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto the parallelograme CF. But as the parallelograme EC, is to the parallelograme CF, so is the line ED to the line DF (by the first of the sixt). Wherefore (by the 10. of the tenth) the line ED is commensurable in length vnto the line DF. But the line ED is rationall and inco<g ref="char:cmbAbbrStroke">̄</g>mensura<g ref="char:EOLhyphen"/>ble in length vnto the line DC, wherefore the line DF is rationall and incommensurable in length vnto the line DC (by the 13. of the tenth). Wherefore the lines CD and DF are rationall commensurable in power onely. But a rectangle figure comprehended vnder ratio<g ref="char:EOLhyphen"/>nall right lines commensurable in power onely, is (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>1. of the tenth) irrationall, and the line that containeth it in power is irrationall, and is called a mediall line. Wherefore the line that containeth in power that which is comprehended vnder the lines CD and DF is a me<g ref="char:EOLhyphen"/>diall line. But the line B containeth in power the parallelograme which is comprehended vn<g ref="char:EOLhyphen"/>der
<pb facs="tcp:7062:279"/>
the lines CD and DF Wherefore the line B is a mediall line. A right line therfore com<g ref="char:EOLhyphen"/>mensurable to a mediall line, is also a mediall line: which was required to be proued.</p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest that a superficies commensurable vnto a mediall su<g ref="char:EOLhyphen"/>perficies, is also a mediall superficies.</hi>
                  </p>
                  <p>
                     <hi>For the lines which contain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> in power those superficieces are commensurable in power, of which the one is a mediall line (by the definitio<g ref="char:cmbAbbrStroke">̄</g> of a mediall line in the</hi> 21. <hi>of this tenth): wherefore the other also is a mediall line by this</hi> 23. <hi>propositio<g ref="char:cmbAbbrStroke">̄</g>. And as it was sayd of rationall lines so also is it to be sayd o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> mediall lines, namely, that a li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e commensurable to a mediall line, is also a mediall line, a line I say which is commensurable vnto a mediall line, whether it be commensurable in length, and also in po<g ref="char:EOLhyphen"/>wer, or ells in power onely. For vniuersally it is true, that lines commensurable in length, are also com<g ref="char:EOLhyphen"/>mensurable in power. Now if vnto a mediall line there be a line commensurable in power, if it be com<g ref="char:EOLhyphen"/>mensurable in length, the<g ref="char:cmbAbbrStroke">̄</g> are those lines called mediall lines commensurable in length &amp; in power. But if they be commensurable in power onely, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y are called mediall lines commensurable in power onely.</hi>
                  </p>
                  <p>There are also other right lines incommensurable in length to the mediall line, and commensura<g ref="char:EOLhyphen"/>ble in power onely to the same: and these lines are also called mediall, for that they are commensu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>a<g ref="char:EOLhyphen"/>ble in power to the mediall line. And in a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> mu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h as they are mediall lines, they are commensurable in power the one to the other. But being compared the one to the other, they may be commensurable ei<g ref="char:EOLhyphen"/>ther in length, and the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>efo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e in power, or ells in power onely. And then if they be commensurable in length, they are called also mediall lines commensu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>able in length, and so consequently they are vnder<g ref="char:EOLhyphen"/>standed to be commensurable in power. But i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> they be commensurable in power onely, yet notwith<g ref="char:EOLhyphen"/>standing they also are called mediall lines commensurable in power onely.</p>
                  <p>
                     <hi>Flussates</hi> after this proposition teacheth how to come to the vnderstanding of me<g ref="char:EOLhyphen"/>diall superficieces and lines, by surd numbers, after this maner. Namely to expresse the mediall superficieces by the rootes of numbers which are not square numbers: and the lines co<g ref="char:cmbAbbrStroke">̄</g>taining in power such medial superficieces, by the rootes of rootes of numbers not square. Mediall lines also commensurable, are expressed by the rootes of rootes of like s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>perficial numbers, but yet not square, but such as haue that proportion that the squares of square numbers haue. For the rootes of those numbers and the rootes of rootes are in proportion as numbers are, namely, if the squares be proportionall the sides also shalbe proportionall (by the 22. of the sixt). But mediall lines incommensu<g ref="char:EOLhyphen"/>rable in power, are the rootes of rootes of numbers, which haue not that proportion, that square numbers haue. For their rootes are the powers of mediall lines, which are incommensurable (by the 9. of the tenth). But mediall lines commensurable in power onely, are the rootes of rootes of numbers, which haue that proportion that simple square numbers haue, and not which the squares of squares haue. For the rootes (which are the powers of the mediall lines) are comme<g ref="char:cmbAbbrStroke">̄</g>surable, but the rootes of rootes (which expresse the sayd mediall lines) are incommensurable.</p>
                  <p>Wherefore there may be found out infinite mediall lines incommensurable in pow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r, by comparing infinite vnlike playne numbers the one to the other. For vnlike playne numbers, which haue not the proportion of square numbers, doo make the rootes which expresse the superficieces of mediall lines inco<g ref="char:cmbAbbrStroke">̄</g>mensurable (by the 9. of the tenth). And therefore the mediall lines containing in power those superficieces are inco<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLunhyphen"/>mensurable in length. For lines incommensurable in power, are alwayes incommensu<g ref="char:EOLhyphen"/>rable in length (by the corrollary of the 9. of the tenth).</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 21. Theoreme. The 24. Proposition. A rectangle parallelogramme comprehended vnder mediall lines co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in length, is a mediall rectangle parallelogramme.</head>
               <p>
                  <pb n="250" facs="tcp:7062:279"/>SVppose that the rectangle parallelogramme AG, be comprehended vnder these mediall right lines AB and BC, which let be commensurable in length.<note place="margin">Construction.</note> Then I say, that AC is a mediall rectangle parallelogramme. Describe (by the 46. of the first) vpon the line AB a square AD.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the square AD is a me<g ref="char:EOLhyphen"/>diall superficies. And <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as the line AB is com<g ref="char:EOLhyphen"/>mensurabl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
in length vnto the line BC, and the line AB is equall vnto the line BD, therefore the line BD is commensurable in length vnto the line BC. But <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the line DB is to the line BC, so is the square DA to the parallelogramme AC (by the first of the sixt). Wher<g ref="char:EOLhyphen"/>fore (by the 10. of the tenth) the square DA is commen<g ref="char:EOLhyphen"/>surable vnto the parallelogramme AC. But the square DA is mediall, for that it is described vpon a mediall line. Wherefore AC also is a mediall parallelogra<g ref="char:cmbAbbrStroke">̄</g>me (by the former Corollary). A rectangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp;c: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 22<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Theoreme. The 25. Proposition. A rectangle parallelogramme comprehended vnder mediall right lines commensurable in power onely, is either rationall, or mediall.</head>
               <p>A Mediall line being geuen, there may be found an other line commensura<g ref="char:EOLhyphen"/>ble vnto it, in power onely (by the 11. of this booke) as was taught there tou<g ref="char:EOLhyphen"/>ching rationall lines. Now then suppose that the rectangle parallelogramme AC be comprehended vnder these mediall right lines AB &amp; BC. Then I say, that the parallelogramme AC is either rationall, or mediall.<note place="margin">Construction.</note> Describe (by the 46. of the first) vpon the lines AB and BC, their squares AD and BE. Wherefore either of these squares AD and BE is mediall (by the 21. of the tenth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Let there be put a rationall line FG. And vpon the line FG, let there be applied a rectangle parallelogramme GH e<g ref="char:EOLhyphen"/>quall to the square AD, and making the breadth the line FH. (How to do this was taught in the 22. of this booke). And vpon the line HM, apply a rectangle parallelogramme MK, equall to the parallelogramme AC, and making in breadth the line HK: (to do this ye must take a fourth line proportionall with the lines HM, AB &amp; BC (by the 12. of the sixt) which fourth line let be HK: wherefore (by the 26. of the sixt) that which is contayned vnder the extremes HM and HK, is equall to the parallelogramme contayned vnder the meanes AB and BC). And more<g ref="char:EOLhyphen"/>ouer
<figure/>
vpon the line KN apply a rectangle paral<g ref="char:EOLhyphen"/>lelogramme NL equall to the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BE, and making in breadth the line KL. Wherfore the lines FH, HK &amp; K, are in one and the selfe same right line. (For those parallelogrammes so applied vpo<g ref="char:cmbAbbrStroke">̄</g> the lines FG, HM, and KN, are rectangle, and the angles EHM and KHM are equall to two right angles, for they are right angles: wherefore the lines FH and HK are in one right line, by the 14. of the first: So also may be sayd of the angles HKN and LKN). And
<pb facs="tcp:7062:280"/>
forasmuch as either of these squares AD and BE is mediall, and the square AD is equall to the parallelogramme GH, and the square BE to the parallelogramme NL: wherefore either of these parallelogrammes GH and NL is mediall. And they are applyed vpon a ra<g ref="char:EOLhyphen"/>tionall line, namely, FG. Wherefore (by the 22. of the tenth) either of these right lines F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>H and KL is a rationall line, and in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line FG. And forasmuch as the square AD is commensurable to the square BE (for the lines AB and BC are suppo<g ref="char:EOLhyphen"/>sed to be commensurable in power) therefore the parallelogramme GH is commensurable to the parallelogramme NL (for they are equall vnto the sayd squares). But as the paralle<g ref="char:EOLhyphen"/>logramme GH is to the parallelogramme NL, so (by the first of the sixt) is the line FH to the line KL. Wherefore (by the 10. of the tenth) the line FH is commensurable in length to the line KL. Wherefore these right lines FH and KL, are rationall commensurable in length the one to the other (commensurable in length, I say, the one to the other, for vnto the line FG, by reason of which they are rationall, they are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length as it hath bene proued). Wherefore the parallelogramme contayned vnder the lines FH and KL, is rationall (by the 19. of the tenth).</p>
               <p>
                  <note place="margin">In these fewe words consisteth the full de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> of the corol<g ref="char:EOLhyphen"/>lary fol<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ow<g ref="char:EOLhyphen"/>ing.</note>
                  <q>And forasmuch as the line DB is equall to the line BA, and the line XB to the line BC: there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore as the line DB is to the line BC, so is the line AB to the line BX. But as the line DB is to the line BC, so (by the first of the sixt) is the square DA to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AC. And as the line AB is to the line BX, so is the parallelogramme AC to the square CX. Wherefore as the square DA is to the parallelogramme AC, so is the parallelogramme AC to the square CX. But the square AD is equall to the parallelogramme GH, and to the pa<g ref="char:EOLhyphen"/>rallelogramme AC is the parallelogramme MK also equall, and to the parallelogramme NL is equall the square BE:</q>
Wherefore as the parallelogramme GH is to the parallelo<g ref="char:EOLhyphen"/>gramme MK, so is the parallelogramme MK to the parallelogramme NL. Wherefore (by the first of the sixt) as
<figure/>
the line FH is to the line HK, so is the line HK to the line KL. Wherefore (by the 17. of the sixt) the paralle<g ref="char:EOLhyphen"/>logra<g ref="char:cmbAbbrStroke">̄</g>me contayned vn<g ref="char:EOLhyphen"/>der the lines FH and KL, is equall to the square of the line HK. But the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contayned vnder the lines FH and KL, is rationall, as hath before bene proued. Wherefore the square of the line HK is also rationall. Wherefore also the line HK is rationall.</p>
               <p>And now if the line HK be commensurable in length vnto the line HM, that is, vnto the line FG,<note place="margin">Note.</note> which is equall to the line HM, then (by the 19. of the tenth) the parallelogramme NH is rationall. But if it be incommensurable in length vnto the line FG, then the lines HK and HM are rationall commensurable in power onely. And so shall the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me HN be mediall. Wherefore the parallelogramme HN is either rationall, or mediall. But the parallelogramme HN is equall to the parallelogramme AG. Wherefore the parallelo<g ref="char:EOLhyphen"/>gramme AC is either rationall, or mediall. A rectangle parallelogramme therefore com<g ref="char:EOLhyphen"/>prehended vnder mediall right lines commensurable in power onely, is either rationall, or mediall: which was required to be demonstrated.</p>
               <p>
                  <hi>How to finde mediall lines commensurable in power onely contayning a rationall parallelogramme, and also other mediall lines commensurable in power contayning a
<pb n="251" facs="tcp:7062:280"/>
mediall parallelogramme, shall afterward be taught in the 27. and 28. Propositions of thys booke.</hi>
               </p>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest, that a rectangle parallelogramme contayned vnder two right lines, is the meane proportionall betwene the squares of the sayd lines.</hi>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note> As it was manifest (by the first of the sixt) that that which is contayned vnder the lines AB and BC, is the meane proportionall betwene the squares AD and CX. This Corollary is put after the 53. Proposition of this booke as an Assumpt, and there demonstrated, which there in his place you shall finde. But because it followeth of this Proposi<g ref="char:EOLhyphen"/>tion so euidently and briefly without farther demonstration, I thought it not amisse here by the way to note it.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 23. Theoreme. The 26. Proposition. A mediall superficies excedeth not a mediall superficies, by a rationall su<g ref="char:EOLhyphen"/>perficies.</head>
               <p>FOr if it be possible, let AB being a mediall superficies, exceede AC being also a mediall superficies, by DB being a rationall superficies. And let there be put a rationall right line EF.<note place="margin">Construction.</note> And vpon the line EF apply a rectangle parallelo<g ref="char:EOLhyphen"/>gramme FH, equall vnto the mediall superficies AB, whose other side let be EH: and from the parallelogramme FH take away the parallelogramme FG, equall vnto the mediall superficies AC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Wherefore (by the third common sentence) the re<g ref="char:EOLhyphen"/>sidue BD is equall to the residue KH. But (by supposition) the superficies DB is rationall. Wherfore the superficies KH is also rationall. And forasmuch as either of these superficieces AB and AC is mediall, and AB is equall vnto FH,
<figure/>
&amp; AC vnto FG: therefore either of these superficieces FH and FG is mediall: and they are applyed vpon the rationall line EF. Wherefore (by the 22. of the tenth) either of these lines HE and EG is rationall &amp; incom<g ref="char:EOLhyphen"/>mensurable in length vnto the line EF. And forasmuch as the superficies DB is rationall, and the superficies KH is equall vnto it: therefore KH is also rationall: and it is applied vpo<g ref="char:cmbAbbrStroke">̄</g> the rationall line EF (for it is ap<g ref="char:EOLhyphen"/>plied vpon the line GK, which is equall to the line EF). Wherefore (by the 20. of the tenth) the line GH is ratio<g ref="char:EOLhyphen"/>nall and commensurable in length vnto the line GK. But the line GK is equall to the line EF. Wherfore the line GH is rationall and commensurable in length vnto the line EF. But the line EG is ra<g ref="char:EOLhyphen"/>tionall and incommensurable in length to the line EF. Wherefore (by the 13. of the tenth) the line EG is incommensurable in length vnto the line GH. And as the line EG is to the line GH, so is the square of the line EG to the parallelogramme contayned vnder the lines EG and GH (by the Assumpt put before the 21. of the tenth). Wherefore (by the 10. of the tenth) the square of the line EG is incommensurable vnto the parallelogramme contayned vnder the lines EG and GH. But vnto the square of the line EG are commensurable the squares of the lines EG and GH, for either of them is rationall, as hath before bene proued. Wherefore the squares of the lines EG and GH are incommensurable vnto the parallelo<g ref="char:EOLhyphen"/>gramme contayned vnder the lines EG and GH. But vnto the parallelogramme contayned vnder the lines EG and GH, is commensurable that which is contayned vnder the lines FG and GH twise (for they are in proportion the one to the other as number is to number,
<pb facs="tcp:7062:281"/>
namely, as vnitie is to the number 2, or as 2. is to 4: and therefore (by the 6. of this booke) they are commensurable). Wherefore (by the 13. of the tenth) the squares of the lines EG and GH are incommensurable vnto that which is contayned vnder the lines EG and GH twise. (This is more brie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ly concluded by the corollary
<figure/>
of the 13. of the tenth). But the squares of the lines EG and GH together with that which is contayned vnder the lines EG and GH twise are equall to the square of the line EH (by the 4. of the second). Wherefore the square of the line EH is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to the squares of the lines EG and GH (by the 16. of the tenth). But the squares of the lines FG &amp; GH are rationall. Wher<g ref="char:EOLhyphen"/>fore the square of the line EH is irrationall. Wherefore the line also EH is irrationall. But it hath before bene proued to be rationall: which is impossible. Wherefore a mediall superficies exceedeth not a mediall superficies by a rationall superficies: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 4. Probleme. The 27. Proposition. To finde out mediall lines commensurable in power onely, contayning a rationall parallelogramme.</head>
               <p>LEt there be put two rationall lines commensurable in power onely, namely, A and B.<note place="margin">Construction.</note> And (by the <hi>13.</hi> of the six) take the meane proportionall betwene the lines A and B, and let the same line be C. And as the line A is to the line B, so (by the <hi>12.</hi> of the sixt) let the line C be to the line D.<note place="margin">Demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>a<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as A and B are rationall lines commensurable in power onely, therfore (by the <hi>21.</hi> of the tenth) that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines A and B, that is, the square of the line C. For the square of the line C is equall to the parallelogramme contayned vnder the lines A an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> B (by the <hi>17.</hi> of the sixth) is mediall, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>herfore C also is a mediall line. And for that
<figure/>
as the line A is to the line B, so is the line C to the line D, therfore as the square of the line A is to the square of the lyne B, so is the square of the line C to the square of the line D (by the 22. of the sixth). But the squares of the lines A and B are commensurable, for the li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s A and B a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e supposed to be rationall comme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable in power onely. Wherefore also the squares of the lines C and D are commensurable (by the 10. of the tenth) wherfore the lines C and D are commensu<g ref="char:EOLhyphen"/>rable in power onely. And C is a mediall line. Wherfore (by the 23. of the tenth) D also is a mediall line. Wherfore C and D are mediall lynes commensurable in power onely. Now also I say that they contayne a rationall parallelogramme. For for that as the line A is to the line B, so is the line C to the line D: therfore alternately also (by the 16. of the fift) as the line A is to the line C, so is the lyne B to the lyne D. But as the lyne A is to the lyne C, so is the line C to the lyne B: wherfore as the line C is to the line B, so is the line B to the lyne D. Wherfore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines C and D is equal to the square of the line B. But the square of the lyne B is rationall. Wherfore the parallelograme which is contayned vnder the lynes C and D is also rationall. Wherfore there are found out mediall lines commensura<g ref="char:EOLhyphen"/>bl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <pb n="252" facs="tcp:7062:281"/>
in pow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ly contayning a rationall parallelogramme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> required to be done.</p>
            </div>
            <div type="part">
               <head>The 5. Probleme. The 28. Proposition. To finde out mediall right lynes commensurable in power onely, contayning a mediall parallelogramme.</head>
               <p>LEt there be put three rationall right lines commensurable in power only, name<g ref="char:EOLhyphen"/>ly, A, B, and C,<note place="margin">Construction.</note> and (by the 13. of the sixt) take the meane proportional betwene the lines A and B, &amp; let th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> same be D. And as the line B is to the line C, so (by the 12. of the sixt) let the line D be to the line E. And forasmuch as the lines A and B are rationall commensurable in power onely,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore (by the 21. of the tenth) that which is contained vnder the lines A
<figure/>
and B, that is the square of the line D, is mediall. Wherfore D is a mediall line. And forasmuch as the lines B and C are commensurable in power onely; and as the line B is to the line C, so is the line D to the line E: wherfore the lines D and E are commensurable in power onely (by the corollary of the tenth of this booke) but D is a mediall line. Wherefore E also is a mediall line (by the 23. of this booke.) Wherfore D &amp; E are mediall lines commensurable in power onely. I say also that they containe a mediall parallelograme. For for that as the line B is to the line C, so is the line D to the line E: therfore alternately (by the 16 of the fift) as the line B is to the line D, so is the line C to y<hi rend="sup">e</hi> line E. But as the lyne B is to the line D, so is the line D to the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by conuerse proportion (which is proued by the corollary of the fourth of the fifth) Wherfore as the line D is to the line A, so is the line C to the line E. Wherfore that which is contained vnder the lines A &amp; C, is (by the 16. of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) equall to that which is contayned vnder the lines D &amp; E. But that which is contained vnder the lines A and C is medial (by the 21. of the tenth.) Wherfore that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines D and E is mediall. Wherfore there are found out mediall lines commensu<g ref="char:EOLhyphen"/>rable in power onely, containing a mediall superficies: which was required to be done.</p>
               <div type="section">
                  <head>An Assumpt.</head>
                  <p>To finde out two square numbers, which added together make a square number.</p>
                  <p>Let there be put two like superficiall numbers AB and BC (which how to finde out, hath bene taught after the 9. proposition of this booke) And let them both be either euen numbers or odde. And let the greater number be AB. And forasmuch as if from any euen number be taken away an euen number, or fro<g ref="char:cmbAbbrStroke">̄</g>
                     <figure/>
an odde number be taken away an odde num<g ref="char:EOLhyphen"/>ber, the residue shall be euen (by the 24. and 26 of the ninth). If therfore from AB being an euen number be taken away BC an euen num<g ref="char:EOLhyphen"/>ber, or from AB being an odde number be taken away BC being also odde: the residue AC shall be euen. Deuide the number AC into two equall partes in D: wherefore the number which is produced of AB into BC together with the square number of CD, is (by the sixt of the second, as Barlaam demonstrateth it in numbers) equall to the square number of BD.
<pb facs="tcp:7062:282"/>
But that which is produced of AB into BC is a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. For it was proued (by the first of the ninth) that if two like plaine numbers multiplieng the one the other, produce any nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber, the number produced shal be a square number. Wherfore there are found out two square numbers, the one being the square number which is produced of AB into BC, and the other the square number produced of CD, which added together make a square number, namely, the square number produced of BD multiplied into himselfe, forasmuch as they were demo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>strated equall to it.</p>
               </div>
               <div type="section">
                  <head>
                     <note n="*" place="margin">A Corollary.</note>A Corollary.</head>
                  <p>
                     <note place="margin">To finde out two square n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>mbers ex<g ref="char:EOLhyphen"/>ceeding the one the other by a square <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>umber.</note>And hereby it is manifest, that there are found out two square numbers, namely, the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the square number of BD, and the other the square number of CD, so that that numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r wherin thone excedeth the other, the number (I say) which is produced of AB into BC, is also a square number: namely, when A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp;
<figure/>
BC are like playne numbers. But when they are not like playne numbers, then are there found out two square numbers, the square number of BD, and the square number of DC, whose excesse, that is, the number wherby the greater excedeth the lesse, namely, that which is produced of AB into BC, is not a square number.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>To finde out two square numbers which added together make not a square number.</p>
                  <p>
                     <note place="margin">An Assumpt.</note>Let AB and BC be like playne numbers, so that (by the first of the ninth) that which is produced of AB into BC is a square number, and let AC be an euen number. And deuide C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> into two equall par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es in D. Now by that which hath before bene sayd in the former as<g ref="char:EOLhyphen"/>sumpt, it is manifest that the square number produced of AB into BC, together with the square number of CD, is equall to the square number of BD. Take away from CD vnitie DE. Wherfore that which is produced of AB into BC together with the square of CE is lesse then the square number of
<figure/>
BD. Now then I say that the square num<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er produced of AB into BC added to the square number of CE, make not a square number. For if they do make a square number, then that square number which they make, is either greater the<g ref="char:cmbAbbrStroke">̄</g> the square number of BE, or equall vnto it, or lesse then it. First, greater it cannot be, for it is already proued that the square number produced of AB into BC, together with the square number of CE, is lesse then the square number of BD. But betwene the square number of BD, and the square number of BE, there is no meane square number. For the number BD excedeth the number BE onely by vnitie: which vnitie can by no meanes be deuided into numbers. Or if the number produced of AB into BC together with the square of the nu<g ref="char:cmbAbbrStroke">̄</g>ber CE, should be greater then the square of the number BE, then should the selfe same number produced of AB into BC together with the square of the number CE, be equall to the square of the number BD, the contrary wherof is already proued. Wherfore if it be possible, let that which is produced of AB into BC together with the square number of the number CE be equall to the square number of BE. And let GA be double to vnitie DE, that is, let it be the num<g ref="char:EOLhyphen"/>ber two. Now forasmuch as the whole number AC is by supposition double to the whole number CD, of which the number AG is double to vnitie DE, therfore (by the 7. of the se<g ref="char:EOLhyphen"/>uenth)
<pb n="253" facs="tcp:7062:282"/>
the residue, namely, the number GC is double to the residue, namely, to the number EC. Wherfore the number GC is deuided into two equall partes in E. Wherefore that which is produced of GB into BC together with the square number of CE is equall to the square nu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ber of BE. But that which is produced of AB into BC, together with the square number of CE, is supposed to be equal to the square number of BE wherfore that which is produced of GB into BC together with the square number of CE is equall to that which is produced of AB into BC, together with the square number of CE. Wherefore taking away the square number of CE, which is common to them both, the number AB shall be equall to the num<g ref="char:EOLhyphen"/>ber GB, namely, the greater to the lesse, which is impossible. Wherfore that which is produced of AB into BC together with the square number of CE, is not equall to the square number of BE, I say also that that which is produced of AB into BC together with the square num<g ref="char:EOLunhyphen"/>ber of CE is not lesse then the square number of BE. For if it be possible, the<g ref="char:cmbAbbrStroke">̄</g> shall it be equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l to some square number lesse then the square number of BE. Wherfore let the number produ<g ref="char:EOLhyphen"/>ced of AB into BC together with the square of the number CE be equal to the square num<g ref="char:EOLhyphen"/>ber of BF. And let the number HA be double to the number DF. The<g ref="char:cmbAbbrStroke">̄</g> also it followeth that the number HC is double to the number CF, so that HC also is deuided into two equall partes in F, and therfore also the number which is produced of HD into BC, together with the square number of FC, is equall to the square number of the number BF. But by supposi<g ref="char:EOLhyphen"/>tion, the number which is produced of AB into BC together with the square number of CE is equall to the square number of BF. Wherfore it followeth that the number produced of AB into BC together with the square number of CE, is equall to that which is produced of HB into BC together with the square number CF, which is impossible. For if it should be e<g ref="char:EOLhyphen"/>quall, then forasmuch as the square of CF is lesse then the square of CE, the number produ<g ref="char:EOLhyphen"/>ced of HB into BC should be greater then th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> number produced of AB into BC. And so al<g ref="char:EOLhyphen"/>so should the number HB be greater then the number AB, when yet it is lesse then it. Wher<g ref="char:EOLhyphen"/>fore the number produced of AB into BC together with the square number of CE, is not lesse then the square nu<g ref="char:cmbAbbrStroke">̄</g>ber of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> E. And it is also proued that it cannot be equall to the square number of BE, neither greater then it. Wherfore that which is produced of AB into BC added to the square number of CE, maketh not a square number. And although it be possible to demonstrate this many other wayes, yet this semeth to vs suffici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, least the matter beyng ouer long, should seeme to much tedious.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 6. Probleme. The 29. Proposition. To finde out two such rationall right lynes commensurable in power on<g ref="char:EOLhyphen"/>ly, that the greater shall be in power more then the lesse, by the square of a right line commensurable in length vnto the greater.</head>
               <p>LEt there be put a rational line AB, and take al<g ref="char:EOLhyphen"/>so
<figure/>
two such square numbers CD and DE,<note place="margin">Construction.</note> that their excesse CE be not a square number (by the corolary of the first assumpt of the 28. of the tenth) And vpon the line AB describe a semicircle AFB. And by the corollary of the 6. of the tenth, as the number DC is to the number CE, so let the square of the lyne BA be to the square of the line AF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And draw a line from F to B. Now for that as the square of the line BA is to the square of the line AF, so is the number CD to the num<g ref="char:EOLhyphen"/>ber
<pb facs="tcp:7062:283"/>
CE, therfore the square of the line BA hath to the square of the line AF, that propor<g ref="char:EOLhyphen"/>tion that the nu<g ref="char:cmbAbbrStroke">̄</g>ber CD hath to the number CE. Wherfore the square of the line BA is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>surable to the square of the line AF (by the 6. of the te<g ref="char:cmbAbbrStroke">̄</g>th). But the square of the line AB is rational. Wherfore also the square of the line AF is rational. Wherfore also the line AF is rationall. And forasmuch as the number CD hath not vnto the number CE that proporti<g ref="char:EOLhyphen"/>on that a square number hath to a square number, therfore neither also hath the square of the line AB to the square of the line AF that proportion
<figure/>
that a square number hath to a square number. Wherfore (by the 9. of the te<g ref="char:cmbAbbrStroke">̄</g>th) the line AB is vnto the line AF in<g ref="char:EOLhyphen"/>commensurable in length. Wherfore the lines AF and AB are rationall commensurable in power onely. And for that as the number DC is to the number CE, so is the square of the line AB to the square of the line AF: ther<g ref="char:EOLhyphen"/>fore by conuersion or euerse proportio<g ref="char:cmbAbbrStroke">̄</g> which is demonstra<g ref="char:EOLhyphen"/>ted (by the corollary of the 19. of the fifth) as the number CD is to the number DE, so is the square of the line AB to the square of the line BF, which is the excesse of the square of the line AB aboue the square of the line AF (by the assumpt put before the 14. of this booke). But the number CD hath to the number DE that proportion that a square number hath to a square number: wherfore the square of the line AB hath to the square of the line BF, that proportion that a square num<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er hath to a square number. Wherefore (by the 9. of the tenth) the line AB is commensurable in length vnto the line BF. And (by the 47. of the first) the square of the line AB is equall to the squares of the lines AF and FB. Wherfore the line AB is in power more then the line AF by the square of the line BF which is commensurable in length vn<g ref="char:EOLhyphen"/>to the line AB. Wherefore there are found out two such rationall lines commensurable in power onely, namely, AB and AF, so that the greater line AB is in power more then the lesse line AF, by the square of the line FB, which is commensurable in length vnto the line AB: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶The 7. Theoreme. The 30. Proposition. To finde out two such rationall lines commensurable in power onely,<note place="margin">Montaureus maketh this an Assumpt: as the Grecke text seemeth to do likewise but without a cause.</note> that the greater shalbe in power more then the lesse by the square of a right line incommensurable in length to the greater.</head>
               <p>LEt there be put a rationall line <hi>AB,</hi> and take also (by the 2. assumpt of the 28. of the tenth) two square numbers <hi>CE</hi> and <hi>ED,</hi> which being added together make not a square number, and let the numbers <hi>CE</hi> and <hi>ED</hi> added together make the number <hi>CD.</hi>
                  <note place="margin">Construction.</note> And vpon the line <hi>AB</hi> describe a sencircle <hi>AFB.</hi> And (by the corollary of the 6. of the tenth) as the number <hi>DC</hi> is to the
<figure/>
number <hi>CE,</hi> so let the square of the line <hi>AB</hi> be to the square of the line <hi>AF,</hi> and draw a line from <hi>F</hi> to <hi>B.</hi> And we may in like sort,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> as we did in the former proposition, proue that the lines <hi>BA</hi> and <hi>AF</hi> are rationall commensurable in power one<g ref="char:EOLhyphen"/>ly. And for that as the number <hi>DC</hi> is to the number <hi>CE,</hi> so is the square of the line <hi>AB</hi> to the square of the line <hi>AF</hi>: therefore by conuersion (by the corollary of the 19. of the fifte) as the number <hi>CD</hi> is to the number <hi>DE,</hi> so is the square of the line <hi>AB</hi> to the square to the line <hi>FB.</hi> But the number <hi>C
<pb n="254" facs="tcp:7062:283"/>
D</hi> hath not to the number <hi>DE</hi> that proportion that a squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mbe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th to a square number. Wherefore neither also the square of the line <hi>AB</hi> hath to the square of the line <hi>BF</hi> that proportion that a square number hath to a square number. Wherefore the line <hi>AB</hi> is (by the 9 of the tenth) incommensurable in length to the line <hi>BF.</hi> And the line <hi>AB</hi> is in power more then the line <hi>AF</hi> by the square of the right line <hi>BF,</hi> which is incommen<g ref="char:EOLhyphen"/>surable in length vnto the line <hi>AB.</hi> Wherfore the lines <hi>AB</hi> and <hi>AF</hi> are rationall commen<g ref="char:EOLhyphen"/>surable in power onely. And the line <hi>AB</hi> is in power more then the line <hi>AF</hi> by the square of the line <hi>FB</hi> which is commensurable in length vnto the line <hi>AB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be done.</p>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>
                     <q>If there be two right lines hauing betwene them selues any proportion: as the one right line is to the other, so is the parallelograme contained vnder both the right lines to the square of the lesse of those two lines.</q>
                  </p>
                  <p>Suppose that these two right <hi>AB</hi> and <hi>BC</hi> be
<figure/>
in some certaine proportion.<note place="margin">This Assumpt setteth fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>th nothing <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ls but that which the first o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>t <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>et<g ref="char:EOLhyphen"/>teth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>orth, and therefore in s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>me exam<g ref="char:EOLhyphen"/>plars it is not founde.</note> Then I say that as the line <hi>AB</hi> is to the line <hi>BC,</hi> so is the parallelograme contained vnder <hi>AB</hi> and <hi>BC</hi> to the square of <hi>BC.</hi> Describe the square of the line <hi>BC</hi> and let the same be <hi>CD,</hi> and make perfect the parallelograme <hi>AD</hi> now it is manifest that as the line <hi>AB</hi> is to the line <hi>BC,</hi> so is the parallelograme <hi>AD</hi> to the parallelograme or square <hi>BE</hi> (by the first of the sixt). But the parallelograme <hi>AD</hi> is that which is bontained vnder the lines <hi>AB</hi> and <hi>BC,</hi> for the line <hi>BC</hi> is equall to the line <hi>BD</hi> and the parallelograme <hi>BE</hi> is the square of the line <hi>BC.</hi> Wherefore as the line <hi>AB</hi> is to the line <hi>BC</hi> so is the parallelograme coutained vnder the lines <hi>AB</hi> and <hi>BC</hi> to the square of the line <hi>BC,</hi> which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 8. Probleme. The 31. Proposition. To finde out two mediall lines commensurable in power onely, comprehen<g ref="char:EOLhyphen"/>ding a rationall superficies, so that the greater shall be in power more then the lesse by the square of a line commensurable in length vnto the greater.</head>
               <p>LEt there be taken (by the 29. of the tenth) two rationall lines commensurable in pow<g ref="char:EOLhyphen"/>er onely <hi>A</hi> and <hi>B,</hi>
                  <note place="margin">Construction.</note> so that let the line <hi>A</hi> being the greater be in power more then the line <hi>B,</hi> being the lesse by the square of a line commensurable in length vnto the line <hi>A</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And let the square of the line <hi>C</hi> be equall to the parallelograme contained vnder the lines <hi>A</hi> and <hi>B</hi> which is done by finding out the meane proportionall line, namely the line <hi>C</hi> be<g ref="char:EOLhyphen"/>twene the lines <hi>A</hi> and <hi>B</hi> (by the 13. of the sixt). Now the parallelograme contained vnder the lines <hi>A</hi> and <hi>B</hi> is mediall (by the 21. of this booke). Wherefore (by the corollary of the 23. of the tenth) the square also of the line <hi>C</hi> is mediall. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore the line <hi>C</hi> also is mediall: Vnto the square of the line <hi>B</hi> let the parallelograme contained vnder the lines <hi>C</hi> and <hi>D</hi> be equall (by finding out a third line proportionall) namely the line <hi>D</hi> to the two lines <hi>C</hi> and <hi>B</hi> (by the 11. of the sixt). But the square of the line <hi>B</hi> is rationall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the parallelograme con<g ref="char:EOLhyphen"/>tained vnder the line <hi>C</hi> and <hi>D</hi> is rationall. And for that as the line <hi>A</hi> is to the line <hi>B,</hi>
                  <pb facs="tcp:7062:284"/>
so is the parallelograme contained vnder the lines <hi>A</hi>
                  <figure/>
and <hi>B</hi> to the square of the line <hi>B</hi> (by the assumpt go<g ref="char:EOLhyphen"/>ing before). But vnto the parallelograme contained vnder the lines <hi>A</hi> and <hi>B</hi> is equall the square of the line <hi>C,</hi> and vnto the square of the line <hi>B</hi> is equal the parallelograme contained vnder the lines <hi>C</hi> and <hi>D,</hi> as it hath now bene proued: therefore as the line <hi>A</hi> is to the line <hi>B,</hi> so is the square of the line <hi>C</hi> to the pa<g ref="char:EOLhyphen"/>rallelograme contained vnder the lines <hi>C, D.</hi> But as the square of the line <hi>C</hi> is to that which is contayned vnder the lines <hi>C</hi> and <hi>D,</hi> so is the line <hi>C</hi> to the line <hi>D.</hi> Wherefore as the line <hi>A</hi> is to the line <hi>B,</hi> so is the line <hi>C</hi> to the line <hi>D.</hi> But (by supposition) the line <hi>A</hi> is commensura<g ref="char:EOLhyphen"/>ble vnto the line <hi>B</hi> in power onely. Wherefore (by the <hi>11.</hi> of the tenth) the line <hi>C</hi> also is vnto the line <hi>D</hi> commensurable in power onely. But the line <hi>C</hi> is mediall. Wherefore by the 23<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth) the line <hi>D</hi> also is mediall. And for that as the line <hi>A</hi> is to the line <hi>B,</hi> so is the line <hi>C</hi> to the line <hi>D</hi>: but the line <hi>A</hi> is in power more then the line <hi>B,</hi> by the square of a line com<g ref="char:EOLhyphen"/>mensurable in length vnto the line <hi>A</hi> (by supposition). Wherefore the line <hi>C</hi> also is in power more then the line <hi>D</hi> by the square of a line commens<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rable in length vnto the line <hi>C.</hi> Wherefore there are found out two mediall lines <hi>C</hi> and <hi>D</hi> commensurable in power one<g ref="char:EOLhyphen"/>ly comprehending a rationall superfi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ies, and the line <hi>C</hi> is in power more then the line <hi>D,</hi> by the square of a line commensurable in length vnto the line <hi>C.</hi> And in like sort may be found out two mediall lines commensurable in power onely contayning a rationall superficies, so that the greater shalbe in power more the<g ref="char:cmbAbbrStroke">̄</g> the lesse by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the greater, namely, when the line <hi>A</hi> is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line <hi>B</hi> by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensura<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>le in length vnto the line <hi>A,</hi> which to do is taught by the 30. of this booke.</p>
               <p>
                  <hi>The selfe same construction remaining, that part of this proposition fro<g ref="char:cmbAbbrStroke">̄</g> these wordes.</hi> And for that as the line A is to the line B, <hi>to these wordes,</hi> But (by supposition) the line A is commensurable vnto the line B, <hi>may more easely be demonstrated after this maner. The lines C, B, D, are in continuall proportion by the second part of the 17. of the sixt. But the lines A, C, D are also in continuall proportion by the same. Wherefore by the 11. of the fifth, as the line A is to the line C, so is the line B to the line D. Wherfore alternately as the line A is to the line B so is the line C to the line D. &amp;c. which was required to be doone.</hi>
               </p>
               <div type="section">
                  <head>¶An assumpt.</head>
                  <p>
                     <q>If there be three right lines hauing betwene them selues any proportion: as the first is to the third, so is the parallelograme contained vnder the first and the second, to the parallelograme contained vnder the second and the third.</q>
                  </p>
                  <p>Suppose that these three lines <hi>AB, B
<figure/>
C,</hi> and <hi>CD</hi> be in some certayne proporti<g ref="char:EOLhyphen"/>on. Then I say that as the line <hi>AB</hi> is to the line <hi>CD,</hi> so is the parallelograme contay<g ref="char:EOLhyphen"/>ned vnder the lines <hi>AB</hi> and <hi>BC</hi> to the parallelograme contayned vnder the lines <hi>BC</hi> and <hi>CD.</hi>
                     <note place="margin">Construction.</note> From the point <hi>A</hi> raise vp vnto the line <hi>AB</hi> a perpendicular line <hi>AE,</hi> and let <hi>AE</hi> be equall to the line <hi>BC</hi>: and by the poynt <hi>E</hi> draw vnto the line <hi>AD</hi> a pa<g ref="char:EOLhyphen"/>rallel line <hi>EK</hi>: and by euery one of the poyntes <hi>B, C,</hi> and <hi>D</hi> draw vnto the line <hi>AE</hi> parallel lines <hi>BF, CH,</hi> and <hi>DK.</hi> And for that as the line <hi>AB</hi> is to the line <hi>BC</hi> so is the parallelo<g ref="char:EOLhyphen"/>grame <hi>AF</hi> to the parallelograme <hi>BH</hi> (by the first of the sixt):<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and as the line <hi>BC</hi> is to the
<pb n="255" facs="tcp:7062:284"/>
lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>CD,</hi> so is the parallelograme <hi>BH</hi> to the parallelograme <hi>CK.</hi> Wherefore of equalitie as the line <hi>AB</hi> is to the line <hi>CD,</hi> so is the parallelograme <hi>AF</hi> to the parallelograme <hi>CK.</hi> But the parallelograme <hi>AF</hi> is that which is contayned vnder the lines <hi>AB</hi> and <hi>BC,</hi> for the line <hi>AE</hi> is put equall to the line <hi>BC.</hi> And the parallelograme <hi>CK</hi> is that which is contained vnder the lines <hi>BC</hi> and <hi>CD,</hi> for the line <hi>BC</hi> is equal to the line <hi>CH,</hi> for that the line <hi>CH</hi> is equall to the line <hi>AE</hi> (by the 34. of the first). If therefore there be three right lines ha<g ref="char:EOLhyphen"/>uing betwene them selues any proportion: as the first is to the third, so is the parallelograme contained vnder the first and the second, to the parallelogramme co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the second and the third: which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 9. Probleme. The 32. Proposition. To finde out two mediall lines commensurable in power onely, comprehen<g ref="char:EOLhyphen"/>ding a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, so that the greater shall be in power more then the lesse, by the square of a line commensurable in length vnto the greater.</head>
               <p>LEt there be taken three rationall lines co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely, A, B, C, so, that (by the 29. of the tenth) let the line A be in power more then the line C, by the square of a line commensurable in length vnto the line A.<note place="margin">Construction.</note> And vnto the parallelogramme contayned vnder the lines A &amp; B, let the square of the line D be equall. But that which is contayned vnder the lines A and B is mediall. Wherefore (by the Corollary of the 23. of the tenth) the square of the line D also is mediall. Wherefore the line D also is mediall. And vnto that which is contayned vnder the lines B and C, let be equall that which is contayned vnder the lines D and E (which is done by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inding out a fourth line proportionall vnto the lines D, B, C, which let be the line E).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for that (by the As<g ref="char:EOLhyphen"/>sumpt
<figure/>
going before) as that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines A and B is to that which is contayned vnder the lines B and C, so is the line A to the line C. But vnto that which is contayned vnder the lines A &amp; B, is equall the square of the line D, and vnto that which is contayned vnder the lines B &amp; C, is equall that which is contayned vnder the lines D and E. Wherefore as the line A is to the line C, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the square of the line D, to that which is contayned vnder the lines D and E. But as the square of the line D is to that which is contayned vnder the lines D and E, so is the line D to the line E (by the Assumpt put before the 22. of the tenth). Wherefore as the line A is to the line C, so is the line D to the line E. But the line A is vnto the line C commensurable in power onely. Wherefore the line D is vnto the line E commensurable in power onely. But D is a mediall line. Wherefore (by the 23. of the tenth) E also is a mediall line. And for that as the line A is to the line C, so is the line D to the line E, and the line A is in power more then the line C, by the square of a line commensurable in length vnto the line A. Wherefore (by the 14. of the tenth) D is in power more then E, by the square of a line commensurable in length vnto the line D. I say moreouer that that which is contayned vnder the lines D and E is mediall. For forasmuch as that which is contayned vnder the lines B &amp; C, is equall to that which is contayned vnder the lines D and E: but that which is contayned vnder the lines B and C is mediall. Wherefore that which is contayned vnder the lines D and E is al<g ref="char:EOLhyphen"/>so mediall. Wherefore there are found out two mediall lines D and E, co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely, comprehending a mediall superficies, so that the greater is in power more then the lesse, by the square of a line commensurable in length to the greater: which was required to be
<pb facs="tcp:7062:285"/>
done. And thus i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> eu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t, how in like sorte may be founde out two mediall lines comm<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>surable in power. onely,<note place="margin">A Corollary.</note> contayning a mediall superfic<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es, so that the greater shall be in power more the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lesse, by the square of a line incommensurable in length vnto the greater. When the line A is in power more then the line C, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line A: as the thirteth teacheth vs.</p>
               <div type="section">
                  <head>
                     <hi>1.</hi> ¶An Assumpt.</head>
                  <p>Suppose that there be a rectangle triangle ABC, hauing the angle BAC a right angle. And (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y the 12. of the first) from the poynt A to the right line BC, a perpendicular line being drawen AD: then I say first, that the parallelogramme contayned vnder the lines <milestone type="tcpmilestone" unit="unspecified" n="1"/> C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and BD, is equall to the square of the line BA. Secondly I say, that the parallelo<g ref="char:EOLhyphen"/>gramme <milestone type="tcpmilestone" unit="unspecified" n="2"/> contay<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed vnder the lines BC and CD, is equall to the square of the line CA. Thirdly I say, that the parallelogramme contayned vnder the lines BD and DC, is equall to <milestone type="tcpmilestone" unit="unspecified" n="3"/> the square of the line AD. And fourthly I say, that the parallelogramme contayned vnder <milestone type="tcpmilestone" unit="unspecified" n="4"/> the lines BC &amp; AD, is equall to the parallelogramme co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines BA &amp; AC.</p>
                  <p>As tou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hing <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he first, that the parallelogramme contayned vnder the lines CB and BD, is equall to the square of the line AB, is thus proued.
<figure/>
For foras<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uch as in the rectangle triangle BAC, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>om the right angle vnto the base is drawen a per<g ref="char:EOLhyphen"/>pendicular l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ne AD, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ore (by the 8. of the sixt) the triangles ABD and ADC, are like to the whole <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ria<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gle ABC, and are also like the one to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he other. A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d for that the triangle ABC is like to the triangle ADE, therefore both the triangles are equiangle by the d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>finitio<g ref="char:cmbAbbrStroke">̄</g> of like figures. Wher<g ref="char:EOLhyphen"/>fore (by the 4. of the sixt) as the line CB is to the line BA, so is the line AB to the line BD. <note n="*" place="margin">A Corollary added by I. Dee.</note> Wherefore (by the 17. of the sixt) the parallelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines BC &amp; BD, is equall to the, square of the line AB.<note n="*" place="margin">
                        <hi>Therefore if you deuide the square of the side AB by the side BC: the quotient will be BD. Which maketh DC also knowen: by ei<g ref="char:EOLhyphen"/>the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of which (by the 47. of the first) the per<g ref="char:EOLhyphen"/>pendicular AD is knowen: which considera<g ref="char:EOLhyphen"/>tion in practise, is very profitable, where the three sides AB, AC, and BC, are knowen or geuen.</hi>
                     </note>
                  </p>
                  <p>As touching the second, that the parellelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and CD, is equall to the square of the line AC is by the selfe same reason proued. For the triangle ABC is like to the triangle ADC. Wherefore as the line BC is to the line AC,<note place="margin">I. Dee</note> so is the line, AC to the line DC. <note n="*" place="margin">The second Corollary.</note> Wherefore the parallelogramme contained vnder the lines BC and CD, is equall to the square of the line AC.<note n="*" place="margin">
                        <hi>Therefore if you deuide the square of the side AC, by the side BC, the portion DC, will be the product. &amp;c. as in the former Corol<g ref="char:EOLhyphen"/>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ry.</hi>
                     </note> As tou<g ref="char:EOLhyphen"/>ching the third, that the parallelogramme contained vnder the lines BD and DC, is equall to the square of the line DA, is thus proued. For, forasmuch as if in a rectangle triangle be drawn fro<g ref="char:cmbAbbrStroke">̄</g> the right angle to the base a perpendicular line, the perpendicular so drawen is the meane proportional betwene the segme<g ref="char:cmbAbbrStroke">̄</g>ts of the base (by the corollary of the 8. of the sixt): ther<g ref="char:EOLunhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore as the line BD is to the line DA, so is th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> line AD to the line DC. Wherefore (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the sixt) the parallelogramme contayned vnder the lines BD and DC, is equall to the square of the line DA. As touching the fourth, that the parallelogramme contained vnder the lines BC and AD, is equall to the parallelogramme contained vnder the lines BA and AC, is thus proued. For forasmuch as (as we haue already declared) the triangle ABC is like, and therefore equiangle, to the triangle ABD, therefore as the line BC is to
<pb n="256" facs="tcp:7062:285"/>
the line AC,<note place="margin">I. D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e</note> so is the line BA to the line AD (by the 4. of the sixt).<note n="*" place="margin">The thirde Corollary.</note> But if there be foure right lines proportionall, that which is contained vnder the first and the last, is equall to that which is contained vnder the two meanes (by the 16. of the sixt). Wherefore that which is contained vnder the lines BC and AD, is equall to that which is contayned vnder the lines BA and AC.<note n="*" place="margin">
                        <hi>Therfore if the parallelogramme of BA, and AC, be deuided by BC, the product will geue the p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ndicular D A. These three Co<g ref="char:EOLhyphen"/>rollaryes in practise Logisticall and Geometri<g ref="char:EOLhyphen"/>call are profitable.</hi>
                     </note>
                  </p>
                  <p>I say moreouer, that if there be made a parallelogramme complete,<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration of this fourth part of the determinatio<g ref="char:cmbAbbrStroke">̄</g>.</note> contained vnder the lines BC and AD, which let be EC: and if likewise be made complete the parallelogramme contained vnder the lines BA and AC, which let be AF, it may by an other way be proued that the parallelogramme EC is equall to the parallelogramme AF. For, forasmuch as ei<g ref="char:EOLhyphen"/>ther of them is double to the triangle ACB (by the 41. of the first): and thinges which are double to one and the selfe same thing, are equall the one to the other. Wherefore that which is contained vnder the lines BC and AD, is equall to that which is contained vnder the lines BA and AC.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>2.</hi> ¶An Assumpt.</head>
                  <p>
                     <q>If a right line be deuided into two vnequall partes: as the greater part is to the lesse,<note place="margin">An Assumpt.</note> so is the parallelogramme contayned vnder the whole line and the greater part, to the parallelogramme contayned vnder the whole line and the lesse part.</q>
                  </p>
                  <p>Deuide the right line AB into two vnequall partes in the point E: And let AE be the greater part. Then I say, that as the line AE is to the line EB, so is the parallelogramme contained vnder the lines BA and AE to the parallelogramme contained vnder the lines BA &amp; BE. Describe the square of the line AB, and let the
<figure/>
same be ACDB. And from the point E draw vnto either of these lines AC and DE a parallell line EF. Now it is manifest, that as the line AE is to the line EB, so is the pa<g ref="char:EOLhyphen"/>rallelogramme AF to the parallelogramme BF (by the first of the sixt). But the parallelogramme AF is contayned vn<g ref="char:EOLhyphen"/>der the lines BA and AE (for the line AC is equall to the line AB) and the parallelogramme BF is contained vn<g ref="char:EOLhyphen"/>der the lines AB and BE (for the line DB is equall to the line AB). Wherefore as the line AE is to the line EB, so is the parallelogramme contained vnder the lines BA and AE, to the parallelogramme contained vnder the lines AB and BE: which was requi<g ref="char:EOLhyphen"/>red to be demonstrated.</p>
                  <p>
                     <hi>This Assumpt differeth litle from the first Proposition of the sixt booke.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>
                     <hi>3.</hi> ¶An Assumpt.</head>
                  <p>
                     <q>If there be two vnequall right lines, and if the lesse be deuided into two e<g ref="char:EOLhyphen"/>quall partes: the parallelogramme contained vnder the two vnequall lines, is double to the parallelogramme contained vnder the greater line &amp; halfe of the lesse line.</q>
                  </p>
                  <p>
                     <pb facs="tcp:7062:286"/>Suppose that there be two vnequall right lines AB and BC, of which le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> AB be the greater, and deuide the line BC into two equall partes in the point D. Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n I say, that the pa<g ref="char:EOLhyphen"/>rallelogramme contained vnder the lines AB &amp; BC, is double to the parallelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines AB and BD. From the point B raise vp vpon the right line BC, a perpendicular line BE, and let BE be equall to the line BA. And drawing from the point C and D, the lines CG and DF parallels and equall
<figure/>
to BE: and then drawing the right line GFE, the figure is complete. N<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> for that a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he line DB is to the line DC, so is the parallelogramme BF to the parallelogramme DG (by the 1. of the sixt): ther<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore by composition of proportion, as the whole line BC is to the line DC, so is the parallelogramme BG to the parallelogramme DG (by the 18. of the fift). But the line BC is double to the line DC. Where<g ref="char:EOLhyphen"/>fore the parallelogramme BG is double to the pa<g ref="char:EOLhyphen"/>rallelogramme DG. But the parall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>logramme BG is contained vnder the lines AB and BC) for the line AB is equall to the line BE (and the parallelo<g ref="char:EOLhyphen"/>gramme DG is contayned vnder the lines AB and BD) for the line BD is equall to the line DC, and the line AB to the line DF: which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 10. Probleme. The 33. Proposition. To <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inde out two right lines incommensurable in power, whose squares ad<g ref="char:EOLhyphen"/>ded together make a rationall superficies, and the parallelogramme contai<g ref="char:EOLhyphen"/>ned vnder them make a mediall superficies.</head>
               <p>TAke by the 30. of the tenth, two rationall right lines commensurable in power onely, namely, AB and BC, so that let the line AB, being the greater, be in power more then the line BC being the lesse,<note place="margin">Construction.</note> by the square of a line incommen<g ref="char:EOLhyphen"/>surable in length vnto the line AB. And by the 10. of the first, deuide the line BC into two equall partes in the point D. And vpon the line AB apply a parallelogramme equall to the square
<figure/>
either of the line BD or of the line DC, and wanting in figure by a square, by the 28. of the sixth, and let that pa<g ref="char:EOLhyphen"/>rallelogramme be that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AE and EB. And vpon the line AB describe a se<g ref="char:EOLhyphen"/>micircle AFB. And by the 11. of the first, from the point E, raise vp vnto the line AB, a perpendiculer line EF, cutting the circumference in the point F. And draw lines from A to F,<note place="margin">De<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>onstra<g ref="char:EOLhyphen"/>tion.</note> and from F to B. And forasmuch as there are two vnequall right lines AB and BC, and the line AB is in power more then the line BC, by the square of a line incommensurable in l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ngth vnto AB, and vpon the line AB is applied a parallelograme equall to the fourth part o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the square of the line BC, that is, to the square of the halfe of the line BC, and wanting in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure by a square, and the said parallelogramme is that which is contained vnder the lines AE and EB. wherfore by the 2. part of the 18. of the tenth, the line AE is incomme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable in length vnto the line EB. But as the line AE is to the line EB, so is the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines BA and AE, to the parallelogramme contayned vnder the lines AB and BE by the second assumpt before put. And that which is contained vnder
<pb n="257" facs="tcp:7062:286"/>
the line BA and AE is equall to the square of the line AF, by the second part of the first as<g ref="char:EOLhyphen"/>sumpt before put. And that which is contained vnder the lines AB and BE is by the first part of the same assumpt, equall to the square of the line BF. Wherfore the square of the line AF is incomm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nsurable to the square of the line BF. Wherfore the lines AF and BF are incommensurable in power. And forasmuch as AB is a rationall line (by supposition) ther<g ref="char:EOLhyphen"/>fore (by the 7 definition of the tenth) the square of the line AB is rationall.<note place="margin">The first part of the d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ter<g ref="char:EOLhyphen"/>mination con<g ref="char:EOLhyphen"/>cluded.</note> Wherefore also the squares of the lines AF and FB added together make a rationall superficies. For (by the 47. of the first) they are equal to the square of the line AB. Again forasmuch as (by the third part of the first assumpt going before) that which is contained vnder the lines AE and EB, is equall to the square of the line EF. But by supposition that which is contained vnder the lines AE and EB is equall to the square of the line BD. Wherfore the line FE is equall to the line BD. Wherfore the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BG is double to the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> E. Wherfore (by the third assumpt going before) that which is contained vnder the lines AB and BC, is double to that which is contained vnder the lines AB and EF. But that which is contained vnder the lines AB and BC, is by supposition mediall. Wherfore (by the corollary of the 23. of the tenth) that which is contained vnder the lines AB and EF is also mediall, but that whiche is con<g ref="char:EOLhyphen"/>tayned vnder the lines AB and EF,<note place="margin">The second part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</note> is (by the last parte of the first assumpt goyng be<g ref="char:EOLhyphen"/>fore) equall to that which is contained vnder the lines AF and FB. Wherefore that which is contained vnder the lines AF &amp; FB is a mediall superficies. And it is proued, that that which is composed of the squares of the lines AF and FB added together is rationall.<note place="margin">The totall conclusion.</note> Wherfore there are found out two right lines AF and FB incommensurable in power, whose squares added together, make a rationall superficies, and the parallelogramme contained vn<g ref="char:EOLhyphen"/>der them, is a mediall superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 11. Probleme. The 34. Proposition. To finde out two right lines inc<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mensurable in power, whose squares ad<g ref="char:EOLhyphen"/>ded together make a mediall superficies, and the parallelogramme contay<g ref="char:EOLhyphen"/>ned vnder them, make a rationall superficies.</head>
               <p>TAke (by the 31. of the tenth) two mediall lines AB and BC, commensurable in power onely, comprehending a rationall superficies,<note place="margin">Construction.</note> so that let the line AB be in power more then the line BC by the square of a line incommensurable in length vnto the line AB. And describe vpon the line AB a semicircle ADB. And by the 10. of the first, deuide the line BC
<figure/>
vnto two equall partes in the point E. And (by the 28. of the sixt) vpon the line AB apply a parallelogramme equall to the square of the line BE, and wantyng in figure by a square, and let that paral<g ref="char:EOLhyphen"/>lelogramme be that which is contayned vnder the lines AF and FB. Wherfore the line AF is incommensurable in length vnto the line FB (by the 2. part of the 18. of the tenth.) And from the point F vnto the right line AB, raise vp (by the 11. of the first) a per<g ref="char:EOLhyphen"/>pendiculer line FD, and draw lines from A to D, and from D to B.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AF is incommensurable vnto the line FB: but (by the second assumpt going before the 33. of the tenth) as the line AF is to the line FB, so is the parallelogramme contayned vnder the lines BA and AF, to the parallelogramme contained vnder the lines BA and BF, wher<g ref="char:EOLhyphen"/>fore (by the tenth of the tenth) that which is contained vnder the lines BA and AF is in<g ref="char:EOLhyphen"/>commensurable to that which is contayned vnder the lines AB and BF: but that which is
<pb facs="tcp:7062:287"/>
contained vnder the lines BA and AF is equall to the square of the line AD, and that which is contained vnder the lines AB and BF is also equall to the square of the line DB (by the second part of the first assumpt going before the 33. of the te<g ref="char:cmbAbbrStroke">̄</g>th) wherfore the square of the line AD is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to the
<figure/>
square of the line DB. Wherefore the lines AD and DB are incommensura<g ref="char:EOLhyphen"/>ble in power. And forasmuch as the square of the line AB is mediall, there<g ref="char:EOLhyphen"/>fore also the superficies made of the squares of the lines AD and DB ad<g ref="char:EOLhyphen"/>ded together is mediall.<note place="margin">The first part of the deter<g ref="char:EOLhyphen"/>mination con<g ref="char:EOLhyphen"/>cluded.</note> For the squares of the lines AD and DB are (by the 47. of the first) equall to the square of the line AB. And forasmuch as the line BC is double to the line FD (as it was proued in the pro<g ref="char:EOLhyphen"/>position going before) therefore the parallelogramme contained vnder the lines AB and and BC is double to the parallelogramme contained vnder the lines AB and FD (by the third assumpt going before the 33. proposition) wherefore it is also commensurable vnto it (by the sixt of the tenth) But that which is contained vnder the lines AB and BC is suppo<g ref="char:EOLhyphen"/>sed to be rationall. Wherfore that which is contained vnder the lines AB and FD is also ra<g ref="char:EOLhyphen"/>tionall. But that which is contained vnder the lines AB and FD, is equall to that which is contained vnder the lines AD and DB (by the last part of the first assumpt going before the 33. of the tenth) Wherfore that which is contayned vnder the lines AD and DB is also ra<g ref="char:EOLhyphen"/>tionall.<note place="margin">
                     <p>The second part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</p>
                     <p>The totall conclusion.</p>
                  </note> Wherefore there are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ound out two right lines AD and DB incommensurable in power, whose squares added together, make a mediall superficies, and the parallelogramme co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned vnder them, make a rationall superficies: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 12. Probleme. The 35. Proposition. To finde out two right lines incommensurable in power, whose squares ad<g ref="char:EOLhyphen"/>ded together, make a mediall superficies, and the parallelogramme contai<g ref="char:EOLhyphen"/>ned vnder them, make also a mediall superficies, which parallelogramme moreouer, shall be incommensurable to the superficies made of the squares of those lines added together.</head>
               <p>TAke (by the 32. of the tenth) two mediall lines AB and BC commensurable in power onely,<note place="margin">Construction.</note> comprehending a mediall superficies, so that let the line AB be in power more then the line BC by the square of a line incommensurable in length vnto the line AB. And vpon the line AB describe
<figure/>
a semicircle ADB, and let the rest of the construction be as it was in the two former propositions. And forasmuch as (by the 2 part of the 18. of the tenth) the line AF is in<g ref="char:EOLhyphen"/>commensurable in length vnto the line FB,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore the line AD is incommensurable in power vnto the line DB (by that which was demonstrated in the pro<g ref="char:EOLhyphen"/>positio<g ref="char:cmbAbbrStroke">̄</g> going before). And forasmuch as the square of the line AB is mediall, therefore that also which is composed of the squares of the lines AD and DB (which squares are equall to the square of the line AB by the 47. of the first) is mediall.<note place="margin">The first part concluded.</note> And forasmuch as that which is
<pb n="258" facs="tcp:7062:287"/>
contained vnder the lines AF and FB, is equall to either of the squares of the lines EB and FD, for by supposition the parallelogramme contained vnder the lines AF and FB is e<g ref="char:EOLhyphen"/>quall to the square of the line EB, and the same parallelogramme is equall to the square of the line DF (by the third part of the first assumpt going before the 33. of the te<g ref="char:cmbAbbrStroke">̄</g>th). Wherfore the line BE is equall to the line DF. Wherfore the line BC is double to the line FD. Where<g ref="char:EOLhyphen"/>fore that which is contained vnder the lines AB and BC is double to that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AB and FD. Wherfore they are commensurable by the sixt of this boke: but that which is contained vnder the lines AB and BC is mediall by supposition. Wherfore also that which is contained vnder the lines AB and FD is mediall (by the corollary of the 23, of the tenth) but that which is contained vnder the lines AB and FD, is (by the fourth part of the first assumpt going before the 33. of the tenth) equall to that which is contained vnder the lines AD and DB:<note place="margin">The second part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</note> wherfore that which is contained vnder the lines AD and DB is also mediall. And forasmuch as the line AB is incommensurable in length vnto the line BC. But the line BC is commensurable in length vnto the line BE. Wherfore (by the 13<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth) the line AB is incommensurable in length vnto the line BE. Wherefore the square of the line AB is incommensurable to that which is contained vnder the lines AB and BE (by the first of the sixt and 10. of this booke) But vnto the square of the line AB are equall the squares of the lines AD and DB added together (by the 47. of the first): and vnto that which is contayned vnder the lines AB and BE, is equall that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AB and FD, that is, which is contained vnder AD and DB. For the parallelogramme contained vnder the lines AB and FD is equall to the parallelogramme contained vnder the lines AD and DB (by the last part of the first assumpt going before the 33. of this tenth booke).<note place="margin">The third part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</note> Wherfore that which is composed of the squares of the lines AD and DB is incommensurable to that which is contained vnder the lines AB and DB. Where<g ref="char:EOLhyphen"/>fore there are found out two right lines AD and DB incommensurable in power, whose squares added together, make a mediall superficies,<note place="margin">The totall con<g ref="char:EOLunhyphen"/>clusion.</note> and the parallelogramme contayned vn<g ref="char:EOLhyphen"/>der them, make also a mediall superficies, which parallelogramme moreouer is incommensu<g ref="char:EOLhyphen"/>rable to the superficies composed of the squares of those lines added together, which was re<g ref="char:EOLhyphen"/>quired to be done.</p>
            </div>
            <div type="part">
               <head>The beginning of the Senaries by Composition. ¶ The 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Theoreme. The 36. Proposition. If two rationall lines commensurable in power onely be added together:<note place="margin">
                     <hi>The first Se<g ref="char:EOLhyphen"/>nary by com<g ref="char:EOLhyphen"/>position.</hi>
                  </note> the whole line is irrationall, and is called a binomium, or a binomiall line.</head>
               <p>
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>
                  <pb facs="tcp:7062:288"/>
                  <hi>B and BC is incommensurable to the square of the
<figure/>
line BC. But vnto the parallelograme contained vnder the lines AB and BC is commensurable the parallelograme contained vnder AB and BC twise (by the 6. of the tenth): wherefore that which is contained vnder AB and BC twise is in<g ref="char:EOLhyphen"/>commensurable to the square of the line BC (by the 13 of the tenth). But vnto the square of the line BC is commensurable that which is composed of the squares of the lines AB and BC (by the 15. of the tenth), for by supposition the lines AB and BC are commensurable in power onely. Wherefore (by the 13. of the tenth) that which is composed of the squares of the lines AB and BC added together is incommensurable to that which is contained vnder the lines AB and BC twise. Wherefore (by the 16. of the tenth) that which is contained vn<g ref="char:EOLhyphen"/>der AB and BC twise together with the squares of the lines AB and BC, which (by the 4. of the second) is equall to the square of the whole line AC, is incommensurable to that which is composed of the squares of AB and BC added together. But that which is composed of the squares of AB and BC added together is rationall, for it is commensurable to either of the squares of the lines AB and BG of which either of them is rationall by supposition: wherfore the square of the line AC is (by the 10. definition of the tenth) irrationall. Wherefore the line AC also is irrationall, and is called a binomiall line.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>Diffinition of a binomiall line.</hi>
                  </note>This proposition sheweth the generation and production of the second kinde of ir<g ref="char:EOLhyphen"/>rationall lines which is called a binomium, or a binomial line. The definition whereof is fully gathered out of this proposition, and that thus.</p>
               <p>
                  <hi>A binomium or a binomiall line, is an irrationall line composed of two rationall lines commensu<g ref="char:EOLhyphen"/>rable the one to the other in power onely.</hi> And it is called a binomium, that is, hauing two names, because it is made of two such lines as of his partes which are onely commensu<g ref="char:EOLhyphen"/>rable in power and not in length: and therefore ech part or line, or at the least the one of them, as touching length, is vncertaine and vnknowne. Wherefore being ioyned to<g ref="char:EOLhyphen"/>gether their quantitie cannot be expressed by any one number or name, but ech part remayneth to be seuerally named in such sort as it may. And of these binomiall lines there are sixe seuerall kindes,<note place="margin">
                     <hi>Sixe kindes of binomiall lines.</hi>
                  </note> the first binomiall, the second, the third, the fourth, the fifth, and the sixt, of what nature and condition ech of these is shalbe knowne by their definitious which are afterward set in their due place.</p>
            </div>
            <div type="part">
               <head>¶The 25. Theoreme. The 37. Proposition. If two mediall lines commensurable in power onely containing a rationall superficies, be added together: the whole line is irrationall, and is called a first bimediall line.</head>
               <p>LEt these two mediall lines <hi>AB</hi> and <hi>BC</hi> being commensurable in power onely, and contayning a rationall superficies (the 27. of the tenth teacheth to finde out two such lines) be composed. Then I say that the whole line <hi>AC</hi> is irrationall. For as <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> sayd in the proposition next going before that which is composed of the squares of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <hi>AB</hi> and <hi>BC</hi> is incomme<g ref="char:cmbAbbrStroke">̄</g>surable to that which is contai<g ref="char:EOLhyphen"/>ned
<figure/>
vnder the lines <hi>AB</hi> and <hi>BC</hi> twis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore (by the 16. of the tenth) that which is composed of the squares of the lines <hi>AB</hi> and <hi>BC</hi> together with that which is con<g ref="char:EOLhyphen"/>tained vnder the lines <hi>AB</hi> and <hi>BC</hi> twise,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> that is, the square of the line <hi>AC</hi> is incommensurable to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise. But that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> commen<g ref="char:EOLhyphen"/>surable to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> once (by the 6. of the tenth)
<pb n="259" facs="tcp:7062:288"/>
wherefore the square of the whole line <hi>AC,</hi> is (by the 13<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth) inc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mmensurable <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o that which is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> once. But by supposition the lines <hi>AB</hi> and <hi>BC</hi> comprehend a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. Wherefore the square of the whole line <hi>AC</hi> is irrationall: wherefore also the line <hi>AC</hi> is irrationall. And it is called a first bimediall line.</p>
               <p>
                  <hi>The third irrational line which is called a first bimediall line, is sh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wed by this pro<g ref="char:EOLhyphen"/>position, and the definition thereof is by it made manifest,</hi>
                  <note place="margin">Diffinition of a first bimedi<g ref="char:EOLhyphen"/>all line.</note> 
                  <hi>which is this.</hi> A first bimediall line, is an irrationall line, which is composed of two mediall lines commensurable in power onely con<g ref="char:EOLhyphen"/>tayning a rationall parallelograme. <hi>It is called a first bimediall line, by cause the two mediall lines or partes whereof it is composed contayne a rationall superficies, which is prefer<g ref="char:EOLhyphen"/>red before an irrationall.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 26. Theoreme. The 38. Proposition. If two mediall lines commensurable in power onely contayning a mediall superficies, be added together: the whole line is irrationall, and is called a second bimediall line.</head>
               <p>LEt these two medial lines <hi>AB</hi> and <hi>BC</hi> being commensurable in power one<g ref="char:EOLhyphen"/>ly, and contayning a mediall superficies (the 28. of the tenth teacheth to find<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> out two such lines) be added together.<note place="margin">Construction.</note> Then I say that the whole line <hi>AC</hi> is irrational. Take a rationall line <hi>DE.</hi> And (by the 44. of the first) vpon the line <hi>DE</hi> apply the parallelograme <hi>DF,</hi> equal to the square of the line <hi>AC.</hi> whose other side let be the line <hi>DG.</hi> And forasmuch as the square of the line <hi>AC</hi> is (by the 4. of the second) equall to that which
<figure/>
is composed of the squares of the lines <hi>AB</hi> and <hi>BC,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> to<g ref="char:EOLhyphen"/>gether with that which is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> twise: but the square of the line <hi>AC</hi> is equall to the parallelograme <hi>DF.</hi> Wherefore the parallelograme <hi>DF</hi> is equall to that which is composed of the squares of the lines <hi>AB</hi> and <hi>BC</hi> together with that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise. Now then a<g ref="char:EOLhyphen"/>gayne (by the 44, of the first) vpon the line <hi>DE</hi> apply the parallelograme <hi>EH</hi> equall to the squares of the lines <hi>AB</hi> and <hi>BC.</hi> Wherefore the parallelograme remayning, namely, <hi>HF,</hi> is equall to that which is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> twise. And forasmuch as either of these lines <hi>AB</hi> and <hi>BC</hi> is mediall, therefore the squares of the lines <hi>AB</hi> and <hi>BC</hi> are al<g ref="char:EOLhyphen"/>so mediall. And that which is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> twise is (by the corolla<g ref="char:EOLhyphen"/>ry of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>4. of the tenth) mediall. For by the 6. of this booke it is comme<g ref="char:cmbAbbrStroke">̄</g>surable <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> that <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> once, which is by supposition medial. <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> squares of the lines. <hi>AB</hi> and <hi>BC</hi> is equall the parallelograme <hi>EH,</hi> and vnto that <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise is equall the parallelograme <hi>HE</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> either of these parallelogrames <hi>HE</hi> and <hi>AF</hi> is mediall: and they are applyed vpon the ratio<g ref="char:EOLhyphen"/>nall line <hi>ED.</hi> Wherefore (by the 22. of the tenth) either of these lines <hi>DH</hi> and <hi>HG</hi> is a ra<g ref="char:EOLhyphen"/>tionall line, and inc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mmensurable in length vnto the line <hi>DE.</hi> And forasmuch as (by suppo<g ref="char:EOLhyphen"/>sition) the line <hi>AB</hi> is incommensurable in length vnto the line <hi>BC.</hi> But as the line <hi>AB</hi> is to the line <hi>BC</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the square of the line <hi>AB</hi> to the parallelograme which is contayned vn<g ref="char:EOLhyphen"/>der the lines <hi>AB</hi> and <hi>BC</hi> (by the first of the sixt). Wherefore (by the 10 of this book<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) the square of the line <hi>AB</hi> is incommensurable to the parallelograme contayned vnder the lines
<pb facs="tcp:7062:289"/>
                  <hi>AB</hi> and <hi>BC.</hi> But to the square of the line <hi>AB</hi> is com<g ref="char:EOLhyphen"/>mens<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>able
<figure/>
that which is composed of the squares of the lines <hi>AB</hi> and <hi>BC</hi> (by the <hi>15.</hi> of the tenth). For the squares of the lines <hi>AB</hi> and <hi>BC</hi> are commensurable (when as the lines <hi>AB</hi> and <hi>BC</hi> are put to be commen<g ref="char:EOLhyphen"/>surable in power onely). And to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> is commensurable that which is contained vnder the lines <hi>AB</hi> and <hi>BC</hi> twise (by the 6 of the tenth) wherefore that which is composed of the squares of the lines <hi>AB</hi> and <hi>BC</hi> is incommensu<g ref="char:EOLhyphen"/>rable to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise. But to the squares of the lines <hi>AB</hi> and <hi>BC</hi> is equall the parallelograme <hi>EH.</hi> And to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise is equall the parallelograme <hi>FH.</hi> Wherfore the parallelograme <hi>FH</hi> is incommensurable to the parallelograme <hi>HE.</hi> Wherfore the line <hi>DH</hi> is incommensurable in length to the line <hi>HG</hi> (by the 1 of the sixt and 10 of this booke). And it is proued that they are rationall lines. Wherefore the lines <hi>DH</hi> &amp; <hi>HG</hi> are rationall commensurable in power onely. Wherefore (by the 36. of the tenth) the whole line <hi>DG</hi> is irrationall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And the line <hi>DE</hi> is rationall. But a rectangle super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies compre<g ref="char:EOLhyphen"/>hended vnder a rationall line and an irrationall line is (by the corollary added after the <hi>21</hi> of the tenth) irrationall. Wherefore the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies <hi>DF</hi> is irrationall. And the line also which containeth it in power is irrational. But the line <hi>AC</hi> containeth in power the superficies <hi>DF.</hi> Wherefore the line <hi>AC</hi> is irrationall. And it is called a second bimediall line.</p>
               <p>
                  <hi>This Proposition sheweth the generation of the fourth irrationall line, called a second bimediall line.</hi>
                  <note place="margin">Diffinition of a second bime<g ref="char:EOLunhyphen"/>diall line.</note> 
                  <hi>The definition wherof is euident by this Proposition, which is thus.</hi> A second bimediall line is an irrationall line, which is made of two mediall lines commensurable in power onely ioyned together, which comprehend a mediall superficies. <hi>And it is called a second bi<g ref="char:EOLhyphen"/>mediall, because the two mediall lines of which it is composed, co<g ref="char:cmbAbbrStroke">̄</g>taine a mediall super<g ref="char:EOLhyphen"/>ficies, and not a rationall. Now a mediall is by nature &amp; in knowledge after a rationall.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 27. Theoreme. The 39. Proposition. If two right lines inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power be added together, hauing that which is composed of the squares of them rationall, and the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contayned vnder them mediall: the whole right line is irrationall, and is called a greater line.</head>
               <p>LEt t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ese two right lines AB and BC being incommensurable in power onely, and making that which is required in the Proposition (The 33. of the tenth teacheth to finde out two such lines) be added together. Then I say, that the whole line AC is irrationall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as (by supposition) the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines AB and BC is mediall, therefore the pa<g ref="char:EOLhyphen"/>rallelogramme contained twise vnder the lines AB and BC is mediall. (For that which is contained vnder AB and BC twise, is commensurable to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder AB and BC once (by the 6. of the tenth). Wherefore (by the Corollary of the 23. of the tenth) that which is contained vnder AB &amp; BC twise, is mediall). But by supposition that which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> composed of the squares of the lines AB and BC, is rationall. Wherefore that which is contained vnder the lines AB and BC twise is incommensurable to that which is composed
<pb n="260" facs="tcp:7062:289"/>
of the squares of the lines AB and BC. Wher<g ref="char:EOLhyphen"/>fore
<figure/>
(by the 16. of the tenth) that which is com<g ref="char:EOLhyphen"/>posed of the squares of the lines AB and BC to<g ref="char:EOLhyphen"/>gether with that which is contayned vnder the lines AB &amp; BC twise, which is (by the 4. of the second) equall to the square of the line AC, is incommensurable to that which is composed of the squares of the lines AB and BC. But that which is composed of the squares of the lines AB and BC, is rationall. Wherefore the square of the whole line AC, is irrationall. Wherefore the line AC also is irrationall. And is called a greater line. And it is called a greater line for that that which is composed of the squares of the lines AB &amp; BC which are rationall, is greater then that which is contayned vnder the lines AB and BC twise, which are mediall. Now it is meete that the name should be geuen according to the propertie of the rationall.</p>
               <div type="section">
                  <head>An Assumpt.</head>
                  <p>And that the superficies composed of the squares of the lines AB and BC,<note place="margin">An Assumpt.</note> is greater then that which is contained vnder the lines AB and BC twise, may thus be proued. First it is manifest, that the lines AB and BC are vnequall: for if they were equall, then the squares of the lines AB and BC should be equall to that which is contained vnder the lines AB and BC twise, so that that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder AB
<figure/>
and BC, should also be rationall: which is contrary to the supposition. Wherefore the lines AB and BC are vnequall. Suppose then that the line AB be the greater, &amp; let the line BD be equall to the line BC. Wherfore (by the <hi>7.</hi> of the second) the squares of the lines AB and BD, are equall to that which is contained vnder the lines AB and BC twise and to the square of the line AD. But the line DB is equall to the line BC. Wherefore the squares of the lines AB and BC, are equall to that which is contained vnder the lines AB and BC twise and to the square of the line AD. Wherefore the squares of the lines AB and BC are greater then that which is contayned vnder the lines AB and BC twise, by the square of the line AD: which was required to be proued.</p>
                  <p>
                     <hi>This Proposition teacheth the production of the fift irrationall line, which is called a greater line: which is by the sense of this Proposition thus defined.</hi>
                  </p>
                  <p>A greater line is an irrationall line, which is composed of two right lines which are incommensu<g ref="char:EOLhyphen"/>rable in power,<note place="margin">Diffinition of a greater line.</note> the squares of which added together, make a rationall superficies, and the parallelo<g ref="char:EOLhyphen"/>gramme which they containe, is mediall. <hi>It is therefore called a greater line, as</hi> Theon <hi>sayth, because the squares of the two lines of which it is composed, added together being ra<g ref="char:EOLhyphen"/>tionall, are greater then the mediall superficies contained vnder them twise. And it is conuenient that the denomination be taken of the proprietie of the rationall part, ra<g ref="char:EOLhyphen"/>ther then of the mediall part.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶The 28. Theoreme. The 40. Proposition. If two right lines inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power be added together, hauing that which is made of the squares of them added together mediall, and the pa<g ref="char:EOLhyphen"/>rallelogramme contayned vnder them rationall: the whole right line is ir<g ref="char:EOLhyphen"/>rationall, and is called a line contayning in power a rationall and a mediall superficies.</head>
               <p>
                  <pb facs="tcp:7062:290"/>LEt these two right lines AB and BC being incommensurable in power, &amp; making that which is required in the Proposition be added together. (The 34. of the tenth teacheth to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inde out two such lines). Then I say, that the whole line AC is irra<g ref="char:EOLhyphen"/>tionall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as that which is composed of the squares of the lines AB and BC is mediall, but that which is contained vnder the lines AB and BC twise is ratio<g ref="char:EOLhyphen"/>nall, therefore that which is composed of the squares of the lines AB and BC, is incommen<g ref="char:EOLhyphen"/>surable to that which is contained vnder the lines AB and BC twise. Wherefore (by the 16. of the tenth) that which is composed of the squares of the lines AB and BC together with that which is contayned vnder the lines AB and BC
<figure/>
twise, which (by the 4. of the second) is the square of the whole line AG, is incommensurable to that which is contained vnder the lines AB and BC twise. But that which is contained vnder the lines AB and BC twise, is rationall, for that which is contained vnder the lines AB and BC once is put to be rationall. Wherefore the square of the whole line AC is irrationall. Wherefore the line AC also is irrationall. And it is therefore called a line containing in power a rationall and a me<g ref="char:EOLhyphen"/>diall superficies, because the power thereof contayneth two superficieces, whereof the one is ra<g ref="char:EOLhyphen"/>tionall, namely, that which is contained vnder the two partes, and the other mediall, namely, that which is made of the squares of the partes added together. And by reason that the ra<g ref="char:EOLhyphen"/>tionall is in order of nature &amp; of knowledge before the mediall, therfore the first part of the denomination is taken of it, and the second part is taken of the mediall which was required to be proued.</p>
               <p>
                  <hi>In this Proposition is taught the generation of the sixt irrationall line, which is cal<g ref="char:EOLhyphen"/>led a line whose power is rationall and mediall. The definition of which is gathered of thys Proposition after this maner.</hi>
               </p>
               <p>A line whose power is rationall and mediall, is an irrationall line which is made of two right lines incommensurable in power added together,<note place="margin">Diffinition of a line whose power is rati<g ref="char:EOLhyphen"/>onall and mediall.</note> whose squares added together make a mediall superficies, but that supersicies which they containe is rationall. <hi>The reason of the name is before set forth in the Proposition.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 29. Theoreme. The 41. Proposition. If two right lines incommensurable in power be added together, hauyng that which is composed of the squares of them added together mediall, and the parallelogramme contayned vnder them mediall, and also incommen<g ref="char:EOLhyphen"/>surable to that which is composed of the squares of them added together<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the whole right line is irrationall, and is called a line contayning in power two medials.</head>
               <p>LEt these two right lines AB and BC being incommensurable in power, and hauing that which is composed of the squares of the lines AB and BC mediall, and the pa<g ref="char:EOLhyphen"/>rallelogramme which is contayned vnder the lines AB and BC mediall, and also incommensurable to that which is composed of the squares of the lines AB and BC added together,<note place="margin">Construction.</note> be added together (the 35. of the tenth teacheth to finde out two such lines) Then I say that the whole line AC is irrationall. Take a rationall line DE, and (by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he 44. of the first) vpon the line DE apply the parallelogramme DF equall to the squares of the lines AB and BC. And vpon the line GF, which is equall to the line DE, apply the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me GH equall to that which is contained vnder the lines AB and BC tw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se. Wherefore the
<pb n="261" facs="tcp:7062:290"/>
whole parallelogramme DH is equall to the square of the line AC (by the 4. of the second.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note>) And forasmuch as that which is composed of the squares of the lines AB and BC is mediall and is equall to the parrllelogramme DF: therfore DF also is mediall (by that which was said in the 38. proposition of this booke). And it is
<figure/>
applied vpon the rationall line DE. Wherefore the line DG is rationall and incommensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth vnto y<hi rend="sup">e</hi> line DE (by the 22. of the te<g ref="char:cmbAbbrStroke">̄</g>th). And by the same reason the lyne GK is rational &amp; inco<g ref="char:cmbAbbrStroke">̄</g>mesura<g ref="char:EOLhyphen"/>ble in length vnto the line GF, that is, vnto the line DE. And forasmuch as that which is composed of y<hi rend="sup">e</hi> squares of the lines AB &amp; BC added together, is by supposition inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AB and BC twise, therfore also the parallelogramme DF is incommensurable vnto the parallelogramme GH. Wherfore also the line DG is incommensurable vnto the line GK (by the first of the sixt) and by the tenth of the tenth. But it is now proued that they are rationall. Wherfore the lines DG and GK are rationall commensurable in power onely. Wherfore (by the 36. of the tenth) the whole line DK is rationall, and is called a binomiall line, but the line DE is irrationall. Wherfore the parallelogramme DH is irrationall (by the corollary added after the 22. propositio<g ref="char:cmbAbbrStroke">̄</g> of the tenth). Wherfore also the line which containeth it in power is irrationall: but the line AC containeth it in power. Wherfore the line AC is irrationall, and is called a line contayning in power two medials. It is called a line containing in power two medials, for that it contay<g ref="char:EOLhyphen"/>neth in power two mediall supersicieces; one of which is composed of the squares of the lynes AB and BC added together; and the other is that which is contained vnder the lines AB &amp; BC twise: which was required to be demonstrated.</p>
               <p>
                  <hi>In this proposition is taught the nature of the 7. kinde of irrationall lines which is called a line whose power is two medials. The definition whereof is taken of this pro<g ref="char:EOLhyphen"/>position after this maner.</hi>
               </p>
               <p>A line whose power is two medials, is an irrationall line which is composed of two right lines in<g ref="char:EOLhyphen"/>commensurable in power,<note place="margin">Diffinition of a li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e contai<g ref="char:EOLhyphen"/>ning in power two medials.</note> the squares of which added together, make a mediall superficies, and that which is contained vnder them is also mediall, and moreouer it is incommensurable to that which is composed of the two squares added together.</p>
               <p>
                  <hi>The reason why this line is called a line whose power is two medials, was before in the ende of the demonstration declared.</hi>
               </p>
               <p>And that the said irrationall lines are deuided one way onely, that is, in one point onely, into the right lines of which they are composed, and which make euery one of the kindes of those irrationall lines, shall straight way be demonstrated: but first will we demonstrate two assumptes here following.</p>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>Take a right line and let the same be AB, and deuide it into two vnequall partes in the point C,<note place="margin">An Assumpt.</note> and againe deuide the same line AB into two other vnequal partes, in an other point namely, in D, and let the line AC (by supposition) be greater then the line DB. Then I say
<pb facs="tcp:7062:291"/>
that the squares of the lines AC and BC added together, are greater then the squares of the lines AD and DB added together. Deuide the line AB (by the 10. of the first) into two e<g ref="char:EOLhyphen"/>quall partes in the point E. And forasmuch as the line AC is greater then the line DB, take away the line DC which is common to them both: wherfore the residue AD is greater then the residue CB, but the line AE is equall to the line EB. Wherfore the line DE is lesse then the line EC. Wherfore the pointes C and D are not
<figure/>
equally distant from the point E, which is the point of the section into two equall partes. And foras<g ref="char:EOLhyphen"/>much as (by the 5. of the second) that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines AC and CB together with the square of the line EC is equall to the square of the line EB. And by the same reason that which is contayned vnder the lynes AD and DB together with the square of the line DE, is also equall to the self same square of the line EB: wherfore that which is contained vnder the lines AC and CB together with the square of the line EC is equall to that which is contained vnder the lines AD and DB to<g ref="char:EOLhyphen"/>gether with the square of the line DE: of which the square of the line DE is lesse then the square of the line EC (for it was proued that the line DE is lesse then the line EC). Wher<g ref="char:EOLhyphen"/>fore the parallelogramme remayning, contayned vnder the lines AC and CB is lesse the<g ref="char:cmbAbbrStroke">̄</g> the parallelogramme remayning contayned vnder the lines AD and DB. Wherfore also that which is contayned vnder the lines AC and CB twise is lesse then that which is contayned vnder the lines AD and DB twise. But (by the fourth of the second) the square of the whole line AB is equall to that which is composed of the squares of the lines AC and CB toge<g ref="char:EOLhyphen"/>ther with that which is contained vnder the lynes AC and CB twise, and by the same reason the square of the whole line AB is equall to that which is composed of the squares of the lines AD and DB together with that which is contayned vnder the lynes AD and DB twise: wherfore that which is composed of the squares of the lynes AC and CB together with that which is contayned vnder the lynes AC and CB twise, is equall to that which is composed of the squares of the lynes AD and DB, together with that which is con<g ref="char:EOLhyphen"/>tayned vnder the lynes AD and DB twise. But it is already proued that that which is con<g ref="char:EOLhyphen"/>tayned vnder the lynes AC and CB twise, is lesse then that which is contayned vnder the lines AD &amp; DB twise. Wherfore the residue, namely, that which is composed of the squares of the lines AC and CB is greater then the residue, namely, then that which is composed of the squares of the lines AD and DB: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>A rationall superficies exceedeth a rationall superficies, by a rationall su<g ref="char:EOLhyphen"/>perficies.</q>
                  </p>
                  <p>Let AD be a rationall superficies, and let it exceede AF
<figure/>
being also a rationall superficies by the superficies ED. Then I say that the superficies ED is also rationall. For the paral<g ref="char:EOLhyphen"/>lelogramme AD is commensurable to the parallelogramme AF, for that either of them is rationall. Wherefore (by the second part of the 15. of the tenth) the parallelogramme AF is commensurable to the parallelogramme ED. But the the parallelogramme AF is rationall. Wherfore also the pa<g ref="char:EOLhyphen"/>rallelogramme ED is rationall.</p>
               </div>
            </div>
            <div type="part">
               <pb n="262" facs="tcp:7062:291"/>
               <head>¶The 30. Theoreme. The 42. Proposition. A binomiall line is in one point onely deuided into his names.<note place="margin">The second Se<g ref="char:EOLhyphen"/>nary by com<g ref="char:EOLhyphen"/>position.</note>
               </head>
               <p>SVppose that AB be a binomiall line, and in the point G let it be deuided into his names, that is, into the lines wherof the whole line AB is composed. Where<g ref="char:EOLhyphen"/>fore these lines AC and CB are rationall commensurable in power onely. Now I say that the line AB cannot in any other point besides C be deuided into two ra<g ref="char:EOLhyphen"/>tionall lines commensurable in power onely.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For
<figure/>
if it be possible, let it be deuided in the point D, so that let the lines AD and DE b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> rationall commensurable in power onely. First this manifest, that neith<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> of these poin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es C and D de<g ref="char:EOLhyphen"/>uideth the right line AB into two equall partes. Otherwise the lines AC and CB should be rationall commensurable in length and so likewise should the lines AD and DB be. For e<g ref="char:EOLhyphen"/>uery line measureth it selfe, and any other line equall to it selfe. Moreouer the line DB is either one and the same with the line AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that is, is equall to the line AC, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> els it is greater<g ref="char:EOLhyphen"/>then the line AC, either els it is lesse then it. If DB be equall to the line AC, then putting the line DB vpon the line AC eche endes of the one, shall agree with eche endes of the other. Wherfore putting the point B vpon the point A, the point D also shall fall vpon the point C, and the line AD which is the rest of the line AC, shall also be equall to the line CB, which is the rest of the line DB. Wherfore the line AB is deuided into his names in the point C. And so also shal the line AB being deuided in the point D be deuided in the self <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame point that the self same line AB was before deuided in the point C, which is co<g ref="char:cmbAbbrStroke">̄</g>trary to the suppositio<g ref="char:cmbAbbrStroke">̄</g>. For by suppositio<g ref="char:cmbAbbrStroke">̄</g> it was deuided in sundry pointes, namely, in C &amp; D. But if the line DB be greater<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the the line AC, let the line AB be de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ided into two equal partes in the point E. Wherfore the points C &amp; D shal not equally be distant fro<g ref="char:cmbAbbrStroke">̄</g> the point E (Now (by the first assupt going before this propositio<g ref="char:cmbAbbrStroke">̄</g>) that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of y<hi rend="sup">e</hi> lines AD &amp; DB is greater the<g ref="char:cmbAbbrStroke">̄</g> that which is composed of the squares of the lines AC &amp; CB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But that which is composed of the squares of the lines AD &amp; DB together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AD &amp; DB twise, is equall to that which is composed of the squares of the lines AC &amp; CB together with that which is contained vnder the lines AC and CB twise, for either of them is equall to the square of the whole line AB (by the 4. of the second) wherefore how much that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of the lines AD and DB added together is greater then that which is composed of the squares of the lines AC and CB added together, so much is that which is contained vnder the lines AC and CB twise greater then that which is contained vnder the lines AD and DB twise. But that which is composed of the squares of the lines AD and DB excedeth that which is composed of the squares of the lines AC and CB by a rationall su<g ref="char:EOLhyphen"/>perficies (by the 2. assumpt going before this proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) For that which is composed of the squares of the lines AD and DB is rationall, and so also is that which is composed of the squares of the lines AC and CB: for the lines AD and DB are put to be rationall commen<g ref="char:EOLhyphen"/>surable in power onely, and so likewise are the lines AC and CB. Wherfore also that which is contained vnder the lines AC and CB twise, exceedeth that which is contained vnder the lines AD &amp; DB twise by a rational superficies, whe<g ref="char:cmbAbbrStroke">̄</g> yet notwithsta<g ref="char:cmbAbbrStroke">̄</g>ding they are both medial superficieces (by the 21. of the tenth) which (by the 26. of the same) is impossible. And if the line DB be lesse then the line AC, we may by the like demonstration proue the selfe same im<g ref="char:EOLhyphen"/>possibilitie. Wherfore a binomiall line is in one point onely deuided into his names: Which was required to be demonstrated.</p>
               <div type="section">
                  <pb facs="tcp:7062:292"/>
                  <head>
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>ollary added by <hi>Flussates.</hi>
                  </head>
                  <p>
                     <hi>Two ration <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>surable in power onely being added together cannot be equall to two other rationall line <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> in power onely added together.</hi>
                     <note place="margin">
                        <hi>A Corollary.</hi>
                     </note> For either of them should make a binomia<gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> so should a binomiall line be deuided into his names in moe poyntes then on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ch by this proposition is proued to be impossible. The like shall follow in the fiue <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> irrationall lines as touching their two names.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 31. Probleme. The 43. Proposition. A first bimediall line is in one poynt onely deuided into his names.</head>
               <p>SVppose that <hi>AB</hi> be a first bimediall line, and let it be deuided into his partes in the point <hi>C,</hi> so that let the lines <hi>AC</hi> and <hi>CB</hi> be mediall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely, and containing a rationall superficies. Then I say that the line <hi>AB</hi> can not be deuided into his names in any other poynt then in <hi>C.</hi> For if it be possible let it be deuided into his names in the poynt <hi>D,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e.</note> so that let
<figure/>
                  <hi>AD</hi> &amp; <hi>DB</hi> be mediall lines commensurable in power onely, comprehending a rationall superficies. Now for<g ref="char:EOLhyphen"/>asmuch as how much that which is contayned vnder the lines <hi>AD</hi> and <hi>DB</hi> twise di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ferreth from that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> twise, so much differreth that which is composed of the squares of the lines <hi>AD</hi> and <hi>DB</hi> from that which is composed of the squares of the lines <hi>AC</hi> and <hi>CB</hi>: but that which is contayned vnder the lines <hi>AD</hi> and <hi>DB</hi> twise differreth from that which is contay<g ref="char:EOLunhyphen"/>ned vnder the lines <hi>AC</hi> and <hi>CB</hi> twise, by a rationall superficies (by the second assumpt go<g ref="char:EOLhyphen"/>ing before the 41. of the tenth). For either of those superficieces is rationall. Wherefore that which is composed of the squares of the lines <hi>AC</hi> and <hi>CB</hi> differeth from that which is com<g ref="char:EOLunhyphen"/>posed of the squares of the lines <hi>AD</hi> and <hi>DB</hi> by a rationall superficies, when yet they are both mediall superficieces: which is impossible. Wherefore a first bimediall line is in one poynt onely deuided into his names: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 32. Theoreme. The 44. Proposition. A second bimediall line is in one poynt onely deuided into his names.</head>
               <p>SVppose that the line <hi>AB</hi> being a second bimediall line, be deuided into hys names in the poynt <hi>C</hi>: so that let the lines <hi>AC</hi> and <hi>CB</hi> be mediall lines com<g ref="char:EOLhyphen"/>mensurable in power onely, comprehending a mediall superficies. It is manifest that the poynt <hi>C</hi> deuideth not the whole line <hi>AB</hi> into two equall partes.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ie.</note> For the lines <hi>AC</hi> and <hi>CB</hi> are not commensurable in length the one to the other. Now I say that the line <hi>AB</hi> cannot be deuided into his names in any other poynt but onely in <hi>C.</hi> For if it be possible, let it be deuided into his names in the poynt <hi>D,</hi> so that let not the line <hi>AC</hi> be one and the same, that is, let it not be equall, with the line <hi>DB.</hi> But let it be greater then it. Now it is manifest (by the first assumpt going before the 42. proposition of this booke) that the squares of the lines <hi>AC</hi> and <hi>CB</hi> are greater then the squares of the lines <hi>AD</hi> and <hi>DB.</hi> And also that the lines <hi>AD</hi> and <hi>DB</hi> are mediall lines commensurable in power onely, comprehending a mediall supersicies. Take a rationall line <hi>EF.</hi> And (by the 44. of the first) vpon the line <hi>EF</hi> apply a rectangle parallelograme <hi>EK</hi> equall to the square of the line <hi>AB.</hi> From which pa<g ref="char:EOLhyphen"/>rallelograme take away the parallelograme <hi>EG</hi> equall to the squares of the lines <hi>AC</hi> and <hi>CB</hi> Wherefore the residue, namely, the parallelograme <hi>HK</hi> is equall to that which is contai<g ref="char:EOLhyphen"/>ned
<pb n="263" facs="tcp:7062:292"/>
vnder the lines <hi>AC</hi> and <hi>CB</hi> twise. Agayne from the the parallelograme <hi>EK</hi> take away the parallelograme <hi>EL</hi> equall to the squares of the lines <hi>AD</hi> and <hi>DB</hi> which are lesse then the squares of the lines <hi>AC</hi> and <hi>CB.</hi> Wherefore the re<g ref="char:EOLhyphen"/>sidue,
<figure/>
namely, the parallelograme <hi>MK</hi> is equall to that which is contayned vnder the lines <hi>AD</hi> and <hi>DB</hi> twise. And forasmuch as the squares of the lines <hi>AC</hi> and <hi>CB</hi> are mediall, therefore the parallelograme <hi>EG</hi> also is me<g ref="char:EOLhyphen"/>diall.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an imp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ssibi<g ref="char:EOLhyphen"/>litie.</note> And it is applyed vpon the rationall line <hi>EF</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> where<g ref="char:EOLhyphen"/>fore the line <hi>EH</hi> is rationall and incommensurable in length to the line <hi>EF.</hi> And by the same reason, the parallelograme <hi>HK</hi> is mediall (for that which is e<g ref="char:EOLhyphen"/>quall vnto it, namely, that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> twise is mediall) therefore the line <hi>HN</hi> is also rationall and incommensurable in length vn<g ref="char:EOLhyphen"/>to the line <hi>EF.</hi> And forasmuch as the lines <hi>AC</hi> and <hi>CB</hi> are mediall lines commensurable in power onely, therefore the line <hi>AC</hi> is incommensurable in length vnto the line <hi>CB.</hi> But as the line <hi>AC</hi> is to the line <hi>CB,</hi> so is the square of the line <hi>AC</hi> to that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> (by the 1. of the sixt). Wherefore the square of the line <hi>AC</hi> is incommensurable to that which is contayned vnder the lines <hi>AC</hi> and <hi>CB.</hi> But (by the 16. of the tenth) vnto the square of the line <hi>AC</hi> are commensurable the squares of the lines <hi>AC</hi> and <hi>CB</hi> added to<g ref="char:EOLhyphen"/>gether, for the lines <hi>AC</hi> and <hi>CB</hi> are commensurable in power onely. And vnto that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> is commensurable that which is contayned vn<g ref="char:EOLhyphen"/>der the lines <hi>AC</hi> and <hi>CB</hi> twise. Wherefore that which is composed of the squares of the lines <hi>AC</hi> and <hi>CB,</hi> is incommensurable to that which is contained vnder the lines <hi>AC</hi> and <hi>CB</hi> twise. But to the squares of the lines <hi>AC</hi> and <hi>CB</hi> is equall the parallelograme <hi>EG,</hi> and to that which is contained vnder the lines <hi>AC</hi> and <hi>CB</hi> twise is equall the para<g ref="char:EOLhyphen"/>lelograme <hi>HK.</hi> Wherfore the parallelograme <hi>EG</hi> is incommensurable to the parallelograme <hi>HK.</hi> Wherefore also the line <hi>EH</hi> is incommensurable in length to the line <hi>HN.</hi> And the lines <hi>EH</hi> and <hi>HN</hi> are rationall. Wherefore they are rationall commensurable in power one<g ref="char:EOLhyphen"/>ly: but if two rationall lines commensurable in power onely be added together, the whole line is irrationall, and is called a binomiall line (by the 36. of the tenth). Wherefore the binomiall line <hi>EN</hi> is in the poynt <hi>H</hi> deuided into his names. And by the same reason also may it be proued that the lines <hi>EM</hi> and <hi>MN</hi> are rationall lines commensurable in power onely. Wher<g ref="char:EOLunhyphen"/>fore <hi>EN</hi> being a binomiall line is deuided into his names in sundry poyntes, namely, in <hi>H</hi> and <hi>M,</hi> neither is the line <hi>EH</hi> one and the same, that is, equal with <hi>MN.</hi> For the squares of the lines <hi>AC</hi> and <hi>CB</hi> are greater then the squares of the lines <hi>BD</hi> and <hi>AD</hi> (by the <hi>1.</hi> as<g ref="char:EOLhyphen"/>sumpt put after the <hi>41.</hi> of the tenth). But the squares of the lines <hi>AD</hi> and <hi>DB</hi> are greater then that which is contayned vnder the lines <hi>AD</hi> and <hi>DB</hi> twise (by the assumpt put after the 39. of the tenth). Wherefore the squares of the lines <hi>AC</hi> and <hi>CB,</hi> that is, the parallelo<g ref="char:EOLhyphen"/>grame <hi>EG</hi> is much greater then that which is contained vnder the lines <hi>AD</hi> and <hi>DB</hi> twise, that is then the parallelograme <hi>MK.</hi> Wherfore (by the first of the sixt) the line <hi>EH</hi> is greater then the line <hi>MN.</hi> Wherefore <hi>EH</hi> is not one and the same with <hi>MN.</hi> Wherefore a binomiall line is in two sundry poyntes deuided into his names. Which is impossible. The selfe same absurditie also will follow if the line <hi>AC</hi> be supposed to be lesse then the line <hi>DB.</hi> A second binomiall line therefore is not deuided into his names in sundry poyntes. Wherefore it is deuided in one onely: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 33. Theoreme. The 45. Proposition. A greater line is in one poynt onely deuided into his names.</head>
               <p>
                  <pb facs="tcp:7062:293"/>LEt <hi>AB</hi> being a greater line be deuided into his names in the poynt <hi>C,</hi> so that let the lines <hi>AC</hi> and <hi>CB</hi> be rationall incommensurable in power, hauing that which is composed of the squares of the lines <hi>AC</hi> and <hi>CB</hi> rationall, and that which is con<g ref="char:EOLhyphen"/>tained vnder the lines <hi>AC</hi> and <hi>CB</hi> mediall.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>e.</note> Then I say that the line <hi>AB</hi> can not in any o<g ref="char:EOLhyphen"/>ther poynt then in <hi>C</hi> be deuided into his names. For
<figure/>
if it be possible, let it be deuided into his names in the poynt <hi>D,</hi> so that let <hi>AD</hi> and <hi>DB</hi> be lines incomme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable in power, hauing that which is composed of the squares of the lines <hi>AD</hi> and <hi>DB</hi> rationall, and that which is contayned vnder the lines <hi>AD</hi> and <hi>DB</hi> mediall. Now forasmuch as how much the squares of the lines <hi>AC</hi> and <hi>CB</hi> di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>er from the squares of the lines <hi>AD</hi> and <hi>DB,</hi> so much differeth that which is contained vnder the lines <hi>AD</hi> and <hi>DB</hi> twise from that which is contained vnder the lines <hi>AC</hi> and <hi>CB</hi> twise, by those thinges which haue be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e sayd in the demonstration of the 42. propositi<g ref="char:EOLhyphen"/>on. But the squares of the lines <hi>AC</hi> and <hi>CB</hi> exceede the squares of the lines <hi>AD</hi> and <hi>DB</hi> by a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies (for they are either of them rationall). Wherfore that which is containe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>der the lines <hi>AD</hi> and <hi>DB</hi> twise exceedeth that which is contained vnder the lines <hi>AC</hi> and <hi>CB</hi> twise by a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies: when as either of them is a mediall super<g ref="char:EOLhyphen"/>s<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>es. Which is impossible (by the 26. of the tenth). Wherefore a greater line is in one poynt on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ly deuided into his names: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 34. Theoreme. The 46. Proposition. A line contayning in power a rationall and a mediall, is in one point one<g ref="char:EOLhyphen"/>ly deuided into his names.</head>
               <p>LEt AB being a line containing in power a rationall and a mediall, be deui<g ref="char:EOLhyphen"/>ded into his names in the point C, so that let the lines AC &amp; CB be incom<g ref="char:EOLhyphen"/>mensurable in power, hauing that which is composed of the squares of the lines AC and CB mediall, and that which is contained vnder the lines AC and CB rationall.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> Then I say, that the line AB can not in any other point be deuided into his names but onely in the point C. For if it be possi<g ref="char:EOLhyphen"/>ble, let it be deuided into his names in the point D,
<figure/>
so that let the lines AD and DB be incommensura<g ref="char:EOLhyphen"/>ble in power, hauing that which is composed of the squares of the lines AD and DB mediall, and that which is contayned vnder the lines AD and DB rationall. Now forasmuch as how much that which is contained vnder the lines AD &amp; DB twise differeth from that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AC and CB twise, so much differ the squares of the lines AC &amp; CB added together from the squares of the lines AD and DB added together. But that which is contayned vnder the lines AC and CB twise excedeth that which is contained vnder the lines AD and DB twise by a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies (for either of them is rationall). Wher<g ref="char:EOLhyphen"/>fore also the squares of the lines AC and CB added together, exceede the squares of the lines AD and DB added together by a rationall superficies, when yet ech of them is a mediall su<g ref="char:EOLhyphen"/>perficies: which is impossible. Wherefore a line containing in power a rationall and a medi<g ref="char:EOLhyphen"/>all, is in one point onely deuided into his names: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <pb n="264" facs="tcp:7062:293"/>
               <head>¶The 35. Theoreme. The 47. Proposition. A line contayning in power two medials, is in one point onely deuided into his names.</head>
               <p>SVppose that AB being a line containing in power two medialls, be deuided into his names in the point C, so that let the lines AC and CB be incommensurable in power, hauing that which is composed of the squares of the lines AC &amp; CB mediall, and that also which is contained vnder the lines AC and CB mediall, and moreouer incommensurable <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o that which is composed of the squares of the lines AC and CB. Then I say,<note place="margin">Construction.</note> that the line AB can in no other point
<figure/>
be deuided into his names but onely in the point C. For if it be possible, let it be deuided into his names in the point D, so that let not the line AC be one and the same, that is, equall with the line DB: but by supposition let the line AC be the greater. And take a rationall line EF. And (by the 43. of the first) vpon the line EF apply a rectangle paralle<g ref="char:EOLhyphen"/>logra<g ref="char:cmbAbbrStroke">̄</g>me EG equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of the lines AC and CB: and likewise vpon the line HC, which is equall to the line EF, apply the parallelogramme HK equall to that which is contained vnder the lines AC and CB twise. Wherefore the whole parallelogramme EK is equall to the square of the line AB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Againe vpon the same line EF describe the paralle<g ref="char:EOLhyphen"/>logramme EL equall to the squares of the lines AD and DB. Wherefore the residue, name<g ref="char:EOLhyphen"/>ly, that which is contayned vnder the lines AD and DB twise, is equall to the parallelo<g ref="char:EOLhyphen"/>gramme remaining, namely, to MK. And forasmuch as that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of the lines AC and CB, is (by supposition) mediall, therefore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me EG which is equall vnto it, is also mediall: and it is applied vpon the rationall line EF. Wherefore (by the 22. of the tenth) the line HE is rationall and incommensurable in length vnto the line EF. And by the same reason also the line HN is rationall and incommensurable in length to the same line EF. And forasmuch as that which is composed of the squares of the lines AC and CB is incommensurable to that which is contained vnder the lines AC and CB twise (for it is supposed to be incommensurable to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AC and CB once): therefore the parallelogramme EG is incommensurable to the paralle<g ref="char:EOLhyphen"/>logramme H <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Wherefore the line EH also is incommensurable in length to the line HN, and they are rationall lines: wherfore the lines EH and HN are rationall commensurable in power onely. Wherefore the whole line EN is a binomiall line, and is deuided into his names in the point H. And in like sort may we proue, that the same binomiall line EN is de<g ref="char:EOLhyphen"/>uided into his names in the point M, and that the line EH is not one and the same that is e<g ref="char:EOLhyphen"/>quall with the line MN, as it was proued in the end of the demonstration of the 44. of this booke. Wherefore a binomiall line is deuided into his names in two sundry pointes: which is impossible (by the 42. of the tenth). Wherefore a line containing in power two medials, is not in sundry pointes deuided into his names. Wherefore it is deuided in one point onely: which was required to be demonstrated.</p>
               <div type="section">
                  <pb facs="tcp:7062:294"/>
                  <head>¶Second Definitions.</head>
                  <p>IT was shewed before that of binomiall lines there were sixe kindes,<note place="margin">
                        <hi>Sixe kindes of binomiall lines.</hi>
                     </note> the definitions of all which are here now set, and are called second definitio<g ref="char:cmbAbbrStroke">̄</g>s. All binomiall lines, as all other kindes of irrationall lines, are co<g ref="char:cmbAbbrStroke">̄</g>ceaued, co<g ref="char:cmbAbbrStroke">̄</g>sidered, and perfectly vnderstanded onely in respecte of a rationall line (whose partes as before is taught, are certayne, and knowen, and may be distinctly expressed by number) vnto which line they are compared. Thys rational<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> line must ye euer haue before your eyes, in all these definitions, so shall they all be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asie inough.</p>
                  <p>
                     <note place="margin">
                        <hi>A binomiall line co<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>ste<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>h of two pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s.</hi>
                     </note>A binomiall line (ye know) is made of two partes or names, wherof the one is greater then the other. Wherfore the power or square also of the one is greater then the power or square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the other. The three first kindes of binomiall lines, namely, the first, the se<g ref="char:EOLhyphen"/>con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, &amp; the third, are produced, when the square of the greater name or part of a bino<g ref="char:EOLhyphen"/>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cedeth the square of the lesse name or part, by the square of a line which is com<g ref="char:EOLhyphen"/>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nsurable in length to it, namely, to the greater. The three last kindes, namely, the fourth, the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t, and the sixt, are produced, when the square of the greater name or part <gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>edeth the square of the lesse name or part, by the square of a line incommensurable in length vnto it, that is, to the greater part.</p>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>Firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> d<gap reason="illegible" resp="#PDCC" extent="3 letters">
                                 <desc>•••</desc>
                              </gap>ini<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>n.</hi>
                        </note>A first binomiall line is, whose square of the greater part exceedeth the square of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e lesse part <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y the square of a line commensurable in length to the greater part, and the greater part is also commensurable in length to t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e rationall line first set.</head>
                     <p>As l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t the ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ll line first set be AB: whose partes are distinctly knowen: suppose also that the line CE be a binomiall line, whose names or partes let be CD and DE. And let the square of the line CD the greater part excede the square of the line DE the lesse part by the square of the line FG: which line FG, let b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e commen<g ref="char:EOLhyphen"/>su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>able
<figure/>
in length to the line CD, which is the greater part of the binomiall line. And moreouer let the line CD the greater pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t be com<g ref="char:EOLunhyphen"/>mensurble in length to the rationall line first set, name<g ref="char:EOLhyphen"/>ly, to AB. So by this d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ini<g ref="char:EOLhyphen"/>tion the binomiall line CE is a first binomiall line.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>Secon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> diffi<g ref="char:EOLhyphen"/>nition.</hi>
                        </note>A second binomiall line is, when the square of the greater part exceedeth the square of the lesse part by the square of a line commensurable in length vnto it, and the lesse part is commensurable in length to the rationall line first set.</head>
                     <p>As (supposing euer the rationall line) let CE be a binomiall line deuided in the poynt D. The square of whose greater part CD let exceede the square of the lesse part DE by the square of the line FG, which
<figure/>
line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>G let be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line CD t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e gr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ater p<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the binomiall line. And let also the line DE the lesse part of the binomiall line be commensu<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>able in l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ngth to the rationall line first set AB. So by this definition the binomiall line CE is a second binomiall line.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>Third <gap reason="illegible" resp="#PDCC" extent="6 letters">
                                 <desc>••••••</desc>
                              </gap>i<g ref="char:EOLhyphen"/>tion.</hi>
                        </note>A third binomiall line is, when the square of the greater part excedeth the
<pb n="265" facs="tcp:7062:294"/>
square of the lesse part, by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto it. And neither part is commensurable in length to the rationall line geue<g ref="char:cmbAbbrStroke">̄</g>.</head>
                     <p>As suppose the line CE to be a binomiall line: whose partes are ioyned together in the poynt D: and let the square of the line CD the greater part exceede the square of the lesse part DE by the square of the line FG, and
<figure/>
let the line FG be commensurable in length to the line CD the greater part of the binomiall. Moreouer, let neither the greater part CD, nor the lesse part DE, be commensurable in length to the rationall line AB, then is the line CE by this definition a third binomiall line.</p>
                  </div>
                  <div type="subsection">
                     <head>A fourth binomiall line is,<note place="margin">
                           <hi>Fourth diffi<g ref="char:EOLhyphen"/>nition.</hi>
                        </note> when the square of the greater part exceedeth the square of the lesse by the square of a line incommensurable in length vn<g ref="char:EOLhyphen"/>to the greater part. And the greater is also commensurable in length to the rationall line.</head>
                     <p>As let the line CE be a binomiall line, whose partes let be CD &amp; DE, &amp; let the square of the line CD the greater part
<figure/>
exceede the square of the line DE the lesse, by the square of the line FG. And let the line FG be incommensu<g ref="char:EOLhyphen"/>rable in length to the line CD the greater. Let also the line CD the greater part be commensurable in length vnto the rat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>onall line AB. Then by this definition the line CE is a fourth binomiall line.</p>
                  </div>
                  <div type="subsection">
                     <head>A fift binomiall line is,<note place="margin">
                           <hi>Fifth dif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> when the square of the greater part exceedeth the square of the lesse part, by the square of a line incommensurable vnto it in length. And the lesse part also is commensurable in length to the rationall line geuen.</head>
                     <p>As suppose that CE be a binomiall line, whose greater part let be CD, and let the lesse part be DE. And let the square of the line CD excede the square of the line DE by the square of the line FG, which let be inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vn<g ref="char:EOLhyphen"/>to
<figure/>
the line CD the greater part of the binomiall line. And let the line DE the second part of the binomi<g ref="char:EOLhyphen"/>all line be commensurable in length vnto the rationall line AB. So is the line CE by this definition a fift bi<g ref="char:EOLhyphen"/>nomiall line.</p>
                  </div>
                  <div type="subsection">
                     <head>A sixt binomiall line is,<note place="margin">
                           <hi>Sixth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> when the square of the greater part exceedeth the square of the lesse, by the square of a line incommensurable in length vnto it. And neither part is commensurable in length to the rationall line geuen.</head>
                     <p>As let the line CE be a binomiall line, deuided into his names in the point D. The square of whose greater part CD let exceede the square of the lesse part DE by the square of the line FG, and let the line FG be incommensurable in length to the line CD the greater part of the binomiall line. Let also
<pb facs="tcp:7062:295"/>
n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ither CD the greater part, nor DE
<figure/>
the lesse part be commensurable in length to the rationall line AB. And so by this definition the line CE is a sixt binomiall line. So ye see that by these definitions, &amp; their examples, and declarations, all the kindes of binomiall lines are made very playne.</p>
                     <p>This is to be noted that here is nothing spoken of those lines, both whose portions a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e com<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ensurable in length vnto the rationall line first set, for that such lines cannot be binomiall lines. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>or binomiall lines are composed of two rationall lines commensu<g ref="char:EOLhyphen"/>rable in Power onely (by the 36. of this booke). But lines both whose portions are comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nsurable in length to the rationall line first set are not binomiall lines. For that the partes of such lines should by the 12. of this booke be commensurable in length the one to the other. And so should they not be such lines as are required to the compositi<g ref="char:EOLhyphen"/>on of a binomiall line. Moreouer such lines should not be irrationall but rationall, for that they are commensurable t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ch of the parts whereof they are co<g ref="char:cmbAbbrStroke">̄</g>posed (by the 15. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> this booke). And therefore they should be rationall for that the lines which compos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> them are rationall.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 13. Probleme. The 48. proposition. To finde out a first binomiall line.</head>
               <p>
                  <note place="margin">The third Se<g ref="char:EOLhyphen"/>nary by com<g ref="char:EOLhyphen"/>position.</note>TAke two numbers <hi>AC</hi> and <hi>CB,</hi> &amp; let them be such, that the number which is made of them both added together, namely, <hi>AB,</hi> haue vnto one of them, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, vnto <hi>BC</hi> that proportion that a square number hath to a square numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut vnto the other, namely, vnto <hi>CA</hi> let it not haue that proportion that a square number hath to a square number (such as is euery square num<g ref="char:EOLhyphen"/>ber which may be deuided into a square number and into a number not square).<note place="margin">Construction.</note> Take also a certayne rationall line, and let the
<figure/>
same be <hi>D.</hi> And vnto the line <hi>D</hi> let the line <hi>EF</hi> be commensurable in length. Wherefore the line <hi>EF</hi> is ra<g ref="char:EOLunhyphen"/>tionall. And as the number <hi>AB</hi> is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>AC,</hi> so let the square of the line <hi>EF</hi> be to the square of an other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e, namely, of <hi>FG</hi> (by the co<g ref="char:EOLhyphen"/>rollary of the sixt of y<hi rend="sup">e</hi> te<g ref="char:cmbAbbrStroke">̄</g>th). Wher<g ref="char:EOLhyphen"/>fore the square of the line <hi>EF</hi> hath to the square of the line <hi>FG</hi> that proportion that number hath to number. Wherefore the square of the line <hi>EF</hi> is commensurable to the square of the line <hi>FG</hi> (by the 6. of this booke) And the line <hi>EF</hi> is rationall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n.</note> Wherefore the line <hi>FG</hi> also is rationall. And forasmuch as the number <hi>AB</hi> hath not to the number <hi>AC,</hi> that proportion that a square number hath to a square number, neither shal the square of the line <hi>EF</hi> haue to the square of the line <hi>FG</hi> that proportion that a square number hath to a square number. Wherefore the line <hi>EF</hi> is incom<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>FG</hi> (by the 9. of this booke). Wherefore the lines <hi>EF</hi> and <hi>FG</hi> are rationall commensurable in power onely. Wherefore the whole line <hi>EG</hi> is a binomi<g ref="char:EOLhyphen"/>all line (by the 36. of the tenth). I say also that it is a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst binomiall line. For for that as the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <hi>BA</hi> is to the number <hi>AC,</hi> so is the square of the line <hi>EF</hi> to the square of the line <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>G</hi>: but the number <hi>BA</hi> is greater then the number <hi>AC</hi>: wherefore the square of the line <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>F</hi> is also greater then the square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line <hi>FG.</hi> Vnto the square of the line <hi>EF</hi> let the squares of the lines <hi>FG</hi> and <hi>H</hi> be equall (which how to finde out is taught in the assumpt put
<pb n="266" facs="tcp:7062:295"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ft<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r the 13. of the t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nth). And f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r that as th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> number <hi>BA</hi> is to the number <hi>AC,</hi> so is the square of the line <hi>EF</hi> to the square of the line <hi>FG</hi>: therefore (by co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uersion or e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ersion of proportion (by the corollary of the 19. of the fift) as the number <hi>AB</hi> is to the number <hi>BC,</hi> so is the square of the line <hi>EF</hi> to the square of the line <hi>H.</hi> But the number <hi>AB</hi> hath to the num<g ref="char:EOLunhyphen"/>ber <hi>BC</hi> that proportion that a square number hath to a square number. Wherefore also the square of the line <hi>EF</hi> hath to the square of the line <hi>H</hi> that proportion that a square number hath to a square number. Wherefore the line <hi>EF</hi> is commensurable in length to the line <hi>H</hi> (by the 9. of this booke. Wherefore the line <hi>EF</hi> is in power more then the line <hi>FG</hi> by the square of a line commensurable in length to the line <hi>EF.</hi> And the lines <hi>EF</hi> and <hi>FG</hi> are ra<g ref="char:EOLhyphen"/>tionall commensurable in power onely. And the line <hi>EF</hi> is commensurable in length to the rationall line <hi>D.</hi> Wherefore the line <hi>EG</hi> is a first binomiall line: which was required to be doone.</p>
            </div>
            <div type="part">
               <head>The 14. Probleme. The 49. Proposition. To finde out a second binomiall line.</head>
               <p>TAke two numbers <hi>AC</hi> and <hi>CB,</hi> and let them be such that the number made of them both added together, namely,<note place="margin">Construction.</note> 
                  <hi>AB</hi> haue vnto <hi>BC</hi> that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square number, and vnto the number <hi>CA</hi> let it not haue that proportion that a square number hath to a square number, as it was declared in the former proposition. Take also a ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ionall line, and let the same be <hi>D,</hi> and vnto the line <hi>D</hi> let the line <hi>FG</hi> be commensurable in length Wherefore <hi>FG</hi> is a rationall line. And as the number <hi>CA</hi> is to the number <hi>AB</hi> so let the square of the line <hi>GF</hi> be to the square of the line <hi>FE</hi> (by the 6. of the tenth), Wherefore the square of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>GF</hi> is commensurable to the square of the line <hi>FE.</hi> Wherfore also <hi>FE</hi> is a rationall line.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the nu<g ref="char:cmbAbbrStroke">̄</g>ber <hi>CA</hi> hath not vnto the number <hi>AB</hi> that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, ther<g ref="char:EOLhyphen"/>fore neither also the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the line <hi>GF</hi>
                  <figure/>
hath to the square of the line <hi>FE</hi> that propor<g ref="char:EOLhyphen"/>tion that a square number hath to a square number. Wherefore the line <hi>GF</hi> is incom<g ref="char:EOLhyphen"/>mensurable in length vnto the line <hi>FE</hi> (by the 9. of the tenth): wherefore the lines <hi>FG</hi> and <hi>FE</hi> are rationall commensurable in po<g ref="char:EOLhyphen"/>wer onely. Wherefore the whole line <hi>EG</hi> is a binomiall line. I say moreouer that the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>EG</hi> is a second binomiall line. For for that by contrary proportion as the number <hi>BA</hi> is to the number <hi>AC,</hi> so is the square of the line <hi>EF</hi> to the square of the line <hi>FG.</hi> But the number <hi>BA</hi> is greater then the number <hi>AC,</hi> wherefore also the square of the line <hi>EF</hi> is greater then the square of the line <hi>FG.</hi> Vnto the square of the line <hi>EF,</hi> let the squares of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s <hi>GF</hi> and <hi>H</hi> be equall. Now by conuersion (by the corollary of the 19. of of the fift) as the number <hi>AB</hi> is to the number <hi>BC,</hi> so is the square of the line <hi>EF</hi> to the square of the line <hi>H.</hi> But the number <hi>AB</hi> hath to the number <hi>BC</hi> that proportion that a square number hath to a square number. Wherefore the square of the line <hi>EF</hi> hath to the square of the line <hi>H,</hi> that proportion that a square num<g ref="char:EOLhyphen"/>ber hath to a square number. Wherefore (by the 9. of the tenth) the line <hi>EF</hi> is commensu<g ref="char:EOLhyphen"/>rable in length vnto the line <hi>H.</hi> Wherefore the line <hi>EF</hi> is in power more then the line <hi>FG</hi> by the square of a line commensurable in length vnto the line <hi>EF</hi>: and the lines
<pb facs="tcp:7062:296"/>
                  <hi>EF</hi> and <hi>FG</hi> are rationall commensurable in power onely, and <hi>FG</hi> being the lesse name <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s commensurable in length vnto the rationall line geuen, namely, to <hi>D.</hi> Wherefore <hi>EG</hi> is a se<g ref="char:EOLhyphen"/>cond binomiall line: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶The 15. Probleme. The 50. Proposition. To finde out a third binomiall line.</head>
               <p>
                  <note place="margin">Construction.</note>TAke two numbers AC and CB, and let them be such that the numb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r made of them both added together, namely, AB, haue to the number BC that proportion that a square number hath to a square number. But to the number AC let it not haue that proportion that a square number hath to a square number, as it was declared in the two for<g ref="char:EOLhyphen"/>mer. And take also some other number that is not a square number, and let the same be D, and let not the number D haue either to the number BA, or to the number AC that propor<g ref="char:EOLhyphen"/>tion that a square number hath to a square num<g ref="char:EOLhyphen"/>ber.
<figure/>
And take a rational line and let the the same be E. And as the number D is to the number AB, so let the square of the line E be to the square of the line FG. Wherfore the square of the line E is co<g ref="char:cmbAbbrStroke">̄</g>mensu<g ref="char:EOLhyphen"/>rable to the square of the line FG, but the line E is rational, wherfore the line FG also is rational. And for that the nu<g ref="char:cmbAbbrStroke">̄</g>ber D hath not to the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB that proportion that a square number hath to a square number, neither also shall the square of the line E haue to the square of the line FG that proportion that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wher<g ref="char:EOLunhyphen"/>fore the line E is incommensurable in length to the line FG (by the 9. of the tenth.) Now againe as the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber AC, so let the square of the line FG be to the square of the line GH. Wherfore the square of the line FG is commensurable to the square of the line GH.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And the line FG is rationall. Wherfore also the line GH is rationall. And for that the number BA hath not to the nu<g ref="char:cmbAbbrStroke">̄</g>ber AC, that pro<g ref="char:EOLhyphen"/>portion that a square number hath to a square number, therefore neither also hath the square of the line FG to the square of the line GH that proportion that a square number hath to a square number. Wherfore the line FG is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line GH. Wherfore the lines FG &amp; GH are rational co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power only. Wherfore the whole line FH is a binomial line. I say moreouer that it is a third binomial line. For for that as the nu<g ref="char:cmbAbbrStroke">̄</g>ber D is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB, so is the square of the line E to the square of the line FG: but as the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB is to the number AC, so is the square of the line FG to the square of the line GH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equalitie (by the 22. of the fift) as the number D is to the number AC, so is the square of the line E to the square of the line GH. But the nu<g ref="char:cmbAbbrStroke">̄</g>ber D hath not to the nu<g ref="char:cmbAbbrStroke">̄</g>ber A. C that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore neither also hath the square of the line E to the square of the line GH that proportio<g ref="char:cmbAbbrStroke">̄</g> tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a square number hath so a square number. Wherfore the line E is incommensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line GH. And for that as the number AB is to the number AC, so is the square of the line FG to the square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line GH, therfore the square of the line FG is greater then the square of the line GH. Vnto the square of the line FG, let the squares of the lines GH and K be equal. Wherfore (by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uerse proportio<g ref="char:cmbAbbrStroke">̄</g> by the corollary of the 19. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift) as the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB is to the number BC, so is the square of y<hi rend="sup">e</hi> line FG to the square of the line K. But the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB hath to the nu<g ref="char:cmbAbbrStroke">̄</g>ber
<pb n="267" facs="tcp:7062:296"/>
BC that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square number. Wherfore also the square of the line FG hath to the square of the line K that proportion that a square number hath to a square number. Wherfore the line FG is commensurable in length to the line K. Wherfore the line FG is in power more then the line GH by the square of a line commensurable in length vnto it. And the lines FG and GH are rationall commensurable in power onely, and neither of them is commensurable in length vnto the rationall line E: wherfore the lyne F<g ref="char:EOLhyphen"/>H is a third binomiall line: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶The 16. Probleme. The 51. Proposition. To finde out a fourth binomiall line.</head>
               <p>TAke two numbers AC and CB,<note place="margin">Const<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uction.</note> &amp; let the<g ref="char:cmbAbbrStroke">̄</g> be such that the nu<g ref="char:cmbAbbrStroke">̄</g>ber made of the<g ref="char:cmbAbbrStroke">̄</g> both added together, namely, AB, haue to neither of the nu<g ref="char:cmbAbbrStroke">̄</g>bers AC and CB that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber (such as is euery square nu<g ref="char:cmbAbbrStroke">̄</g>ber to two nu<g ref="char:cmbAbbrStroke">̄</g>bers not square, which are lesse the<g ref="char:cmbAbbrStroke">̄</g> it &amp; make the said square nu<g ref="char:cmbAbbrStroke">̄</g>ber). And take a rationall line, and let the same be D. And vnto the line D let the line EF be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note>
                  <figure/>
Wherfore EF is a rationall line, and as the num<g ref="char:EOLhyphen"/>ber BA is to the number AC, so let the square of the line EF be to the square of y<hi rend="sup">e</hi> line FG. Where<g ref="char:EOLhyphen"/>fore the square of the line EF is commensura<g ref="char:EOLhyphen"/>ble to the square of the line FG, and the line BF is a rationall line. Wherfore also the line FG is a rationall line. And for that the number BA hath not to the number AC that proportion that a square number hath to a square number, neither also shall the square of the line EF haue to the square of the line FG that proportion that a square number hath to a square number. Wherfore the line EF is incommensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line FG. Wherfore the lines EF and FG are rationall commensurable in power onely. Wherefore the whole line EG is a binomiall line. I say moreouer that it is a fourth binomiall lyne. For for that as the number BA is to the number AC, so is the square of the line EF to the square of the line FG. But the number BA is greater then the number AC. Wherefore also the square of the line EF is greater then the square of the line FG. Vnto the square of the line EF let the squares of the lines FG and H be equall. Wherfore by conuersion (by the corollary of the 19. of the fift) as the number AB is to the number BC, so is the square of the line EF to the square of the line H. But the num<g ref="char:EOLunhyphen"/>ber AB hath not to the number BC that proportion that a square number hath to a square number: therfore neither also hath the square of the line EF to the square of the line H that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore (by the 9. of the te<g ref="char:cmbAbbrStroke">̄</g>th) the line EF is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length vnto the line H. Wherfore the line EF is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line FG by the square of a line incommensurable in length vnto it. And the lynes EF and FG are rationall commensurable in power onely, and the line EF is commensurable in length to the rationall line D. Wherfore the line EG is a fourth binomiall line: which was required to be found out.</p>
            </div>
            <div type="part">
               <head>¶The 17. Probleme. The 52. Proposition. To finde out a fift binomiall lyne.</head>
               <p>
                  <pb facs="tcp:7062:297"/>TAke two numbers AC and CB, and let them be such, that the number AB haue to neither of the numbers AC or CB that proportion that a square num<g ref="char:EOLhyphen"/>ber hath to a square number,<note place="margin">Construction.</note> as in the former proposition. And take a ratio<g ref="char:EOLhyphen"/>nall line and let the same be D. And vnto the line D let the line FG be com<g ref="char:EOLhyphen"/>mensurable in length. Wherfore the line FG is rationall. And as the number CA is to the number AB, so let the square
<figure/>
of the line GF be to the square of the line EF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the square of the line GF is commensurable to the square of the line FE. Wherefore also the line FE is ratio<g ref="char:EOLhyphen"/>nall. An<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for that the number CA hath not to the number AB that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square num<g ref="char:EOLhyphen"/>ber, ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore neither also hath the square of the line GF to the square of the line FE that proportion that a square number hath to a square number. Wherfore (by the 9. of the tenth) the line GF is incommensurable in length to the line FE. Wherfore the lines EF and FG are rationall commensurable in power only. Wherfore the whole line EG is a binomiall line. I say moreouer that it is a fift binomiall line. For for that as the number CA is to the number AB, so is the square of the line GF to the square of the line FE, therfore contrariwise, as the number BA is to the number AC, so is the square of the line EF to the square of the line FG. but the number BA is greater then the number AC. Wherfore also the square of the line EF is greater then the square of the line FG. Vnto the square of the line EF, let the squares of the lines FG and H be equall. Wherfore by conuersio<g ref="char:cmbAbbrStroke">̄</g> (by the corollary of the 19. of the fift) as the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB is to the num<g ref="char:EOLhyphen"/>ber BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the square of the line EF to the square of the line H. But the nu<g ref="char:cmbAbbrStroke">̄</g>ber AB hath not to the number BC that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square number. Wherefore neither also hath the square of the line EF to the square of the line H that proportion that a square number hath to a square number. Wherfore (by the 9. of the tenth) the line EF is in<g ref="char:EOLhyphen"/>commensurable in length to the line H. Wherfore the line EF is in power more then the line FG by the square of a line incommensurable in length vnto it. And the lines EF and FG are rationall commensurable in power onely. And the line FG being the lesse name, is com<g ref="char:EOLhyphen"/>mensurable in length to the rationall line geuen, namely, to D. Wherfore the whole line EG is a fift binomiall line: which was required to be found out.</p>
            </div>
            <div type="part">
               <head>¶The 18. Probleme. The 53. Proposition. To finde out a sixt binomiall line.</head>
               <p>TAke two numbers AC &amp; CB, and let the<g ref="char:cmbAbbrStroke">̄</g> be such that the number which is made of them both added together,<note place="margin">Construction.</note> namely, AB, haue to neither of the numbers AC or CB that proportion that a
<figure/>
square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square number. Take also any other number which is not a square number, and let the same be D. And let not the num<g ref="char:EOLhyphen"/>ber D haue to any one of these num<g ref="char:EOLhyphen"/>bers AB and AC that proportion that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Let there be put moreouer a rationall line,
<pb n="268" facs="tcp:7062:297"/>
and let the same be E. And as the number D is to the number AB, so let the square of the line E be to the square of FG. Wherefore (by the 6. of the tenth) the line E is commensurable in power to the line FG, &amp; the line E is rationall. Wherfore also the line FG is rationall. And for that the number D hath not <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the number AB that proportion that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square number, therefore neither also shall the square of the line E haue to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line E. Againe, as the number BA is to the number AC, so let the square of the line FG be to the square of the line GH.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore (by the 6. of the tenth) the square of the line FG is commensurable to the square of the line GH. And the square of the line FG is rationall. Wherefore the square of the line GH is also ra<g ref="char:EOLhyphen"/>tionall. Wherefore also the line GH is rationall. And for that the number AB hath not to the number AC, that proportion that a square number hath to a square number: there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore neither also hath the square of the line FG to the square of the line GH, that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length to the line GH. Wherefore the lines FG and GH are rationall commensurable in power onely. Wherefore the whole line FH is a binomiall line. I say moreouer, that it is a sixt binomiall line. For for that as the number D is to the number AB, so is the square of the line E to the square of the line FG. And as the number BA is to the number AC, so is the square of the line FG to the square of the line GH. Wherefore of equalitie (by the 22. of the fift) as the number D is to the number AC, so is the square of the line E to the square of the line GH. But the number D hath not to the nu<g ref="char:cmbAbbrStroke">̄</g>ber AC that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line E to the square of the line GH that proportion that a square number hath to a square number. Wher<g ref="char:EOLhyphen"/>fore the line E is incommensurable in length to the line GH. And it is already proued, that the line FG is also incommensurable in length to the line E. Wherefore either of these lines FG and GH is incommensurable in length to the line E. And for that as the number <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>A is to the number AC; so is the square of the line FG to the square of the line GH: therfore the square of the line FG is greater then the square of the line GH. Vnto the square of the line FG, let the squares of the lines GH and K be equall. Wherefore by euersion of propor<g ref="char:EOLhyphen"/>tion, as the number AB is to the number BC, so is the square of the line FG to the square of the line K. But the number AB hath not to the number BC that proportion that a square number hath to a square number. Wherefore neither also hath the square of the line FG to the square of the line K that proportion that a square number hath to a square number. Wherefore the line FG is incommensurable in length vnto the line K. Wherefore the line FG is in power more then the line GH, by the square of a line incommensurable in length to it. And the lines FG and GH are rationall commensurable in power onely. And neither of the lines FG &amp; GH is commensurable in length to the rationall line geuen, namely, to E. Wherefore the line FH is a sixt binomiall line: which was required to be found out.</p>
               <div type="section">
                  <head>¶A Corollary added out of <hi>Flussates.</hi>
                  </head>
                  <p>
                     <hi>By the</hi> 6. <hi>form<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r Proposi<gap reason="illegible" resp="#PDCC" extent="5 letters">
                           <desc>•••••</desc>
                        </gap> it i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> manifest, ho<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> divide any right line geuen into the names of euery one of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> foresayd binomiall lines.</hi>
                     <note place="margin">
                        <hi>A Corollary added by Flussates.</hi>
                     </note> For if it be required to deuide a right line ge<g ref="char:EOLhyphen"/>uen into a first binomiall line, then by the 48<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this booke finde out a first binomiall line. And this right line being so found out deuided into his names, you may by the 10. of the sixt, deuide the right line geuen in like sort. And so in the other fiue following.</p>
                  <p>Although I here note vnto you this Corollary out of <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>,</hi> in very conscience and of gratefull <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> I am enforced to certifie you, that, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> any yeare<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, before the trauailes of <hi>Flussas</hi> (vpo<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>Eu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>li<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>
                     </hi> Geo<g ref="char:EOLhyphen"/>metricall Elementes) were published, the order how to deuide, not onely the 6. Binomiall lines into their names, but also to adde to the 6. Resid<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ls their due partes: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rthermore to deuide all the o<g ref="char:EOLhyphen"/>ther
<pb facs="tcp:7062:298"/>
irrational<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lines (of this tenth booke) into the partes distinct, of which they are composed: with many other straunge conclusions Mathematicall, to the better vnderstanding of this tenth booke and o<g ref="char:EOLhyphen"/>ther Mathematicall bookes, most necessary, were by <hi>M. Iohn Dee</hi> inuented and demonstrated:<note place="margin">
                        <hi>M. Dee his booke called Ty<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ni<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>m Mathemati<g ref="char:EOLhyphen"/>cum.</hi>
                     </note> as in his booke, whose title is <hi>Tyrocinium Mathematicum</hi> (dedicated to <hi>Petru<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> Nonnius,</hi> An. 1559.) may at large appeare. Where also is one new arte, with sundry particular pointes, whereby the Mathematicall Sci<g ref="char:EOLhyphen"/>ences, greatly may be enriched. Which his booke, I hope, God will one day allowe him opportunitie to publishe: with diuers other his Mathematicall and Metaphysicall labours and inuentions.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>Is a right line be deuided into two partes how soeuer: the rectangle paral<g ref="char:EOLhyphen"/>lelogramme contayned vnder both the partes, is the meane proportionall betwene the squares of the same parts. And the rectangle parallelogramme contained vnder the whole line and one of the partes, is the meane pro<g ref="char:EOLhyphen"/>portionall betwene the square of the whole line and the square of the sayd part.</q>
                  </p>
                  <p>
                     <note place="margin">This Assumpt as was before noted, f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ll<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>w<g ref="char:EOLhyphen"/>eth most <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>f<g ref="char:EOLhyphen"/>ly without farther demon<g ref="char:EOLunhyphen"/>stration of the <hi>25.</hi> of this booke.</note>Suppose that there be two squares AB and BC, and let the lines DB and BE so be put that they both make one right line. Wherefore (by the 14. of the first) the lines FB and BG make also both one right line. And make perfect the parallelogramme AC. Then I say, that the rectangle parallelogramme DG is the meane proportionall betwene the squares AB and BC: and moreouer, that the parallelogramme DC is the meane proportionall betwene the squares AC and CB. First the parallelogramme AG is a square. For forasmuch as the line DB is equall to the line BF, and the line BE vnto the line BG, therfore the whole line DE is equall to the whole line FG. But the line DE is equall to either of these lines AH &amp; KC, and the line FG is equall to either of these lines AK and HC (by the 34. of the first).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wher<g ref="char:EOLhyphen"/>fore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AC is equilater, it is also rectan<g ref="char:EOLhyphen"/>gle
<figure/>
(by the 29. of the first). Wherefore (by the 46. of the first) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AC is a square. Now for that as the line FB is to the line BG, so is the line DB to the line BE. But as y<hi rend="sup">e</hi> line FB is to the line BG, so (by the 1. of the sixt) is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AB, which is the square of the line DB, to the parallelogramme DG, and as the line DB is to the line BE, so is the same parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DG to the parallelogramme BC, which is the square of the line BE. Wherefore as the square AB is to the pa<g ref="char:EOLhyphen"/>rallelogramme DG, so is the same parallelogramme DG to the square BC. Wherefore the parallelogramme DG is the meane proportionall betwene the squares AB and BC. I say moreouer, that the paral<g ref="char:EOLhyphen"/>lelogramme DC is the meane proportionall betwene the squares AC and CB. For for that as the line AD is to the line DK, so is the line KG to the line GC (for they are ech equall to eche). Wherefore by composition (by the 18. of the fift) as the line AK is to the line KD, so is the line KC to the line CG. But as the line AK is to the line KD, so is the square of the line AK, which is the square AC, to the parallelogramme co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AK and KD, which is the parallelogramme CD: and as the line KC is to the line CG, so also is the parallelogramme DC to the square of the line GC, which is the square BC. Wherefore as the square AC is to the parallelogramme DC, so is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DC to the square BC. Wherefore the parallelogramme DC is the meane proportionall betwene the squares AC and BC: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <pb n="269" facs="tcp:7062:298"/>
                  <head>¶An Assumpt.</head>
                  <p>
                     <q>Magnitudes that are meane proportionalls betwene the selfe same or e<g ref="char:EOLhyphen"/>quall magnitudes, are also equall the one to the other.</q>
                  </p>
                  <p>Suppose that there be three magnitudes <hi>A, B, C.</hi>
                     <note place="margin">An Assumpt.</note>
                     <figure/>
And as <hi>A</hi> is to <hi>B,</hi> so let <hi>B</hi> be to <hi>C.</hi> And likewise as the same magnitude <hi>A</hi> is to <hi>D,</hi> so let <hi>D</hi> be to the same magnitude <hi>C.</hi> Then I say that <hi>B</hi> and <hi>D</hi> are equall the one the other. For the proportion of <hi>A</hi> vnto <hi>C</hi> is double to that proportion which <hi>A</hi> hath to <hi>B</hi> (by the <hi>10.</hi> definition of the fift) and likewise the selfe same proportion of <hi>A</hi> to <hi>C</hi> is (by the same definition) double to that proportion which <hi>A</hi> hath to <hi>D.</hi> But magnitudes whose equemultiplices are either equall or the selfe same, are also equall. Wherefore as <hi>A</hi> is to <hi>B,</hi> so is <hi>A</hi> to <hi>D.</hi> Wherefore (by the 9. of the fift) <hi>B</hi> and <hi>D</hi> are equall the one to the other. So shall if also be if there be other magnitudes equal to <hi>A</hi> and <hi>C,</hi> namely, <hi>E</hi> and <hi>F,</hi> betwene which let the magnitude <hi>D</hi> be the meane proportionall.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 36. Theoreme. The 54. Proposition. If a superficies be contained vnder a rationall line &amp; a first binomiall line: the line which containeth in power that superficies is an irrationall line, &amp; a binomiall line.<note place="margin">The fourth Se<g ref="char:EOLunhyphen"/>nary by com<g ref="char:EOLhyphen"/>position.</note>
               </head>
               <p>SVppose that the superficies ABCD, be contained vnder the rational line AB, and vnder a first binomial line AD. Then I say that the line which containeth in power the superficies AC, is an irrational line, and a binomial line.<note place="margin">Construction.</note> For foras<g ref="char:EOLhyphen"/>much as the line AD is a first binomial line, it is in one only point deuided into
<figure/>
his names (by the 42. of this tenth): let it be deuided into his names in the point E. And let AE be the greater name. Now it is manifest that the lines AE and ED are rationall c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>m<g ref="char:EOLhyphen"/>mensurable
<pb facs="tcp:7062:299"/>
in power onely, and that the line AE is in power more then the line ED, by the square of a line commensurable in length to the line AE, and moreouer that the line AE is commensurable in length to the rationall line geuen AB by the definition of a first Binomi<g ref="char:EOLhyphen"/>all line set before the 48. proposition of this tenth. Deuide (by the 10. of the first) the line ED into two equall partes in the point F. And forasmuch as the line AE is in power more then the line ED by the square of a line commensurable in length vnto the line AE, therefore if vpon the greater line, namely, vpon the line AE be applied a parallelogramme equall to the fourth part of the square of the lesse line, that is, to the square of the line EF, &amp; wan<g ref="char:EOLhyphen"/>ting in forme by a square, it shall deuide the greater line, namely, AE into two partes còm<g ref="char:EOLhyphen"/>mensurable in length the one to the other (by the second part of the 17. of the tenth.) Apply therfore vpon the line AE a parallelogramme equall to the square of the line EF, and wan<g ref="char:EOLhyphen"/>ting in forme by a square by the 28. of the sixt, and let the same be that which is contained vnder the lines AG and GE. Wherfore the line AG is commensurable in length to the lyne
<figure/>
GE. Draw by the pointes G, E, and F, to either of these lines AB and DG these parallel lines GH, EK, and FL (by the 31. of the first) And (by the 14. of the second) vnto the paralle<g ref="char:EOLhyphen"/>logramme AH describe an equall square SN. And vnto the parallelogramme GK describe (by the same) an equal square NP. And let these lines MN &amp; NX be so put, that they both make one right line. Wherfore (by the 14. of the first) the lines RN and NO make also both one right line.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Make perfect the parallelogramme SP. Wherfore the parallelogramme SP is a square by those thinges which were demonstrated after the determination in the first as<g ref="char:EOLhyphen"/>sumpt going before. And forasmuch as that which is contained vnder the lines AG and GE is equall to the square of the line EF (by construction): therfore as the line AG is to the EF, so is the line EF to the line EG (by the 14. or 17. of the sixt) Wherfore also (by the 1. of the sixt) as the parallelogramme AH is to the parallelogramme EL, so is the parallelo<g ref="char:EOLhyphen"/>gramme EL to the parallelogramme GK. Wherfore the parallelogramme EL is the meane proportionall betwene the parallelogrammes AH and GK. But the parallelogramme AH is equal to the square SN, and the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me GK is equal to the square NP by co<g ref="char:cmbAbbrStroke">̄</g>structi<g ref="char:EOLhyphen"/>on. Wherfore the parallelogramme EL is the meane proportionall betwene the squares SN and NP (by the 7. of the fifth) But (by the first assumpt going before) the parallelogramme MR is the meane proportionall betwene the squares SN and NP. Wherefore the parallelo<g ref="char:EOLhyphen"/>gramme MR is equall to the parallelogramme EL (by the last assumpt going before) But the parallelogramme MR is equal to the parallelogramme OX (by the 43. of the first) and the pa<g ref="char:EOLhyphen"/>rallelogramme EL is equall to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me FC by construction, and by the first of the sixt. Wherfore the whole parallelogramme EC is equall to the two parallelogrammes MR &amp; OX. And the parallelogrammes AH and GK are equall to the squares SN and NP by construction. Wherfore the whole parallelogramme AC is equal to the whole square SP, that
<pb n="270" facs="tcp:7062:299"/>
is, to the square of the line MX.<note place="margin">The first part of this demon<g ref="char:EOLhyphen"/>stration con<g ref="char:EOLhyphen"/>cluded.</note> Wherefore the line MX containeth in power the parallelo<g ref="char:EOLhyphen"/>gramme AC. I say moreouer that the line MX is a binomiall line. For forasmuch as (by the 17. of the tenth) the line AG is commensurable in length to the line EG. Therefore (by the 15. of the tenth) the whole line AE is commensurable in length to either of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se lines AG and GE. But by supposition the line AE is commensurable in length to the line AB. Wherfore (by the 12. of the tenth) either of the lines AG &amp; GE are commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line AB. But the line AB is rationall. Wherefore either of these lines AG and GE is rationall. Wherfore (by the 19. of the tenth) either of these parallelogrammes AH and GK is rationall. Wherfore (by the first of the sixt, and 10. of the tenth) the parallelogramme AH is commensurable to the parallelogramme GK. But the parallelogramme AH is equall to the
<figure/>
square SN,<note place="margin">The seco<g ref="char:cmbAbbrStroke">̄</g>d part of the demon<g ref="char:EOLhyphen"/>stration con<g ref="char:EOLhyphen"/>cluded.</note> and the parallelogramme GK is equall to the square NP Wherfore the squares SN and NP, which are the squares of the lines MN and NX are rationall and commensu<g ref="char:EOLhyphen"/>rable. And forasmuch as (by supposition) the line AE is incommensurable in length to the line ED. But the line AE is commensurable in length to the line AG. And the line DE is commensurable in length to the line EF, for it is double to it by construction. Wherfore (by the 13. of the tenth) the line AG is incommensurable in length to the line EF. Wherfore the parallelogramme AH is incommensurable to the parallelogramme EL. But the parallelo<g ref="char:EOLhyphen"/>gramme AH is equal to the square SN, and the parallelogramme EL is equall to the paral<g ref="char:EOLhyphen"/>lelogramme MR. Wherfore the square SN is incommensurable to the parallelogramme MR. But as the square SN is to the parallelogramme MR, so is the line ON to the line NR by the 1. of the sixt.<note place="margin">The third part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</note> Wherfore the line ON is incommensurable to the line NR. But the line ON is equall to the line MN, and the line NR to the line NX. Wherfore the line MN is in<g ref="char:EOLhyphen"/>commensurable to the line NX. And it is already proued that the squares of the lines MN and NX are rationall and commensurable. Wherefore the lines MN and NX are rationall commensurable in power onely.<note place="margin">The totall con<g ref="char:EOLunhyphen"/>clusion.</note> Wherfore the whole line MX is a binomiall line, and it con<g ref="char:EOLhyphen"/>taineth in power the parallelogramme AC: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 37. Theoreme. The 55. Proposition. If a superficies be comprehended vnder a rationall line and a second binomi<g ref="char:EOLhyphen"/>all line: the line that contayneth in power that superficies is irrationall, and is a first bimediall line.</head>
               <p>SVppose that the superficies ABCD be contayned vnder a rationall line AB, and vnder a second binomiall line AD. Then I say that the line that containeth in power the superficies AC is a first bimediall line. For forasmuch as AD is a second binomiall line, it can in one onely point be deuided into his names, by the
<pb facs="tcp:7062:300"/>
43. of this tenth: let it therefore by supposition be deuided into his names in the poynt E, so that let AE be the greater name. Wherefore the lines AE and ED are rationall commensu<g ref="char:EOLhyphen"/>rable in power onely, and the line AE is in power more then the line ED by the square of a line commensurable in length to AE, and the lesse name, namely, ED is commensurable in length to the line AB by the definition of a second binomiall line, set before the 48. propositio<g ref="char:cmbAbbrStroke">̄</g> of this tenth. Deuide the line ED (by the te<g ref="char:cmbAbbrStroke">̄</g>th of the first) into two equall partes in the poynt F. And (by the 28. of the sixt) vpon the line AE apply a parallelogramme equal to the square of the line EF, and wanting in figure by a square. And let the same parallelogramme be that which is contayned vnder the lines AG and GE. Wherefore (by the second part of the 17. of this tenth) the line AG is commensurable in length to the line GE. And (by the 31. of the first) by the poyntes G, E, F, draw vnto the lines AB and CD these parallel lines, GH, EK, &amp; FL. And (by the 14. of the second) vnto the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AH describe an equall square SN. And to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me GK describe an equall square NP, and let the lines MN &amp; NX be so put that they both make one right line: wherefore (by the 14. of the first) the lines al<g ref="char:EOLhyphen"/>so RN, and NO make both one right line.<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> Make perfect the parallelogramme SP. Now it is manifest (by that which hath bene demo<g ref="char:cmbAbbrStroke">̄</g>strated in the propositio<g ref="char:cmbAbbrStroke">̄</g> next going before) that the parallelog<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame MR is the meane proportionall betwene the squares SN and NP, and is e<g ref="char:EOLhyphen"/>quall to the parallelogramme EL, and that the line MX contayneth in power the superficies
<figure/>
AC. Now resteth to proue that the line MX is a first bimediall line. Forasmuch as the line AE is incommensurable in length to the line ED, and the line ED is commensurable in length to the line AB, therefore (by the 13. of the tenth) the line AE is incommensurable in length to the line AB. And forasmuch as the line AG is commensurable in length to the line GE therefore the whole line AE is (by the 15. of the tenth) commensurable in length to either of these lines AG and GE. But the line AE is rationall: wherefore either of these lines AG and GE is rationall. And forasmuch as the line AE is incommensurable in length to the line AB, but the line AE is commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to either of these lines AG and GE: wherefore either of the lines AG and GE are incommensurable in length to the line AB (by the 13. of the tenth). Wherefore the lines AB, AG, and GE are rationall com<g ref="char:EOLhyphen"/>mensurable in power onely. Wherefore (by the 21. of the tenth) either of these parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes AH and GK is a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. Wherefore also either of these squares SN and NP is a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies by the corollary of the 23. of y<hi rend="sup">e</hi> tenth. Wherfore the lines MN &amp; NX are mediall lines by the 21. of this te<g ref="char:cmbAbbrStroke">̄</g>th.<note place="margin">The first part of this demon<g ref="char:EOLhyphen"/>stration con<g ref="char:EOLhyphen"/>cluded.</note> And forasmuch as the line AG is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line GE, therefore (by the 1. of the sixt and 11. of the tenth) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AH is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the parallelogramme GK that is, the square SN to the square NP, that is, the square of the line MN to the square of the line NX. Wherefore the lines MN and NX are medialls commensurable in power.<note place="margin">The third part co<g ref="char:cmbAbbrStroke">̄</g>cluded.</note> And forasmuch as the line AE is incommensurable
<pb n="271" facs="tcp:7062:300"/>
in length to the line ED, but the line AE is commensurable in length to the line AG, and the line ED is commensurable in length to the line E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, therefore (by the 13. of the tenth) the line AG is incommensurable in length to the line E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the sixt and 11. of the tenth) the parallelograme AH is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to the parallelogramme EL, that is, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> square SN to the parallelogramme MR, that is, the line ON is incommensurable to the line NR, that is the line MN to the line NX. And it is proued that the lines MN and NX are mediall lines commensurabl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in power.<note place="margin">The fourth part co<g ref="char:cmbAbbrStroke">̄</g>clude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> Wherefore the lines MN and NX are medi<g ref="char:EOLhyphen"/>all lines commensurable in power onely. Now I say moreouer that they comprehend a rational superficies. For forasmuch as by supposition the line DE is comme<g ref="char:cmbAbbrStroke">̄</g>surable in length to either of these lines AB and EF, therefore the line FE is commensurable in length to the line EK which is equall to the line AB (by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 12. of the tenth). And either of these lines EF and EK is a rationall line. Wherefore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me EL, that is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me MR, is a rationall superficies (by the 19. of the tenth).<note place="margin">The fift part concluded.</note> But the parallelogramme MR is that which is contayned vnder the lines MN and NX. But if two mediall lines commensurable in power onely and comprehending a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies be added together, the whole line is irrational and is called a first bimediall (by the 37. of the tenth).<note place="margin">The total<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> con<g ref="char:EOLunhyphen"/>clusion.</note> Wherefore the line MX is a first bime<g ref="char:EOLhyphen"/>diall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 38. Theoreme. The 56. Proposition. If a superficies be contayned vnder a rationall line and a third binomiall line: the line that contayneth in power that superficies is irrationall, and is a second bimediall line.</head>
               <p>SVppose that the superficies ABCD be comprehended vnder the rationall line AB and a third binomiall line AD, and let the line AD be supposed to be de<g ref="char:EOLhyphen"/>uided into his names in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> point E, of which let AE be the greater name. Then I say, that the line that containeth in power the superficies AC is irrationall, and is a second bimediall line. Let the same construction of the figures be in this, that was in the two Propositions nex<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> going before. And now forasmuch as the line AD is a third binomi<g ref="char:EOLhyphen"/>all line, therefore these lines AE and ED are rationall commensurable in power onely.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And
<figure/>
the line AE is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line ED by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line AE, and neither of the lines AE nor ED is commensurable in length to the line AB by the definition of a third binomiall line set before the 48. Proposition. As in the for<g ref="char:EOLhyphen"/>mer Propositions it was demonstrated, so also may it in this Proposition be proued, that the
<pb facs="tcp:7062:301"/>
line MX containeth in power the superficies AC, and that the lines MN and NX are medi<g ref="char:EOLhyphen"/>all lines commensurable in power onely. Wherefore the line MX is a bimediall line. Now re<g ref="char:EOLhyphen"/>steth to proue that it is a second bimediall line. Forasmuch as the line DE is (by supposition)
<figure/>
incommensurable in length to the line AB, that is, to the line EK. But the line ED is com<g ref="char:EOLhyphen"/>mensurable in length to the line EF. Wherefore (by the 13. of the tenth) the line EF is in<g ref="char:EOLhyphen"/>commensurable in length to the line EK. And the lines FE and EK are rationall. For by sup<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>osition the line ED is rationall, vnto which the line FE is commensurable. Wherefore the lines FE and EK are rationall lines commensurable in power onely. Wherefore (by the 21. of the tenth) the parallelogramme EL, that is, the parallelogramme MR which is con<g ref="char:EOLhyphen"/>tayned vnder the lines MN and NX is a mediall superficies. Wherefore that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines MN and NX is a mediall superficies. Wherefore the line MX is a second bimediall line (by the 38. Proposition and definition annexed thereto): which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 39. Theoreme. The 57. Proposition. If a superficies be contained vnder a rationall line, and a fourth binomiall line: the line which contayneth in power that superficies is irrationall, and is a greater line.</head>
               <p>SVppose that the superficies AC be comprehended vnder a rationall line AB and a fourth binomiall line AD, &amp; let the binomiall line AD be supposed to be deuided into his names in the point E, so that let the line AE be the greater name. Then I say, that the line which contayneth in power the superficies AC is irrationall, and is a grea<g ref="char:EOLhyphen"/>ter line.<note place="margin">Construction.</note> For, forasmuch as the line AD is a fourth binomiall line, therefore the lines AE and ED are rationall commensurable in power onely. And the line AE is in power more then the line ED by the square of a line incommensurable in length to AE. And the line AE is commensurable in length to the line AB. Deuide (by the 10. of the first) the line DE into two equall partes in the point F. And vpon the line AE apply a parallelogramme equall to the square of EF and wanting in figure by a square: and let the same parallelogramme be that which is contayned vnder the lines AG &amp; GE. Wherefore (by the second part of the 18. of the tenth) the line AG is incommensurable in length to the line EG. Draw vnto the line AB, by the pointes G, E, F, parallell lines GH, EK, and PL, and let the rest of the con<g ref="char:EOLhyphen"/>struction be as it was in the three former Propositions.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now it is manifest, that the line MX contayneth in power the superficies AC. Now resteth to proue that the line MX is an
<pb n="272" facs="tcp:7062:301"/>
irrationall line, and a greater line. Forasmuch as the line AG is incommensurable in length to the line EG, therefore (by the 1. of the sixt, and 11. of the tenth) the parallelogramme <milestone type="tcpmilestone" unit="unspecified" n="1"/> AH is incommensurable to the parallelogramme GK, that is, the square SN to the square NP. Wherefore the lines MN and NK are incommensurable in power. And forasmuch <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>
                  <milestone type="tcpmilestone" unit="unspecified" n="2"/> the line AE is commensurable in length to the rationall line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, therefore the parallelo<g ref="char:EOLhyphen"/>gramme
<figure/>
AK is rationall. And it is equall to the squares of the lines MN and NX. Wher<g ref="char:EOLhyphen"/>fore that which is composed of the squares of the lines MN and NX added together is rati<g ref="char:EOLhyphen"/>onall. And forasmuch as the line ED is incommensurable in length to the line AB, that is, to the line EK, but the line ED is commensurable in length to the line EF, therefore the <milestone type="tcpmilestone" unit="unspecified" n="3"/> line EF is incommensurable in length to the line EK. Wherefore the lines EK and EF are rationall commensurable in power onely. Wherefore (by the 21. of the tenth) the parallelo<g ref="char:EOLhyphen"/>gramme LE, that is, the parallelogramme MR is mediall. And the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me MR <milestone type="tcpmilestone" unit="unspecified" n="4"/> is that which is contayned vnder the lines MN, and NX. Wherefore that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines MN and NX is mediall. And that which is composed of the squares of the lines MN &amp; NX is proued to be rationall, &amp; the line MN is demonstrated to be incom<g ref="char:EOLhyphen"/>mensurable in power to the line NX. But if two lines incommensurable in power be added together, hauing that which is made of the squares of them added together rationall, &amp; that which is vnder them mediall, the whole line is irrationall, and is called a greater line (by the 39. of the tenth). Wherefore the line MX is irrationall, and is a greater line, and it contai<g ref="char:EOLhyphen"/>neth in power the superficies AC: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 40. Theoreme. The 58. Proposition. If a superficies be contained vnder a rationall line and a fift binomiall line: the line which contayneth in power that superficies is irrationall, and is a line contayning in power a rationall and a mediall superficies.</head>
               <p>SVppose that the superficies AG be contayned vnder the rationall line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and vnder a fift binomiall line AD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd let the same lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AD be supposed to be deuided into his names in the poynt E, so that let the line AE be the greater name. Then I say, that the line which contayneth in power the superficies AC is irrationall, and is a line contayning in power a rationall and a mediall superficies.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Let the selfe same constructions be in this, that were in the foure Proposition next going before. And it is manifest, that the line MX contayneth in power the superfici<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> AG. Now testeth to proue that the line MX is a line contayning in power a rationall &amp; a mediall superficies. For<g ref="char:EOLhyphen"/>asmuch
<pb facs="tcp:7062:302"/>
as the line AG is incommensurable in length to the line GE, therefore (by the 1. of the sixt, and 10. of the tenth) the parallelogramme AH is incommensurable to the paralle<g ref="char:EOLhyphen"/>logramme <milestone type="tcpmilestone" unit="unspecified" n="2"/> HE, that is, the square of the line MN to the square of the line NX. Wherefore the lines MN and NX are incommensurable in power. And forasmuch as the line AD i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a fif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> binomiall line, and his lesse name or part is the line ED, therefore the line ED is com<g ref="char:EOLhyphen"/>mensurable
<figure/>
in length to the line AB. But the line AE is incommensurable in length to the line ED. Wherefore (by the 13. of the tenth) the line AB is incommensurable in length to the line AE. Wherefore the lines AB and AE are rationall commensurable in power onely. Wherefore (by the 21. of the tenth) the parallelogramme AK is mediall, that is, that which is composed of the squares of the lines MN &amp; NX added together. And forasmuch <milestone type="tcpmilestone" unit="unspecified" n="3"/> as the line DE is commensurable in length to the line AB, that is, to the line EK, but the line DE is commensurable in length to the line EF, wherefore (by the 12. of the tenth) the line EF is also commensurable in length to the line EK. And the line EK is rationall. Wherefore (by the 19. of the tenth) the parallelogramme EL, that is, the parallelogramme MR, which is contayned vnder the lines MN and NX is rationall. Wherefore the lines <milestone type="tcpmilestone" unit="unspecified" n="4"/> MN and NX are incommensurable in power, hauing that which is composed of the squares of them added together, Mediall, and that which is contayned vnder them, Rationall. Wherefore (by the 40. of the tenth) the whole line MX is a line contayning in power a ra<g ref="char:EOLhyphen"/>tionall and a mediall superficies, and it contayneth in power the superficies AC: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 41. Theoreme. The 59. Proposition. If a superficies be contayned vnder a rationall line, and a sixt binomiall line, the lyne which contayneth in power that superficies, is irrational, &amp; is called a line contayning in power two medials.</head>
               <p>SVppose that the superficies ABCD be contained vnder the rationall line AB, and vnder a sixt binomiall line AD, and let the line AD be supposed to be deuided <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>to his names in the point E, so that let the line AE be the greater name. Then I say that the line that containeth in power the superficies AC is irrationall, and is a line contay<g ref="char:EOLhyphen"/>ning in power two medials.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Let the selfe same constructio<g ref="char:cmbAbbrStroke">̄</g>s be in this, that were in the former propositions. Now it is manifest that the line MX containeth in power the superficies AC, <milestone type="tcpmilestone" unit="unspecified" n="1"/> and that the line MN is incommensurable in power to the line NX. And forasmuch as the <milestone type="tcpmilestone" unit="unspecified" n="2"/> line AE is incommensurable in length to the line AB, therfore the lines AE and AB are
<pb n="273" facs="tcp:7062:302"/>
rationall commensurable in power only. Wherfore (by the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of the tenth) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AK that is, that which is composed of the squares of the lines MN and NX added together <milestone type="tcpmilestone" unit="unspecified" n="3"/> is mediall. Againe forasmuch as the line ED is incommensurable in length to the line AB, therefore also the line EF is inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line EK. Wherfore the lines EF and EK are rationall commensurable in power onely. Wherfore the parallelogramme EL, that is, the parallelogramme MR which is contained vnder the lines MN and NX is me<g ref="char:EOLhyphen"/>diall.
<figure/>
And forasmuch as the line AE is incommensurable in length to the line EF, therfore <milestone type="tcpmilestone" unit="unspecified" n="4"/> the parallelogramme AK is also incommensurable to the parallelogramme EL (by the first of the sixt, and 10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth.) But the parallelogramme AK is equal to that which is com<g ref="char:EOLhyphen"/>posed of the squares of the lines MN and NX added together. And the parallelogramme EL is equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>tai<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>d vnd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r the lines MN and NX. Wherfore that which <milestone type="tcpmilestone" unit="unspecified" n="5"/> is c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>posed of the squares of the lines MN and NY added together, is incommensurable to that which is contained vnder the l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nes MN and NX<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and e<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>her of them, nam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ly, that which is composed of the squares of the lines MN and NX added tog<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>her, and that which is contained v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>der the lines MN and N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is proued mediall; and the lines MN and NX are proued incommensurable in power. Wherfore (by the 41. of the tenth) the whole line MX is a line contayning in power two medials, and it containeth in power the superfices AC: which was required to be d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>onstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ted.</p>
               <div type="section">
                  <head>An A<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>umpt.</head>
                  <p>
                     <hi>If a right line be deuided into two vnequall partes, the squares which are made of the vnequall partes, are greater then the rectangle parallelogramme c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tayned vnder the vnequall partes, twise.</hi>
                  </p>
                  <p>
                     <hi>Suppose that AB be a right line, and let it be <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> point C. And let the line AC be the greater part. <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>B, are greater the<g ref="char:cmbAbbrStroke">̄</g> that which is contained vnder the lines A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and CB twise. D<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>id<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (by the 10. of the first) the line AB into two e<g ref="char:EOLhyphen"/>quall
<figure/>
partes, in the point D. Now forasmuch as the right line AB is deuided into two equall parte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> in the point D, and into two vnequall parte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> in the point C, therfore (by the 5. of the second) that which is contained vnder the lines <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> line CD, is equall to the square of the line AD. <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <pb facs="tcp:7062:303"/>
the lines AC and CB, (omitting the square of the line CD) is lesse then the square of the AD (by the 9. common sentence, and the seuenth of the fifth:) Wherefore that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AC and CB, twise, is lesse then the double of the square of the line AD (that is, the<g ref="char:cmbAbbrStroke">̄</g> twise the square of the line AD)
<figure/>
by</hi> 
                     <note n="*" place="margin">
                        <hi>Looke after the Assumpt concluded at this marke: for plainer ope<g ref="char:EOLhyphen"/>ning of this place.</hi>
                     </note> 
                     <hi>alternate proportio<g ref="char:cmbAbbrStroke">̄</g>, and the 14. of the fift. But the squares of the lines AC and CB are double to the squares of the lines AD and DC (by the 9. of the seco<g ref="char:cmbAbbrStroke">̄</g>d). Therfore the squares of AC and CB are more then double to the square of AD alone, (leauing out the square of DC) by the 8. of the fift. But the parallelogramme contained vnder the lines AC and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>B twise, is proued lesse the<g ref="char:cmbAbbrStroke">̄</g> the double of the square of the line AD. Therfore the same parallelogramme contained vnder the lines AC and CB twise, is much lesse then the squares of the lines AC and CB. If a right line therfore be deuided into two vnequall partes, the squares which are made of the vnequall partes, are greater the<g ref="char:cmbAbbrStroke">̄</g> the rectangle parallelogramme contained vnder the vnequall partes, twise: which was required to be demonstrated.</hi>
                  </p>
                  <p>In numbers I neede not to haue so alleaged, for the 17. of the seuenth had confirmed the doubles to be one to the other, as their singles were, but in our magnitudes, it likewise is true and euident by alternate proportion, thus. As the parallelogramme of the lines AC and CB is to his double, so is the square of the line AD to his double (eche being halfe). Wherfore, alternately, as the parallelogramme is to the square, so is the parallelograme his double to the double of the square. But the parallelograme was proued lesse then th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> square: wherfore his double is lesse then the square his double, by the 14. of the fifth.</p>
                  <p>This Assumpt is in some bookes not read, for that in maner it semeth to be all one with that which was put after the 39. of this booke: but for the diners maner of demon<g ref="char:EOLhyphen"/>strating, it is necessary.<note place="margin">
                        <hi>The vse of this Assumpt is in the next proposition, &amp; other follow<g ref="char:EOLhyphen"/>ing.</hi>
                     </note> For the feare of inuentio<g ref="char:cmbAbbrStroke">̄</g> is therby furthered. And though Zam<g ref="char:EOLhyphen"/>bert did in the demonstration hereof, omitte that which P. Montaureus could not sup<g ref="char:EOLhyphen"/>ply<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> but plainly doubted of the sufficiencie of this proofe, yet M. Dee, by onely allega<g ref="char:EOLhyphen"/>tion of the due places of credite, whose pithe &amp; force<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Theon his wordes do containe, hath restored to the demonstration sufficiently, both light and authoritie, as you may perceiue, and chiefly such may iudge, who can compare this demonstration here (thus furnished) with the Greeke of Theon, or latine translation of Zambert.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 42. Theoreme. The 60. Proposition. <note place="margin">The fift Se<g ref="char:EOLhyphen"/>nary by com<g ref="char:EOLhyphen"/>position.</note>The square of a binomiall line applyed vnto a rationall line, maketh the breadth or other side a first binomiall line.</head>
               <p>SVppose that the line <hi>AB</hi> be a
<figure/>
binomiall line, and let it be supposed to be deuided into his names in the poynt <hi>C,</hi> so that let <hi>AC</hi> be the greater name.<note place="margin">Construction.</note> And take a rationall line <hi>DE</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>4. of the first) vnto the line <hi>DE</hi> apply a rectan<g ref="char:EOLhyphen"/>gle parallelograme <hi>DEFG</hi> equall to the square of the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>line <hi>AB</hi> and making in breadth the line <hi>DG.</hi> Then I say that the line <hi>DG</hi> is a first binomiall line. Vnto the line <hi>DE</hi> apply the parallelograme <hi>DH</hi> e<g ref="char:EOLhyphen"/>quall to the square of the line <hi>AC,</hi> and vnto the line <hi>KH</hi> which is equall to the
<pb n="274" facs="tcp:7062:303"/>
line <hi>DE</hi> apply the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>KL</hi> equall to the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the line <hi>BC.</hi> Wherefore the residue namely, that which is contayned vnder the lines <hi>AC</hi> &amp; <hi>CB</hi> twise is equall to the re<g ref="char:EOLhyphen"/>sidue, namely, to the parallelograme <hi>MF,</hi> by <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the second Deuide (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first) the line <hi>MG</hi> into two equall partes in the poynt <hi>N.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And (by the 31. of the first) draw the line <hi>NX</hi> parallel to either of these lines <hi>ML</hi> and <hi>GF.</hi> Wherefore either of these parallelo<g ref="char:EOLhyphen"/>grammes <hi>MX</hi> and <hi>NF</hi> is equall to that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> once, by the 15. of the fifth. And forasmuch as the line <hi>AB</hi> is a binomiall line, and is deuided into his names in the poynt C therefore the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ines <hi>AC</hi> and <hi>CB</hi> are rationall commensura<g ref="char:EOLhyphen"/>ble in power onely. Wherefore the squares of the lines <hi>AC</hi> and <hi>CB</hi> are rationall, and there<g ref="char:EOLhyphen"/>fore commensurable the one to the other. Wherefore (by the <hi>15.</hi> of the tenth) that which is made of the squares of the lines <hi>AC</hi> and <hi>CB</hi> added together is commensurable to either of the squares of the lines <hi>AC</hi> or <hi>CB</hi>: wherefore that which is made of the squares of the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>AC</hi> and <hi>CB</hi> added together is rationall. And it is equall to the parallelograme <hi>DL</hi> by con<g ref="char:EOLhyphen"/>struction. Wherefore the parallelograme <hi>DL</hi> is rationall. And it is applyed vnto the rational line <hi>DE</hi>: wherefore (by the 20. of the tenth) the line <hi>DM</hi> is rationall, and commensurable in length to the line <hi>DE.</hi> Agayne forasmuch as the lines <hi>AC</hi> and <hi>CB</hi> are rationall commen<g ref="char:EOLhyphen"/>surable in power onely, therefore that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi>
                  <milestone type="tcpmilestone" unit="unspecified" n="1"/> twise, that is, the parallelograme <hi>MF</hi> is mediall by the 21 of the tenth, and it is applyed vnto the rationall line <hi>ML.</hi> Wherefore the line <hi>MG</hi> is rationall and incommensurable in length to the line <hi>ML</hi> (by the 22. of this tenth) that is, to the line <hi>DE.</hi> But the line <hi>MD</hi> is proued rationall and commensurable in length to the line <hi>DE.</hi> Wherefore (by the 13. of the tenth) the <milestone type="tcpmilestone" unit="unspecified" n="2"/> line <hi>DM</hi> is incommensurable in length to the line <hi>MG.</hi> Wherefore the lines <hi>DM &amp; M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> are rationall commensurable in power onely.<note place="margin">Concluded that DG is a binomiall line.</note> Wherefore (by the 36. of the tenth) the whole line <hi>DG</hi> is a binomiall line. Now resteth to proue that it is a first binomiall line. Forasmuch as (by the the assumpt going before the 54. of the tenth) that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> is the meane proportionall betwene the squares of the lines <hi>AC</hi> and <hi>CB,</hi> therefore the parallelograme <hi>MX</hi> is the meane proportionall betwene the parallelogrammes <hi>DH</hi> and <hi>KL.</hi> Wherefore as the parallelogame <hi>DH</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the parallelograme <hi>MX,</hi> so is the parallelograme <hi>MX</hi> to the parallelograme <hi>KL,</hi> that is, as the line <hi>DK</hi> is to the line <hi>MN,</hi> so is the line <hi>MN</hi> to the line <hi>MK.</hi> Wherefore that which is contayned vnder the lines <hi>DK</hi> and <hi>KM</hi> is equall to the square of the line <hi>MN.</hi> And forasmuch as the square of the line <hi>AC</hi> is commensurable to the square of the line <hi>CB,</hi> the parallelograme <hi>DH</hi> is commensu<g ref="char:EOLhyphen"/>rable to the prarallelograme <hi>KL.</hi> Wherefore (by the <hi>1.</hi> of the sixt and 10. of the tenth) the line <milestone type="tcpmilestone" unit="unspecified" n="3"/> 
                  <hi>DE</hi> is co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line <hi>KM.</hi> And forasmuch as the squares of the lines <hi>AC</hi> and <hi>CB</hi> are greater then that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> twise by the assumpt going before this proposition, or by the assumpt after this 39. of the tenth, therefore the parallelograme <hi>DL</hi> is greater then the parallelograme <hi>MF.</hi> Wherefore (by the first of the sixt) the line <hi>DM</hi> is greater then the line <hi>MG.</hi> And that which is contayned vnder the lines <hi>DK</hi> and <hi>KM</hi> is equall to the square of the line <hi>MN,</hi> that is to the fourth part of the square <milestone type="tcpmilestone" unit="unspecified" n="4"/> of the line <hi>MG.</hi> But (by the 17. of the tenth), if there be two vnequall right lines, and if vp<g ref="char:EOLhyphen"/>pon the greater be applyed a parallelograme equall to the fourth part of the square made of the lesse line and wanting in figure by a square, if also the parallelograme thus applyed de<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ide the line whereupon it is applyed into parts commensurable in length, then shall the grea<g ref="char:EOLhyphen"/>ter line be in power more then the lesse by the square of a line commensurable in length to the greater. Wherefore the line <hi>DM</hi> is in power more then the line <hi>MG</hi> by the square of a line commensurable in length vnto the line <hi>DM.</hi> And the lines <hi>DM</hi> and <hi>MG</hi> are proued ra<g ref="char:EOLhyphen"/>tionall <milestone type="tcpmilestone" unit="unspecified" n="5"/> commensurable in power onely. And the line <hi>DM</hi> is proued the greater name and commensurable in length to the rationall line geuen <hi>DE.</hi> Wherefore by the definition of a first binomiall line se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> before the 48. proposition of this booke, the line <hi>DG</hi> is a first binomiall
<pb facs="tcp:7062:304"/>
line: which was required to be proued.</p>
               <p>
                  <hi>This proposition and the fiue following are the conuerses of the sixe former pro<g ref="char:EOLhyphen"/>positions.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 43. Theoreme. The 61. Proposition. The square of a first bimediall line applied to a rationall line, maketh the breadth or other side a second binomiall line.</head>
               <p>SVppose that the line AB be a first bimediall line, and let it be supposed to be de<g ref="char:EOLhyphen"/>uided into his partes in the point C, of which let AC be the greater part. Take also a rationall line DE, and (by the 44. of the first) apply to the line DE the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DF equall to the square of the line AB, &amp; making in breadth the line DG.<note place="margin">Construction.</note> Then I say, that the line DG is a second binomiall line. Let the same con<g ref="char:EOLhyphen"/>structions be in this, that were in the Proposition going before. And forasmuch as the line AB is a first bimediall line,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n.</note> and is deuided into his partes in the point C, therefore (by the 37. of the tenth) the lines AC and CB are
<figure/>
mediall commensurable in power onely, co<g ref="char:cmbAbbrStroke">̄</g>prehending a rationall superficies. Wher<g ref="char:EOLhyphen"/>fore also the squares of the lines AC and CB are mediall. Wherefore the parallelo<g ref="char:EOLhyphen"/>gramme DL is mediall (by the Corollary of the 23. of the tenth) and it is applied vp<g ref="char:EOLhyphen"/>pon the rationall line DE. Wherefore (by the 22. of the tenth) the line MD is rati<g ref="char:EOLhyphen"/>onall and incommensurable in length to the line DE. Againe forasmuch as that <milestone type="tcpmilestone" unit="unspecified" n="1"/> which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AC and CB twise is rationall, therefore also the pa<g ref="char:EOLhyphen"/>rallelogramme MF is rationall, and it is applied vnto the rationall line ML. Wherefore the line MG is rationall and commensura<g ref="char:EOLhyphen"/>ble <milestone type="tcpmilestone" unit="unspecified" n="2"/> in length to the line ML, that is, to the line DE (by the 20. of the tenth). Wherefore the line DM is incommensurable in length to the line MG, and they are both rationall. <milestone type="tcpmilestone" unit="unspecified" n="3"/> Wherefore the lines DM and MG are rationall commensurable in power onely. Wherefore the whole line DG is a binomiall line.<note place="margin">Concluded that DG is a binomiall line.</note> Now resteth to proue that it is a second binomiall line. Forasmuch as the squares of the lines AC and CB are greater then that which is contayned vnder the lines AC and CB twise (by the Assumpt before the 60. of this booke): therefore the parallelogramme DL is greater then the parallelogrrmme MF. Wherefore also (by the first of the sixt) the line DM is greater then the line MG. And forasmuch as the square of <milestone type="tcpmilestone" unit="unspecified" n="4"/> the line AC is commensurable to the square of the line CB, therefore the parallelogramme DH is commensurable to the parallelogramme KL. Wherefore also the line DK is com<g ref="char:EOLhyphen"/>mensurable <milestone type="tcpmilestone" unit="unspecified" n="5"/> in length to the line KM. And that which is contayned vnder the lines DK and KM is equall to the square of the line MN, that is, to the fourth part of the square of the line MG. Wherefore (by the 17. of the tenth) the line DM is in power more then the line MG, by the square of a line commensurable in length vnto the line DM: and the line MG is commensurable in length to the rationall line put, namely, to DE. Wherefore the line DG is a second binomiall line: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb n="275" facs="tcp:7062:304"/>
               <head>¶ The 44. Theoreme. The 62. Proposition. The square of a second bimediall line, applied vnto a rationall line: maketh the breadth or other side therof, a third binomiall lyne.</head>
               <p>SVppose that AB be a second bimediall line, and let AB be supposed to be deuided into his partes in the point C, so that let AC be the greater part. And take a ra<g ref="char:EOLhyphen"/>tionall line DE. And (by the 44. of the first) vnto the line DE apply the paralle<g ref="char:EOLhyphen"/>logramme DF equall to the square of the line AB, and making in breadth the line DG. Then I say that the line DG is a third binomiall line. Let the selfe same construc<g ref="char:EOLhyphen"/>tions be in this that were in the propositions next going before. And forasmuch as the line AB is a second bimediall line,<note place="margin">Construction.</note> and is deuided
<figure/>
into his partes in the point C,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore (by the 38. of the tenth) the lines AC and CB are medials commensurable in power only, comprehe<g ref="char:cmbAbbrStroke">̄</g>ding a mediall superficies. Wher<g ref="char:EOLhyphen"/>fore <note place="margin">†</note> that which is made of the squares of the lines AC and CB added together, is mediall, and it is equall to the parallelo<g ref="char:EOLhyphen"/>gramme DL by construction. Wherefore the parallelogramme DL is mediall, and is applied vnto the rationall line DE, wherfore (by the 22. of the tenth) the line MD is rationall and incommensurable in length to the line DE. And by the lyke reason also <note place="margin">*</note> the line MG is rationall and incommensurable in length to the line ML, that is, to the line DE. Wherfore either of these lines DM and MG is rational, and incommensurable in length to the line DE. And foras<g ref="char:EOLhyphen"/>much as the line AC is incommensurable in length to the line CB, but as the line AC is to the line CB, so (by the assumpt going before the 22. of the tenth) is the square of the line AC to that which is contained vnder the lines AC and CB. Wherfore the square of the line AC is inc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mmmensurable to that which is contayned vnder the lines AC and CB. Wherfore that <note place="margin">‡</note> that which is made of the squares of the lines AC and CB added together, is incommen<g ref="char:EOLhyphen"/>surable to that which is contained vnder the lines AC and CB twise, that is, the parallelo<g ref="char:EOLhyphen"/>gramme DL to the parallelogramme MF. Wherfore (by the first of the sixt, and 10. of the tenth) the line DM is incommensurable in length to the line MG. And they are proued both rationall,<note place="margin">DG, conclu<g ref="char:EOLhyphen"/>ded a binomi<g ref="char:EOLhyphen"/>all line.</note> wherfore the whole line DG is a binomiall line by the definition in the 36. of the tenth. Now resteth to proue that it is a third binomiall line. As in the former propositions, so also in this may we conclude that the line DM is greater then the line MG, and that the line DK is commensurable in length to the line KM. And that that which is contained vnder the lines DK and KM is equall to the square of the line MN. Wherfore the line DM is in power more then the line MG by the square of a line commensurable in length vnto the line DM, and neither of the lines DM nor MG is commensurable in length to the rational line DE. Wherfore (by the definition of a third binomi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll line) the line DG is a third binomiall line: which was required to be proued.</p>
               <div type="section">
                  <head>¶Here follow certaine annotations by M. Dee, made vpon three places in the demonstration, which were not very euident to yong beginners.</head>
                  <p>† (The squares of the lines AC and C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are medials (<gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> taught after the 21<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this tenth) and ther<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore forasmuch as they are (by supposition) comme<g ref="char:cmbAbbrStroke">̄</g>surable th'one to the other: (by the 15. of the te<g ref="char:cmbAbbrStroke">̄</g>th)
<pb facs="tcp:7062:305"/>
the compound of them both is commensurable to ech part. But the partes are medials, therfore (by the co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ollary of the 23. of the tenth) the compound shall be mediall.</p>
                  <p>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For that MX is equall (by construction) to that which is contayned vnder the lines AC and CB, which is proued mediall: therfore (by the corollary of the 23. of this tenth) MX is mediall, and ther<g ref="char:EOLhyphen"/>fore (by the same corollary) his double MF is mediall. And it is applied to a rationall line, ML (beyng equall to D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>) therfore by the 22. of the tenth, the line MG is rationall and incommensurable in length to ML, that is, to DE.</p>
                  <p>‡ Because the compound of the two squares (of the lines AC and C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>) beyng commensurable one to the other, is also to eyther square (by the 15.) commensurable, therfore to the square of AC: But the square of AC is proued incommensurable to that which is contained vnder AC &amp; CB once. Wherfore (by the 13. of the tenth) the compound of the two squares (of the lines AC and CB) is incommensu<g ref="char:EOLhyphen"/>rable, to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AC and C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> once. But to that which is twise contained vnder the same lines AC and CB, the paralleloga<g ref="char:cmbAbbrStroke">̄</g>rme once contayned, is commensurable (for it is as 1. is to 2.) therfore that which is made of the squares of the lines AC and CB is incommensurable to the parallelogramme contained vnder AC and CB twise, by the sayd 13. of this tenth.</p>
               </div>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <note place="margin">A Corollary added by M. Dee.</note>Hereby it is euident, that the squares made of the two partes of a second bi<g ref="char:EOLhyphen"/>mediall line, composed, is a compound mediall, and that the same compound is in<g ref="char:EOLhyphen"/>commensurable to the parallelogramme contayned vnder the two partes of the second bimediall lyne.</p>
                  <p>
                     <hi>The proofe hereof, is in the first and third annotations here before annexed.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 45. Theoreme. The 63. Proposition. The square of a greater line applied vnto a rationall line, maketh the breadth or other side a fourth binomiall line.</head>
               <p>SVppose that the line AB be a greater line, and let it be supposed to be deuided into his partes in the point C, so that let AC be the greater part. And take a rationall line DE. And (by the 44. of the first) vnto the line DE, apply the parallelogramme DF equall to the square of the line AB, and making in breadth the line DG. Then I say, that the line DG is a fourth binomiall line. Let the selfe same construction be in this, that was in the former Propositions.<note place="margin">Construction.</note> And for<g ref="char:EOLhyphen"/>asmuch
<figure/>
as the line AB is a greater line, &amp; is deuided into his partes in the point C:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore the lines AC and CB are incom<g ref="char:EOLhyphen"/>mensurable in power, hauing that which is made of the squares of them added toge<g ref="char:EOLhyphen"/>ther rationall, and the parallelogramme which is contayned vnder them, mediall. Now forasmuch as that which is made of the squares of the lines AC and CB added together is rationall, therefore the paralle<g ref="char:EOLhyphen"/>logramme DL is rationall. Wherefore al<g ref="char:EOLhyphen"/>so the line MD is rationall and commen<g ref="char:EOLhyphen"/>surable <milestone type="tcpmilestone" unit="unspecified" n="1"/> in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line DE (by the 20. of this tenth). Againe forasmuch as that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AC and CB twise is mediall, that is, the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me
<pb n="276" facs="tcp:7062:305"/>
MF, and it is applied vnto the rationall line ML, therefore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>2. of the tenth) the <milestone type="tcpmilestone" unit="unspecified" n="2"/> line MG is rationall and incommensurable in length to the line DE, Therefore (by the 13. of the tenth) the line DM is incommensurable in length to the line MG. Wherefore the lines DM and MG are rationall commensurable in power onely. Wherfore the whole <milestone type="tcpmilestone" unit="unspecified" n="3"/> line DG is a binomiall line. Now resteth to proue, that it is also a fourth binomiall line. Euen as in the former Propositions, so also in this may we conclude, that the line DM is greater then the line MC. And that that which is contayned vnder the lines DK and KM is equall to the square of the line MN. Now forasmuch as the square of the line AC is incommensurable to the square of the line CB, therefore the parallelogramme DH is incommensurable to the parallelogramme KL. Wherefore (by the 1. of the sixt, and 10. of the tenth) the line DK is incommensurable in length to the line KM. But if there be two vnequall right lines, and if vpon the greater be applied a pa<g ref="char:EOLhyphen"/>rallelogramme equall to the fourth part of the square made of the lesse, and wanting in figure <milestone type="tcpmilestone" unit="unspecified" n="4"/> by a square, and if also the parallelogramme thus applied deuide the line wherupon it is ap<g ref="char:EOLhyphen"/>plied into partes incommensurable in length, the greater line s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all be in power more then the lesse, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the greater (by the 18. of the tenth). Wherefore the line DM is in power more then the line MG, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mensurable in length to DM. And the lines DM and MG are proued to be rationall co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mensurable in power onely. And the line DM is commensurable in length to the rationall line geuen DE. Wherefore the line DG is a fourth binomiall line: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 46. Theoreme. The 64. Proposition. The square of a line contayning in power a rationall and a mediall super<g ref="char:EOLhyphen"/>ficies applied to a rationall line, maketh the breadth or other side a fift bi<g ref="char:EOLhyphen"/>nomiall line.</head>
               <p>SVppose that the line AB be a line contayning in power a rationall and a mediall superficies, and let it be supposed to be deuided into his partes in the point C, so that let AC be the greater part, and take a rationall line DE. And (by the 44. of the first) vnto the line DE apply the parallelogramme DF equall to the square of the line AB, and making in breadth
<figure/>
the line DG. Then I say, that the line DG is a fift binomiall line.<note place="margin">Construction.</note> Let the selfe same co<g ref="char:cmbAbbrStroke">̄</g>structi<g ref="char:EOLhyphen"/>on be in this, that was in the former. And for<g ref="char:EOLhyphen"/>asmuch as AB is a line contayning in power a rationall and a mediall superficies,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and is deui<g ref="char:EOLhyphen"/>ded into his partes in the poynt C, therefore the lines AC &amp; CB are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power, hauing that which is made of the squares of the<g ref="char:cmbAbbrStroke">̄</g> added together mediall, and that which is con<g ref="char:EOLhyphen"/>tayned vnder then rationall. Now forasmuch as that which is made of the squares of the lines AC and CB added together is mediall, there<g ref="char:EOLhyphen"/>fore also the parallelogramme DL is mediall. Wherefore (by the 22. of the tenth) the line DM is rationall and incommensurable in length to the line DE. Againe forasmuch as <milestone type="tcpmilestone" unit="unspecified" n="1"/> that which is contayned vnder the lines AC and CB twise, that is, the parallelogramme MF, is rationall, therefore by the 20. the line MG is rationall &amp; co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length <milestone type="tcpmilestone" unit="unspecified" n="2"/>
                  <pb facs="tcp:7062:306"/>
to the line DE. Wherefore (by the 13. of the tenth) the line DM is incommensurable in <milestone type="tcpmilestone" unit="unspecified" n="3"/> length to the line MG. Wherefore the lines DM and MG are rationall commensurable in power onely. Wherefore the whole line DG is a binomiall line. I say moreouer, that it is a fift binomiall. For, as in the former, so al<g ref="char:EOLhyphen"/>so
<figure/>
in this may it be proued, that that which is contayned vnder the lines DK and KM is e<g ref="char:EOLhyphen"/>quall to the square of MN the halfe of the lesse: and that the line DK is incommensurable in length to the line KM. Wherefore (by the 18. of the tenth) the line DM is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line MG by the square of a line incommen<g ref="char:EOLhyphen"/>surable in length to the line DM. And the lines DM and MG are rationall commensurable in power onely, and the lesse line, namely, MG is commensurable in length to the rationall line geuen DE. Wherefore the line DG is a fift binomiall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 47. Theoreme. The 65. Proposition. The square of a line contayning in power two medialls applyed vnto a ra<g ref="char:EOLhyphen"/>tionall line, maketh the breadth or other side a sixt binomiall line.</head>
               <p>SVppose that the line <hi>AB</hi> be a line contayning in power two medialls, and let it be supposed to be deuided into his partes in the poynt <hi>C.</hi> And take a rationall line <hi>DE.</hi> And (by the 44. of the first) vnto the rationall line <hi>DE</hi> apply the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>DF</hi> equall to the square of the line <hi>AB</hi> and making in breadth the line <hi>DG.</hi> Then I say that the line <hi>DG</hi> is a sixt binomiall line.<note place="margin">Construction.</note>
                  <figure/>
Let the selfe same construction be in thys that was in the former. And forasmuch as the line <hi>AB</hi> is a line contayning in power two medialls,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and is diuided into his partes in the poynt <hi>C,</hi> therefore the lines <hi>AC</hi> &amp; <hi>CB</hi> are incommensurable in power ha<g ref="char:EOLhyphen"/>uing that which is made of the squares of them added together mediall, and that which is contayned vnder them, mediall, and moreouer incommensurable to that which is made of the squares of them ad<g ref="char:EOLhyphen"/>ded together. Wherefore by those thinges which haue bene before proued, either of these parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes <hi>DL</hi> and <hi>MF</hi> is mediall, and either of them is applyed vpon the ratio<g ref="char:EOLhyphen"/>nall line <hi>DE.</hi> Wherefore (by the 22. of the tenth) either of these lines <hi>DM</hi> and <hi>MG</hi> is ratio<g ref="char:EOLhyphen"/>nall <milestone type="tcpmilestone" unit="unspecified" n="1"/> and incommensurable in length to the line <hi>DE.</hi> And forasmuch as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hat which is mad<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the squares of the lines <hi>AC</hi> and <hi>CB,</hi> added together is incommensurable to that which is contayned vnder the lines <hi>AC</hi> and <hi>CB</hi> twise, therefore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>DL</hi> is in<g ref="char:EOLhyphen"/>commensurable to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>MF.</hi> Wherefore (by the 1. of the sixt and 10. of the tenth) the line <hi>DM</hi> is incommensurable in length to the line <hi>MG.</hi> Wherefore the lines <milestone type="tcpmilestone" unit="unspecified" n="2"/>
                  <hi>DM</hi> and <hi>MG</hi> are rationall commensurable in power onely. Wherefore the whole line <hi>DG</hi>
                  <pb n="277" facs="tcp:7062:306"/>
is a binomiall line. I say also that is a sixt binomiall line. For euen as in the other proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> it hath bene proued so also in this may it be proued, that that which is contayned vnder the lines <hi>DK</hi> and <hi>KM,</hi> is equall to the square of the line <hi>MN,</hi> and that the line <hi>DK</hi> is incom<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>KM,</hi> and therfore (by the 18. of the tenth) the line <hi>DM</hi> is in power more then the line <hi>MG</hi> by the square of a line incommensurable in length to the line <hi>DM.</hi> And neither of these lines <hi>DM</hi> nor <hi>MG</hi> is commensurable in length to ther an<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>nall line geuen <hi>DE.</hi> Wherefore the line <hi>DG</hi> is a sixt <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>nomiall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 48. Theoreme. The 66. Proposition. A line commensurable in length to a binomiall line, is also a binomiall line of the selfe same order.</head>
               <p>SVppose that the line <hi>AB</hi> be a binomiall line,<note place="margin">The sixt Se<g ref="char:EOLhyphen"/>nary.</note> and vnto the line <hi>AB</hi> let the line <hi>CD</hi> be commensurable in length. Then I say that the line <hi>CD</hi> is a binomiall line and of the selfe same order that the line <hi>AB</hi> is. For forasmuch as <hi>AB</hi> is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> binomiall line let it be deuided into his names in the poynt <hi>E,</hi>
                  <note place="margin">Construction.</note> and let <hi>AE</hi> be the greater name. Wherefore the lines <hi>AE</hi> and <hi>EB</hi> are
<figure/>
rationall commensurable in power onely. And as the line <hi>AB</hi> is to the line <hi>CD,</hi> (so by the 12. of the sixt) let the line <hi>AE</hi> be to the line <hi>CF.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore (by the <hi>19.</hi> of the fift) the residue, namely, the line <hi>EB</hi> is to the residue, namely, to the line <hi>FD,</hi> as the line <hi>AB</hi> is to the line <hi>CD.</hi> But (by supposition) the line <hi>AB</hi> is commensurable in length to the line <hi>CD.</hi> Wherefore (by the 10. of the tenth) the line <hi>AE</hi> is commensurable in length to the line <hi>CF,</hi> and the line <hi>EB</hi> to the line <hi>FD.</hi> And the lines <hi>AE</hi> and <hi>EB</hi> are rationall. Wherefore the lines <hi>CF</hi> and <hi>FD</hi> are also rationall. And for that as the line <hi>AE</hi> is to the line <hi>CF,</hi> so is the line <hi>EB.</hi> to the line <hi>FD,</hi> therefore alternately (by the 16. of the fift) as the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>E</hi> is to the line <hi>EB,</hi> so is the line <hi>CF</hi> to the line <hi>FD.</hi> But the lines <hi>AE</hi> and <hi>EB</hi> are commensura<g ref="char:EOLhyphen"/>ble in power onely, wherefore the lines <hi>CF</hi> and <hi>FD</hi> are also commensurable in power onely, and they are rationall. Wherefore the whole line <hi>CD</hi> is a binomiall line. I say also that it is of the selfe same order of binomiall lines that the line <hi>AB</hi> is. For the line <hi>AE</hi> is in power more then the line <hi>EB</hi> either by the square of a line commensurable in length to the line <hi>AE,</hi> or by the square of a line incommensurable in length to the line <hi>AE.</hi> If the line <hi>AE</hi> be in power more then the line <hi>EB</hi> by the square of a line commensurable in length to the line <hi>AE,</hi> the line also <hi>CF,</hi> by the 14. of the tenth) shalbe in power more then the line <hi>FD</hi> by the square of a line commensurable in length to <hi>CF.</hi> And if the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>AE</hi> be commensurable in length to a rationall line geuen, the line <hi>CF</hi> also shalbe commensurable in length to the same (by the 12. of the tenth). And so either of these lines <hi>AB</hi> and <hi>CD</hi> is a first binomiall line, that <milestone type="tcpmilestone" unit="unspecified" n="1"/> is, they are both of one and the selfe same order. But if the line <hi>EB</hi> be commensurable in length to the rationall line put, the line <hi>FD</hi> also shalbe commensurable in length to the same. And by that meanes agayne the lines <hi>AB</hi> and <hi>CD</hi> are both of one and the selfe same order<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for either of them is a second binomiall line. But if neither of the lines <hi>AE</hi> nor <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> be com<g ref="char:EOLhyphen"/>mensurable <milestone type="tcpmilestone" unit="unspecified" n="2"/> in length to the rationall line put, neither also of these lines <hi>CF</hi> nor <hi>FD</hi> shalbe commensurable in length to the same. And so neither of the lines <hi>AB</hi> and <hi>CD</hi> is a third bi<g ref="char:EOLhyphen"/>nomiall <milestone type="tcpmilestone" unit="unspecified" n="3"/> line. But if the line <hi>AE</hi> be in power more then the line <hi>EB</hi> by the square of a line incommensurable in length to the line <hi>AE,</hi> the line also <hi>CF</hi> shalbe in power more then the line <hi>FD</hi> by the square of a line incommensurable in length to the line <hi>CF,</hi> (by the 14. of
<pb facs="tcp:7062:307"/>
the tenth). And then if the line <hi>AE</hi> be commensurable in length to the rationall line put the line <hi>CF</hi> also shalbe commensurable in length to the same, and so either of the lines <hi>AB,</hi> 
                  <milestone type="tcpmilestone" unit="unspecified" n="4"/> and <hi>CD</hi> shalbe a fourth binomiall line. And if the line <hi>EB</hi> be commensurable in length to the rationall line geuen, the line <hi>FD</hi> also shalbe commensurable in length to the same. And <milestone type="tcpmilestone" unit="unspecified" n="5"/> so either of the lines <hi>AB</hi> and <hi>CD</hi> shalbe a fift binomiall line. But if neither of the lines <hi>AE</hi> nor <hi>EB</hi> be commensurable in length to the rationall line geuen, neither also of the lines <hi>CF</hi> nor <hi>FD</hi> shalbe commensurable in length to the same, and so either of the lines <hi>AB</hi> and <milestone type="tcpmilestone" unit="unspecified" n="6"/>
                  <hi>CD</hi> shalbe a sixt binomiall line. A line, therefore commensurable in length to a binomiall line, is also a binomiall line of the selfe same order: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 49. Theoreme. The 67. Proposition. A line commensurable in length to a bimediall line, is also a bimediall lyne and of the selfe same order.</head>
               <p>SVppose that the line AB be a bimediall line, And vnto the line AB, let the lyne CD be commensurable in length. Then I say that the line CD is a bimediall line, and of the self order that the line AB is. Deuide the line AB into his partes in the point E.<note place="margin">Construction.</note> And forasmuch as the line AB is a bimediall line, and is deuided into his partes in the point E, therfore (by the 37. and 38. of the tenth) the lines AE and EB are medials commensurable in power onely, And (by the 12. of the sixt) as the line AB is to the line CD, so let the line AE be to the line CF.
<figure/>
Wherfore (by the 19. of the fift) the residue,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> name<g ref="char:EOLunhyphen"/>ly, the line EB is to the residue, namely, to the line FD, as the line AB is to the line CD. But the line AB is commensurable in length to the lyne CD. Wherfore the line AE is commensurable in length to the line CF, and the line EB to the line FD. Now the lines AE and EB are me<g ref="char:EOLhyphen"/>diall, wherfore (by the 23. of the tenth) the lines CF and FD are also mediall. And for that as the line AE is to the line EB, so is the line CF to the line FD. But the lines AE and EB are commensurable in power onely, wherfore the lines CF and FD are also commensura<g ref="char:EOLhyphen"/>ble in power onely. And it is proued that they are mediall. Wherfore the lyne CD is a bimedi<g ref="char:EOLhyphen"/>all <milestone type="tcpmilestone" unit="unspecified" n="1"/> line. I say also that it is of the selfe same order that the line AB is. For, for that as the line AE is to the line EB, so is the line CF to the line FD, but as the line CF is to FD, so is the square of the lyne CF to the parallelogramme contained vnder the lynes CF and FD, by the first of the sixt. Therfore as the line AE is to the line EB, so (by the 11. of the fift) is the square of the line CF to the parallelogramme contained vnder the lines CF and FD: but as AE is to EB, so by the 1. of the sixt, is the square of the line AE, to the parallelogramme contained vnder the lines AE and EB, therfore (by the 11. of the fift) as the square of the line AE is to that which is contained vnder the lines AE and EB, so is the square of the line CF to that which is contained vnder the lines CF and FD. Wherfore alternately (by the 16. of the fift) as the square of the line AE is to the square of the line CF, so is that which is contained vnder the lines AE and EB to that which is contained vnder the lines CF &amp; FD. But the square of the line AE is commensurable to the square of the line CF, because AE and CF are commensurable in length. Wherfore that which is contained vnder the lines AE and EB in commensurable to that which is contained vnder the lines CF and FD. If therfore that which is contained vnder the lines AE and EB be rationall, that is, if the line
<pb n="278" facs="tcp:7062:307"/>
AB be a first bimediall line, that also which is contained vnder the lines CF and FD is ra<g ref="char:EOLhyphen"/>tionall. Wherfore also the line CD is a first bimediall line. But if that which is contained vn<g ref="char:EOLhyphen"/>der <milestone type="tcpmilestone" unit="unspecified" n="3"/> the lines AE and EB be mediall, that is, if the line AB be a second bimediall line, that also which is contayned vnder the lines CF and FD is mediall: wherfore also the line CD is a second bimediall line. Wherfore the lines AB and CD are both of one and the selfe same order: which was required to be proued.<milestone type="tcpmilestone" unit="unspecified" n="4"/>
               </p>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Flussates:</hi> but first noted by <hi>P. Monta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>reus.</hi>
                  </head>
                  <p>
                     <q>A line commensurable in power onely to a bimediall line, is also a bimediall line, and of the selfe same order.</q>
                  </p>
                  <p>Suppose that AB be a bimediall line, either a first or a second, wherunto let the line GD be co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in power onely.<note place="margin">
                        <hi>A Coroll<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ry add<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>d by Flussetes.</hi>
                     </note> Take also a rationall line EZ, vpon which (by the 45. of the first) apply a rectan<g ref="char:EOLhyphen"/>gle parallelogramme equall to the square of the line AB, which let be EZFC, and let the rectangle parallelogramme CFIH be equall to the square of the line GD. And forasmuch as vpon the rationall line EZ is applyed a rectangle parallelogramme EF
<figure/>
equall to the square of a first bimediall line, therefore the other side therof, namely, EC, is a second bino<g ref="char:EOLhyphen"/>miall line, by the 61. of this booke. And forasmuch as by supposition the squares of the lines AB &amp; GD are commensurable, therefore the parallelogrammes EF and CI (which are equall vnto them) are also commensurable. And therefore by the 1. of the sixt, the lines EC and CH are commensurable in length. But the line EC is a second binomiall line. Where<g ref="char:EOLhyphen"/>fore the line CH is also a second binomiall line, by the 66. of this booke. And forasmuch as the super<g ref="char:EOLhyphen"/>ficies CI is contayned vnder a rationall line EZ or CF, and a second binomiall line CH, therefore the line which contayneth it in power, namely, the line GD is a first bimediall line, by the 55. of this booke. And so is the line GD in the selfe same order of bi<g ref="char:EOLhyphen"/>mediall lines that the line AB is. The like demonstration also will serue if the line AB be supposed to b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a second bimediall line. For so shall it make the breadth EC a third binomiall line whereunto the line CH shall be commensurable in length, and therefore CH also shall be a third binomiall line, by meanes whereof the line which contayneth in power the superficies CI, namely, the line GD shall also be a second bimediall line. Wherefore a line commensurable either in length, or in power onely to a bimediall line, is also a bimediall line of the selfe same order.</p>
                  <p>But so is it not of necessitie in binomiall lines, for if their powers onely be commensurable, it fol<g ref="char:EOLhyphen"/>loweth not of necessitie that they are binomialls of one and the selfe same order,<note place="margin">
                        <hi>Note.</hi>
                     </note> but they are eche bi<g ref="char:EOLhyphen"/>nomialls eyther of the three first kindes, or of the three last. As for example. Suppose that AB be a first binomiall line, whose greater name let be AG, and vnto AB let the DZ be co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in po<g ref="char:EOLhyphen"/>wer onely. Then I say, that the line DZ is not of the selfe
<figure/>
same order that the line AB is. For if it be possible, let the line DZ be of the selfe same order that the line AB is. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>efore the line DZ may in like sort be deuided as the line AB is, by that which hath bene demonstrated in the 66. Proposition of this booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> let it be so deuided in the poynt E. Wherefore it can not be so deuided in any other poynt, by the 42<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of this booke. And for that the line AB <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> to the line DZ, as the line AG is to the line DE, but the lines AG &amp; DE, namely, the greater names, are com<g ref="char:EOLhyphen"/>mensurable in length the one to the other (by the 10. of this booke) for that they are commensurable in length to <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> and the selfe same rationall line, by the first definition of binomiall lines. Wherefore the lines AB and DZ are commensurable in length, by the 13. of this booke. But by supposition they are commensurable in power onely: which is impossible.</p>
                  <p>
                     <pb facs="tcp:7062:308"/>The selfe same demonstration also will serue, if we suppose the line AB to be a second binomial line: for the lesse names <hi>GB</hi> and EZ being commensurable in length to one and the selfe same ratio<g ref="char:EOLhyphen"/>nall line, shall also be commensurable in length the one to the other. And therefore the lines AB and DZ which are in the selfe same proportion with them, shall also be commensurable in length the one to the other: which is contrary to the supposition. Farther, if the squares of the lines AB and DZ be applyed vnto the rationall line CF, namely, the parallelogrammes CT and HL, they shall make the breadthes CH and HK first binomiall lines, of what order soeuer the lines AB &amp; DZ (whose squares were applyed vnto the rational line) are, (by the 60. of this
<figure/>
booke). Wherefore it is manifest, that vnder a rationall line and a first binomiall line, are confusedly contayned all the powers of binomiall lines (by the 54. of this booke). Wherfore the onely commensuration of the powers doth not of necessitie bryng forth one and the selfe same order of binomiall lines. The selfe same thyng also may be pro<g ref="char:EOLhyphen"/>ued, if the lines AB and DZ be supposed to be a fourth or fifth binomiall line, whose powers onely are conmmensu<g ref="char:EOLhyphen"/>rable, namely, that they shall as the first bring forth bino<g ref="char:EOLhyphen"/>miall lines of diuers orders. Now forasmuch as the pow<g ref="char:EOLhyphen"/>ers of the lines AG and GB, and DE and EZ are com<g ref="char:EOLhyphen"/>mensurable &amp; proportionall, it is manifest, that if the line AG be in power more then the line GB by the square of a line commensurable in length vnto AG, the line DE also shall be in power more then the line EZ by the square of a line commensurable in length vnto the line DE (by the 16. of this booke). And so shall the two lines AB and DZ be eche, of the three first binomiall lines. But if the line AG be in power more then the line GB by the square of a line incommensurable in length vnto the line AG, the line DE shall also be in pow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> then the line EZ by the square of a line incomensurable in length vnto the line DE, by the self<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> same P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oposition. And so shall eche of the lines AB and DZ be of the three last binomiall lines. But why it is not so in the third and sixt binomiall lines, the reason is: For that in them neither of the name<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is commensurable in length to the rationall line put FC.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 50. Theoreme. The 68. Proposition. A line commensurable to a greater line, is also a greater line.</head>
               <p>SVppose that the line AB be a greater line. And vnto the line AB let the line CD be commensurable.<note place="margin">Construction.</note> Then I say that the line CD also is a greater line. Deuide the line AB into his partes in the point E. Wherfore (by the 39. of the tenth) the lines AE and EB are incommensurable in power, hauing that which is made of the squares of them added together rationall, and that which is contained vnder the<g ref="char:cmbAbbrStroke">̄</g> me<g ref="char:EOLhyphen"/>diall. And let the rest of the construction be in this, as it was in the former. And for that as the line AB is to the line CD,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so
<figure/>
is the line AE to the line CF, &amp; th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line EB to the line FD, but the line AB is commensurable to the line CD by suppositio<g ref="char:cmbAbbrStroke">̄</g>. Wher<g ref="char:EOLhyphen"/>fore the line AE is commensura<g ref="char:EOLhyphen"/>ble to the line CF, and the line EB to the line FD. And for that as the line AE is to the line CF, so is the line EB to the line FD. Therfore alternately (by the 16. of the fift) as the line AE is to the line EB, so is the line CF to the line FD. Wherfore by composition also (by the 18. of the fift) as the line AB is to the line EB, so is the line CD to the line FD. Where<g ref="char:EOLhyphen"/>fore (by the 22. of the sixt) as the square of the line AB is to the square of the line EB, so is the square of the line CD to the square of the line FD. And in like sort may we proue that as the square of the line AB is to the square of the line AE, so is the square of the line CD, to the square of the line CF. Wherfore (by the 11. of the fift) as the square of the lyne AB is to the squares of the lines AE and EB, so is the square of the line CD to the squares of the lines CF and FD. Wherfore alternately (by the 16. of the fift) as the square of the line AB
<pb n="279" facs="tcp:7062:308"/>
is to the square of the line CD, so are the squares of the lines AE and EB to the squares of the lines CF and FD. But the square of the line AB is commensurable to the square of the line CD (for the line AB is commensurable to the line CD by suppositio<g ref="char:cmbAbbrStroke">̄</g>). Wherfore also the squares of the lines AE and EB are commensurable to the squares of the lines CF and FD. But the squares of the lines AE and EB are incommensurable, and being added together are rationall. Wherfore the squares of the lines CF and FD are incommensurable, &amp; being added together, are also rationall. And in like sort may we proue that that which is contained vnder the lines AE and EB twise is commensurable to that which is contained vnder the lines CF and FD twise. But that which is contained vnder the lines AE and EB twise, is mediall, wherfore also that which is contained vnder the lines CF and FD twise is medial. Wherfore the lines CF and FD are incommensurable in power, hauing that which is made of the squares of them added together rationall, and that which is contained vnder the<g ref="char:cmbAbbrStroke">̄</g> me<g ref="char:EOLhyphen"/>diall. Wherfore (by the 39. of the tenth) the whole line CD is irrationall, &amp; is called a grea<g ref="char:EOLhyphen"/>ter line. A line therfore commensurable to a greater line, is also a greater line.</p>
               <div type="section">
                  <head>An other demonstration of <hi>Peter Montaureus</hi> to proue the same.</head>
                  <p>Suppose that the line AB be a greater line, and vnto it let the line CD be commensurable any way, that is, either both in length and in power, or els in power onely.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after P. Montaureus.</hi>
                     </note> Then I say that the line CD also is a greater line. Deuide the line AB into his partes in the point E. and let the rest of the construction be in this as it was in the former. And for that as the line AB is to the line CD, so is the line AE to the lyne CF, and the line EB to the line FD,
<figure/>
therfore as the line AE is to the lyne CF, so is the line EB to the line FD, but the line AB is commensurable to the line CD. Wherfore also the lyne AE is commensurable to the lyne CF, and likewise the line EB to the line FD. And for th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> as the line AE is to the line CF, so is the line EB to the line FD, therfore alternately as the lyne AE is to the line EB, so is the line CF to the lyne FD. Wherfore (by the 22. of the sixt) as the square of the lyne AE is to the square of the line EB, so is the square of the line CF to the square of the line FD. Wherfore by composition (by the 18. of the fift) as that which is made of the squares of the lynes A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> added together is to the square of the lyne EB, so is that which is made of the square<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the lyne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and FD added together to the square of the lyne FD. Wherefore by contrary proportion as the square of the line EB is to that which is made of the squares of the lines AE and E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> added together, so is the square of the lyne FD to that which is made of the squares of the lynes CF and FD added together. Wherfore alternately as the square of the line EB is to the square of the lyne FD, so is that which is made of the squares <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nes AE and EB added together to that whiche is made of the squares of the lynes CF and FD added together. But the square of the lyne EB is co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable to the square of the lyne FD, for it hath already bene proued that the lines EB and FD are co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>surable. Wherfore that which is made of the squares of the lines AE &amp; EB added together is com<g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>surable to that which is made of the squares of C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp; FD added together. But that which is made of the squares of the lines AE and EB added together is rationall by suppositio<g ref="char:cmbAbbrStroke">̄</g>. Wherfore that which is made of the squares of the lynes CF and FD added together is also rationall. And as the lyne AE is to the lyne EB, so is the line CF to the lyne FD: But as the lyne AE is to the lyne EB, so is the square of the line A <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> contayned vnder the lynes AE and EB: therfore at the lyne CF is to the lyne FD, so is the square of the lyne AE to the parallelogramme contayned vnder the lines AE and EB &amp; as the lyne CF is to the lyne FD, so is the square of the lyne CF to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contayned vnder the lynes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>F &amp; FD. Wherfore as the square of the lyne AE is to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me con<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>ed vnder the lines AE and EB, so is the square of the lyne CF to the parallelogramme co<g ref="char:cmbAbbrStroke">̄</g>tay<g ref="char:EOLhyphen"/>ned vnder the lynes CF and FD. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s the square of the line AE is to the square of the lyne CF, so is the parallelogramme contained vnder the lynes AE and EB to the parallelogramme <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> vnde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the lines <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>. But the square of the lyne AE is commensurable to the square of the lyne CF, for it is already pr<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>d that the lynes AE and CF are comme<g ref="char:cmbAbbrStroke">̄</g>surable. Wherefore the parallelogramme contayned vnder the lynes AE and EB is commensurable to the parallelogramme contayned vnder the lynes CF and FD. But the parallelogramme contayned vnder the lines AE and EB is mediall by suppo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ition. Wherfore the parallelogramme contayned vnder the lynes CF and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D al<g ref="char:EOLhyphen"/>so is mediall. And (as it hath already bene proued) as the line AE is to the lyne EB, so is the lyne CF to the lyne FD. But the lyne AE was by supposition incommensurable in power to the line EB. Wher<g ref="char:EOLhyphen"/>fore (by the 10. of the tenth) the lyne CF is incommensurable in power to the lyne FD. Wherfore the
<pb facs="tcp:7062:309"/>
lynes CF and FD are incommensurable in power, hauing that which is made of the squares of them added together rationall, and that which is contayned vnder them mediall. Wherfore the whole lyne CD is (by the 39. of the tenth) a greater lyne. Wherfore a lyne commensurable to a greater lyne is also a greater lyne: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>An other more briefe demonstration of the same after Campane.</head>
                  <p>Suppose that A be a greater line, vnto which let the line B be
<figure/>
comme<g ref="char:cmbAbbrStroke">̄</g>surable, either in length and power, or in power onely.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration after Campane.</note> And take a rational line CD. And vpon it apply the superficies C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall to the square of the line A: and also vpo<g ref="char:cmbAbbrStroke">̄</g> the line FE (which is equall to the rationall line CD) apply the parallelogramme FG equall to the square of the line B. And forasmuch as the squares of the two lines A and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are commensurable by supposition, the superficies C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, shalbe commensurable vnto the superficies FG: and therefore by the first of the sixt and tenth of this booke, the line DE is commen<g ref="char:EOLhyphen"/>surable in length to the line GB. And forasmuch as (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>3. of this booke) the line DE is a fourth binomiall line, therefore by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>6. of this booke the line GE is also a fourth binomiall line: where<g ref="char:EOLhyphen"/>fore by the 57. of this booke the line B which contayneth in power the superficies FG is a greater line.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 51. Theoreme. The 69. Proposition. A line commensurable to a line contayning in power a rationall and a me<g ref="char:EOLhyphen"/>diall: is also a line contayning in power a rationall and a mediall.</head>
               <p>SVppose that <hi>AB</hi> be a line contayning in power a rationall and a mediall. And vnto the line <hi>AB</hi> let the line <hi>CD</hi> be commensurable, whether in length and power, or in power onely. The<g ref="char:cmbAbbrStroke">̄</g> I say that the line <hi>CD</hi> is a line co<g ref="char:cmbAbbrStroke">̄</g>tayning in power a rationall &amp; a mediall. Duide the line <hi>AB</hi> into his parts in the poynt <hi>E.</hi>
                  <note place="margin">Construction.</note> Wher<g ref="char:EOLhyphen"/>fore (by the 40. of the tenth) the lines <hi>AE</hi> and <hi>EB</hi>
                  <figure/>
are incommensurable in power, hauing that which is made of the squares of them added together medi<g ref="char:EOLhyphen"/>al, and that which is contayned vnder the<g ref="char:cmbAbbrStroke">̄</g> nationall. Let the same construction be in this that was in the former.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And in like sort we may proue that the lines <hi>CF</hi> and <hi>FD</hi> are incommensurable in power, and that that which is made of the square of the lines <hi>AE</hi> and <hi>EB</hi> is commensurable to that which is made of the squares of the lines <hi>CF</hi> and <hi>FD,</hi> and that that also which is contayned vnder the lines <hi>AE</hi> and <hi>EB</hi> is comme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable to that which is contayned vnder the lines <hi>CF</hi> and <hi>FD.</hi> Wherefore that which is made of the squares of the lines <hi>CF</hi> and <hi>FD</hi> is mediall, and that which is contayned vnder the lines <hi>CF</hi> and <hi>FD</hi> is rationall. Wherefore the whole line <hi>CD</hi> is a line contayning in power a rationall and a mediall: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration of the same after <hi>Campane.</hi>
                  </head>
                  <p>
                     <pb n="280" facs="tcp:7062:309"/>Supose that <hi>AB</hi> be a line contayning in pow<g ref="char:EOLhyphen"/>er
<figure/>
a rationall and a mediall:<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>r Campane.</hi>
                     </note> whereunto let the line <hi>GD</hi> be commensurable either in length and pow<g ref="char:EOLhyphen"/>er, or in power onely. Then I say that the line <hi>G</hi>D is a line contayning in power a rationall and a me<g ref="char:EOLhyphen"/>diall. Take a rational line <hi>EZ,</hi> vpo<g ref="char:cmbAbbrStroke">̄</g> which by the 45. of the first apply a rectangle parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>EZFC</hi> equall to the square of the line <hi>AB</hi>: and vpon the line <hi>CF</hi> (which is equall to the line <hi>EZ</hi>) applye the parallelogramme <hi>FCHI</hi> equall to the square of the line <hi>G</hi>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and let the breadths of the sayd parallelogrammes be the lines <hi>EG</hi> and <hi>CH.</hi> And forasmuch as the line <hi>AB</hi> is commensurable to the line <hi>GD</hi> at the least in power onely, therefore the parallelogrammes <hi>EF</hi> and <hi>FH</hi> (which are equall to their squares) shalbe commensurable. Where<g ref="char:EOLhyphen"/>fore by the 1. of the sixt the right lines <hi>EC</hi> and <hi>CH</hi> are co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth. And forasmuch as the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>EF</hi> (which is equall to the square of the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> which contayneth in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> ratio<g ref="char:EOLhyphen"/>nall and a mediall) is applyed vpon the rationall <hi>EZ,</hi> making in breadth the line <hi>EC,</hi> therefore the line <hi>EC</hi> is a fifth binomiall line (by the 64. of this booke) vnto which line <hi>EC</hi> the line <hi>CH</hi> is co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in length, wherefore by the 66. of this booke the line <hi>CH</hi> is also a fifth binomiall line. And forasmuch as the superficies <hi>CI</hi> is contayned vnder the rationall line <hi>EZ</hi> (that is <hi>CF</hi>) and a fifth binomall line <hi>CH,</hi> therefore the line which contayneth in power the superficies <hi>CI,</hi> which by supposition is the line <hi>GD</hi> is a line contayning in power a rationall and a mediall by the 58. of this booke. A line therefore com<g ref="char:EOLhyphen"/>mensurable to a line contayning in power a rationall and a mediall. &amp;c.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 52. Theoreme. The 70. Proposition. A line commensurable to a line contayning in power two medialls, is also a line contayning in power two medialls.</head>
               <p>SVppose that AB be a line contayning in power two medialls. And vnto the line AB let the line CD be commensurable, whether in length &amp; power, or in power onely. Then I say, that the line CD is a line contayning in power two medialls. Forasmuch as the line AB is a line contayning in power two medialls,<note place="margin">Construction.</note> let it be deuided into his partes in the point E. Wherefore (by the 41. of the tenth) the lines AE and EB are incommensurable in power, hauing that which is made of the squares of them added together mediall, and that also which is contained vnder them mediall, and that which is made of the squares of the lines AE &amp;
<figure/>
EB is incommensurable to that which is contained vnder the lines AE and EB. Let the selfe same construction be in this, that was in the former.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And in like sort may we proue, that the lines CF &amp; FD are incommensurable in power, and that that which is made of the squares of the lines AE and EB added together, is commensurable to that which is made of the squares of the lines CF and FD added together, and that that also which is contained vnder the lines AE and EB is commensurable to that which is con<g ref="char:EOLhyphen"/>tained vnder the lines CF and FD. Wherefore that which is made of the squares of the lines CF and FD is mediall (by the Corollary of the 23. of the tenth): and that which is contayned vnder the lines CF and FD is mediall (by the same Corollary)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and moreouer, that which is made of the squares of the lines CF &amp; FD is incommensurable to that which is contained vnder the lines CF and FD. Wherefore the line CD is a line containing in power two medialls: which was required to be proued.</p>
               <div type="section">
                  <pb facs="tcp:7062:310"/>
                  <head>¶ An Assumpt added by <hi>Montaureus.</hi>
                  </head>
                  <p>
                     <note place="margin">An Assumpt.</note>That that which is made of the squares of the lines CF and FD added together, is in<g ref="char:EOLhyphen"/>commensurable to that which is contained vnder the lines CF and FD is thus proued. For, because as that which is made of the squares of the lines AE and EB added together is to the square of the line AE, so is that which is made of the squares of the lines CF and FD ad<g ref="char:EOLhyphen"/>ded together, to the square of the line CF, as it was proued in the Propositions going before: therefore alternately, as that which is made of the squares of AE and EB added together is to that which is made of the squares of CF and FD added together, so is the square of the line AE to the square of the line CF. But before, namely, in the 68. Proposition, it was pro<g ref="char:EOLhyphen"/>ued, that as the square of the line AE is to the square of the line CF, so is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contained vnder the lines AE and EB to the parallelogramme contained vnder the lines CF and FD. Wherefore as that which is made of the squares of the lines AE and EB is to that which is made of the squares of the lines CF and FD, so is the parallelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines AE and EB to the parallelogramme contained vnder the lines CF and FD. Wherefore alternately, as that which is made of the squares of the lines AE and EB is to the parallelogramme contained vnder the lines AE and EB, so is that which is made of the squares of the lines CF and FD to the parallelogramme contained vnder the lines CF and FD. But by supposition that which is made of the squares of the lines AE and EB, is incomm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nsurable to the parallelogramme contained vnder the lines AE &amp; EB. Wherefore that which is made of the squares of the lines CF and FD added together, is in<g ref="char:EOLhyphen"/>commens<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rable to the parallelogramme contained vnder the lines CF and FD: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>An other demonstration after <hi>Campane.</hi>
                  </head>
                  <p>Suppose that AB be a line contayning in power two medialls: wherunto let the line GD be com<g ref="char:EOLhyphen"/>mensurable either in length, and in power, or in power onely.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Cam<g ref="char:EOLhyphen"/>pan<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.</hi>
                     </note> Then I say, that the line GD is a line co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tayning in power two medialls. Let the same construction be in this, that was in the former. And for<g ref="char:EOLhyphen"/>asmuch as the parallelogramme EF is equall to the
<figure/>
square of the line AB, and is applyed vpon a ratio<g ref="char:EOLhyphen"/>nall line EZ, it maketh the breadth EC a sixt bino<g ref="char:EOLhyphen"/>miall line, by the 65. of this booke. And forasmuch as the parallelogrammes EF &amp; CI (which are equall vnto the squares of the lines AB and GD, which are supposed to be commensurable) are commensurable, therefore the lines EC and CH are commensurable in length, by the first of the sixt. But EC is a sixt bi<g ref="char:EOLhyphen"/>nomiall line: Wherefore CH also is a sixt binomiall line, by the 66. of this booke. And forasmuch as the superficies CI is contayned vnder the rationall line CF and a sixt binomiall line CH, therefore the line which co<g ref="char:cmbAbbrStroke">̄</g>tayneth in power the superficies CI, name<g ref="char:EOLhyphen"/>ly, the line GD is a line contayning in power two medialls, by the 59. of this booke. Wherefore a line commensurable to a line contayning in power two medialls. &amp;c.</p>
               </div>
               <div type="section">
                  <head>An Annotation.</head>
                  <p>If other to hath bene spoken of sixe <hi>Senarys,</hi> of which the first Senary contayneth the pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>duction of irrationall lines by composition: the second, the diuision of them, namely, that those lines are in one poin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> onely deuide<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the third, the finding out of binomiall lines, of the first, I say, the second, the third, the fourth, the fift, and the sixt: after that beginneth the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ourth Senary, containing the difference of irrationall lines betwene them selues. For by the
<pb n="281" facs="tcp:7062:310"/>
nature of euery one of the binomiall lines are demonstrated the differences of irrational lines. The fiueth entre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>teth of the applications of the squares of euery irrational line, namely, what irrationall lines are the breadthes of euery superficies so applied. In the sixt Senary is proued, that any line commensurable to any irrationall line, is also an irrationall line of the same na<g ref="char:EOLhyphen"/>ture. And now shall be spoken of the seuenth Senary, wherein againe are plainly set forth the rest of the differences of the said lines betwene them selues.</p>
                  <p>And the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e is euen in those irrationall lines an arithmeticall proportionalitie.<note place="margin">Note.</note> And that line which is the arithmeticall meane proportionall betwene the partes of any irrationall line, is also an irrationall line of the selfe same kinde. First it is certaine that there is an arith<g ref="char:EOLhyphen"/>meticall proportion betwene those partes. For suppose that the line AB be any of the foresaid irrationall lines, as for example, let it be a binomiall line, &amp; let it be deuided into his names in the point C. And let AC be the greater name, from which take away the line AD equall to the lesse name, namely, to CB. And deuide the line CD into two equall partes in the point E. It is manifest that the line AE is equall to the
<figure/>
line EB. Let the line FG be equall to either of them. It is plaine that how much the line AC dif<g ref="char:EOLhyphen"/>fereth fro<g ref="char:cmbAbbrStroke">̄</g> the line FG, so much the same line FG di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ereth from the line CB: for in eche is the diffe<g ref="char:EOLhyphen"/>rence of the line DE or EC, which is the propertie of arithmeticall proportionalitie. And it is manifest, that the line FG is commensurable in length to the line AB, for it is the halfe thereof. Wherefore (by the 66. of the tenth) the line FG is a binomiall line. And after the selfe same maner may it be proued touching the rest of the irrationall lines.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 53. Theoreme. The 71. Proposition. If two superficieces, namely, a rationall and a mediall superficies be co<g ref="char:cmbAbbrStroke">̄</g>posed together, the line which contayneth in power the whole superficies, is one of these foure irrationall lines, either a binomial line, or a first bimediall lyne, or a greater lyne, or a lyne contayning in power a rationall and a mediall superficies.</head>
               <p>SVppose that there be two superficieces AB and CD,<note place="margin">The seuent<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Senary.</note> of which let the superficies AB be rationall, &amp; the superficies CD mediall. The<g ref="char:cmbAbbrStroke">̄</g> I say that the lyne contay<g ref="char:EOLhyphen"/>ning in power y<hi rend="sup">•</hi> whole superficies AD, is either a binomial line, or a first bimedi<g ref="char:EOLhyphen"/>all line, or a greater line,<note place="margin">Construction.</note> or a line co<g ref="char:cmbAbbrStroke">̄</g>tayning in power a rationall &amp; a mediall su<g ref="char:EOLhyphen"/>perficies. For the superficies AB is either greater or lesse the<g ref="char:cmbAbbrStroke">̄</g> the superficies CD for they ca<g ref="char:cmbAbbrStroke">̄</g> by no meanes be equall, whe<g ref="char:cmbAbbrStroke">̄</g> as the one is rationall, and the other medial). First let it be greater,<note place="margin">
                     <p>Two cases in this propositi<g ref="char:EOLhyphen"/>on, of which ech also hath his two cases.</p>
                     <p>First case.</p>
                  </note> and take a rationall line EF. And (by the 44. of the first) vnto the line EF apply the paral<g ref="char:EOLhyphen"/>lelogramme EG equal to the superficies AB, &amp; ma<g ref="char:EOLhyphen"/>king
<figure/>
in breadth the line EH, and to the same line EF, that is, to the line HG apply the parallelogramme HI equall to the superficies DC, and makyng in breadth the line HK.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the super<g ref="char:EOLhyphen"/>ficies AB is rationall, and is equall to the parallelo<g ref="char:EOLhyphen"/>gramme EG, therfore the parallelogramme EG is also rational, and it is applied vnto the rational line EF, making in breadth the line EH. Wherfore the line EH is rationall and commensurable in length to the line EF (by the 20. of the tenth). Againe for<g ref="char:EOLunhyphen"/>asmuch as the superficies CD is mediall, and is e<g ref="char:EOLhyphen"/>qual
<pb facs="tcp:7062:311"/>
to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me HI, therfore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me HI is also mediall, and is appli<g ref="char:EOLhyphen"/>ed vnto the rationall line EF, that is, vnto the lyne HG making in breadth the line HK. Wherfore the lyne HK is rationall and incommensurable in length to the line EF (by the 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth.) And forasmuch as the superficies CD is mediall, and the superficies AB is ra<g ref="char:EOLhyphen"/>tionall, therfore the superficies AB is incommensurable to the superficies CD. Wherfore also the parallelogramme EG is incommensurable to the parallelogramme HI. But as the paral<g ref="char:EOLhyphen"/>lelogramme GE is to the parallelogramme HI, so (by the 1. of the sixt) is the line EH to the lyne HK. Wherfore (by the 10. of the tenth) the line EH is incommensurable in length to the line HK, and they are both rationall. Wherfore the lines EH and HK are rationall com<g ref="char:EOLunhyphen"/>mensurable in power onely. Wherfore the whole line EK is a binomiall line, and is deuided into his names in the poynt H. And forasmuch as the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies AB is greater then the su<g ref="char:EOLhyphen"/>perficies CD, but the superficies AB is equall to the parallelogramme EG, and the superficies CD to the parallelogramme HI. Wherfore the parallelogramme EG is greater then the pa<g ref="char:EOLhyphen"/>rallelogramme HI. Wherfore the line EH is greater then the line HK. Wherfore the line EH is in power more then the line HK either by the square of a line commensurable in length to the lyne EH, or by the square of a lyne incommensurable in length to the lyne EH.<note place="margin">First part of the first case.</note> First let it be in power more by the square of a lyne co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth vnto the line EH. Now the greater name, namely, EH is commensurable in length to the rational line geuen EF, as it hath already bene proued. Wherfore the whole line
<figure/>
EK is a first binomiall lyne. And the lyne EF is a rationall lyne. But if a superficies be contayned vn<g ref="char:EOLhyphen"/>der a rationall line, and a first binomiall lyne, the lyne that contayneth in power the same superficies, is (by the 54. of the tenth) a binomiall line. Wherefore the lyne containing in power the parallelogramme EI is a binomiall line. Wherefore also the line contai<g ref="char:EOLhyphen"/>ning in power the superficies AD is a binomiall line.</p>
               <p>But now let the lyne EH be in power more then the line HK by the square of a line incommensurable in length to the line EH:<note place="margin">Second part of the first case.</note> now the greater name that is, EH is commensurable in length to the rationall line geuen EF. Wher<g ref="char:EOLhyphen"/>fore the line EK is afourth binomiall line. And the line EF is rationall. But if a superfi<g ref="char:EOLhyphen"/>cies be contained vnder a rationall line and afourth binomiall line, the line that containeth in power the same superficies is (by the 57. of the tenth) irrational, and is a greater line. Wher<g ref="char:EOLunhyphen"/>fore the line which containeth in power the parallelogramme EI is a greater line. Wherefore also the line containing in power the superficies AD is a greater lyne.</p>
               <p>
                  <note place="margin">The second case.</note>But now suppose that the superficies AB which is rationall, be lesse then the superficies CD which is mediall. Wherfore also the parallelogramme EG is lesse then the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me HI. Wherfore also the line EH is lesse then the line HK. Now the line HK is in power more then the lyne EH either by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line HK, or by the square of a lyne incommensurable in length vnto the lyne HK. First let it be in power more by the square of a line commensurable in length vnto HK:<note place="margin">First part of the seco<g ref="char:cmbAbbrStroke">̄</g>d case.</note> now the lesse name, that is EH is commensurable in length to the rationall line geuen EF, as it was before proued. Wherfore the whole line EK is a second binomiall line. And the line EF is a rationall line. But if a superficies be contained vnder a rationall line and a second binomiall lyne, the lyne that contayneth in power the same superficies, is (by the 55. of the tenth) a first bimediall line. Wherfore the line which contayneth in power the parallelograme EI is a first bimediall line. Wherfore also the line that containeth in power the superficies AD is a first bimediall lyne.</p>
               <p>
                  <note place="margin">Second part of the seco<g ref="char:cmbAbbrStroke">̄</g>d case.</note>But now let the line HK be in power more then the line EH, by the square of a line in<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>mensurable
<pb n="282" facs="tcp:7062:311"/>
in length to the lyne HK, now the lesse name, that is, EH is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the rationall lyne geuen EF. Wherfore the whole line EK is a fift binomiall lyne. And the lyne EF is rationall. But if a superficies be contayned vnder a rationall lyne, and a fift binomiall lyne, the line that contayneth in power the same superficies, is (by the 58. of the tenth) a line containing in power a rationall and a mediall. Wherefore the lyne that contay<g ref="char:EOLhyphen"/>neth in power the parallelogramme EI is a line contayning in power a rationall and a medi<g ref="char:EOLhyphen"/>all. Wherfore also the lyne that containeth in power the superficies AD is a lyne contayning in power a rationall and a mediall. If therfore a rationall and a mediall superficies be added together, the lyne which contayneth in power the whole superficies, is one of these foure irrati<g ref="char:EOLhyphen"/>onall lines, namely, either a binomiall line, or a first bimediall line, or a greater lyne, or a lyne contayning in power a rationall and a mediall: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 54. Theoreme. The 72. Proposition. If two mediall superficieces incommensurable the one to the other be com<g ref="char:EOLhyphen"/>posed together: the line contayning in power the whole superficies is one of the two irrationall lines remayning, namely, either a second bimediall line, or a line contayning in power two medialls.</head>
               <p>LEt these two mediall superficieces AB and CD being incommensurable the one to the other be added together. Then I say, that the line which contayneth in power the superficies AD is either a second bimediall line, or a line contayning in power two medialls.<note place="margin">Construction.</note> For the superficies AB is either greater or lesse then the superficies CD (for they can by no meanes be equall, when as they are incom<g ref="char:EOLhyphen"/>mensurable).<note place="margin">
                     <p>Two cases in this Proposi<g ref="char:EOLhyphen"/>tion.</p>
                     <p>The first case.</p>
                  </note> First let the superficies AB be greater then
<figure/>
the superficies CD. And take a rationall line EF. And (by the 44. of the first) vnto the line EF apply the paralle<g ref="char:EOLhyphen"/>logramme EG equall to the superficies AB, and making in breadth the line EH: and vnto the same line EF, that is, to the line HG, apply the parallelogramme HI equall to the superficies CD, &amp; making in breadth the line HK. And forasmuch as either of these superficieces AB &amp; CD is mediall, therefore also either of these parallelogrammes EG and HI is mediall. And they are eche applied to the rationall line EF, making in breadth the lines EH and HK. Wherefore (by the 22. of the tenth) either of these lines EH and HK is rationall and incommensurable in length to the line EF. And forasmuch as the superficies AB is incommensurable to the superficies CD, and the superficies AB is equall to the parallelogramme E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and the superficies CD to the parallelogramme HI: therefore the parallelogramme EG is incommensurable to the pa<g ref="char:EOLhyphen"/>rallelogramme HI. But (by the 1. of the sixt) as the parallelogramme EG is to the paralle<g ref="char:EOLhyphen"/>logramme HI, so is the line EH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the line HK. Wherefore (by the 10. of the tenth) the line EH i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>spans<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> HK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>nes EH <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd HK are rationall commensurable in power onely. Wherfore the whole line EK is a binomiall line. And as in the former Proposition so als<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in this may it be proued, that the line EH is greater then the line HK. Wherefore the line EH is in power more then the line HK, either by the square of a line commensurable in length to the line EH, or by the square of a line incom<g ref="char:EOLhyphen"/>mensurable in length to the line EH. First let it be greater by the square of a line commen<g ref="char:EOLhyphen"/>surable in length vnto the line EH.<note place="margin">The first part of the first case.</note> Now neither of these lines EH and HK is commensu<g ref="char:EOLhyphen"/>rable in length to the rationall line geuen EF. Wherefore the whole line EK is a third bi<g ref="char:EOLhyphen"/>nomiall
<pb facs="tcp:7062:312"/>
line. And the line EF is a rationall line. But if a superficies be contayned vnder a rationall line &amp; a third binomiall line, the line that co<g ref="char:cmbAbbrStroke">̄</g>taineth in power the same superficies, is (by the 56. of the tenth) a second bimediall line. Wherefore the line that containeth in power the superficies EI, that is, the superficies AD, is a second bimediall line.</p>
               <p>
                  <note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> second <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> case.</note>But now suppose that the line EH be in power more then the line HK by the square of a line incommensurable in length to the line EH. And forasmuch as either of these lines EH and HK is incommensurable in length to the rationall line geuen EF, therfore the line EK is a sixt binomiall line. But if a superficies be contained vnder a rationall line and a sixt binomiall line, the line that containeth in power the same superficies, is (by the 59. of the tenth) a line containing in power two medialls. Where<g ref="char:EOLhyphen"/>fore
<figure/>
the line that containeth in power the superficies AD, is a line contayning in power two medialls.<note place="margin">The second case.</note> And after the selfe same maner, if the superficies AB be lesse then the su<g ref="char:EOLhyphen"/>perficies CD, may we proue, that the line that contayneth in power the superficies AD, is either a second bimediall line, or a line containing in power two medialls. If there<g ref="char:EOLhyphen"/>fore two mediall superficieces incommensurable the one to the other be added together, the line contayning in power the whole superficies is one of the two irrationall lines re<g ref="char:EOLhyphen"/>mayning, namely, either a second bimediall line, or a line co<g ref="char:cmbAbbrStroke">̄</g>taining in power two medialls: which was required to be proued.</p>
               <div type="section">
                  <head>¶ A Corollary following of the former Propositions.</head>
                  <p>A binomiall line and the other irrationall lines following it, are neither mediall lines, nor one and the same betwene them selues. For the square of a mediall line applied to a rati<g ref="char:EOLhyphen"/>onall line,<note place="margin">A Corollary</note> maketh the breadth rationall and incommensurale in length to the rationall line, wherunto it is applied (by the 22. of the tenth). The square of a binomiall line applyed to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> rationall line, maketh the breadth a first binomiall line (by the 60. of the tenth). The square of a first bimediall line applied vnto a rationall line, maketh the breadth a second binomiall line (by the 61. of the tenth). The square of a second bimediall line applied vnto a rationall line, maketh the breadth a third binomiall line (by the 62. of the tenth). The square of a greater line applied to a rationall line, maketh the breadth a fourth binomiall line (by the 63. of the tenth). The square of a line containing in power a rationall &amp; a mediall superficies, maketh the breadth a fift binomiall line (by the 64. of the tenth). And the square of a line containing in power two medialls, applied vnto a rationall line, maketh the breadth a sixt binomiall line (by the 65. of the tenth). Seing therefore that these foresaid breadthes differ both from the first breadth, for that it is rationall, and differ also the one from the other, for that they are binomials of diuers orders: it is manifest that those irrationall lines differ also the one from the other.</p>
               </div>
            </div>
            <div type="part">
               <head>Here beginneth the Senaries by substraction. ¶The 55. Theoreme. The 73. Proposition. If from a rationall line be taken away a rationall line commensurable in power onely to the whole line: the residue is an irrationall line, and is called a residuall line.</head>
               <p>
                  <pb n="283" facs="tcp:7062:312"/>SVppose that AB be a rationall line, and from AB take away a rationall line BC commensurable in power onely to the whole line AB.<note place="margin">The first Se<g ref="char:EOLhyphen"/>nary by sub<g ref="char:EOLhyphen"/>straction.</note> Then I say that the line re<g ref="char:EOLhyphen"/>mayning, namely AC is irrationall and is called a residuall line. For forasmuch as the line AB is incommensurable in length vnto the line BC, and (by the as<g ref="char:EOLhyphen"/>sumpt going before the 22. of the tenth) as the line
<figure/>
AB is to the line BC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the square of the line AB to that which is contayned vnder the lines AB and BC: wherefore (by the 10. of the tenth) the square of the line AB is incommensurable to that which is contayned vnder the lines AB and BC. But vnto the square of the line AB are commensurable the squares of the lines AB and BC (by the 15. of the tenth). Wherefore the squares of the lines AB and BC are in<g ref="char:EOLhyphen"/>commensurable to that which is contayned vnder the lines AB and BC. But vnto that which is contayned vnder the lines AB and BC is commensurable that which is contayned vnder the lines AB and BC twise. Wherefore the squares of the lines AB and BC are inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>sura<g ref="char:EOLhyphen"/>ble to that which is contayned vnder the lines AB and BC twise. But the squares of the lines AB and BC are equall to that which is contayned vnder the lines AB and BC twise, and to the square of the line AC (by the 7. of the second). Wherefore that which is contayned vnder the lines AB and BC twise together with the square of the line AC is incommensurable to that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AB and BC twise. Wherefore (by the 2 part of the 16. of the te<g ref="char:cmbAbbrStroke">̄</g>th) that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AB and BC twise, is inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable to the square of the line AC. Wherefore (by the first part of the same) that which is contayned vn<g ref="char:EOLhyphen"/>der the lines AB and BC twise together with the square of the line AC, that is, the squares of the lines AB and BC are incommensurable to the square of the line AC. But the squares of the lines AB and BC are rationall, for the lines AB and BC are put to be rationall: wher<g ref="char:EOLhyphen"/>fore the line AC is irrationall and is called a residuall line: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration after <hi>Campane.</hi>
                  </head>
                  <p>
                     <hi>Campane</hi> demonstrateth this Proposition by a figure more briefly after
<figure/>
this m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ner.<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Cam<g ref="char:EOLhyphen"/>pane.</hi>
                     </note> Let the superficies EG be equall to the squares of the lines AB and BC added together: which shall be rationall (for that the lines AB and BC are supposed to be rationall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely). Fro<g ref="char:cmbAbbrStroke">̄</g> which superficies take away the superficies DF equall to that which is con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ya<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d vnder the lines AB &amp; DC twise, which shall be mediall (by the 21. of this booke). Now by the 7. of the second, the superficies FG is equall to the square of the line AC. And forasmuch as the superficies EG is incommensurable to the superficies DF (for that the one is rati<g ref="char:EOLhyphen"/>onall and the other mediall): therefore (by the 16. of this booke) the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> superficies EG is incommensurable to the superficies FG. Wher<g ref="char:EOLhyphen"/>fore the superficies FG is irrationall. And therefore the line AC which contayneth it in power is irrationall: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>An annotation of <hi>P. Monta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s.</hi>
                  </head>
                  <p>This Theoreme teacheth nothing els but that that portion of the greater name of a binomiall line which remayneth after the taking away of the lesse name from the grea<g ref="char:EOLhyphen"/>ter name is irrationall, which is called a residuall line, that is to say, if from the greater name of a binomiall line, which greater name is a rationall line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely to the lesse name, be taken away the lesse name, which selfe lesse name is also com<g ref="char:EOLhyphen"/>mensurable in power onely to the greater name (which greater name this Theoreme calleth the whole line) the rest of the line which remaineth is irrational, which he calleth
<pb facs="tcp:7062:313"/>
a residuall line. Wherfore all the lines which are intreated in this Theoreme, and in the fiue other which follow are the portions remayning of the greater partes of the whole lines which were intreated of in the 36.37.38.39.40.41. propositio<g ref="char:cmbAbbrStroke">̄</g>s, after the taking away the lesse part from the greater.</p>
                  <p>In this proposition is set forth the nature of the eight kinde of irrationall lines which is called a residuall line, the definition whereof by this proposition is thus.</p>
                  <p>
                     <note place="margin">
                        <hi>Diffinition of the eight irra<g ref="char:EOLhyphen"/>tionall line.</hi>
                     </note>
                     <hi>A residuall line is an irrationall line which remayneth, when from a rationall line geuen, is taken away a rationall line commensurable to the whole line in power onely.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶The 56. Theoreme. The 74. Proposition. If from a mediall line be taken away a mediall line commensurable in power onely to the whole line, and comprehending together with the whole line a rationall superficies: the residue is an irrationall line, and is called a first mediall residuall line.</head>
               <p>SVppose that AB be a mediall line. And
<figure/>
from the line AB take away a mediall line BC commensurable in power onely to the whole line AB and comprehending toge<g ref="char:EOLhyphen"/>ther with the line AB a rationall superficies, that is, let that which is comprehended vnder the lines AB and BC be rationall. Then I say that the line remayning, namely, AC is irratio<g ref="char:EOLhyphen"/>nall and is called a first mediall residuall line.<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>monstra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> For forasmuch as the lines AB and BC are me<g ref="char:EOLhyphen"/>diall, therefore also the squares of the lines AB and BC are mediall. But that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines AB and BC twise is rationall. Wherefore that which is composed of the squares of the lines AB and BC, that is, that which is contayned vnder AB and BC twise together with the square of the line AC is incommensurable to that which is contained vnder the lines AB and BC twise. Wherefore (by the second part the 16. of the tenth) that which is contayned vnder the lines AB and BC twise is incommensurable to the square of the line AC. But that which is contayned vnder the lines AB and BC twise it rationall, wherefore the square of the line AC is irrationall. Wherefore also the line AC is irrationall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and is called a first mediall residuall line. This first mediall residuall line is also that part of the greater part of a first bimediall line, which remayneth after the taking away of the lesse part from the greater, wherof it hath also his name, and is called a first mediall residuall lines which was required to be proued.</p>
               <p>
                  <hi>Out of this proposition is taken the definition of the ninth kinde of irrationall lines, which is called a first residuall mediall line the difinition whereof is thus.</hi>
               </p>
               <p>
                  <note place="margin">Diffinition of <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inth irra<g ref="char:EOLhyphen"/>tionall line.</note>A first residuall mediall line is an irrationall line which remayneth, when from a mediall line is taken away a mediall line commensurable to the whole in power onely, and the part taken away and the whole line contayne a mediall superficies.</p>
               <div type="section">
                  <head>An other demonstration after Campane.</head>
                  <p>Let the line DE be rationall, vpon which apply the superficies DF equall to that which is cont<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<g ref="char:EOLhyphen"/>ned
<pb n="284" facs="tcp:7062:313"/>
vnder the lynes AB and BC twise, and let the superficies GE be e<g ref="char:EOLhyphen"/>qual
<figure/>
to that which is composed of the squares of the lynes AB and BC:<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>ter Campane.</hi>
                     </note> wherfore by the 7. of the second, the superficies FG is equal to the square of the lyne AC. And forasmuch as (by supposition) the superficies EG is mediall, therfore (by the 22. of the tenth) the lyne DG is rationall co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in power onely to the rational lyne DE. And forasmuch as by sup<g ref="char:EOLhyphen"/>position the superficies EH is rational, therfore by the 20. of the tenth, the line DH is rational commensurable in length vnto the rationall line DE. Wherfore the lynes DG and DH are rationall commensurable in power only (by the assumpt put before the 13. of this boke). Wherfore by the 73 of this boke, the lyne GH is a residuall lyne, and is therefore irrationall. Wherfore (by the corollary of the 21. of this boke) the superficies FG is irrational. And therfore the line AC which co<g ref="char:cmbAbbrStroke">̄</g>tayneth it in power is irra<g ref="char:EOLhyphen"/>tionall, and is called a first medial residuall lyne.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 57. Theoreme. The 75. Proposition. If from a mediall lyne be taken away a mediall lyne commensurable in power only to the whole lyne, and comprehending together with the whole lyne a mediall superficies, the residue is an irrationall lyne, and is called a second mediall residuall lyne.</head>
               <p>SVppose that AB be a mediall line, and from AB take away a mediall line CB commensurable in power onely to the whole line AB, and comprehending toge<g ref="char:EOLhyphen"/>ther with the whole line AB a mediall superficies, namely, the parallelogramme contained vnder the lines AB and BC.<note place="margin">Construction.</note> Then I
<figure/>
say that the residue, namely; the line AC is irrationall, and is called a second mediall residuall line. Take a rationall line DI, and (by the 44. of the first) vnto the line DI apply the pa<g ref="char:EOLhyphen"/>rallelogramme DE equall to the squares of the lines AB &amp; BC, and making in bredth the line DG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And vnto the same line DI apply the parallelogramme DH equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AB &amp; BC twise, and makyng in breadth the line DF. Now the parallelogramme DH is lesse then the parallelogramme DE, for that also the square of the lines AB and BC are greater then that which is contained vnder the lines AB and BC twise, by the square of the line AC by the 7. of the second. Wherfore the parallelogramme remayning, namely, FE, is equal to the square of the line AC. And forasmuch as the squares of the lines AB and BC are medi<g ref="char:EOLhyphen"/>all, therfore also the parallelogramme DE is mediall, and is applied to the rationall line DI, making in bredth the line DG. Wherfore (by the 22. of the tenth) the line DG is rational and incommensurable in length to the line DI. Againe forasmuch as that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AB and BC is mediall, therfore also that which is contained vnder the lines AB and BC twise is mediall, but that which is contained vnder the lines AB and BC twise is equall to the parallelogramme DH. Wherfore the parallelogramme DH is mediall and is applied to the rationall line DI making in breadth the line DF. Wherfore the line DF is rationall and incommensurable in length to the line DI. And forasmuch as the lines AB and BC are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely, therfore the line AB is incomme<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line BC. Wherfore (by the assumpt going before the 22. of the tenth, and by the 10. of the tenth) the square of the line AB is incommensurable to that which is contained vnder the lines AB and BC. But vnto the square of the line AB are commensurable the squares of AB and BC (by the 15. of the tenth) And vnto that which is contained vnder the lines
<pb facs="tcp:7062:314"/>
AB and BC is comme<g ref="char:cmbAbbrStroke">̄</g>surable to that which is contained vnder the lines AB and BC twise. Wherfore the squares of the lines AB and BC are incommensurable to that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AB and BC twise. But vnto the squares of the lines AB and BC is equal the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DE, and to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AB and BC twise, is equall the parallelogramme DH. Wherefore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me DE is inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable to the parallelogramme DH. But as the parallelogramme DE is to the parallelogramme DH, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line GD to the line DF. Wherfore the line GD is incom<g ref="char:EOLhyphen"/>mensurable
<figure/>
in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line DF. And either of the<g ref="char:cmbAbbrStroke">̄</g> is ra<g ref="char:EOLhyphen"/>tionall. Wherfore the lines GD and DF are rationall com<g ref="char:EOLhyphen"/>mensurable in power onely. Wherfore the line FG is a residu<g ref="char:EOLhyphen"/>all line (by the 73. proposition of the tenth) And the line DE is a rationall line. but a supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ficies comprehended vnder a rationall line, and an irrationall line is irrationall (by the 21 of the te<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> and the line which containeth in power the same super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies is irrationall (by the assumpt going before the same) Wherfore the parallelograme FE is irrationall. But the line AC containeth in power the parallelogramme FE. Wherfore the line AC is an irrationall line and is called a second mediall residuall line. And this second mediall residuall line is that part of the greater part of a bimediall line which remayneth after the taking away of the lesse part from the greater: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstrtion more briefe after <hi>Campane.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>mon<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion af<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>er Cam<g ref="char:EOLhyphen"/>pane.</hi>
                     </note>Suppose that <hi>AB</hi> be a mediall line, from which take away the me<g ref="char:EOLhyphen"/>diall
<figure/>
line <hi>GB</hi> commensurable vnto the whole line <hi>AB</hi> in power onely and contayning with it <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> mediall superficies, namely, that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines <hi>AB</hi> and <hi>BG.</hi> Then I say that the residue <hi>AG</hi> is an i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rationall line, and is called a second mediall residua<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l line. Take a ratio<g ref="char:EOLhyphen"/>nall line <hi>DC,</hi> vpon which apply a parallelogramme equall to that which is composed of the squares of the lines <hi>AB</hi> and <hi>BG,</hi> which by the 45. of the first let be <hi>DCEI.</hi> Agayne let the parallelogramme <hi>ZFEI</hi> be equall to that which is contayned vnder the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>AB</hi> and <hi>BG</hi> twise. Wherfore the superficies remaining <hi>DF</hi> is equal to the square of the line <hi>AG</hi> by the 7. of the second. (For that which is contayned vnder the lines <hi>AB</hi> and <hi>BG</hi> twise together with the square of the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is equall to that which is composed of the squares of the lines <hi>AB</hi> and B<hi>G</hi>). And forasmuch as the squares of the lines <hi>AB</hi> and <hi>BG</hi> are mediall, for that they are descri<g ref="char:EOLhyphen"/>bed of medial lines: the parllelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>DE</hi> which is equal vnto the<g ref="char:cmbAbbrStroke">̄</g>, shall also be mediall. And foras<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uch as that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines <hi>AB</hi> and <hi>BG</hi> is by supposition mediall, therfore the superficie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>Z<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> which is double vnto it is also mediall. But the mediall superficies <hi>DE</hi> excedeth not the mediall supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ficies <hi>ZE</hi> by a rational superficies (by the 26. of this booke). Wherfore the excesse, namely, the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies <hi>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is irrationall, vnto which the square of the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is equall: wherefore the square of the line <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> is ir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ationall, and therefore the line <hi>AG</hi> which contayneth it in power is irratio<g ref="char:EOLhyphen"/>nall by the assumpt put before the 21. of this booke, and is called a second mediall residuall line.</p>
                  <p>This proposition setteth forth the nature of the tenth kinde of irrational lines, which is called a second residuall mediall line, which is thus defined.</p>
                  <p>
                     <note place="margin">
                        <hi>Diffinition of the tenth ir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>a<g ref="char:EOLunhyphen"/>tionall line.</hi>
                     </note>
                     <hi>A second residual<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> mediall lyne is an irrationall lyne which remayneth, when from a medial line is taken away a mediall lyne commensurable to the whole in power onely, and the part taken away &amp; the whole lyne contayne a mediall superficies.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶The 58. Theoreme. The 76. Proposition. I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rom a right line be taken away a right line incommensurable in power to the whole, and if that which is made of the squares of the whole line and of the line taken away added together be rationall, and the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me
<pb n="285" facs="tcp:7062:314"/>
contained vnder the same lines mediall: the line remayning is irrationall, and is called a lesse line.</head>
               <p>SVppose that AB be a right line, and from the right line AB take away a right line BC incommensurable in power to the whole line, namely, to AB, and let that which is composed of the squares of the lines AB and BC be rationall, and let the parallelogramme contayned vnder the same lines AB and BC be mediall. Then I say, that the line remayning, namely, the
<figure/>
line AC is irrationall, &amp; is called a lesse line. For forasmuch as that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of the lines AB and BC is rationall,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and that which is contayned vnder the lines AB and BC twise is mediall, therefore that which is composed of the squares of the lines AB and BC is incommensurable to that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AB and BC twise. Wherefore the squares of the lines AB and BC, are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to the square of AC, as it was said in the 73. Proposition. But that which is made of the squares of the lines AB and BC is rationall. Wherefore the square of the line AC is irrationall: wherefore also the line AC is irrationall: and is called a lesse line. And is therefore so called, for that it is that portio<g ref="char:cmbAbbrStroke">̄</g> of the greater part of a greater line, which remai<g ref="char:EOLhyphen"/>neth after the taking away of the lesse part fro<g ref="char:cmbAbbrStroke">̄</g> the greater: which was required to be proued.</p>
               <p>
                  <hi>In thys Proposition is contayned the definition of the eleuenth kinde of irratio<g ref="char:EOLhyphen"/>nall lines, which is called a lesse line, whose definition is thus.</hi>
               </p>
               <p>A lesse line is an irrationall line which remayneth,<note place="margin">Diffinition of the eleue<g ref="char:cmbAbbrStroke">̄</g>th ir<g ref="char:EOLhyphen"/>rationall line.</note> whe<g ref="char:cmbAbbrStroke">̄</g> from a right line is taken away a right line incommensurable in power to the whole, and the square of the whole line, &amp; the square of the part taken away added together, make a rationall superficies, and the parallelogramme contayned of them is mediall.</p>
               <p>
                  <hi>This Proposition may after</hi> Campanes <hi>way be demonstrated, if you remember well the order &amp; po<g ref="char:EOLhyphen"/>sitions which he in the three former Propositions vsed.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 19. Theoreme. The 77. Proposition. If from a right line be taken away a right line incommensurable in power to the whole line, and if that which is made of the squares of the whole line and of the line taken away added together be mediall, and the paral<g ref="char:EOLhyphen"/>lelogramme contained vnder the same lines rationall: the line remaining is irrationall, and is called a line making with a rationall superficies the whole superficies mediall.</head>
               <p>SVppose that AB be a right line, and from the right line AB take away a right line BC incommensurable in power to the whole line AB, and let that which is made of the squares of the lines AB and BC added together, be mediall, and the parallelogramme contained vnder the same lines rationall. Then I say, that the line remayning, namely, the line AC, is irratio<g ref="char:EOLhyphen"/>nall,
<figure/>
and is called a line making with a rationall superficies the whole superficies mediall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For for<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>smuch as that which is made of the squares of the lines AB and BC added together is mediall, and that which is contained vnder the lines
<pb facs="tcp:7062:315"/>
AB and BC twise is rationall, therefore that which is made of the squares of the lines AB and BC added together, is incommensurable to that which is contained vnder the lines AB and BC twise. Wherefore (by the 16. of the tenth) the residue, namely, the square of the line AC is incommensurable to that which is contained vnder the lines AB and BC twise. But that which is contained vnder the lines AB and BC twise is rationall. Wherfore the square of the line AC is irrationall. Wherefore also the line AC is irrationall: and is called a line making with a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies the whole su<g ref="char:EOLhyphen"/>perficies
<figure/>
mediall: and is therfore so called for that that which is made of the squares of the lines AB and BC added together is mediall, &amp; is a certaine whole superficies, part whereof is that which is contained vnder the lines AB &amp; BC, which is a rationall superficies. For the squares of the lines AB and BC, are equall to that which is contained vnder the lines AB and BC twise, and to the square of the line AC (by the 7. of the second). Or it is therefore so called for that the square thereof added to a rationall su<g ref="char:EOLhyphen"/>per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, maketh the whole superficies mediall, as shall be proued by the 109. Proposition of this booke: which was required to be proued.</p>
               <p>
                  <hi>In this Proposition is declared the nature of the twelueth kind of irrationall lines, which is called a line making with a rationall superficies the whole superficies mediall, whose definition is thus.</hi>
               </p>
               <p>
                  <note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ition of the twelueth ir<g ref="char:EOLunhyphen"/>ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ionall line.</note>A line making with a rationall superficies the whole superficies mediall, is an irrationall line which remaineth, whe<g ref="char:cmbAbbrStroke">̄</g> fro<g ref="char:cmbAbbrStroke">̄</g> a right line is taken away a right line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power to the whole line, and the square of the whole line &amp; the square of the part taken away added together make a me<g ref="char:EOLhyphen"/>diall superficies, and the parallelogramme contained of them is rationall.</p>
               <p>
                  <hi>This Proposition also may after</hi> Campanes <hi>way be demonstrated, obseruing the former caution.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 60. Theoreme. The 78. Proposition. If from a right line be taken away a right line incommensurable in power to the whole line, and if that which is made of the squares of the whole line and of the line taken away added together be medial, and the parallelo<g ref="char:EOLunhyphen"/>gramme contayned vnder the same lines be also mediall, and incommen<g ref="char:EOLhyphen"/>surable to that which is made of the squares of the sayd lines added toge<g ref="char:EOLhyphen"/>ther: the line remayning is irrationall, and is called a line making with a mediall superficies the whole superficies mediall.</head>
               <p>SVppose that <hi>AB</hi> be a right line,
<figure/>
and from <hi>AB</hi> take away a right line <hi>BC</hi> incommensurable in power to the whole line <hi>AB.</hi> And let that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added together be mediall, and let the parallelograme contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> be also medial, and let that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added together be in<g ref="char:EOLhyphen"/>commensurable to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC.</hi> The<g ref="char:cmbAbbrStroke">̄</g> I say that the line remayning, namely, the line <hi>AC</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <pb n="286" facs="tcp:7062:315"/>
irrationall and is called a line making with a mediall superficies the whole superficies medial. Take a rationall line <hi>DI.</hi> And (by the 44. of the first) vnto the line <hi>DI</hi> apply the parallelo<g ref="char:EOLhyphen"/>gramme <hi>DE</hi> equall to that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added to<g ref="char:EOLhyphen"/>gether,<note place="margin">Construction.</note> and making in breadth the line <hi>DG.</hi> And vnto the same line <hi>DI</hi> apply the paralle<g ref="char:EOLhyphen"/>logramme <hi>DH</hi> equall to that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise and making in breadth the line <hi>DF.</hi> Wherefore the parallelogramme remayning, namely the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>FE,</hi> is equall to the square of the line <hi>AC.</hi> Wherefore the line <hi>AC</hi> contay<g ref="char:EOLhyphen"/>neth in power the parallelograme <hi>FE.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added together is mediall, and is equall to the parallelogramme <hi>DE,</hi> therefore also the parallelogramme <hi>DE</hi> is mediall. And the parallelogramme <hi>DE</hi> is applyed to the rationall line <hi>DI</hi> making in breadth the line <hi>DG.</hi> Wherfore (by the 22. of the tenth) the line <hi>DG</hi> is rationall and incommensurable in length to the line <hi>DI.</hi> Agayne forasmuch as that which is contayned vnder the lines <hi>AB</hi> and <hi>BC</hi> twise is mediall and is equall to the parallelogramme <hi>DH,</hi> therefore the parallelogramme <hi>DH</hi> is mediall. And the parallelograme <hi>DH</hi> is applyed vnto the rationall line <hi>DI</hi> making in breadth the line <hi>DF,</hi> wherefore the line <hi>DF</hi> is rationall and incommensurable in length to the line <hi>DI.</hi> And for<g ref="char:EOLhyphen"/>asmuch as that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added together is in<g ref="char:EOLhyphen"/>commensurable to that which contayned vnder the lines <hi>AB</hi> &amp; <hi>BC</hi> twise, ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>DE</hi> is incommensurable to the parallelogramme <hi>DH.</hi> But as the parallelo<g ref="char:EOLhyphen"/>gramme <hi>DE</hi> is to the parallelogramme <hi>DH,</hi> so (by the first of the sixt) is the line <hi>DG</hi> to the line <hi>DF</hi>: wherfore the line <hi>DG</hi> is incommensurable in length to the line <hi>DF.</hi> And they are both rationall lines. Wherefore the lines <hi>DG</hi> and <hi>DF</hi> are rationall commensurable in pow<g ref="char:EOLhyphen"/>er onely, where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the line <hi>FG</hi> is a residuall line by the 73. of this booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But the line <hi>FH</hi> is rationall for that it is equall vnto the line <hi>DI.</hi> But a rectangle parallelogramme contayned vnder a rationall line and an irrationall line is irrationall, and the line also that contayneth in power the same parallelogramme is irrationall (by the 21. of the tenth). But the line <hi>CA</hi> contayneth in power the parallelogramme <hi>FE.</hi> Wherefore the line <hi>AC</hi> is irrationall and is called a line making with a mediall superficies the whole superficies mediall. And is therfore so called, for that that which is made of the squares of the lines <hi>AB</hi> and <hi>BC</hi> added together is mediall, &amp; is a certayne whole superficies, part whereof is that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines <hi>AB</hi> and <hi>BC,</hi> which is also mediall: you shall also by the 110. proposition of this booke vnderstand an other cause why it is so called.</p>
               <p>
                  <hi>This proposition may thus more briefely be demonstrated: forasmuch as that which is composed of the squares of the lines AB and BC is mediall, and that also which is contayned vnder them is me<g ref="char:EOLhyphen"/>diall, therefore the parallelogramm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s DE and DH which are equall vnto them are mediall: but a me<g ref="char:EOLhyphen"/>diall superficies exceedeth not a mediall superficies by a rationall superficies. Wherefore the superficies FE which is the excesse of the mediall superficies DE aboue the mediall superficies DH is irrational. And therefore the line AC which contayneth it in power is irrationall. &amp;c.</hi>
               </p>
               <p>
                  <hi>In this proposition is shewed the conditio<g ref="char:cmbAbbrStroke">̄</g> and nature of the thirtenth and last kinde of irrationall lines, which is called a line making with a mediall superficies the whole superficies mediall, whose definition is thus.</hi>
               </p>
               <p>A line making with a mediall superficies the whole superficies mediall is an irrationall line which remayneth,<note place="margin">Diffinition of the thirtenth and last irra<g ref="char:EOLhyphen"/>tionall line.</note> when from a right li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e is taken away a right line incommensurable in power to the whole line, and the squares of the whole line and of the line taken away added together make a mediall superficies, and the parallelogramme contayned of the<g ref="char:cmbAbbrStroke">̄</g> is also a mediall superficies, moreouer the squares of them are incommensurable to the parallelogramme contayned of them.</p>
               <div type="section">
                  <head>An assumpt of <hi>Campane.</hi>
                  </head>
                  <p>
                     <pb facs="tcp:7062:316"/>
                     <hi>If there be fower qua<g ref="char:cmbAbbrStroke">̄</g>tities, &amp; if the difference of the first to the second, be as the difference of the third to the fourth, then alternately, as the difference of the first is to the third, so is the difference of the second to the fourth.</hi>
                  </p>
                  <p>This is to be vnderstand of qua<g ref="char:cmbAbbrStroke">̄</g>tities in like sort referred the one to the other, that is if the first be greater then the second,<note place="margin">
                        <hi>An Assumpt of Campane.</hi>
                     </note> the third ought to be greater then the fourth and if the first be lesse then the se<g ref="char:EOLhyphen"/>cond, the third ought to be lesse then the fourth: and is also to be vnderstand in arithmeticiall propor<g ref="char:EOLhyphen"/>tionality. As for example let the difference of A be vnto B as the difference of C is to D.<note place="margin">
                        <p>
                           <hi>I. Dee.</hi>
                        </p>
                        <p>
                           <hi>Though Cam<g ref="char:EOLhyphen"/>panes lemma be true, ye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> the maner of de<g ref="char:EOLhyphen"/>monstrating it, (narrowly considered) is not artificiall.</hi>
                        </p>
                     </note> Then I say that as the difference of A is to C, so is the difference of B to D. For (by this common se<g ref="char:cmbAbbrStroke">̄</g>tence, the difference of the extreames is composed of the differences of the ex<g ref="char:EOLhyphen"/>treames
<figure/>
to the meanes), the difference of A to C is compo<g ref="char:EOLhyphen"/>sed of the difference of A to B and of the difference of B to C. And (by the same common sentence) the difference of B to D is composed of the difference of B to C, and of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he difference of C to D. And forasmuch as (by supposition) the difference of A to B is as the difference of C to D, and the difference of B to C is common to them both. Where<g ref="char:EOLhyphen"/>fore it followeth, that as the difference of A is to C, so is the difference of B to D: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 61. Theoreme. The 79. Proposition. Vnto a residual line can be ioyned one onely right lyne rational, and com<g ref="char:EOLhyphen"/>mensurable in power onely to the whole lyne.</head>
               <p>
                  <note place="margin">Second Sena<g ref="char:EOLhyphen"/>ry.</note>LEt AB be a residuall line, and vnto it let the line BC be supposed to be ioyned, so that let the lines AC and BC be rationall commensurable in power onely. Then I say that vnto the line AB cannot be ioyned any other rationall line commensura<g ref="char:EOLhyphen"/>ble in power onely to the whole line.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if it be possible, let BD be such a line ad<g ref="char:EOLhyphen"/>ded vnto it. Wherfore the lines AD and DB
<figure/>
are rationall commensurable in power onely. And forasmuch as how much the squares of the lines AD and DB do exceede that which is contained vnder the lines AD and DB twise, so much also do the squares of the lines AC and CB exceede that which is contained vnder the lines AC and CB twise, for the excesse of eche is one and the same, namely, the square of the line AB (by the <hi>7.</hi> of the second.) Wher<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore alternately (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ormer assumpt of Campanus) how much the squares of the lines AD and DB do exceede the squares of the lines AC &amp; CB, so much also excedeth that which is contayned vnder the lines AD and DB twise, that which is contained vnder the lines AC and CB twise. But that which is made of the squares of the lines AD and DB added together, exceedeth that which is made of the squares of the lynes AC and CB added together by a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies: (for they are either of them rational). Where<g ref="char:EOLhyphen"/>fore that which is contained vnder the lines AD and DB twise, exceedeth that which is contained vnder the lines AC and CB twise by a rationall superficies. But that which is contained vnder the lines AD and DB twise, is mediall, for it is commensurable to that which is contained vnder the lines AD and DB once, which superficies is mediall (by the 21. of the tenth) and by the same reason also that which is contained vnder the lynes AC and CB twise is mediall. Wherfore a mediall superficies differeth from a mediall superficies by a rationall superficies, which (by the 26. of the tenth) is impossible. Wherfore vnto the lyne AB cannot be ioyned any other rationall line besides BC commensurable in power onely, to the whole line. Wherfore vnto a residuall line can be ioyned one onely right line rationall and commensurable in power onely to the whole lyne: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <pb n="287" facs="tcp:7062:316"/>
               <head>¶ The 62. Theoreme. The 80. Proposition. Vnto a first medial residuall line can be ioyned one onely mediall right lyne, commensurable in power onely to the whole lyne, and comprehendyng wyth the whole lyne a rationall superficies.</head>
               <p>SVppose that AB be a first mediall residuall line, &amp; vnto AB ioyne the lyne BC, so that let the lynes AC and BC be mediall commensurable in power onely, &amp; let that which is contained vnder the lines AC and BC be rationall. Then I say that vnto the lyne AB cannot be ioyned any other mediall line commensurable in power onely to the whole lyne, and comprehending together with the whole lyne a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if it be possible let the line BD be such a line. Wherfore the lynes AD and DB are mediall com<g ref="char:EOLhyphen"/>mensurable in power onely, and that which is con<g ref="char:EOLhyphen"/>tayned
<figure/>
vnder the lynes AD and DB is rational. And forasmuch as how much the squares of the lynes AD and DB exceede that which is contay<g ref="char:EOLhyphen"/>ned vnder the lynes AD and DB twise, so much also exceede the squares of the lynes AC &amp; BC, that which is contayned vnder the lynes AC and CB twise (for the excesse of eche is one and the same, namely, the square of the lyne AB). Wherfore alternately (as it was sayd in the former proposition) how much the squares of the lynes AD and DB exceede the squares of the lines AC and CB, so much also that which is contained vnder the lines AD and DB twise, excedeth that which is contained vnder the lines AC and CB twise. But that which is contained vnder the lines AD and DB twise, excedeth that which is contained vnder the lines AC and CB twise by a rationall superficies, for they are either of them a rationall su<g ref="char:EOLhyphen"/>persicies. Wherfore that which is made of the squares of the lines AD &amp; DB excedeth that which is made of the squares of the lines AC &amp; CB by a rationall superficies, which (by the 26. of the tenth) is impossible. For they are either of them mediall (for those foure lines were put to be mediall.) Wherfore vnto a first mediall residuall line can be ioyned onely one right mediall line commensurable in power onely to the whole line, and comprehending with the whole line a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 63. Theoreme. The 81. Proposition. Vnto a second mediall residuall line can be ioyned onely one mediall right line, commensurable in power onely to the whole line, and comprehending with the whole line a mediall superficies.</head>
               <p>SVppose that AB be a seco<g ref="char:cmbAbbrStroke">̄</g>d mediall re<g ref="char:EOLhyphen"/>sidual
<figure/>
line, &amp; vnto the line AB ioyne the line BC, so that let the lines AC and CB be mediall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power one<g ref="char:EOLhyphen"/>ly, and let that which is comprehended vnder the lines AC and CB be mediall. Then I say, that vnto the line AB can not be ioyned any o<g ref="char:EOLhyphen"/>ther mediall right line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely to the whole line, and comprehending to<g ref="char:EOLhyphen"/>gether with the whole line a mediall superficies. For if it be possible, let the line BD be such a line. Wherefore the lines AD &amp; DB are me<g ref="char:EOLhyphen"/>diall commensurable in power onely, and that
<pb facs="tcp:7062:317"/>
which is contained vnder the lines AD and DB is also mediall. Take a rationall line EF.<note place="margin">Construction.</note> And (by the 44. of the first) vnto the line EF apply the parallelogramme EG equall to the squares of the lines AC and CB, and making in breadth the line EM: and from that parallelogramme EG take away the parallelogramme HG e<g ref="char:EOLhyphen"/>quall to that which is contained vnder AC and CB twise, and making in breadth the line HM. Wherefore the parallelogramme remayning, namely, EL, is (by the 7. of the second) equall to the square of the line AB. Wherefore the line AB containeth in power the parallelograme EL. Againe, vnto the line EF apply (by the 44. of the first) the parallelogramme EI equall to the squares of the lines AD and DB, and making in breadth the line EN. But the squares of the lines AD and DB are equall to that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AD and DB twise, and to the square of the line AB.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> Wherefore the parallelogramme EI is equall to that which is contained vnder the lines AD &amp; DB twise, and to the square of the line AB. But the parallelogramme EL is equall to the square of the line AB. Wherefore the parallelogramme remaining, namely, HI, is equall to that which is contained vnder the lines AD and DB twise. And forasmuch as the lines AC and CB are <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>iall, there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the squares also of the lines AC and CB are mediall: and they are equall to the parallelogramme EG: wherefore the parallelogramme EG is (by that which was spo<g ref="char:EOLhyphen"/>ken in the 75. Proposition) mediall: and it is applied vnto the rationall line EF, making in breadth the line EM. Wherefore (by the 22. of the tenth) the line EM is rationall, and in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ommensurable in length to the line EF. Againe, forasmuch as that which is contayned vnder the lines AC and CB is mediall, therefore (by the Corollary of the 23. of the tenth) that which is contained vnder the lines AC and CB twise is also mediall: and it is equall to the parallelogramme HG: wherefore also the parallelogramme HG is mediall, and is ap<g ref="char:EOLhyphen"/>plyed to the rationall line EF, making in breadth the line HM. Wherefore (by the 22 of the tenth) the line HM is rationall, and incommensurable in length to the line EF. And forasmuch as the lines AC and CB are commensurable in power onely, therefore the line AC is incommensurable in length to the line CB. But as the line AC is to the line CB, so (by the Assumpt going before the 22. of the tenth) is the square of the line AC to that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines AC &amp; CB. Wherefore (by the 10. of the tenth) the square of the line AC is incommensurable to that which is con<g ref="char:EOLhyphen"/>tained
<figure/>
vnder the lines AC and CB. But vnto the square of the line AC are commensurable the squares of AC &amp; CB, and vnto that which is contained vnder the lines AC and CB, is commensurable that which is contained vnder the lines AC and CB twise. Wherefore the squares of the lines AC &amp; CB are incommen<g ref="char:EOLhyphen"/>surable to that which is contained vnder the lines AC and CB twise. But vnto the squares of the lines AC and CB is equall the parallelo<g ref="char:EOLhyphen"/>gramme EG, and vnto that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AC &amp; CB twise, is equall the parallelogramme GH. Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme EG is incommensurable to the parallelogramme HG. But as the parallelo<g ref="char:EOLhyphen"/>gramme EG is to the parallelogrmme HG, so is the line EM to the line HM. Wherefore the line EM is incommensurable in length to the line HM. And they are both rationall lines. Wherefore the lines EM and MH are rationall commensurable in power onely. Wherefore the line EH is a residuall line, and vnto it is ioyned a rationall line HM commensurable in power onely to the whole line EM. In like sort also may it be proued, that vnto the line EH is ioyned the line HN, being also rationall, and commensurable in power onely to the whole
<pb n="288" facs="tcp:7062:317"/>
line EN. Wherefore vnto a residuall line is ioyned <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>re then one onely line commensura<g ref="char:EOLhyphen"/>ble in power onely to the whole line: which (by the 79. of the tenth) is impossible. Wherefore vnto a second mediall residuall line can be ioyned onely one mediall right line commensura<g ref="char:EOLhyphen"/>ble in power onely to the whole line, and comprehending with the whole line a mediall super<g ref="char:EOLhyphen"/>ficies: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 64. Theoreme. The 82. Proposition. Vnto a lesse line can be ioyned onely one right line incommensurable in power to the whole lyne, and making together with the whole lyne that which is made of their squares added together rationall, and that which is contayned vnder them mediall.</head>
               <p>SVppose that AB be a lesse line, and to AB ioyne the line BC, so that let BC be such a line as is required in the Theoreme. Wherfore the lines AC and CB are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power, hauing that which is made of the squares of them ad<g ref="char:EOLhyphen"/>ded together rationall, and that which is contained vnder them mediall. Then I say that vnto AB cannot be ioyned any other
<figure/>
such right line.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie.</note> For if it be possible, l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t the lyne BD be such a line. Wherfore the lines AD &amp; DB are incommensurable in power, hauing that which is made of the squares of them added together, rationall, and that which is con<g ref="char:EOLhyphen"/>tained vnder them mediall. And for that how much the squares of the lines AD and DB excede the squares of the lines AC and CB, so much that which is contained vnder the lines AD and DB twise, excedeth that which is contained vnder the lines AC and CB twise (by those things which were spoken in the 79. proposition) But that which is made of the squares of the lines AD and DB added together excedeth that which is made of the squares of the lines AC and CB added together by a rationall superficies, for they are either of them ratio<g ref="char:EOLhyphen"/>nall by supposition. Wherfore that which is contained vnder the lines AD and DB twise, excedeth that which is contained vnder the lines AC and CB twise by a rationall superfi<g ref="char:EOLhyphen"/>cies: which (by the 26. of the tenth) is impossible, for either of them is mediall by supposition. Wherfore vnto a lesse line can be ioyned onely one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together rationall, and that which is contained vnder them mediall: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 65. Theoreme. The 83. Proposition. Vnto a line making with a rationall superficies the whole superficies me<g ref="char:EOLhyphen"/>diall, can be ioyned onely one right lyne incommensurable in power to the whole lyne, and making together with the whole line that which is made of their squares added together mediall, and that which is contained vnder them rationall.</head>
               <p>SVppose that AB be a line making with a rationall superficies the whole superficies mediall, and vnto it let the line BC be ioyned, so that let BC be such a line as is re<g ref="char:EOLhyphen"/>quired in the Theoreme. Wherfore the lines AC and CB are incommensurable in
<pb facs="tcp:7062:318"/>
power, ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ing that which is made of the squares
<figure/>
of the lines AC and CB added together mediall, and that which is contained vnder the lynes AC and CB rationall.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibilitie.</note> Then I say that vnto the lyne AB cannot be ioyned any other such line. For if it be possible, let the line BD be such a line. Wherfore the lines AD and DB are incommensurable in power, hauing that which is made of the squares of the lines AD and DB added together mediall, and that which is contained vnder the lines AD and DB rationall. Now for that how much the squares of the lines AD and DB exceede the squares of the lines AC and CB, so much that which is contained vnder the lines AD and DB twise exceedeth that which is contayned vnder the lynes AC and CB twise, by that which was spoken in the 79. proposition. But that which is contained vnder the lines AD and DB twise, exceedeth that which is contained vnder the lines AC and CB twise by a rationall superficies, for they are either of them ratio<g ref="char:EOLhyphen"/>nall by suppositio<g ref="char:cmbAbbrStroke">̄</g>. Wherfore that which is made of the squares of the lines AD and DB ad<g ref="char:EOLhyphen"/>ded together, excedeth that which is made of the squares of the lines AC and CB added to<g ref="char:EOLhyphen"/>gether by a rationall superficies, which by the 26. of the tenth, is impossible, for they are either of them mediall by supposition. Wherfore vnto the line AB cannot be ioyned any other lyne besides BC, making that which is required in the proposition. Wherfore vnto a line making with a rationall superficies the whole superficies mediall can be ioyned onely one right line in<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power to the whole line, and making together with the whole line that which is made of their squa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es added together mediall, and that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder them ra<g ref="char:EOLhyphen"/>tionall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 66. Theoreme. The 84. Proposition. Vnto a line making with a mediall superficies the whole superficies medial, can be ioyned onely one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together mediall, and that which is contained vnder them mediall, and moreouer making that which is made of the squares of them added together incommensurable to that which is contayned vn<g ref="char:EOLhyphen"/>der them.</head>
               <p>SVppose that AB be a line making with a mediall superficies the whole superficies mediall, and vnto it let the line BC be ioyned, so that let BC be such a line as is required in the Theoreme. Wherefore the lines AC and CB are incommensu<g ref="char:EOLhyphen"/>rable in power, hauing that which is made of the squares of the lines AC and CB added together mediall, &amp; that which
<figure/>
is contained vnder the lines AC and CB mediall, and moreouer that which is made of the squares of the lines AC and CB is incommensurable so that which is contained vnder the lines AC and CB. Then I say, that vnto the line AB can be ioyned no other such line. For if it be possible, let BD be such a line. Wherefore the lines AD and DB are in<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power, hauing that which is made of the squares of the lines AD and
<pb n="289" facs="tcp:7062:318"/>
DB added together mediall, and that which is contained vnder the lines AD and DB mediall, and moreouer that which is made of the squares of the lines AD and and DB added together, is incommensurable to that which is contained vnder the lines AD and DB. Take a rationall line EF.<note place="margin">Construction.</note> And (by the 44. of the first) vnto the line EF apply the parallelogramme EG equall to the squares of the lines AC and CB, and making in breadth the line EM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and from the parallelogramme EG take away the pa<g ref="char:EOLhyphen"/>rallelogramme HG equall to that which is contained vnder the lines AC &amp; CB twise, and making in breadth the line HM. Wherefore the residue, namely, the square of the line AB is equall to the parallelogramme EL (by the 7. of the second). Wherefore the line AB con<g ref="char:EOLhyphen"/>tayneth in power the parallelogramme EL. Againe (by the 44. of the first) vnto the line EF apply the parallelogramme E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equall to the squares of the lines AD and DB, and ma<g ref="char:EOLhyphen"/>king in breadth the line EN. But the square of the line AB is equall to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me EL. Wherefore the residue, namely, the parallelogramme HI is equall to that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AD and DB twise.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> an abjurd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>
                  </note> And forasmuch as that which is made of the squares of the lines AC and CB is mediall, and is equall to the parallelogramme EG, there<g ref="char:EOLhyphen"/>fore also the parallelogramme EG is mediall. And it is applied vnto the rationall line EF, making in breadth the line EM. Wherefore (by the 22. of the tenth) the line EM is ratio<g ref="char:EOLhyphen"/>nall and incommensurable in length to the line EF. Againe, forasmuch as that which is contained vnder the lines AC and CB twise is mediall, and is equall to the parallelogramme HG. Wherefore the parallelogramme H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G is mediall, which parallelogramme HG is appli<g ref="char:EOLhyphen"/>ed to the rationall line EF, making in breadth the line HM. Wherefore the line HM is ra<g ref="char:EOLhyphen"/>tionall and incommensurable in length to the line EF. And forasmuch as the squares of the lines AC and CB are incommensurable to that which is contained vnder the lines AC and CB twise, therefore the parallelogramme EG is incommensurable to the parallelogramme HG. Wherefore the line EM is incommensurable in length to the line MH, and they are both rationall. Wherefore the lines EM and MH are rationall commensurable in power onely, Wherefore EH is a residuall line. And the line ioyned vnto it is HM. And in like sort may we proue, that the line EH is a residuall line, and that the line HN is ioyned vnto it. Wherefore vnto a residuall line is ioyned two sundry lines, being eche commensurable in power onely so the whole line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which (by the 79. of the tenth) is impossible. Wherefore vn<g ref="char:EOLhyphen"/>to the line AB can not be ioyned any other right line besides the line BC, which shall be incommensurable in power to the whole line, &amp; haue together with the whole line that which is made of their squares added together mediall, and that which is contained vnder them mediall, and moreouer incommensurable to that which is made of their squares added toge<g ref="char:EOLhyphen"/>ther. Wherefore vnto a line making with a mediall superficies the whole superficies mediall, can be ioyned onely one right line incommensurable in power to the whole line, and making together with the whole line that which is made of their squares added together mediall, and that which is contained vnder them mediall, and moreouer making that which is made of the squares of them added together incommensurable to that which is contained vnder the<g ref="char:cmbAbbrStroke">̄</g> which was required to be proued.</p>
               <div type="section">
                  <head>¶ Third Definitions.</head>
                  <p>AS of binomiall lynes, there are 6. diuers kindes, so also of residuall lynes which are correspondent vnto them and depend of them (for a residuall line is nothing els (as was before said) but that which remayneth whe<g ref="char:cmbAbbrStroke">̄</g> the lesse part of a binomiall line is taken from the greater part or name ther<g ref="char:EOLhyphen"/>of,) there are likewise sixe seuerall kindes. All which are knowne and con<g ref="char:EOLhyphen"/>sidered
<pb facs="tcp:7062:319"/>
in comparison to a rationall line set forth &amp; appointed, and these residual line haue the selfe same order of production that the binomials had.<note place="margin">
                        <hi>Sixe kindes of re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>iduall lines.</hi>
                     </note> For as the three first kindes of binomiall lines, namely, the first, second, and third, were produced when the square of the greater part of the part of the binomiall excedeth the square of the lesse part thereof by the square of a line commensurable vnto it in length: so in likewise, the first three kindes of residuall lines, namely, the first, second, and third, are produced, when the square of the whole, namely, of that which is made of the residuall line, and the line ioy<g ref="char:EOLhyphen"/>ned vnto it added together, excedeth the square of the line ioyned to the residuall, by the square of a lyne which is commensurable vnto it in length. And as the three last kindes of binomials, namely, the fourth, fifth, and sixth were produced when the square of the greater part excedeth the square of the lesse, by the square of a line incommen<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>rable in length vnto it, euen so the three last kyndes of residuall lynes are produced whe<g ref="char:cmbAbbrStroke">̄</g> the square of the whole excedeth the square of the lyne adioined, by the square of a line incommensurable vnto it in length. As ye may perceiue by their definitions following.</p>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>First diffini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note>A first residuall line is, when the square of the whole excedeth the square of the lyne adioyned, by the square of a lyne commensurable vnto it in le<g ref="char:cmbAbbrStroke">̄</g>gth, and also the whole is commensurable in length to the rationall line first set.</head>
                     <p>As let AB be a ratio<g ref="char:EOLhyphen"/>nall
<figure/>
line, whose partes are certaine, distinct, &amp; to be expressed by num<g ref="char:EOLhyphen"/>ber. And let the residual lyne be CD, and let the line ioyned vnto it be EC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and let the whole being composed of the residuall CD, and the line adioy<g ref="char:EOLhyphen"/>ned EC, be the line ED: let moreouer the square of the whole line ED, excede the square of the line adioyned EC by the square of the line F, which line F let be commensurable in length to the whole lyne ED, and let the whole line ED be also commensurable in length to the rationall line AB: then is the residuall lyne CD by this definition a first residuall line.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>Second diffi<g ref="char:EOLhyphen"/>nition.</hi>
                        </note>A second residual line is, when the square of the whole excedeth the square of the line adioyned, by the square of a line commensurable vnto it in le<g ref="char:cmbAbbrStroke">̄</g>gth, and also the line adioyned is commensurable in length to the rationall lyne.</head>
                     <p>As suppose the line CD to be a residuall, and let the line adioy<g ref="char:EOLhyphen"/>ned
<figure/>
vnto it be EC, and the whole made of them both, let be the line ED: &amp; let the square of ED the whole line excede the square of the lyne adioyned EC, by the square of the lyne F, and let the lyne F be commensurable in length to the whole lyne ED, moreo<g ref="char:EOLhyphen"/>uer let the line adioyned EC be commensurable in length to the rationall line AB: then by this definition, the residuall line CD is a second residuall line.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">
                           <hi>Third diffini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note>A third residuall line is, when the square of the whole excedeth the square of the lyne adioyned, by the square of a line commensurable vnto it in le<g ref="char:cmbAbbrStroke">̄</g>gth and neither the whole line, nor the line adioyned is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the rationall lyne.</head>
                     <p>As (the former supposition standing) suppose that the square of the
<figure/>
whole lyne ED exceede the square of the lyne adioyned EC by the square of the lyne F, and let the lyne F be commensurable in length to the whole lyne EC, and let neither the whole lyne ED, nor the line adioyned EC be commensurable in length to the rationall lyne AB, then by this definition the residuall lyne CD is a third residu<g ref="char:EOLhyphen"/>all lyne.</p>
                  </div>
                  <div type="subsection">
                     <pb n="290" facs="tcp:7062:319"/>
                     <head>A fourth residuall line is,<note place="margin">
                           <hi>Fourth diffi<g ref="char:EOLhyphen"/>nition.</hi>
                        </note> when the square of the whole lyne excedeth the square of the lyne adioyned, by the square of a lyne incommensurable vnto it in length, and the whole lyne is also commensurable in length to the ra<g ref="char:EOLhyphen"/>tionall lyne.</head>
                     <p>As the residuall lyne
<figure/>
beyng as before CD, &amp; the lyne adioyned EC, and the whole ED, let the square of the whole lyne ED exceede the square of the line adioyned EC by the square of the lyne F, and let the lyne F be incommensurable in length to the whole line ED, and let ED the whole line be commensurable in length to the rationall lyne AB, then is the residuall line CD by this declaration a fourth residuall lyne.</p>
                  </div>
                  <div type="subsection">
                     <head>A fiueth residuall line is,<note place="margin">
                           <hi>Fifth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> when the square of the whole lyne exceedeth the square of the lyne adioyned, by the square of a lyne incommensurable vnto it in length, and the lyne adioyned is commensurable in length to the ra<g ref="char:EOLhyphen"/>tionall lyne.</head>
                     <p>As the residuall line beyng CD, the lyne adioyned EC, and
<figure/>
the whole lyne ED, let the square of the whole lyne ED exceede the square of the line adioyned EC by the square of the lyne F, and let the line F be incommensurable in length to the whole lyne ED, and let also EC the lyne adioy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ed be commensurable in length to the rationall line AE, then shall the residuall CD be by this defini<g ref="char:EOLhyphen"/>tion a fifth residuall lyne.</p>
                  </div>
                  <div type="subsection">
                     <head>A sixth residuall line is when the square of the whole line,<note place="margin">
                           <hi>Sixth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> exceedeth the square of the line adioyned, by the square of a line incommensurable vnto it in length, and neither the whole line nor the line adioyned is commensu<g ref="char:EOLhyphen"/>rable in length to the rationall line.</head>
                     <p>As suppose the residuall line to be CD, and the lyne adioyned to
<figure/>
be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>C, and the whole lyne composed of them let be ED, and let the square of the whole lyne <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>D exceede the square of the line adioyned by the square of the lyne F, which line F let be incommensurable in length to the whole lyne ED: moreouer let neither the whole lyne CD nor the line adioyned EC, be comm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nsurable in length to the rationall line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, then shall the residuall lyne CD be by this explica<g ref="char:EOLhyphen"/>tion a sixt residuall lyne, and the last.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 19. Probleme. The 85. Proposition.
To finde out a first residuall line.</head>
               <p>TAke a rationall line and let the same be <hi>A,</hi>
                  <note place="margin">Third Sena<g ref="char:EOLhyphen"/>ry.</note> and vnto it let the line <hi>BG</hi> be commen<g ref="char:EOLhyphen"/>surable in length. Wherefore the line <hi>BG</hi> also is rationall. And take two square numbers <hi>DE</hi> and <hi>EF</hi> which let be such,<note place="margin">Construction.</note> that the excesse of the greater, namely, of
<pb facs="tcp:7062:320"/>
                  <hi>DE,</hi> aboue the lesse <hi>EF</hi> (which excesse let be the
<figure/>
number <hi>DF</hi>) be no square number (by the corollary of the first assumpt of the 28. of the tenth). Wherfore the number <hi>ED</hi> hath not to the number <hi>DF</hi> that proportion that a square number hath to a square number (by the 24. of the eight). And as the number <hi>ED</hi> is to the number <hi>DF,</hi> so let the square of the line <hi>BG,</hi> be to the square of the line <hi>GC</hi> (by the corrollary of the sixt of the tenth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> Wherefore the square of the line <hi>BG</hi> is comme<g ref="char:cmbAbbrStroke">̄</g>surable to the square of the line <hi>GC.</hi> But the square of the line <hi>BG</hi> is rationall, wherefore also the square of the line <hi>GC</hi> is rationall. Wherefore the line <hi>GC</hi> is also rationall. And forasmuch as the number <hi>ED</hi> hath not to the number <hi>DF,</hi> that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber, therfore neither also hath y<hi rend="sup">e</hi> square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line <hi>BG</hi> to the square of the line <hi>GC</hi> that proportion that a square number hath to a square number. Wherfore (by the 9. of the tenth) the line <hi>BG</hi> is incommensurable in length to the line <hi>GC.</hi> And they are both rationall. Wherefore the lines <hi>BG</hi> and <hi>GC</hi> are rational commensurable in power onely. Wherefore the line <hi>BC</hi> is a residuall line. I say moreouer that it is a first residuall line. For forasmuch as the square of the line <hi>BG</hi> is greater then the square of the line <hi>GC</hi> (that it is greater it is manifest, for by supposition the square of the line <hi>BG</hi> is to the square of the line <hi>GC,</hi> as the greater number, namely, <hi>ED</hi> is to the num<g ref="char:EOLhyphen"/>ber <hi>DF</hi>) vnto the square of the line <hi>BG</hi> let the squares of the lines <hi>GC</hi> and <hi>H</hi> be equall. And for that as the number <hi>DE</hi> is to the number <hi>DF,</hi> so is the square of the line <hi>BG</hi> to the square of the line <hi>GC,</hi> therefore by conuersion of proportion (by the corollary of the 9. of the fifth) as the number <hi>DE</hi> is to the number <hi>EF,</hi> so is the square of the line <hi>BG</hi> to the square of the line <hi>H.</hi> But the number <hi>DE</hi> hath to the number <hi>EF</hi> that proportion that a square number hath to a square number, for either of them is a square number, wherefore also the square of the line <hi>BG</hi> hath to the square of the line <hi>H</hi> that proportion that a square num<g ref="char:EOLhyphen"/>be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hath to a square number. Wherefore the line <hi>GB</hi> is commensurable in length to the line <hi>H.</hi> Wherefore the line <hi>GB</hi> is in power more then the line <hi>GC</hi> by the square of a line com<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>GB</hi>: and the whole line, namely, <hi>GB</hi> is commensurable in length to the rationall line <hi>A.</hi> Wherefore the line <hi>BC</hi> is a first residuall line. Wherefore there is founde out a first residuall line which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Probleme. The 86. Proposition. To finde out a second residuall line.</head>
               <p>
                  <note place="margin">Construction.</note>TAke a rationall line, and let the same be A, and vnto it let the line GC be com<g ref="char:EOLhyphen"/>mensurable in length. And take two square numbers DE and EF, and let them be such that the excesse of the greater, namely, DF, be no square number. And as the number DF is to the number DE, so let the square of the line GC be to the square of the line GB. Wherefore both the squares are commensu<g ref="char:EOLhyphen"/>rable.
<figure/>
And forasmuch as the square of the line GC is rationall,<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stra<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</note> therefore the square of the line BG is also rationall: Wherefore also the line BG is rationall. And forasmuch as the squares of the lines BG &amp; GC haue not that proportion the one to the other that a square number hath to a square number, therefore the lines BG and GC are incommensurable in length, and they are both ratio<g ref="char:EOLhyphen"/>nall. Wherefore the lines BG and GC are rationall commensura<g ref="char:EOLhyphen"/>ble
<pb n="291" facs="tcp:7062:320"/>
in power onely. Wherefore the line BC is a residuall line. I say moreouer, that it is a se<g ref="char:EOLhyphen"/>cond residual line. For forasmuch as the square of the line BG is greater the<g ref="char:cmbAbbrStroke">̄</g> the square of the line GC, vnto the square of the line BG let the squares of the lines G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp; H be equall. And for that as the number DE is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber DF, so is the square of the line GB to the square of the line GC, therefore (by conuersion of proportion) as the number DE is to the number EF, so is the square of the line BG to the square of the line H. But either of these numbers DE and EF is a square number. Wherefore the line GB is commensurable in length to the line H. Wherefore the line BG is in power more then the line GC, by the square of a line commensurable in length to the line BG: and the line GC that is ioyned to the residuall line is commensurable in length to the rationall line A. Wherefore the line BC is a second resi<g ref="char:EOLhyphen"/>duall line. Wherefore there is found out a second residuall line: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 21. Probleme. The 87. Proposition. To finde out a third residuall line.</head>
               <p>TAke rationall line, &amp; let the same be A: and take three numbers E, B, C,<note place="margin">Construction.</note> and CD, not hauing the one to the other that proportion that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square number: and let the number BC haue to the number BD that proportion that a square number hath to a square number. And let the num<g ref="char:EOLhyphen"/>ber BC be greater then the number CD. And as the number E is to the num<g ref="char:EOLhyphen"/>ber BC, so let the square of the line A be to the square of
<figure/>
the line FG: and as the number BC is to the number CD, so let the square of the line FG be to the square of the line HG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the square of the line A is com<g ref="char:EOLhyphen"/>mensurable to the square of the line FG. But the square of the line A is rationall. Wherefore also the square of the line FG is rationall: wherefore the line FG is also rationall. And forasmuch as the number E hath not to the number BC that proportion that a square number<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hath to a square number, therefore neither also hath the square of the line A to the square of the line FG that proportion that a square number hath to a square number. Wherefore the line A is incom<g ref="char:EOLhyphen"/>mensurable in length to the line FG. Againe for that as the number BC is to the number CD, so is the square of the line FG to the square of the line HG, therefore the square of the line FG is commensurable to the square of the line HG. But the square of the line FG is ra<g ref="char:EOLhyphen"/>tionall. Wherefore also the square of the line HG is rationall. Wherefore also the line HG is rationall. And for that the number BC hath not to the number CD, that proportion that a square number hath to a square number, therefore neither also hath the square of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G to the square of the line HG, that proportion that a square nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to a square num<g ref="char:EOLhyphen"/>ber. Wherefore the line FG is incommensurable in length to the line HG: and they are both rationall. Wherefore the lines FG &amp; HG are rationall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power onely. Wher<g ref="char:EOLhyphen"/>fore the line FH is a residuall line. I say moreouer, that it is a third residuall line. For for that as the number E is to the number BC, so is the square of the line A to the square of the line FG: and as the number <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>C is to the number CD, so is the square of the line FG to the square of the line HG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore by equalitie of proportion, as the number E is to the num<g ref="char:EOLhyphen"/>ber CD, so is the square of the line A to the square of the line HG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> but the number E hath <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> to the number CD that proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that a square num<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>r hath to a square number, ther<g ref="char:EOLhyphen"/>fore
<pb facs="tcp:7062:321"/>
neither also hath the square of the line A to the square of the line HG that proportion that a square number hath to a square number, therefore the line A is incommensurable in length to the line HG. Wherefore neither of the lines FG and HG is commensurable in length to the rationall line A. And forasmuch as the
<figure/>
square of the line FG is greater then the square of the line HG (that the line FG is greater then the line HG it is mani<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>est, for by supposition the number BC is grea<g ref="char:EOLhyphen"/>ter then the number CD) vnto the square of the line FG let the squares of the lines HG &amp; K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be equall. And for that as the nu<g ref="char:cmbAbbrStroke">̄</g>ber BC is to the number CD, so is the square of the line FG to the square of the line H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>er<g ref="char:EOLhyphen"/>fore (by conuersion of proportion) as the number BC is to the number BD, so is the square of the line FG to the square of the line K. But the nu<g ref="char:cmbAbbrStroke">̄</g>ber BC hath to the num<g ref="char:EOLhyphen"/>ber BD that proportion that a square number hath to a square number. Wherefore the square of the line FG hath to the square of the line K that proportion that a square number hath to a square number. Wherefore the line FG is commensurable in length to the line K. Wherefore the line FG is in power more then the line HG, by the square of a line commen<g ref="char:EOLhyphen"/>surable in length to the line FG, and neither of the lines FG and GH is commensurable in length to the rationall line A: when yet notwithstanding either of the lines FG and GH is rationall. Wherefore the line FH is a third residuall line. Wherefore there is found out a third residuall line: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 22. Probleme. The 88. Proposition. To finde out a fourth residuall line.</head>
               <p>TAke a rationall line and let the same be <hi>A</hi>: and vnto it let the line <hi>BG</hi> be commensurable in length. Wherefore the line <hi>BG</hi> is rationall. And take two numbers <hi>DF</hi> and <hi>FE,</hi>
                  <note place="margin">Construction.</note> and let them be such that the whole number, namely, <hi>DE</hi> haue to neither of the numbers <hi>DF</hi> and <hi>FE</hi> that proportion that a square number hath to a square
<figure/>
number. And as the number <hi>DE</hi> is to the number <hi>EF,</hi> so let the square of the line <hi>BG</hi> be to the square of the line <hi>GC</hi>: wherefore the square of the line <hi>BG</hi> is com<g ref="char:EOLhyphen"/>mensurable to the square of the line <hi>GC,</hi> wherefore also the square of the line <hi>GC</hi> is rationall, and the line <hi>GC</hi> is also rationall. And for that the number <hi>DE</hi> hath not the number <hi>EF</hi> that proportion that a square number hath to a square number,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion</note> therefore the line <hi>BG</hi> is incom<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>GC.</hi> And they are both rationall: wherefore the line <hi>BC</hi> is a residuall line. I say moreouer that it is a fourth residuall line. For forasmuch as the square of the line <hi>BG</hi> is greater then the square of the line <hi>GC,</hi> vnto the square of the line <hi>BG</hi> let the squares of the lines <hi>CG</hi> and <hi>H</hi> be equall. And for that as the number <hi>DE</hi> is to the number <hi>EF,</hi> so is the square of the line <hi>BG</hi> to the square of the line <hi>GC,</hi> therefore by con<g ref="char:EOLhyphen"/>uersion of proportion as the number <hi>DE</hi> is to the number <hi>DF,</hi> so is the square of the line <hi>BG</hi> to the square of the line <hi>H.</hi> But the numbers <hi>DE</hi> and <hi>DF</hi> haue not the one to the other that porportion that a square number hath to a square number. Wherefore the line <hi>BG</hi> is incommensurable in length to the line <hi>H.</hi> Wherefore the line <hi>BG</hi> is in power more then
<pb n="292" facs="tcp:7062:321"/>
the line <hi>GC</hi> by the square of a line incommensurable in length so the line <hi>BG</hi>: and the whole line <hi>BG</hi> is commensurable in length to the rationall line <hi>A,</hi> Wherefore the line <hi>BC</hi> is a fourth residuall line. Wherefore there is founde out a fourth residuall line: which was re<g ref="char:EOLhyphen"/>quired to be doone.</p>
            </div>
            <div type="part">
               <head>¶ The 23. Probleme. The 89. Proposition. To finde out a fift residuall lyne.</head>
               <p>TAke a rational line and let the same be A,<note place="margin">Construction.</note> and vnto it let the line CG be commensurable in length. Wherefore the line CG is rationall. And take two numbers DF and FE, which let be such, that the number DE haue to neither of these numbers DF nor FE that proportion that a square number hath to a square number. And as the number FE is to the number DE, so let the square of the line CG be to the
<figure/>
square of the line BG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the square of the line CG is commensurable to the square of the line BG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore the square of the line BG is rationall, and the line BG is also rational. But the numbers DE and EF haue not that proportion the one to the other that a square number hath to a square nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore the lines BG and GC are ra<g ref="char:EOLhyphen"/>tionall commensurable in power onely. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore the line BC is a residuall line. I say moreouer that it is a fift resi<g ref="char:EOLhyphen"/>duall line. For forasmuch as the square of the line BG is greater then the square of the line GC, vnto the square of the line BG let the squares of the lines GC and H be equal. Now ther<g ref="char:EOLhyphen"/>fore for that as the number DE is to the number EF, so is the square of the line BG to the square of the line GC, therfore by conuersion of proportion, at the number DE is to the nu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ber DF, so is the square of the line BG to the square of the line H. But the numbers DE &amp; DF haue not that proportion the one to the other that a square number hath to a square number. Wherefore the line BG is incommensurable in length to the line H. Wherefore the line BG is in power more then the line CG by the square of a line incommensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line BG, and the line CG which is ioyned to the residual line is comme<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rationall line A. Wherefore the line BC is a fift residuall line. Wherefore there is found out a fift residuall line: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 24. Probleme. The 90. Proposition. To finde out a sixth residuall line.</head>
               <p>TAke a rational line and let the same be A,<note place="margin">Construction.</note> And
<figure/>
take three numbers E, BC, and CD, not ha<g ref="char:EOLhyphen"/>uing the one to the other that proportion that a square number hath to a square number. And let not the number BC haue to the number BD that proporti<g ref="char:EOLhyphen"/>on that a square number hath to a square number. And let the number BC be greater then the number CD, &amp; as the number E is to the number BC, so let the square of the line A be to the square of the lyne FG. And as the number BC is to the number CD, so let the square
<pb facs="tcp:7062:322"/>
of the line FG be to the square of the line GH. Now therfore for that as the number E is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber BC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> so is the square of the line A to the square of the line FG, therfore the square of the line A is commensurable to the square of the line F G. Wherfore the square of the line FG is rationall, and the line FG is also rationall. And for that the number E hath not to the number BC that proportion that a square number hath to a square number, therfore the line A is incommensurable in length to the line FG. Againe for that as the number BC is to the number CD, so is the square of the line FG to the square of the line GH, therefore the square of the line FG is commensurable to the square of the line GH. But the square of the line FG is rationall, wherfore the square also of the line GH is rationall, wherfore the line GH is also rationall. And for that the number B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hath not to the number CD that propor<g ref="char:EOLhyphen"/>tion that a square number hath to a square number, therfore the line FG is incommensura<g ref="char:EOLhyphen"/>ble in length to the line GH, and they are both rationall. Wherefore the lines FG and GH are rationall commensurable in power onely. Wherfore
<figure/>
the lyne FH is a residuall line. I say moreouer that it is a sixt residuall line. For for that as the number E is to the number BC, so is the square of the line A to the square of the line FG, and as the number BC is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber CD, so is the square of the line FG to the square of the line GH, therefore by equalitie of proportion as the number E is to the number CD, so is the square of the line A to the square of the line GH. But the num<g ref="char:EOLhyphen"/>ber E hath not to the number CD that proportion that a square number hath to a square number. Wherefore the line A is in<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>mmensurable in length to the line GH, and neither of these lines FG nor G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is commensurable in length to the rationall line A. And forasmuch as the square of the line FG is greater then the square of the line GH, vnto the square of the line FG let the the squares of the lines GH and K be equall. Now therfore for that as the number B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the number CD, so is the square of the line FG to the square of the line GH, therefore by conuersion of proportion as the number BC is to the number BD, so is the square of the line FG to the square of the line K. But the number BC hath not to the number BD that pro<g ref="char:EOLhyphen"/>portion that a square number hath to a square number, therfore the line FG is incommen<g ref="char:EOLhyphen"/>surable in length to the line K. Wherfore the line FG is in power more then the lyne GH by the square of a line incommensurable in length to the line FG, and neither of the lines FG nor GH is commensurable in length to the rationall line A. Wherfore the line FH is a sixt residual line. Wherfore there is found out a sixt residuall line: which was required to be done.</p>
               <p>
                  <note place="margin">An other more redie way to finde out the sixe re<g ref="char:EOLhyphen"/>siduall lines.</note>There is also a certayne other redier way to finde out euery one of the forsayd sixe residu<g ref="char:EOLhyphen"/>all lines which is after this maner. Suppose that it were required to finde out a first residuall line: Take a first binomiall line <hi>AC,</hi> &amp; let the greater name
<figure/>
thereof be <hi>AB.</hi> And vnto the line <hi>BC</hi> let the line <hi>BD</hi> be equall. Wherefore the lines <hi>AB</hi> and <hi>BC,</hi> that is the lines <hi>AB</hi> and <hi>BD</hi> are rationall commensurable in power onely, and the line <hi>AB</hi> is in power more then the line <hi>BC,</hi> that is, then the line <hi>BD</hi> by the square of a line commensurable in length to the line <hi>AB.</hi> And the line <hi>AB</hi> is commensurable in length to the rationall line geuen. For the line <hi>AC</hi> is put to be a first binomiall line. Where<g ref="char:EOLhyphen"/>fore the line <hi>AD</hi> is a first residual line. And in like maner may ye finde out a second, a third, a fourth, a fift, and a sixt residuall line, if ye take for eche a binomiall line of the same order.</p>
            </div>
            <div type="part">
               <pb n="293" facs="tcp:7062:322"/>
               <head>¶The 67. Theoreme. The 91. Proposition. If a superficies be contayned vnder a rationall line &amp; a first residuall line: the line which contayneth in power that superficies, is a residuall line.</head>
               <p>SVppose that there be a rectangle superficies <hi>AB</hi> contayned vnder a rationall line <hi>AC</hi> and a first residuall line <hi>AD.</hi>
                  <note place="margin">Fourth Sena<g ref="char:EOLhyphen"/>ry.</note> Then I say that the line which contayneth in power the superficies <hi>AB</hi> is a residuall line. For forasmuch as <hi>AD</hi> is a first re<g ref="char:EOLhyphen"/>siduall line,<note place="margin">The <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the Con<g ref="char:EOLhyphen"/>struction.</note> let the line ioyned vnto it be <hi>DG</hi> (by the line ioyned vnto it vnder<g ref="char:EOLhyphen"/>stand
<figure/>
such a line as was spoken of in the end of the 79. pro<g ref="char:EOLhyphen"/>position). Wherefore the lines <hi>AG</hi> and <hi>GD</hi> are rationall co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in pow<g ref="char:EOLhyphen"/>er only, &amp; the whole line <hi>AG</hi> is co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in length to the rationall line <hi>AC,</hi> and the line <hi>AG</hi> is in power more then the line <hi>GD</hi> by the square of a line commensurable in length vnto <hi>AG,</hi> by the definition of a first residuall line. Deuide the line <hi>GD</hi> into two equall partes in the poynt <hi>E.</hi> And vpon the line <hi>AG</hi> apply a parallelogramme equall to the square of the line <hi>EG</hi> and wanting in figure by a square, and let the sayd parallelogramme be that which is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned vnder the lines <hi>AF</hi> and <hi>FG.</hi>
                  <note place="margin">The first part of the demon<g ref="char:EOLhyphen"/>stration.</note> Wherefore the line <hi>AF</hi> is commensurable in length to the line <hi>FG</hi> (by the <hi>17.</hi> of the tenth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And by the poyntes <hi>E, F</hi> and <hi>G,</hi> draw vnto the line <hi>AC</hi> these parallel lines <hi>EH, FI,</hi> and <hi>GK.</hi> And make perfect the parallelograme <hi>AK.</hi> And for<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>as much as the line <hi>AF</hi> is commensurable in length to the line <hi>FG,</hi> therefore also the whole line <hi>AG</hi> is commensurable i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> length to either of the lines <hi>AF</hi> and <hi>FG</hi> (by the <hi>15.</hi> of the tenth). But the line <hi>AG</hi> is commensurable in length to the line <hi>AC.</hi> Wherefore either of the lines <hi>AF</hi> and <hi>FG</hi> is commensurable in length to the line <hi>AC.</hi> But the line <hi>AC</hi> is ra<g ref="char:EOLhyphen"/>tionall, wherefore either of the lines <hi>AF</hi> and <hi>FG</hi> is also rationall. Wherefore (by the 19. of the tenth) either of the parallelogrammes <hi>AI</hi> and <hi>FK</hi> is also rationall.<note place="margin">Note. AI and FK concluded ra<g ref="char:EOLhyphen"/>tional paralle<g ref="char:EOLhyphen"/>logramme.</note> And forasmuch as the line <hi>DE</hi> is comme<g ref="char:cmbAbbrStroke">̄</g>surable in length to the line <hi>EG,</hi> therfore also (by the 15. of the tenth) the line <hi>DG</hi> is commensurable in length to either of the lines <hi>DE</hi> and <hi>EG.</hi> But the line <hi>DG</hi> is rationall, wherefore either of the lines <hi>DE</hi> and <hi>EG</hi> is rationall, and the selfe same line <hi>DG</hi> is incommensurable in length to the line <hi>AC</hi> (by the definition of a first residuall line, or by the 13. of the tenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>). For the line <hi>DG</hi> is incommensurable in length to the line <hi>AG,</hi> which line <hi>AG</hi> is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line <hi>AC</hi>:<note place="margin">
                     <p>Note.</p>
                     <p>DH and FK parallelo<g ref="char:EOLhyphen"/>grammes me<g ref="char:EOLhyphen"/>diall.</p>
                  </note> wherfore either of the lines <hi>DE</hi> and <hi>EG</hi> is rationall and incommensurable in length to the line <hi>AC</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore (by the 21. of the tenth) either of these parallelogrammes <hi>DH</hi> and <hi>EK</hi> is mediall. Vnto the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>AI</hi> let the square <hi>LM</hi> be equall, and vnto the parallelogramme <hi>FK</hi> let the square <hi>NX</hi> be equall, being taken away from the square <hi>LM</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ing the angle <hi>LOM</hi> common to them both.<note place="margin">Second part of the construc<g ref="char:EOLhyphen"/>tion.</note> (And to doo this, there must be founde out the meane proportionall be<g ref="char:EOLhyphen"/>twene the lines <hi>FI</hi> and <hi>FG.</hi> For the square of the meane proportionall is equall to the pa<g ref="char:EOLhyphen"/>rallelogramme contayned vnder the lines <hi>FI</hi> and <hi>FG.</hi> And from the line <hi>LO</hi> cut of a line equall to the meane proportionall so founde out, and descri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the square thereof). Wherefore both the squares <hi>LM</hi> and <hi>NX</hi> are about one and the selfe same diameter (by the 20. of the
<pb facs="tcp:7062:323"/>
sixth) let their diameter be <hi>OR</hi> and describe the figure as it is h<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t forth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Now then for<g ref="char:EOLhyphen"/>asmuch as the parallelogramme contayned vnder the lines <hi>AF</hi> &amp; <hi>FG</hi> is equal to the square of the line <hi>EG,</hi>
                  <note place="margin">Second part of the demonstra<g ref="char:EOLunhyphen"/>tion.</note> therefore (by the 17. of the sixth) as the line <hi>AF</hi> is to the line <hi>EG,</hi> so is the line <hi>EG</hi> to the line <hi>FG.</hi> But as the line <hi>AF</hi> is to the line <hi>EG,</hi> so is th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> parallelogramme <hi>AI</hi> to the parallelo<g ref="char:EOLhyphen"/>gramme
<figure/>
                  <hi>EK.</hi> And as the line <hi>EG</hi> is to the line <hi>FG,</hi> so is the parallelogramme <hi>EK</hi> to the parallelo<g ref="char:EOLhyphen"/>gramme <hi>FK.</hi> Wher<g ref="char:EOLhyphen"/>fore betwene the pa<g ref="char:EOLhyphen"/>grammes <hi>AI</hi> and <hi>FK</hi> the parallelo<g ref="char:EOLhyphen"/>gramme <hi>EK</hi> is the meane proportionall. But (by the second part of the assumpt going before the 54. of the tenth) be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wene the squares <hi>LM</hi> and <hi>NX</hi> the parallelogramme <hi>MN</hi> is the meane proportionall. And vnto the paral<g ref="char:EOLhyphen"/>lelogramme <hi>AI</hi> is equall the square <hi>LM,</hi> and vnto the parallelogramme <hi>FK</hi> is equall the square <hi>NX</hi> by construction. Wherefore the parallelogramme <hi>MN</hi> is equall to the parallelo<g ref="char:EOLhyphen"/>gramme <hi>EK</hi> (by the 2. assumpt going before the 54. of the tenth). But the parallelogramme <hi>EK</hi> is (by the first of the sixth) equall to the parallelogramme <hi>DH,</hi> and the parallelogramme <hi>MN</hi> is (by the 43. of the first) equall to the parallelogramme <hi>LX.</hi> Wherefore the whole pa<g ref="char:EOLhyphen"/>rallelogramme <hi>DK</hi> is equall to the gnomon <hi>VTZ</hi> (which gnomon consisteth of those pa<g ref="char:EOLhyphen"/>rallelogrammes by which ye see in the figure passeth a portion of a circle greater then a se<g ref="char:EOLhyphen"/>micircle) and moreouer to the square <hi>NX</hi>: and the parallelogramme <hi>AK</hi> is equall to the squares <hi>LM</hi> and <hi>NX</hi> by construction: and it is now proued, that the parallelogramme <hi>DK</hi> is equall to the gnomo<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>VTZ,</hi> and moreouer to the square <hi>NX.</hi> Wherfore the residue name<g ref="char:EOLhyphen"/>ly the parallelogramme <hi>AB</hi> is equall to the square <hi>SQ</hi> which is the square of the line <hi>LN.</hi>
                  <note place="margin">LN, is the onely li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hat we sought &amp; consider.</note> Wherefore the square of the line <hi>LN</hi> is equall to the parallelogramme <hi>AB.</hi> Wherefore the line <hi>LN</hi> contayneth in power the parallelogramme <hi>AB.</hi> I say moreouer that the line <hi>LN</hi> is a residuall line. For forasmuch as either of these parallelogrammes <hi>AI</hi> and <hi>FK</hi> is rationall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> as it is before sayd, therefore the squares <hi>LM</hi> and <hi>NX</hi> which are equall vnto them, that is, the squares of the lines <hi>LO</hi> and <hi>ON</hi> are rationall. Wherefore the lines <hi>LO</hi> and <hi>ON</hi> are also rationall. Agayne forasmuch as the parallelogramme <hi>DH</hi> that is <hi>LX</hi> is mediall, there<g ref="char:EOLhyphen"/>fore the parallelogramme <hi>LX</hi> is incommensurable to the square <hi>NX.</hi> Wherefore (by the 1. of the sixth, and 10. of the tenth) the line <hi>LO</hi> is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line <hi>ON</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and they are both rationall. Wherefore they are lines rationall commensurable in power onely. Wherefore <hi>LN</hi> is a residuall line by the definition, and it contayneth in power the parallelo<g ref="char:EOLhyphen"/>parallelogramme <hi>AB.</hi> If therefore a superficies be contayned vnder a rationall line and a first residual line, the line which contayneth in power that superficies<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is a residuall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 68. Theoreme. The 92. Proposition. If a superficies be contained vnder a rationall line and a second residuall line: the line which containeth in power that superficies, is a first mediall residuall line.</head>
               <p>
                  <pb n="295" facs="tcp:7062:323"/>SVppose that AB be a superficies contained vnder <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ra<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>onall line AC, and a se<g ref="char:EOLhyphen"/>cond residuall li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> AD.<note place="margin">First part of the construc<g ref="char:EOLhyphen"/>tion.</note> Then I say<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that the line th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> containeth in power the superficies AB is a first mediall residuall line. For let the line ioyned to the line AD be DG. Wherefore the lines AG and GD are rationall commensurable in power onely, and the
<figure/>
line that is ioyned to the residual line, namely, the line DG is co<g ref="char:cmbAbbrStroke">̄</g>mensura<g ref="char:EOLhyphen"/>ble in length to the rati<g ref="char:EOLhyphen"/>onall line AC: and the line AG is in power more then the line DG, by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line AG. Deuide the line DG into two e<g ref="char:EOLhyphen"/>quall partes in the point E. And vnto the line AG apply a parallelogramme equall to the fourth part of the square of the line DG, that is, equall to the square of the line EG, and wanting in figure by a square, and let that parallelogramme be that which is contained vn<g ref="char:EOLhyphen"/>der the lines AF and FG. Wherefore (by the 1<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> of the tenth) the line AF is commensura<g ref="char:EOLhyphen"/>ble in length to the line FG. And by the pointes E, F, and G, draw vnto the line AC these parallell lines EH, FI, and GK: and forasmuch as the line AF is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line FG,<note place="margin">The first part of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> demon<g ref="char:EOLhyphen"/>stration.</note> therefore the whole line AG is commensurable in length to either of these lines AF and FG. But the line AG is rationall and incommensurable in length to the line AC. Wherefore either of these lines AF and FG are rationall and incommensurable in length to the line AC. Wherefore either of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se parallelogrammes AI and FK is (by the 21. of the tenth) mediall.<note place="margin">AI and FK concluded pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>mes mediall.</note> Againe, forasmuch as the line DE is commensurable in length to the line EG, therefore the line DG is commensurable in length to either of these lines DE and EG. But the line DG is commensurable in length to the rationall line AC. Wherefore either of these lines DE and EG is rationall and commensurable in length to the line AC. Wherefore (by the 19. of the tenth) either of these parallelogrammes DH and EK is rati<g ref="char:EOLhyphen"/>onall.<note place="margin">DH &amp; EK, rationall.</note> Vnto the parallelogramme AI describe an equall square LM,<note place="margin">The second part of the construction.</note> and vnto the paralle<g ref="char:EOLhyphen"/>logramme FK let the square NX be equall, as in the Proposition going before. Wherefore the squares LM and NX are both about one and the same diameter.<note place="margin">The second part of the de<g ref="char:EOLhyphen"/>monstration.</note> Let the diameter be OR, and describe the figure as is in the former Proposition expressed. Now therefore for<g ref="char:EOLhyphen"/>asmuch as the parallelogrammes AI and FK are mediall are <note n="*" place="margin">Analytically the pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e hereof followeth amo<g ref="char:cmbAbbrStroke">̄</g>g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> other thinges.</note> commensurable the one to the other, and the squares of the lines LO &amp; ON which are equall to those parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes, are mediall, therefore the lines LO and ON are also mediall commensurable in power.
<q>(And it is manifest, that the lines LO and ON are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power, for their squares are commensurable, and those squares, namely, the squares of the lines LO &amp; ON are com<g ref="char:EOLhyphen"/>mensurable, for they are equall to the parallelogrammes AI and FK, which are commensu<g ref="char:EOLhyphen"/>rable the one to the other: and that those parallelogrammes AI and FK are commensurable the one to the other, hereby it is manifest, for that it was before proued that the lines AF and FG are commensurable in length. Wherefore (by the 1. of the sixt, and 10. of the tenth) the parallelogrammes AI and FK are commensurable the one to the other.</q>
Wherefore it is now manifest by the way of resolution, that the lines LO &amp; ON are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And forasmuch as the parallelogramme contained vnder the lines AF and FG is equall to the square of the line EG, therefore as the line AF is to the line EG, so is the line EG to the line FG. But as the line AF is to the line EG, so is the parallelogramme AI to the pa<g ref="char:EOLhyphen"/>rallelogramme EK: and as the line EG is to the line FG, so is the parallelogramme EK to
<pb facs="tcp:7062:324"/>
the parallelogramme FK. Wherefore the parallelogramme <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>K is the mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proportionall betwene the parallelogrammes AI and FK: and the parallelogramme MN i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> also the meane proportionall be<g ref="char:EOLhyphen"/>twene
<figure/>
the squares LM &amp; NX: and the paral<g ref="char:EOLhyphen"/>lelogramme AI is equall to the square LM: and the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me FK is equall to the square NX. Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme MN is equall to the parallelo<g ref="char:EOLhyphen"/>gramme EK. But the parallelogramme EK is equall to the parallelogramme DH: and the parallelogramme LX is equall to the parallelo<g ref="char:EOLhyphen"/>gramme MN. Wherefore the whole parallelogramme DK is equall to the Gnomon VTZ, and to the square NX. Wherefore the residu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, namely, the parallelogramme AB is equall to the square SQ, that is, to the square of the line LN. Wherefore the line LN containeth in power the superficies AB.<note place="margin">The line LN found, which is the princi<g ref="char:EOLhyphen"/>pall drift of all the former discourse.</note> I say moreouer, that the line LN is a first mediall residuall line. For forasmuch as the parallelogramme EK is rationall, and is equall to the parallelo<g ref="char:EOLhyphen"/>gramme MN, that is, LX, therefore LX, that is, the parallelogramme contained vnder the lines LO and ON is rationall. But the square NX is mediall, for it is already proued that the parallelogramme FK which is equall to the square NX, is mediall. Wherefore the pa<g ref="char:EOLhyphen"/>rallelogramme LX is incommensurable to the square NX. But as the parallelogramme LX is to the square NX, so is the line LO to the line ON (by the 1. of the sixt). Wherefore (by the 10. of the tenth) the lines LO and ON are incommensurable in length. And it is alrea<g ref="char:EOLhyphen"/>dy proued, that they are mediall commensurable in power. Wherefore the lines LO &amp; ON are mediall commensurable in power onely containing a rationall superficies. Wherefore the line LN is a first mediall residuall line, and containeth in power the superficies AB, which is contained vnder a rationall line and a second residuall line. If therefore a superficies be contained vnder a rationall line and a second residuall line: the line which containeth in power that superficies, is a first mediall residuall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 69. Theoreme. The 93. Proposition. If a superficies be contained vnder a rationall line and a third residuall line: the line that containeth in power that superficies is a second mediall residuall line.</head>
               <p>SVppose that AB be a superficies contained vnder a rationall line AC, &amp; a third residuall line AD. Then I say, that the line which containeth in power the su<g ref="char:EOLhyphen"/>perficies AB is a second mediall residuall line. Let the line ioyned vnto AD, be DG. Wherefore the lines AG and GD are rationall commensurable in power onely,<note place="margin">The first part of the Con<g ref="char:EOLhyphen"/>struction.</note> and neither of the lines AG nor GD is commensurable in length to the rationall line AC, and the whole line AC is in power more then the line GD, by the square of a line com<g ref="char:EOLhyphen"/>mensurable in length to the line AG. Let the rest of the construction be as it was in the for<g ref="char:EOLhyphen"/>mer Propositions.<note place="margin">The fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>st part of the demon<g ref="char:EOLhyphen"/>stration.</note> Wherefore the lines AF and FG are commensurable in length: and the parallelogramme AI is commensurable to the parallelogramme FK. And forasmuch as the lines AF and FG are commensurable in length, therefore the whole line AG is commensu<g ref="char:EOLhyphen"/>rable
<pb n="295" facs="tcp:7062:324"/>
in length to either of these lines AF and FG. B<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AG is rationall and in<g ref="char:EOLhyphen"/>commensurable in length to the line AG. Wherefore either of these lines AF and FG is ra<g ref="char:EOLhyphen"/>tionall and incommensurable in length to the line AG.<note place="margin">
                     <p>Note.</p>
                     <p>AI and FK mediall.</p>
                  </note> Wherefore (by the 21. of the tenth) either of these parallelogrammes AI and FK is mediall. Again<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, forasmuch as the line DB is commensurable in length to the line EG, therefore also the whole line DG is commensu<g ref="char:EOLhyphen"/>rable in length to either of these lines DE and EG. But the line DG is rationall commen<g ref="char:EOLhyphen"/>surable in power on<g ref="char:EOLhyphen"/>ly
<figure/>
to the line AC. Wherfore also either of the lines DE and EG is rationall and commensurable in power only to the line AC.<note place="margin">
                     <p>Note.</p>
                     <p>DH and EK mediall.</p>
                  </note> Wherefore ei<g ref="char:EOLhyphen"/>ther of these paralle<g ref="char:EOLhyphen"/>logrammes DH and EK is mediall. A<g ref="char:EOLhyphen"/>gaine forasmuch as the lines AG and DG are commensurable in power onely, therefore they are incommensura<g ref="char:EOLhyphen"/>ble in length. But the line AG is commensurable in length to the line AF: and the line DG is commensurable in length to the line GE. Wherefore the line AF is incommensurable in length to the line EG. But as the line AF is to the line EG, so is the parallelogramme AI to the parallelogramme EK.<note place="margin">
                     <p>Note.</p>
                     <p>AI incommen<g ref="char:EOLhyphen"/>surable to EK. Second part of the Constructio<g ref="char:cmbAbbrStroke">̄</g>.</p>
                  </note> Wherefore the parallelogramme AI is incommensurable to the parallelogramme EK. Vnto the parallelogramme AI describe an equall square LM: and vnto the parallelogramme FK describe an equall square NX: and describe the figure as you did in the former Proposition. Now forasmuch as the parallelogramme contained vnder the lines AF and FG is equall to the square of the line EG, therefore as the line AF is to the line EG, so is the line EG to the line FG. But as the line AF is to the line EG, so is the pa<g ref="char:EOLhyphen"/>rallelogramme AI to the parallelogramme EK: and as the line EG is to the line FG, so is the parallelogramme EK to the parallelogramme FK. Wherefore as the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AI is to the parallelogramme EK, so is the parallelogramme EK to the parallelogramme FK. Wherefore the parallelogramme EK is the meane proportionall betwene the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes AI and FK. But the parallelogramme MN is the meane proportionall betwene the squares LM and NX. Wherefore the parallelogramme EK is equall to the parallelogramme MN. Wherefore the whole parallelogramme DK is equall to the Gnomon VTZ, &amp; to the square NX. And the parallelogramme AK is equall to the squares LM and NX. Wherefore the residue, namely, the parallelogramme AB is equall to the square QS, that is,<note place="margin">The principall line, LN fou<g ref="char:cmbAbbrStroke">̄</g>de.</note> to the square of the line LN. Wherefore the line LN containeth in power the superficies AB. I say more<g ref="char:EOLhyphen"/>ouer, that the line LN is a second mediall residuall line. For for that as it is proued, the pa<g ref="char:EOLhyphen"/>rallelogramme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AI and FK are mediall, therefore the squares that are equall vnto them, namely, the squares of the lines LO and ON are also mediall. Wherefore either of these lines LO and ON is mediall. And forasmuch as the parallelogramme AI is <note n="*" place="margin">Because the lines AF and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>G are proued commensurable in length.</note> commensu<g ref="char:EOLhyphen"/>rable to the parallelogramme FK, therefore the squares that are equall to them, namely, the squares of the lines LO and ON are also commensurable. Againe, forasmuch as it is proued, that the parallelogramme AI is incommensurable to the parallelogramme EK, therfore the square LM is incommensurable to the parallelogramme MN, that is, the square of the line LO to the parallelogramme contained vnder the lines LO &amp; ON. Wherfore also the <note n="*" place="margin">By the first o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the sixth and tenth of the tenth.</note> line LO is incommensurable in length to the line ON. Wherefore the lines LO and ON are mediall commensurable in power onely. I say moreouer, that they containe a mediall super<g ref="char:EOLhyphen"/>ficies. For forasmuch as it is proued, that the parallelogramme EK is mediall, therefore the
<pb facs="tcp:7062:325"/>
parallelogramme which is equall vnto it, namely, the parallelogramme contained vnder the lines LO and ON is also mediall. Wherefore the line LN is a second mediall residuall line, and containeth in power the superficies AB. Wherefore the line that containeth in power the superficies AB is a second mediall residuall line. If therefore a superficies be contained vnder a rationall line and a third residuall line, the line that containeth i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> power that superficies, is a second mediall residuall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>The 70. Theoreme. The 94. Proposition. If a superficies be contayned vnder a rationall lyne, and a fourth residuall lyne: the lyne which contayneth in power that superficies, is a lesse lyne.</head>
               <p>SVppose that there be a superficies AB contained vnder a rationall line AC, and a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ourth residuall line AD. Then I say that the line which containeth in power the superficies AB is a lesse line. For let the line ioyned vnto it be DG. Where<g ref="char:EOLhyphen"/>fore the lines AG and DG are rationall commensurable in power only, and the line AG is in power more then the line DG, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line AG, and the line AG is commensurable in length to the line AC. Deuide the line DG into two equall partes in the point E.<note place="margin">The first part of the con<g ref="char:EOLhyphen"/>struction.</note> And vnto the line AG apply a parallelo<g ref="char:EOLhyphen"/>gramme equal to the
<figure/>
square of the line EG, and wanting in fi<g ref="char:EOLhyphen"/>gure by a square, and let that parallelo<g ref="char:EOLhyphen"/>grame be that which is contayned vnder the lines AF and FG. Wherfore (by the 18. of the tenth) the line AF is inco<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in length to the line FG. Draw by the pointes E, F, &amp; G, vnto the lines AC and DB these parallel lines EH, FI, and GK.<note place="margin">The first part of the demon<g ref="char:EOLhyphen"/>stration.</note> Now forasmuch as the line AG is rational, and commensurable in length to the line AC, therfore the whole parallelogramme AK is (by the 19. of the tenth) rational.<note place="margin">
                     <p>Note.</p>
                     <p>AK rational.</p>
                  </note> Againe forasmuch as the line DG is incommensurable in length to the line AC (for if the line DG were commensurable in length to the line AC, then forasmuch as the line AG is commensurable in length to the same line AC, the lines AG and DG should be commensu<g ref="char:EOLhyphen"/>rable in length the one to the other, when yet they are put to be commensurable in power one<g ref="char:EOLhyphen"/>ly) and both these lines AC and DG are rationall.<note place="margin">
                     <p>Note.</p>
                     <p>DK mediall. AI and FK incommensu<g ref="char:EOLhyphen"/>rable.</p>
                     <p>The second part of the construction.</p>
                     <p>The second part of the demonstration.</p>
                  </note> Wherfore the parallelograme DK is me<g ref="char:EOLhyphen"/>diall. Againe forasmuch as the line AF is incommensurable in length to the line FG, ther<g ref="char:EOLhyphen"/>fore the parallelogramme AI is incommensurable to the parallelogramme FK. Vnto the pa<g ref="char:EOLhyphen"/>rallelogramme AI describe an equal square LM, and vnto the parallelograme FK describe an equall square NX, and let the angle LOM be common to both those squares. Wherefore the squares LM and NX are about one and the selfe same diameter. Let their diameter be OR, and describe the figure. And forasmuch as the parallelograme contained vnder the lines AF and FG is equall to the square of the line EG, therfore proportionally as the line AF is to the line EG, so is the line EG to the line FG, but as the line AF is to the line EG, so is the parallelogramme AI to the parallelogramme EK (by the 1. of the sixt) And as the line EG is to the line FG, so is the parallelogramme EK to the parallelogramme FK. Wherfore
<pb n="296" facs="tcp:7062:325"/>
the parallelogramme EK is the meane proportionall betwene the parallelogrammes AI and FK: wherfore as i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> was said in the former propositions, the parallelogramme MN is equal to the parallelogramme EK: but the parallelogramme DH is equall to the parallelograme EK, and the parallelogramme MN to the parallelogramme LX. Wherfore the whole pa<g ref="char:EOLhyphen"/>rallelogramme DK is equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll to the gnomon VTZ, and to the square NX. Wherfore the re<g ref="char:EOLhyphen"/>sidue, namely, the parallelogramme AB is equall to the residue, namely, to the square SQ,<note place="margin">LN (the chiefe line of this theoreme) founde.</note> that is to the square of the line LN. I say moreouer that LN is that irrationall line which is called a lesse line. For forasmuch as the parallelogramme AK is rationall, and is equall to the squares of the lines LO and ON, therfore that which is made of the squares of the lines LO and ON added together is rationall. Againe forasmuch as the parallelogramme DK is me<g ref="char:EOLhyphen"/>diall, and is equall to that which is contained vnder the lines LO and ON twise, therefore that which is contained vnder the lynes LO and ON twise, is also mediall. And forasmuch as the parallelogramme AI is incommensurable to the parallelogramme FK, therefore the squares which are equall vnto them, namely, the squares of the lines LO and ON are incom<g ref="char:EOLhyphen"/>mensurable the one to the other. Wherfore the lines LO and ON are incommensurable in power, hauyng that which is made of their squares added together rationall, and that which is contained vnder them twise mediall, which is commensurable to that which is contayned vnder them once. Wherfore that which is contained vnder them once is also mediall. Wher<g ref="char:EOLhyphen"/>fore LN is that irrationall line which is called a lesse line, and it containeth in power the su<g ref="char:EOLhyphen"/>perficies AB. If therfore a superficies be contained vnder a rationall line and a fourth resi<g ref="char:EOLhyphen"/>duall line, the line which containeth in power that superficies is a lesse line: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 71. Theoreme. The 95. Proposition. If a superficies be contained vnder a rationall line and a fift residual line: the line that co<g ref="char:cmbAbbrStroke">̄</g>tayneth in power the same superficies, is a line making with a rationall superficies, the whole superficies mediall.</head>
               <p>SVppose that there be a superficies AB contained vnder a rationall line AG and a fift residuall line AD. The<g ref="char:cmbAbbrStroke">̄</g> I say that the line that co<g ref="char:cmbAbbrStroke">̄</g>taineth in power y<hi rend="sup">e</hi> sup<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<g ref="char:EOLhyphen"/>ficies AB, is a line making with a rationall superficies the whole superficies me<g ref="char:EOLhyphen"/>diall. For vnto the line AD let the line DG be ioyned, which shal be co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>a<g ref="char:EOLhyphen"/>ble in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rational line AC. And let the rest of the constructio<g ref="char:cmbAbbrStroke">̄</g> be as in the propositio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> next going before.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AG is inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line AC and they are both rationall, therfore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AK is medial. Againe forasmuch as the line DG is rationall
<figure/>
and commensurable in length to the lyne AC, theref<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re the parallelo<g ref="char:EOLhyphen"/>gramme DK is ratio<g ref="char:EOLhyphen"/>nall. Vnto the parallel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>gramme AI describe an equall square LM; and vnto the parallelograme <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> describe an equall square N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and as in <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> proposition next go<g ref="char:EOLhyphen"/>ing
<pb facs="tcp:7062:326"/>
before, so also in this may we proue, that the line LN containeth in power the superficies AB.<note place="margin">The line LN.</note> I say moreouer that that line LN is a line making with a rationall superficies the whole superficies mediall. For forasmuch as the parallelogramme AK is mediall, therefore that which is equall vnto it, namely, that which is made of the squares of the lines LO and ON added together is also mediall. Againe forasmuch as the parallelogramme DK is ra<g ref="char:EOLhyphen"/>tionall, therfore that which is equall vnto it, namely, that which is contained vnder the lines LO and ON twise, is also rationall. And forasmuch as the line AF is incommensurable in length to the line FG, therfore (by the 1. of the sixt, &amp; 10. of the tenth) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is incommensurable to the parallelogramme FK, wherfore also the square of the lyne LO is incommensurable to the square of the line ON. Wherfore the lines LO and ON are in<g ref="char:EOLhyphen"/>commensurable in power hauing that which is made of their squares added together medi<g ref="char:EOLhyphen"/>all, and that which is contayned vnder them twise rationall. Wherfore the line LN is that irrationall line which is called a lyne making with a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies the whole superfi<g ref="char:EOLhyphen"/>cies mediall, and it contayneth in power the superficies AB. Wherfore the line contayning in power the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies AB, is a line making with a rationall superficies the whole superficies mediall. If therfore a superficies be contayned vnder a rationall lyne &amp; a fift residuall line, the line that contayneth in power the same superficies, is a line makyng with a rationall su<g ref="char:EOLhyphen"/>perficies the whole super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies mediall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 72. Theoreme. The 96. Proposition. If a superficies be contayned vnder a rationall line and a sixth residuall line, the line which contayneth in power the same superficies is a line ma<g ref="char:EOLhyphen"/>king with a mediall superficies the whole superficies mediall.</head>
               <p>SVppose that AB be a superficies contayned vnder a rationall line AC, &amp; a sixt residuall line AD. Then I say shal the line which contayneth in power the su<g ref="char:EOLhyphen"/>perficies AB, is a line making with a medial superficies the whole superficies me<g ref="char:EOLhyphen"/>diall. For vnto the line AD let the line DG be ioyned.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And let the rest be as in the propositions go<g ref="char:EOLhyphen"/>ing
<figure/>
before. And for<g ref="char:EOLhyphen"/>asmuch as the line AF is incommensu<g ref="char:EOLhyphen"/>rable in length to the line FG, there<g ref="char:EOLhyphen"/>fore the parallelo<g ref="char:EOLhyphen"/>gramme AI is in<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the parallelogramme FK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And forasmuch as the lines AG and AC are rationall commensurable in power onely, therefore the parallelogramme AK is me<g ref="char:EOLhyphen"/>diall, and in like maner the parallelogramme DK is mediall. Now forasmuch as the lines AG and GD are commensurable in power onely, therefore they are incommensurable in length the one to the other. But as the line AG is to the line GD, so is the parallelogramme AK to the parallelogramme DK, therefore the parallelogramme AK is incommensurable to the parallelogramme DK. Describe the like figure that was described in the former propo<g ref="char:EOLhyphen"/>sitions, and we may in like sort proue that the line LN contayneth in power the superficies AB. I say moreouer that it is a line making w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th a mediall superficies the whole superficies
<pb n="297" facs="tcp:7062:326"/>
mediall. For the parallelograme AK is mediall, wherefore that which is equal vn<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> it, name<g ref="char:EOLhyphen"/>ly, that which is made of the sq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ares of the lines LO and ON added together is also mediall. And forasmuch as the parallelogramme <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>K is mediall, therefore that which is equall vnto <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t, namely, that which is contayned vnder the lines LO and ON twise is also mediall. And forasmuch as the parallogramme AK is incommensurable to the parallelogramme DK, therefore the squares of the lines LO and ON are incommensurable to that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines LO and ON twise. And forasmuch as the parallelogramme AI is in<g ref="char:EOLhyphen"/>commensurable to the parallelogramme FK, therefore also the square of the line LO is in<g ref="char:EOLhyphen"/>commensurable to the square of the line ON. Wherefore the lines LO and ON are incom<g ref="char:EOLhyphen"/>mensurable in power, hauing that which is made of the squares of the lines LO and ON me<g ref="char:EOLhyphen"/>diall, and that which is contayned vnder them twise mediall, and moreouer that which is made of the squares of them is incommensurable to that<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which is contayned vnder them twise. Wherfore the line LN is that irrationall line which is called a line making with a me<g ref="char:EOLhyphen"/>diall superficies the whole superficies mediall, and it contayneth in power the superficies AB. Wherefore the line which contayneth in power the superficies AB is a line making with a me<g ref="char:EOLhyphen"/>diall superficies the whole superficies mediall. If therefore a superficies be contayned vnder a rationall line and a sixth residuall line, the line which contayneth in power the same superfi<g ref="char:EOLhyphen"/>cies is a line making which a mediall superficies the whole superficies mediall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 73. Theoreme. The 97. Proposition. The square of a residuall line applyed vnto a rationall line,<note place="margin">The fiueth Senary.</note> maketh the breadth or other side a first re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iduall line.</head>
               <p>SVppose that AB be a residuall line,<note place="margin">These sixe propositions following are the conu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rses of the sixe for<g ref="char:EOLhyphen"/>mer proposi<g ref="char:EOLhyphen"/>tions.</note> and
<figure/>
let CD be a rationall line. And vnto the line CD apply the parallelogramme CE equall to the square of the line AB, and making the breadth the line CF. Then I say that the line CF is a first residuall line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">Construction.</note> For vnto the line AB let the line co<g ref="char:cmbAbbrStroke">̄</g>ueniently ioyned be supposed to be B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (which selfe line is also called a line ioyned, as we declared in the end of the 79. proposition). Wherefore the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ines AG and GB are rationall <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>mensurable in power onely. And vnto the line CD apply the pa<g ref="char:EOLhyphen"/>rallelogramme CH equall to the square of the line AG, and vnto the line KH (which is equall to the line CD) apply the parallelogramme KL equall to the square of the line BG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the whole parallelogramme CL is equall to the squares of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and GB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And the parallelogr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CE is equall to the square of the line AB, wherefore the parallelogramme remayning, namely, the parallelogramme FL is e<g ref="char:EOLhyphen"/>quall to that which is contayned vnder the lines AG and GB twise. For (by the 7. of the se<g ref="char:EOLhyphen"/>cond) the squares of the lines AG and GB are equall to that which is contayned vnder the lines AG &amp; GB twise, and to the square of the line AB. Deuide the line FM into two equall partes in the point N. And by the poynt N draw vnto the line CD a parallell line NX. Wher<g ref="char:EOLhyphen"/>fore either of the parallelogrammes FX and NL is equall to that which is contayned vn<g ref="char:EOLhyphen"/>der the lines AG and GB once. And forasmuch as the squares of the lines AG and GB are rationall, vnto which squares the parallelogramme CL is equall, therefore the parallelo<g ref="char:EOLhyphen"/>gramme CL also is rationall: wherefore the line GM is rationall and<note n="*" place="margin">By the <hi>20.</hi> of the tenth.</note> commensurable in
<pb facs="tcp:7062:327"/>
length to the line <hi>CD.</hi> Agayne forasmuch as that which is contayned vnder the lines <hi>AG</hi> and <hi>GB</hi> twise, is <note n="**" place="margin">By the <hi>21.</hi> of the tenth.</note> mediall, therefore the parallelogramme equall vnto it, namely the paral<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>elogramme <hi>FL</hi> is also mediall. Wherefore the line <hi>FM</hi> is rationall and <note n="*" place="margin">By the <hi>22.</hi> of the tenth.</note> incommen<g ref="char:EOLhyphen"/>surable in length to the line <hi>CD.</hi> And forasmuch as the squares of the lines <hi>AG</hi> and <hi>GB</hi> are rationall, and that which is contayned vnder the lines <hi>AG</hi> and <hi>GB</hi> twise, is mediall, therefore the squares of the lines <hi>AG</hi> and <hi>GB</hi> are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable to that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines <hi>AG</hi> and <hi>GB</hi> twise. But vnto the squares of the lines <hi>AG</hi> and <hi>GB</hi> is equall the parallelogramme <hi>CL,</hi> and to that which is contayned vnder the lines <hi>AG</hi> and <hi>GB</hi> twise is equall the parallelogramme <hi>FL,</hi> wherefore the parallelogramme <hi>CL</hi> is incom<g ref="char:EOLhyphen"/>mensurable to the parallelogramme <hi>FL.</hi> Wherefore also the line <hi>CM</hi> is incommensurable in length to the line <hi>FM</hi>: and they are both rationall. Wherefore the lines <hi>CM</hi> and <hi>FM</hi> are rationall commensurable in power onely: and therefore the line <hi>CF</hi> is a residuall line (by the 73. of the tenth). I say moreouer that it is a first residuall line.<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>F co<g ref="char:cmbAbbrStroke">̄</g>cluded a residual line.</note> For forasmuch as that which is contayned vnder the lines <hi>AG</hi> and <hi>GB</hi> is the
<figure/>
meane proportionall betwene the squares of the lines <hi>AG</hi> and <hi>GB</hi> (by the assumpt going before the 54. of the tenth). And vnto the square of the line <hi>AG</hi> is equall the parallelogramme <hi>CH,</hi> and vnto that which is contayned vnder the lines <hi>AG</hi> and <hi>GB</hi> is equall the parallelogramme <hi>NL,</hi> and vnto the square of the line <hi>GB</hi> is equall the parallelograme <hi>KL.</hi> Wherefore the parallelogramme <hi>NL</hi> is the meane proportionall betwene the parallelogrammes <hi>CH</hi> and <hi>KL.</hi> Wherefore as <hi>CH</hi> is to <hi>NL,</hi> so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>NL</hi> to <hi>KL.</hi> But as <hi>CH</hi> is to <hi>NL,</hi> so is the line <hi>CK</hi> to the line <hi>NM</hi> &amp; as <hi>NL</hi> is to <hi>KL,</hi> so is the line <hi>NM</hi> to the line <hi>KM.</hi> Wherfore as the line <hi>CK</hi> is to the line <hi>NM,</hi> so is the line <hi>NM</hi> to the line <hi>KM.</hi> Wherfore the parallelogramme contayned vnder the lines <hi>CK</hi> and <hi>KM</hi> is equall to the square of the line <hi>NM,</hi> that is to the fourth part of the square of the line <hi>FM.</hi> And forasmuch as the square of the line <hi>AG</hi> is commensurable to the square of the line <hi>GB,</hi> therefore the parallelogramme <hi>CH</hi> is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the parallelogramme <hi>KL.</hi> But as <hi>CH</hi> is to <hi>KL,</hi> so is the line <hi>CK</hi> to the line <hi>KM</hi>: wherefore the line <hi>CK</hi> is commensura<g ref="char:EOLhyphen"/>ble in length to the line <hi>KM.</hi> Wherefore (by the 17. of the tenth) the line <hi>CM</hi> is in power more then the line <hi>FM</hi> by the square of a line commensurable in length to the line <hi>CM.</hi> But the line <hi>CM</hi> is commensurable in length to the rationall line <hi>CD.</hi> Wherefore the line <hi>CF</hi> is a f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst residuall line. Wherefore the square of a residuall line applyed vnto a rationall line, maketh the breadth or other side a first residuall line: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
            </div>
            <div type="part">
               <head>¶The <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>4. Theoreme. The 98. Proposition. The square of a first mediall residuall line applied to a rationall line, ma<g ref="char:EOLhyphen"/>keth the breadth or other side a second residuall line.</head>
               <p>SVppose that AB be a first mediall residual line, and let CD be a rationall line. And vnto the line CD apply the parallelogramme CE equall to the square of the lyne AB,<note place="margin">Construction.</note> and making in breadth the line CF. Then I say that the line CF is a second
<pb n="298" facs="tcp:7062:327"/>
residuall line. For vnto the line AB, let the lyne
<figure/>
co<g ref="char:cmbAbbrStroke">̄</g>ueniently ioyned be supposed to be BG. Where<g ref="char:EOLhyphen"/>fore the lines AG and BG are mediall commen<g ref="char:EOLhyphen"/>surable in power onely, comprehending a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. And vnto the line CD apply the pa<g ref="char:EOLhyphen"/>rallelogramme CH equal to the square of the line AG, and making in bredth the line CK, and vn<g ref="char:EOLhyphen"/>the line KH (which is equall to the line CD) ap<g ref="char:EOLhyphen"/>ply the parallelogramme KL equall to the square of the line GB and making in breadth the line KM.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the whole parallelogramme CL is e<g ref="char:EOLhyphen"/>quall to both the squares of the lines AG and GB which are mediall &amp; commensurable the one to the other. Wherfore the parallelogrammes CH and KL are mediall and commensurable the one to the other. Wherfore (by the 15. of the tenth) the whole parallelogramme CL is commensurable to either of these parallelogrammes CH and KL. Wherfore (by the corollary of the 23. of the tenth) the whole parallelogramme CL is also mediall. Wherefore (by the 22. of the tenth) the line CM is rationall and incom<g ref="char:EOLhyphen"/>mensurable in length to the line CD. And forasmuch as the parallelogramme CL is equal to the squares of the lines AG and GB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the squares of the lines AG and GB are equall to that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AG and GB twise, together with the square of the line AB (by the 7. of the second): and vnto the square of the line AB is equall the parallelo<g ref="char:EOLhyphen"/>gramme CE. Wherfore the residue, namely, that which is contained vnder the lines AG and GB twise, is equal to the residue, namely, to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me FL. But that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AG &amp; GB twise is rational. Wherfore the parallelogramme FL is also rationall. Wherfore the line FM is rationall and co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line CD (by the 20. of the tenth.) Now forasmuch as the parallelogramme CL is mediall, and the paral<g ref="char:EOLhyphen"/>lelogramme FL is rational, therfore they are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable the one to the other. Wherfore also the lyne CM is incommensurable in length to the lyne FM, and they are both rationall. Wherfore the line CF is a residuall line.<note place="margin">CF concl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ded a residuall line.</note> I say moreouer that it is a second residuall lyne. For deuide the line FM into two equall partes in the point N, from which point draw vnto the line CD a parallel line NX. Wherfore either of these parallelogrammes FX and NL is e<g ref="char:EOLhyphen"/>quall to the parallelogramme contained vnder the lines AG and GB. And forasmuch as the parallelogramme contained vnder the lines AG and GB is the meane proportionall betwene the squares of the lines AG and BG. Therefore the parallelogramme NL is the meane pro<g ref="char:EOLhyphen"/>portionall betwene the parallelogrammes CH and KL. But as CH is to NL, so is the line CK to the line NM, and as NL is to KL, so is the line NM to the line KM. Wherfore as the line CK is to the lyne NM, so is the lyne NM to the line KM. Wherfore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contayned vnder the lines CK and KM is equall to the square of the line NM, that is, to the fourth part of the square of the lyne FM. But the parallelogramme CH is commensura<g ref="char:EOLhyphen"/>ble to the parallelogramme KL. Wherfore also the lyne CK is commensurable in length to the lyne KM. Wherfore (by the 17. of the tenth) the line CM is in power more then the line FM, by the square of a line commensurable in length to the line CM. And the line FM which is the line conueniently ioyned, is commensurable in length to the rationall lyne CD. Wherfore the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CF is a second residuall lyne. Wherfore the square of a first mediall resi<g ref="char:EOLhyphen"/>dual line applied to a rationall line, maketh the breadth or other side a second residuall lyne: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:328"/>
               <head>¶The 75. Theoreme. The 99. Proposition. The square of a second mediall residuall line applied vnto a rationall line, maketh the breadth or other side a third residuall line.</head>
               <p>SVppose that AB be a second mediall residuall line, and let CD be a rationall line. And vnto the line CD apply the parallelogramme CE equall to the square of the line AB, and making in breadth the line CF.<note place="margin">Construction.</note> Then I say, that the line CF is a third residuall line. For vnto the line AB let the line conueniently ioyned be supposed to be BG. Wherefore the lines AG &amp; GB are mediall commensurable in power onely containing a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. And let the rest of the construction be as in the Proposition next going before.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the line CM is rationall and
<figure/>
incommensurable in length to the rationall line CD. And either of the parallelogrammes FX and NL is equall to that which is contained vnder the lines AG and GB. But that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines AG &amp; GB is mediall. Wher<g ref="char:EOLhyphen"/>fore that which is contained vnder the lines AG and GB twise is also mediall. Wherfore the whole parallelogramme FL is also mediall. Wherefore the line FM is rationall and incommensurable in length to the line CD. And forasmuch as the lines AG and GB are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length, therefore also the square of the line AG is incommensurable to the parallelogramme contay<g ref="char:EOLhyphen"/>ned vnder the lines AG and GB. But vnto the square of the line AG are commensurable the squares of the lines AG and GB: and vnto the parallelogramme contained vnder the lines AG and GB, is commensurable that which is contained vnder the lines AG and GB twise. Wherefore the squares of the lines AG and GB are incommensurable to that which is contained vnder the lines AG and GB twise. Wherefore the parallelogrammes which are equall vnto them, namely, the parallelogrammes CL and FL are incommensurable the one to the other. Wherefore also the line CM is incommensurable in length to the line FM: and they are both rationall. Wherefore the line CF is a residuall line.<note place="margin">CF concluded a residual line.</note> I say moreouer, that it is a third residuall line. For forasmuch as the square of the line AG, that is, the parallelo<g ref="char:EOLhyphen"/>gramme CH is commensurable to the square of the line BG, that is, to the parallelogramme KL, therefore the line CK is commensurable in length to the line KM. And in like sort as in the former Proposition, so also in this may we proue, that the parallelogramme contayned vnder the lines CK and KM, is equall to the square of the line NM, that is, to the fourth part of the square of the line FM. Wherefore the line CM is in power more then the line FM, by the square of a line commensurable in length to the line CM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and neither of the lines CM nor FM is commensurable in length to the rationall line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D. Wherefore the line CF is a third residual line. Wherfore the square of a second mediall residual line applied vnto a rationall line, maketh the breadth or other side a third residuall line: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 76. Theoreme. The 100. Proposition. The square of a lesse line applied vnto a rationall line, maketh the breadth or other side a fourth residuall line.</head>
               <p>
                  <pb n="299" facs="tcp:7062:328"/>SVppose that AB be a lesse line, and let CD be a rationall line. And vnto the line CD apply the parallelogramme CE equall to the square of the line AB, and ma<g ref="char:EOLhyphen"/>king in breadth the line CF. Then I say, that the line CF is a fourth residuall line. For vnto the line AB let the line conueniently ioyned be supposed to be BG.<note place="margin">Construction.</note> Wherefore the lines AG &amp; GB are incommensurable in power, hauing that which is made of their squares added together rationall, and that which is contained vnder them mediall. And let the rest of the construction be as in the Propositions going before. Wherefore the whole parallelo<g ref="char:EOLhyphen"/>gramme CL is rationall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the line CM
<figure/>
is also rationall, and commensurable in length to the line CD. And forasmuch as that which is con<g ref="char:EOLhyphen"/>tained vnder the lines AG and GB twise is medi<g ref="char:EOLhyphen"/>all, therefore the parallelogramme which is equall vnto it, namely, the parallelogramme FL, is also mediall. Wherefore the line FM is rationall and incommensurable in length to the line CD. But the line CM is commensurable in length to the line CD. Wherefore (by the 13. of the tenth) the line CM is incommensurable in length to the line FM: and they are both rationall. Wherefore the lines CM and FM are rationall commensurable in power onely.<note place="margin">CF proued a residuall line.</note> Wherefore the line CF is a residuall line. I say moreouer, that it is a fourth residuall line. For forasmuch as the lines AG and GB are incommensurable in power, therefore the squares of them, that is, the pa<g ref="char:EOLhyphen"/>rallelogrammes, which are equall vnto them, namely, the parallelogrammes CH and KL, are incommensurable the one to the other. Wherefore also the line CK is incommensurable in length to the line KM. And in like sort may we proue, that the parallelogramme contay<g ref="char:EOLhyphen"/>ned vnder the lines CK and KM, is equall to the square of the line NM, that is, to the fourth part of the square of the line FM. Wherefore (by the 18. of the tenth) the line CM is in power more then the line FM, by the square of a line incommensurable in length to the line CM. And the whole line CM is commensurable in length to the rationall line CD. Wherefore the line CF is a fourth residuall line. Wherefore the square of a lesse line applied vnto a rationall line, maketh the breadth or other side a fourth residuall line: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 77. Theoreme. The 101. Proposition. The square of a lyne making with a rationall superficies the whole super<g ref="char:EOLhyphen"/>ficies mediall applied vnto a rational line, maketh the breadth or other side a fift residuall lyne.</head>
               <p>SVppose that AB be a line making with a rationall superficies the whole superficies mediall, and let CD be a rationall line. And vnto the line CD apply the parallelo<g ref="char:EOLhyphen"/>gramme CE equall to the square of the line AB, and making in breadth the line CF. Then I say that the line CF is a fift residuall line. For vnto the line AB let the line con<g ref="char:EOLhyphen"/>ueniently ioyned be supposed to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e BG. Wherfore the lines AG and GB are incommensura<g ref="char:EOLhyphen"/>ble in power, hauing that which is made of their squares added together mediall, and that which is contained vnder them rationall. Let the rest of the construction be in this as it was in the former propositions. Wherfore the whole parallelogramme CL is mediall. Wherefore the line CM is rationall and incommensurable in length to the line CD. And either of the
<pb facs="tcp:7062:329"/>
parallelogramme FX &amp; NL is rationall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wher<g ref="char:EOLhyphen"/>for<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
the whole parallelogramme FL is also ratio<g ref="char:EOLhyphen"/>nall. Wherfore also the line FM is rationall and commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line CD. And for<g ref="char:EOLhyphen"/>asmuch as the parallelogramme CL is mediall, and the parallelogramme FL is rationall, there<g ref="char:EOLhyphen"/>fore CL and FL are incommensurable the one to the other, and the line CM is incommensurable in length to the line FM, and they are both ratio<g ref="char:EOLhyphen"/>nall. Wherfore the lines CM and FM are ratio<g ref="char:EOLhyphen"/>nall commensurable in power onely. Wherfore the lyne GF is a residuall line. I say moreouer that it is a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t residual line.<note place="margin">CF proued a residuall line.</note> For we may in like sort proue, that the parallelograme contained vnder the lines CK and KM, is equall to the square of the line NM, that is, to the fourth part of the square of the lyne FM. And forasmuch as the square of the line AG, that is, the paral<g ref="char:EOLhyphen"/>lelogramme CH is incommensurable to the square of the line BG, that is to the parallelo<g ref="char:EOLhyphen"/>gramme KL, therfore the line CK is incommensurable in length to the line KM. Wherfore (by the 18. of the tenth) the line CM is in power more then the line FM, by the square of a line incommensurable in length to the line CM. And the line conueniently ioyned, namely, the line FM is commensurable in length to the rationall line CD. Wherfore the line CF is a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t residuall line. Wherfore the line CF is a fift residuall line. Wherfore the square of a line making with a rationall superficies the whole superficies medial, applied vnto a rational line maketh the breadth or other side a fift residuall lyne: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 78. Theoreme. The 102. Proposition. The square of a lyne making with a mediall superficies, the whole superfi<g ref="char:EOLhyphen"/>cies mediall applied to a rationall line, maketh the breadth or other side, a sixt residuall line.</head>
               <p>SVppose that AB be a line making with a mediall superficies, the whole superficies mediall,<note place="margin">Construction.</note> and let CD be a rationall line. And vnto the line CD apply the paral<g ref="char:EOLhyphen"/>lelogramme CE equall to the square of the line AB and making in breadth the line CF. Then I say that the line CF is a sixt residual line. For vnto the line AB let the line conueniently ioyned be BG. Wherfore the lines AG and BG are incommen<g ref="char:EOLhyphen"/>surable in power hauing that which is made of their squares added together mediall, &amp; that which is contained vnder them mediall, and moreouer that which is made of their squares added together is incommensurable to that which
<figure/>
is contained vnder them. Let the rest of the con<g ref="char:EOLhyphen"/>struction be in this, as it was in the propositio<g ref="char:cmbAbbrStroke">̄</g>s go<g ref="char:EOLhyphen"/>ing before.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the whole parallelogramme CL is mediall, (for it is equall to that which is made of the squares of the lines AG &amp; GB added together, which is supposed to be mediall). Where<g ref="char:EOLhyphen"/>fore the line CM is rationall and incommensura<g ref="char:EOLhyphen"/>ble in length to the line CD: and in like manner the parallelogramme FL is mediall. Wherfore al<g ref="char:EOLhyphen"/>so the line FM is rationall and incommensurable <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n length to the line CD. And forasmuch as that
<pb n="300" facs="tcp:7062:329"/>
which is made of the squares of the lines AG and GB added together, is incommensurable to that which is contained vnder the lines AG and GB twise, therefore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes equall to them, namely, the parallelogrammes CL and FL are incommensurable the one to the other. Wherfore also the lines GM and FM are incommensurable in length, and they are both rational. Wherfore they are rationall co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power only: Wherfore the line CF is a residuall line. I say moreouer that it is a sixt residuall line.<note place="margin">CF <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>roued <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> residuall.</note> Let the rest of the demonstra<g ref="char:EOLhyphen"/>tion be as it was in the former propositions. And forasmuch as the lines AG and BG are in<g ref="char:EOLhyphen"/>commensurable in power, therfore their squares, that is, the parallelogrammes which are e<g ref="char:EOLhyphen"/>quall vnto them, namely, the parallelogrammes CH and KL are incommensurable the one to the other. Wherfore also the line CK is incommensurable in length to the line KM. Wher<g ref="char:EOLhyphen"/>fore (by the 18. of the tenth) the line CM is in power more then the line FM by the square of a line incommensurable in length to the line CM. And neither of the lines CM nor FM is commensurable in length to the rationall line CD. Wherfore the line CF is a sixt residuall line. Wherfore the square of a line making with a medial superficies the whole superficies me<g ref="char:EOLhyphen"/>diall applied to a rationall line, maketh the breadth or other side a sixt residuall line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 79. Theoreme. The 103. Proposition. A line commensurable in length to a residuall line: is it selfe also a residuall line of the selfe same order.</head>
               <p>SVppose that AB be a residuall line, vnto which let the line CD be commensu<g ref="char:EOLhyphen"/>rable in length.<note place="margin">The sixt Se<g ref="char:EOLhyphen"/>nary.</note> Then I say, that the line CD is also a residuall line, and of the selfe same order of residuall lines that the line AB is. For forasmuch as the line AB is a residuall line,<note place="margin">Construction.</note> let the line conueniently ioyned vnto it be supposed to be BE. Wherefore the lines AE and BE are ra<g ref="char:EOLhyphen"/>tionall
<figure/>
commensurable in power onely. As the line AB is to the line CD, so (by the 12. of the sixt) let the line BE be to the line DF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore (by the 12. of the fift) as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. Wherefore as the line AB is to the line CD, so is the whole line AE to the whole line CF, and the line BE to the line DF. Wherefore (by the 10. of the tenth) the line AE is commensurable in length to the line CF, and the line BE to the line DF. But the line AE is rationall. Wherefore the line CF is also rationall. And in like sort the line DF is rationall, for that the line BE, to whom it is commensurable, is also rationall. And for that as the line BE is to the line AE, so is the line DF to the line CF. But the lines BE and AE are commensurable in power onely: Where<g ref="char:EOLhyphen"/>fore the lines CD and DF are commensurable in power onely. Wherefore the line CD is a residuall line.<note place="margin">CD co<g ref="char:cmbAbbrStroke">̄</g>cluded a residuall line.</note> I say moreouer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that it is a residuall line of the selfe same order that the line AB is. For for that as we haue before said, as the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the line CF, so is the line BE to the line DF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore alternately, as the line AE is to the line BE, so is the line CF to the line DF. But the line AE is in power more then the line EB, either by the square of a line commensurable in length to the line AE, or by the square of a line incommensurable in length to the line AE. If AE be in power <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> then BE, by the square of a line com<g ref="char:EOLhyphen"/>mensurable in length to AE, then the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shall also (by the 14. of the tenth) be in power more then the line DF, by the square of a line commensurable in length to the line CF, and so if the line AE be commensurable in length to the rationall line put, forasmuch as the line
<pb facs="tcp:7062:330"/>
AE is commens<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rable in length to the line CF, therefore (by the 12. of the tenth) the line CF shall also be commensurable in length to the same rationall line. Wherefore either of the <milestone type="tcpmilestone" unit="unspecified" n="1"/> lines AB and CD is a first residuall line. And if the line BE be commensurable in length to the rationall line put, forasmuch as the line BE is commensurable in length to the line DF, therefore the line DF shall also be commensurable in length to the rationall line put<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <milestone type="tcpmilestone" unit="unspecified" n="2"/> and then either of the lines AB and CD is a second residuall line. And if neither of the lines AE nor BE be commensurable in length to the ra<g ref="char:EOLhyphen"/>tionall
<figure/>
line put, then neither of the lines CF nor DF shall be commensurable in length to the same rationall line (by the 13. of the te<g ref="char:cmbAbbrStroke">̄</g>th). And so either of the lines AB &amp; CD is a third residuall line. But <milestone type="tcpmilestone" unit="unspecified" n="3"/> if the line AE be in power more the<g ref="char:cmbAbbrStroke">̄</g> the line BE, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line AE, the line CF shall in like sort (by the 14. of the tenth) be in power more then the line DF, by the square of a line incommensurable in length to the line CF: and then if the line AE be commensurable in length to the rationall line, the line CF shall also in like sort be commensurable in length to the same rationall line: and so either of the lines AB and CD <milestone type="tcpmilestone" unit="unspecified" n="4"/> is a fourth residuall line. And if the line BE be co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rationall line, the line DF shall also be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the same line: and so either of the lines AB &amp; <milestone type="tcpmilestone" unit="unspecified" n="5"/> CD is a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t residuall line. And if neither of the lines AE nor BE be commensurable in length to the rationall line, in like sort neither of the lines CF nor DF shall be co<g ref="char:cmbAbbrStroke">̄</g>mensurable <milestone type="tcpmilestone" unit="unspecified" n="6"/> in le<g ref="char:cmbAbbrStroke">̄</g>gth to the same rational line. And so either of the lines AB &amp; CD is a sixt residual line. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the line CD is a residuall line of the selfe same order that the line AB is. A line therfore commensurable in length to a residuall line, is it selfe also a residuall line of the selfe same order: which was required to be proued.</p>
               <p>
                  <hi>As before touching binomiall lines, so also touching residuall lines, this is to be noted, that a line commensurable in length to a residuall line, is alwayes a residuall line of the selfe same order that the residuall line is, vnto whom it is co<g ref="char:cmbAbbrStroke">̄</g>mensurable,</hi>
                  <note place="margin">Note.</note> 
                  <hi>as hath before in this 103. propositio<g ref="char:cmbAbbrStroke">̄</g> bene proued. But if a line be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in power only to a residuall line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> then followeth it not, yea it is impossible, that that line should be a residuall of the self same order that the residual line, is vnto whom it is commensurable in power onely. Howbeit those two lines shall of necessitie be both either of the three first orders of resid<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ll lines, or of the three last orders: which is not hard to proue, if ye marke diligently the former demonstration, and that which was spoken of binomi<g ref="char:EOLhyphen"/>all lines as touching this matter.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 80. Theoreme. The 104. Proposition. A line commensurable to a mediall residuall line, is it selfe also a medial re<g ref="char:EOLhyphen"/>siduall line, and of the selfe same order.</head>
               <p>SVppose that <hi>AB</hi> be a mediall residuall line, vnto whome let the line <hi>CD</hi> be commensurable in length and in power, or in power onely. Then I say that <hi>CD</hi> is also a mediall residuall line, and of the selfe same order. For forasmuch as the line <hi>AB</hi> is a mediall residuall line,<note place="margin">Construction.</note> let the line conueniently ioyned vnto i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <hi>BE</hi>: wherefore the lines <hi>AE</hi> and <hi>BE</hi> are mediall com<g ref="char:EOLhyphen"/>mensurable
<figure/>
in power onely. As <hi>AB</hi> is to <hi>CD,</hi> so by the 22. of the sixth) let <hi>BE</hi> be to <hi>DF.</hi> And in like sort as in the former so also in this may we proue,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> that the line <hi>AE</hi> is commensurable in length and in power, or in power one<g ref="char:EOLhyphen"/>ly
<pb n="301" facs="tcp:7062:330"/>
vnto the line <hi>CF,</hi> &amp; the line <hi>BE</hi> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the line <hi>DF.</hi> Wherefore (by the 23. of the tenth <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> line <hi>CF</hi> is a mediall line, and the line <hi>DF</hi> is also a mediall line, for that it is commensura<g ref="char:EOLhyphen"/>ble to the mediall line <hi>BE.</hi> And in like sort the lines <hi>CF</hi> and <hi>DF</hi> are commensurable in power onely: for that they haue the selfe same proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, that the lines <hi>AE</hi> and <hi>EB</hi> haue, which are commensurable in power onely. Wherefore the line <hi>CD</hi> is a mediall residuall line.<note place="margin">CD proued a mediall.</note> I say moreouer that it is of the selfe same order that the line <hi>AB</hi> is. For for that as the line <hi>AE</hi> is to the line <hi>BE,</hi> so is the line <hi>CF</hi> to the line <hi>DF.</hi> But as the line <hi>AE</hi> is to the line <hi>BE,</hi> so is the square of the line <hi>AE</hi> to the parallelogramme contayned vnder the lines <hi>AE</hi> and <hi>BE</hi> (by the first of the sixth): and as the line <hi>CF</hi> is to the line <hi>DF,</hi> so is the square of the line <hi>CF</hi> to the parallelogramme contayned vnder the lines <hi>CF</hi> and <hi>DF.</hi> Wherefore as the square of the line <hi>AE</hi> is to the parallelogramme contayned vnder the lines <hi>AE</hi> and <hi>BE,</hi> so is the square of the line <hi>CF</hi> to the parallelogramme contayned vnder the lines <hi>CF</hi> and <hi>DF.</hi> Wherefore alternately as the square of the line <hi>AE</hi> is to the square of the line <hi>CF,</hi> so is the parallelogramme contayned vnder the lines <hi>AE</hi> and <hi>BE,</hi> to the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ines <hi>CF</hi> and <hi>DF.</hi> But the square of the line <hi>AE</hi> is commen<g ref="char:EOLhyphen"/>surable to the square of the line <hi>CF</hi> (for the line <hi>AE</hi> is commensurable to the line <hi>CF</hi>). Wherefore also the parallelogramme contayned vnder the lines <hi>AE</hi> and <hi>BE,</hi> is commensu<g ref="char:EOLhyphen"/>rable to the parallelogramme contayned vnder the lines <hi>CF</hi> and <hi>DF.</hi> Wherefore if the pa<g ref="char:EOLhyphen"/>rallelogramme contayned vnder the lines <hi>AE</hi> and <hi>EB</hi> be rationall, the parallelogramme also contayned vnder the lines <hi>CF</hi> and <hi>FD</hi> shall be rationall. And then either of the lines <hi>AB</hi> and <hi>CD</hi> is a first mediall residuall line. But if the parallelogramme contayned vnder <milestone type="tcpmilestone" unit="unspecified" n="1"/> the lines <hi>AE</hi> and <hi>BE</hi> be mediall, the parallelogramme also contayned vnder the lines <hi>CF</hi> and <hi>FD</hi> shall be also mediall (by the corollary of the 23. of the te<g ref="char:cmbAbbrStroke">̄</g>th): and so either of the lines <hi>AB</hi> and <hi>CD</hi> is a second mediall residuall line. Wherefore the line <hi>CD</hi> is a mediall residu<g ref="char:EOLhyphen"/>all <milestone type="tcpmilestone" unit="unspecified" n="2"/> line of the selfe same order that the line <hi>AB</hi> is. A line therefore commensurable to a me<g ref="char:EOLhyphen"/>diall residuall line, is it selfe also a mediall residuall line of the selfe same order: which was re<g ref="char:EOLhyphen"/>quired to be demonstrated-</p>
               <p>
                  <hi>This Theoreme is vnderstanded generally, that whether a line be commensurable in length &amp; in power, or in power onely to a mediall residuall line, it is it selfe also a me<g ref="char:EOLhyphen"/>diall residuall line, and of the selfe same order, which thing also is to be vnderstanded of the three Theoremes which follow.</hi>
               </p>
               <div type="section">
                  <head>An other demonstration after <hi>Campane.</hi>
                  </head>
                  <p>Suppose that <hi>A</hi> be a mediall residuall line, vnto
<figure/>
whome let the line <hi>B</hi> be commensurable in length, or in power onely. And take a rationall line <hi>CD,</hi> vnto which apply the parallelogramme <hi>CE</hi> equall to the square of the line <hi>A,</hi>
                     <note place="margin">
                        <hi>Construction.</hi>
                     </note> and vnto the line <hi>FE</hi> (which is equall to the line <hi>CD</hi>) apply the parallelogramme <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> equall to the square of the line <hi>B.</hi> Now then the parallelogrammes <hi>CE</hi> and <hi>FG</hi> shall be commensurable,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> for that the lines <hi>A, B</hi> are commensurable in power: wherefore by the 1. of the sixth and 10. of this booke, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> lines <hi>DE</hi> and <hi>FG</hi> are commensurable in length. Now then if <hi>A</hi> be a first mediall residuall line, then is the line <hi>DE</hi> a second resi<g ref="char:EOLhyphen"/>duall line by the 98. of this booke: and if the line <hi>A</hi> be a s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cond mediall residuall line, then is the line <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a third residuall line by the 99. of this booke. But if <hi>DE</hi> be a se<g ref="char:EOLhyphen"/>cond residuall line, <hi>G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> also shall be a second residuall line (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>03. of this boke). And if <hi>DE</hi> be a third re<g ref="char:EOLhyphen"/>siduall
<pb facs="tcp:7062:331"/>
line, <hi>GE</hi> also shall (by the same) be also a third residuall line. Wherefore it followeth by the 9<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and 93. of this booke, that <hi>B</hi> is either a first medial residuall line or a second mediall residuall line, accor<g ref="char:EOLunhyphen"/>ding as the line <hi>A</hi> is supposed to be: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 81. Theoreme. The 105. Proposition. A line commensurable to a lesse line: is it selfe also a lesse line.</head>
               <p>
                  <note place="margin">Construction.</note>SVppose that AB be a lesse line, vnto whom let the line CD be commensurable. Then I say, that the line CD is also a lesse line. For let the same construction be in this, that was in the former Propositions.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the lines AE and EB are incommensurable in power, therefore (by the 22.
<figure/>
of the sixt, and 10. of the tenth) the lines CF &amp; FD are incommensurable in power. Againe (by the 22. of the sixt) as the square of the line AE is to the square of the line BE, so is the square of the line CF to the square of the line DF. Wher<g ref="char:EOLhyphen"/>fore by co<g ref="char:cmbAbbrStroke">̄</g>position as the squares of the lines AE and BE are to the square of the line BE, so are the squares of the lines CF and DF, to the square of the line DF: and alternately, as the squares of the lines AE and BE are to the squares of the lines CF and DF, so is the square of the line BE to the square of the line DF. But the square of the line BE is commensurable to the square of the line DF (for the lines BE and DF are commensurable). Wherefore that which is made of the squares of the lines AE and BE added together, is commensurable to that which is made of the squares of the lines CF and DF added together. But that which is made of the squares of the lines AE and BE added together, is rationall. Wherefore that which is made of the squares of the lines CF and DF added together, is also rationall. A<g ref="char:EOLhyphen"/>gaine, for that as the square of the line AE is to the parallelogramme contained vnder the lines AE and BE, so is the square of the line CF to the parallelogramme contained vnder the lines CF and DF (as we declared in the Proposition next going before): therefore al<g ref="char:EOLhyphen"/>ternately, as the square of the line AE is to the square of the line CF, so is the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines AE and BE, to the parallelogramme contained vnder the lines CF and DF. But the square of the line AE is commensurable to the square of the line CF, for the lines AE &amp; CF are commensurable. Wherefore the parallelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines AE and BE, is commensurable to the parallelogramme con<g ref="char:EOLhyphen"/>tained vnder the lines CF and DF. But the parallelogramme contained vnder the lines AE and BE is mediall. Wherefore the parallelogramme contained vnder the lines CF and DF is also mediall. Wherefore the lines CF and DF are incommensurable in power, hauing that which is made of their squares added together rationall, and the parallelogramme contai<g ref="char:EOLhyphen"/>ned vnder them mediall. Wherefore the line CD is a lesse line. A line therefore commensu<g ref="char:EOLhyphen"/>rable to a lesse line, is it selfe also a lesse line: which was required to be proued.</p>
               <div type="section">
                  <head>An other demonstration.</head>
                  <p>Suppose that A be a lesse line, and vnto A let the line B be commensurable whether in length and power, or in power onely. Then I say that B is a lesse line. Take a rationall line CD.<note place="margin">Construction.</note> And vnto the line CD apply (by the 44 of the first) the parallelogramme CE equall to the square of the line A, and making in bredth the line CF. Wherefore (by the 100. propo<g ref="char:EOLhyphen"/>sition)
<pb n="302" facs="tcp:7062:331"/>
the line CF is a fourth residuall line. Vnto the line FE apply
<figure/>
(by the same) the parallelogramme EH equall to the square of the line B, and making in breadth the line FH. Now forasmuch as the line A is commensurable to the line B,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore also the square of the line A is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the square of the line B. But vnto the square of the line A is equall the parallelogramme CE, &amp; vnto the square of the line B is equal the parallelogramme EH. Wherfore the parallelogramme CE is commensurable to the parallelogramme EH. But as the parallelogramme CE is to the parallelogramme EH, so is the line CF to the line FH. Wherfore the line CF is commen<g ref="char:EOLhyphen"/>surable in length to the line FH. But the line CF is a fourth resi<g ref="char:EOLhyphen"/>duall line. Wherfore the line FH is also a fourth residuall line (by the 103. of the tenth): and the line FE is rationall. But if a superfi<g ref="char:EOLhyphen"/>cies be contained vnder a rationall line, and a fourth residuall lyne, the line that containeth in power that superficies is (by the 94. of the tenth) a lesse lyne. But the line B containeth in power the superficies EH. Wherfore the line B is a lesse line: which was required to be pro<g ref="char:EOLhyphen"/>ued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 82. Theoreme. The 106. Proposition. A line commensurable to a lyne making with a rationall superficies the whole superficies mediall, is it selfe also a lyne making with a rationall su<g ref="char:EOLhyphen"/>perficies the whole superficies mediall.</head>
               <p>SVppose that AB be a line making with a rationall superficies the whole superfi<g ref="char:EOLhyphen"/>cies mediall, vnto whom let the line CD be commensurable. Then I say that the line CD is a line making with a rationall superficies the whole superficies me<g ref="char:EOLhyphen"/>diall.<note place="margin">Construction.</note> Vnto the line AB let the line conueniently ioyned be BE. Wherefore the lines AE and EB are incommensurable in power, hauing
<figure/>
that which is made of their squares added together me<g ref="char:EOLhyphen"/>diall, and the parallelogramme contained vnder them ra<g ref="char:EOLhyphen"/>tionall. Let the construction be in this as it was in the for<g ref="char:EOLhyphen"/>mer propositions.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And in like sort may we proue that as the line AE is to the line BE, so is the line CF to the line DF, and that that which is made of the squares of the lines AE and BE added together is commensurable to that which is made of the squares of the lines CF and DF added together, and that that which is contained vn<g ref="char:EOLhyphen"/>der the lynes AE and EB, is in like sort commensurable to that which is contained vnder the lines CF and DF. Wherfore also the lines CF and DF are commensurable in power, hauing that which is made of their squares added together mediall, and that which is contai<g ref="char:EOLhyphen"/>ned vnder them rationall. Wherfore the line CD is a lyne making with a rationall superfi<g ref="char:EOLhyphen"/>cies the whole superficies mediall. Wherfore a line commensurable to a line making with a rationall superficies the whole superficies mediall, is it selfe also a lyne making with a rational superficies the whole superficies mediall: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration.</head>
                  <p>Suppose that A be a line making with a rationall superficies the whole superficies mediall,
<pb facs="tcp:7062:332"/>
and vnto it let the lyne B be commensurable either in length and in power, or in power on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ly. Then I say that B is a lyne making with a rationall superficies the
<figure/>
whole superficies mediall.<note place="margin">Construction.</note> Take a rational line CD, and vnto the line CD apply the parallelogramme CE equal to the square of the line A and makyng in breadth the lyne GF. Wherfore (by the 101. propo<g ref="char:EOLhyphen"/>sition) the lyne CF is a fift residuall lyne. Againe vnto the line FE apply the parallelogramme FG equall to the square of the line B, and makyng in breadth the lyne FH. Now forasmuch as the line A is co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>mensurable to the lyne B,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therfore the square of the lyne A is comme<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>surable to the square of the line B. But vnto the square of the lyne A is equall the parallelogramme CE, and vnto the square of the line B is equall the parallelogramme FG. Wherfore the parallelogramme CE is commensurable to the parallelogramme FG. Wherefore the line CF is also commensurable in length to the line FH. But the line CF is a fift residuall line. Wherfore also the line FH is a fift residual line. And the line FE is rationall. But if a supersicies be contayned vn<g ref="char:EOLhyphen"/>der a rationall line and a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ift residuall lyne, the lyne that contayneth in power that superficies, is (by the 95. of the tenth) a lyne making with a rationall superficies the whole super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies mediall. But the lyne B containeth in power the parallelogramme FG. Wherfore the lyne B is a lyne making with a rationall superficies, the whole superficies medi<g ref="char:EOLhyphen"/>all<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 83. Theoreme. The 107. Proposition. A line co<g ref="char:cmbAbbrStroke">̄</g>mensurable to a line, making with a mediall superficies, the whole superficies mediall, is it selfe also a line making with a mediall superficies the whole superficies mediall.</head>
               <p>SVppose that AB be a line making with a mediall superficies the whole superfi<g ref="char:EOLhyphen"/>cies mediall, vnto whome let the line CD be commensurable. Then I say that the line CD is also a line making with a mediall superficies the whole superfi<g ref="char:EOLhyphen"/>cies mediall. For vnto the line AB let the line coueniently ioyned be BE. And let the rest of the construction be in this as it was in the former propositions.<note place="margin">Construction.</note> Wherefore the lines AE and BE are inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in power, hauing that which is made of their squares added together mediall,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and that which is contained vnder them also mediall, and moreouer that which is made of their squares added together
<figure/>
is incommensurable to that which is contained vn<g ref="char:EOLhyphen"/>der them. But the lines AE and BE (as we haue be<g ref="char:EOLhyphen"/>fore proued) are commensurable to the lines CF &amp; DF, and that which is made of the squares of the lines AE and BE added together, is commensurable to that which is made of the squares of the lines CF and FD added together, and the parallelogramme contained vnder the lines AE and BE is commensurable to the parallelogramme contained vnder the lines CF and DF. Wherfore the lines CF and DF are incommensurable in power hauing that which is made of their squares added together mediall, and that which is contained vnder them also mediall, and moreouer that which is made of their squares added together, is incommensu<g ref="char:EOLhyphen"/>rable to that which is contained vnder them. Wherfore the line CD is a line making with a mediall superficies the whole superficies mediall. A line therefore commensurable to a lyne
<pb n="303" facs="tcp:7062:332"/>
making with a mediall superficies the whole superficies mediall, is it selfe also a line makyng with a mediall superficies the whole superficies mediall: which was required to be proued.</p>
               <p>
                  <hi>This proposition may also be an other way demonstrated, as the three former propositions were. If vpon a rationall line you apply parallelogrammes equall to the squares of the lines AB and CD, the breadthes of which parallelogrames shall be eche a sixth residual line, aud therfore the lines which co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tayne them in power, namely, the lines AB and CD shall be both such lines as is required in the pro<g ref="char:EOLhyphen"/>position, which is easy to conclud<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> marking the orde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the demonstration in the three former pro<g ref="char:EOLhyphen"/>positions.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 84. Theoreme. The 108. Proposition. If from a rationall superficies be taken away a medialt superficies, the line which containeth in power the superficies remayning, is one of these two ir<g ref="char:EOLhyphen"/>rationall lines, namely, either a residuall line, or a lesse line.</head>
               <p>SVppose that BC be a rationall superficies,<note place="margin">Seuenth Se<g ref="char:EOLhyphen"/>nary.</note> and from it take away a mediall su<g ref="char:EOLhyphen"/>perficies, namely, BD. Then I say, that the line which containeth in power the superficies remayning, namely, the superficies EC, is one of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hese two irrationall lines, namely, either a residuall line, or a lesse line. Take a rationall line FG. And vpon FG describe (by the 44. of the first) a rectangle parallelogramme GH equall to the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies BC.<note place="margin">Constraction.</note> And from the parallelogramme GH take away the parallelogramme GK equall to the superficies BD.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore (by the third common
<figure/>
sentence) the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies remayning, namely, EC, is equall to the parallelogramme remayning, namely, to LH. And foras<g ref="char:EOLhyphen"/>much as BC is rationall, and BD is mediall, and BC is equall to the parallelogramme GH, and BD to the parallelogramme GK: therefore GH is rationall, and GK is mediall: and the parallelogramme GH is applied vnto the rationall line FG. Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. of the tenth) the line FH is rationall and commensurable in length to the line FG. And the paral<g ref="char:EOLhyphen"/>lelogramme GK is also applied vnto the ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ionall line FG. Wherefore (by the 22. of the tenth) the line FK is rationall and incommensurable in length to the line FG. Wherefore (by the Assumpt of the 12. of the tenth) the line FH is incommen<g ref="char:EOLhyphen"/>surable in length to the line FK. And they are both rationall. Wherefore the lines FH and FK are rationall commensurable in power onely. Wherefore the line KH is a residuall line: and the line conueniently <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ed vnto it is KF. Now the line FH is in power more th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n the line KF, either by the square of a line commensurable in lengt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the line FH, or by the square of a line incommensurable in length to the line FH. First let it be in power more then the line FK, by the square of a line commensurable in length to the line FH, and the whole line FH is commensurable in length to the rationall line put, namely, to FG. Wherefore the line KH is a first residuall line. But if a superficies be contained vnder a rationall line, and a first residuall line, the line that containeth in power that superficies, is (by the 91. of the tenth) a residuall line. Wherefore the line which containeth in power LH, that is, the super<g ref="char:EOLhyphen"/>ficies EC, is a residuall line. But if the line HF be in power more then the line FK, by the square of a line incommensurable in length to the line FH, and the whole line FH is com<g ref="char:EOLhyphen"/>mensurable in length to the rationall line geuen FG. Wherefore the line KH is a fourth residuall line. But a line containing in power a superficies contained vnder a rationall line and a fourth residuall line, as a lesse line (by the 94. of the tenth). Wherefore the line that containeth in power the superficies LH, that is, the superficies EC, is a lesse line. If there<g ref="char:EOLhyphen"/>fore from a rationall superficies be taken away a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, the line which containeth
<pb facs="tcp:7062:333"/>
in power the superficies remayning, is one of these two irrationall lines, namely, either a resi<g ref="char:EOLhyphen"/>duall line, or a lesse line: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 85. Theoreme. The 109. Proposition. If from a mediall superficies be taken away a rationall superficies, the line which contayneth in power the superficies remayning is one of these two ir<g ref="char:EOLhyphen"/>rationall lines, namely either a first mediall residuall line, or a line ma<g ref="char:EOLhyphen"/>king with a rationall superficies the whole superficies mediall.</head>
               <p>SVppose that <hi>BC</hi> be a
<figure/>
mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies and from it take away a rati<g ref="char:EOLunhyphen"/>onal super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, namely, <hi>BD.</hi> Then I say that the line which contayneth in power the superficies remayning, namely, the superficies <hi>EC</hi> is one of these two irrationall lines, either a first mediall residuall line, or a line making with a rational superficies the whole superficies mediall. Take a rationall line <hi>FG,</hi> and let the rest of the con<g ref="char:EOLhyphen"/>struction be in this as it was in the former proposition.<note place="margin">Construction.</note> Wherefore it followeth that the line <hi>F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> is rationall and incommensurable in length to the line <hi>FG</hi> (by the 22. of the tenth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And that the line <hi>KF</hi> is (by the 20. of the tenth) rationall and commensurable in length to the line <hi>FG.</hi> Wherefore the lines <hi>FH</hi> and <hi>FK</hi> are rationall commensurable in power onely. Wherefore <hi>KH</hi> is a residuall line. And the line conueniently ioyned vnto it is <hi>FK.</hi> Now the line <hi>FH</hi> is in power more then the line <hi>FK,</hi> either by the square of a line commensurable in <milestone type="tcpmilestone" unit="unspecified" n="1"/> length to the line <hi>FH,</hi> or by the square of a line incommensurable in length vnto it. If the line <hi>FH</hi> be in power more then the line <hi>FK</hi> by the square of a line commensurable in length to the line <hi>FH</hi> and the line co<g ref="char:cmbAbbrStroke">̄</g>ueniently ioyned vnto it, namely, <hi>FK,</hi> is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rationall line <hi>FG.</hi> Wherefore the line <hi>KH</hi> is a second residuall line. And the line <milestone type="tcpmilestone" unit="unspecified" n="2"/>
                  <hi>FG</hi> is a rationall line. But a line contayning in power a superficies comprehended vnder a ra<g ref="char:EOLhyphen"/>tionall line and a second residuall line is (by the 92. of the tenth) a first mediall residuall line. Wherefore the line thus contayneth in power the superficies <hi>LH,</hi> that is, the superficies <hi>CE</hi> is a first mediall residuall lne. But if the line <hi>HF</hi> be in power more then the line <hi>FK</hi> by the square of a line incommensurable in length to the line <hi>FH,</hi> and the line conueniently ioyned, namely, the line <hi>FK</hi> is commensurable in length to the rationall line put, namely, to <hi>FG</hi>: wherefore the line <hi>KH</hi> is a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift residuall line. Wherefore (by the 95. of the tenth) the line that contayneth in power the superficies <hi>LH,</hi> that is, the superficies <hi>EC,</hi> is a line making with a rationall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, the whole superficies mediall which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 86. Theoreme. The 110. Proposition. If from a mediall superficies be taken away a mediall superficies incom<g ref="char:EOLhyphen"/>mensurable to the whole superficies, the line which containeth in power the superficies which remaineth, is one of these two irrationall lines, namely, either a second mediall residuall line, or a line making with a mediall su<g ref="char:EOLhyphen"/>perficies the whole superficies mediall.</head>
               <p>
                  <pb n="304" facs="tcp:7062:333"/>AS in the former descriptions, s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> also <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ake away fro<g ref="char:cmbAbbrStroke">̄</g> the mediall superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> mediall superficies BD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D be incomm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nsur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ble to the whole superficies BC. Then I say, that the line which containeth in power the su<g ref="char:EOLhyphen"/>per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies EC, is one of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se two irrationall lines, namely, either a second me<g ref="char:EOLhyphen"/>diall residuall line, or a li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e making with a mediall superfi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>es the whole super<g ref="char:EOLhyphen"/>ficies mediall.<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stra<g ref="char:EOLhyphen"/>tio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> For forasmuch as either of these superficie<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s
<figure/>
BC and BD is mediall, and BC is incommensurable to BD, it followeth (by the 22. of the tenth) tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> either of these lines FH and FK is rationall and incomm<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>surable in length to the line FG. And forasmuch as the superficies BC is incom<g ref="char:EOLhyphen"/>mensurable to the superficies BD, that is, the superficies GH, to the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies GK, therefore (by the first of the sixt, &amp; 10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the tenth) the line FH is incommensurable in length to the line FK. Wherfore the lines HF and FK are rational co<g ref="char:cmbAbbrStroke">̄</g>men<g ref="char:EOLhyphen"/>surable in power onely. Wherefore (by the 73. of the tenth) the line KH is a residuall line, and the line co<g ref="char:cmbAbbrStroke">̄</g>ueniently ioyned vnto it is FK. Now the line HF is in power more then the line FK, either by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line HF, or by the square of a line incommensurable in length vnto it. If the line HF be in power more then the line FK, by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line FH,<milestone type="tcpmilestone" unit="unspecified" n="1"/> and neither of the lines HF nor FK i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> commensur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ble to the rationall line put FG. Wher<g ref="char:EOLhyphen"/>fore the line KH is a third residuall. But the line CF, that is, the line KL, is rationall. And a rectangle super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies contained vnder a rationall line and a third residuall line, is irratio<g ref="char:EOLhyphen"/>nall, and the line which containeth in power that superficies, is (by the 93. of the tenth) a se<g ref="char:EOLhyphen"/>cond mediall residuall line. Wherefore the line that containeth in power the superficies LH, that is, the superficies EC is a second mediall residuall line. But if the line HF be in power more then the line FK, by the square of a line incomme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>surable in length to the line FH,<milestone type="tcpmilestone" unit="unspecified" n="2"/> and neither of the lines HF nor FK is commensurable in length to the line FG. Wherefore the line HK is a sixt residuall line. But a line containing in power a super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies contained vn<g ref="char:EOLhyphen"/>der a rationall line and a sixt residuall line, is (by the 96. of the tenth) a line making with a mediall superficies the whole superficies mediall. Wherefore the line that containeth in power the superficies LH, that is, the superficies EC, is a line making with a mediall super<g ref="char:EOLhyphen"/>ficies the whole superficies mediall. If therefore from a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies be taken away a mediall superficies, incommensurable to the whole superficies, the line that containeth in power the superficies which remaineth, is one of the two irrationall lines remaining, namely, either a second mediall residuall line, or a line making with a mediall superficies the whole superficies mediall: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 87. Theoreme. The 111. Proposition. A residuall line, is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ot one and the same with a binomiall lyne.</head>
               <p>SVppose that AB be a residuall line. Then I say that AB is not one and the same with a binomiall line. For if it be possible let it be a binomiall line. And take a ra<g ref="char:EOLhyphen"/>tionall line DC.<note place="margin">Construction.</note> And (by the 44. of the first) vnto the line CD apply a rectangle parallelogramme CE equall to the square of the line AB, and making in bredth the line DE.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> And forasmuch as AB is a residuall line, therfore (by the 97. of the tenth) the line DE is a first residuall line. Let the line co<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>niently ioyned vnto it be E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Wherfore the lines D
<pb facs="tcp:7062:334"/>
F and FE are rationall commensurable in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n<g ref="char:EOLhyphen"/>ely,
<figure/>
and the line DF is in power more then the line FE by the square of a line commensurable in length to the line DE &amp; the line DF, is co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rational line put DC. Again forasmuch as AB is by position a binomiall line, therefore (by the 60. of the tenth) the line DE is a first binomiall line. De<g ref="char:EOLhyphen"/>uide it into his names in the point G. And let DG be the greater name. Wherfore the lines DG and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E are rationall commensurable in power onely. And the line DG is in power more then the line GE by the square of a line commensurable in length to the lyne DG, and the line DG is commensurable in length to the rationall line put DC. Wherefore the line DF is commensurable in length to the line DG. Wherfore (by the 13. of the tenth) the whole line DF is commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line remaining, namely, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the line GF. And forasmuch as the line DF is co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable to the line FG, but the line FD is rationall. Wherfore the line FG is also rationall. And forasmuch as the line FD is commensurable in length to the line FG, but the line DF is incommensurable in length to the line FE. Wher<g ref="char:EOLhyphen"/>fore the line FG is incommensurable in length to the line FE (by the 13. of the tenth) and they are both rationall lines. Wherfore the lines GF and FE are rationall commensurable in power onely. Wherfore (by the 73. of the tenth) the line EG is a residuall line, but it is also rationall (as before hath bene proued): which is impossible, namely, that one &amp; the same line should be both rationall and irrationall. Wherfore a residuall line is not one and the same with a binomiall line, that is, is not a binomiall line: which was required to be demonstrated.</p>
               <div type="section">
                  <head>
                     <note place="margin">A Corollary.</note>¶ A Corollary.</head>
                  <p>A residuall lyne and the other fiue irrationall lynes following it, are neither mediall lines, nor one and the same betwene themselues<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> that is, one is vtterly of a diuers kinde fro<g ref="char:cmbAbbrStroke">̄</g> an other. For the square of a mediall line applied to a rationall line, maketh the breadth rationall and incommensurable in length to the rationall lyne, whereun<g ref="char:EOLhyphen"/>to it is applied (by the 22. of the tenth) The square of a residuall line applied to a rationall line, maketh the breadth a first residuall line (by the 97. of the tenth). The square of a first mediall residuall line applied to a rationall line, maketh the breadth a second residuall lyne (by the 98. of the tenth) The square of a second mediall residuall line applied vnto a ratio<g ref="char:EOLhyphen"/>nall line, maketh the breadth a third residuall line (by the 99. of the tenth) The square of a lesse line applied to a rationall line, maketh the breadth a fourth residuall line (by the 100. of the tenth) The square of a line making with a rationall superficies the whole superficies mediall applied to a rationall line, maketh the breadth a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ift residuall line (by the 101. of the tenth) And the square of a line making with a mediall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies the whole superficies medi<g ref="char:EOLhyphen"/>all applied to a rationall line, maketh the breadth a sixt residuall line (by the 102. of the te<g ref="char:cmbAbbrStroke">̄</g>th) Now forasmuch as these foresaid sides which are the breadthes differ both from the first breadth, sor that it is rational, and differ also the one fro<g ref="char:cmbAbbrStroke">̄</g> the other, for that they are residuals of diuers orders and kindes, it is manifest that those irrationall lines differ also the one from the other. And forasmuch as it hath bene proued in the 111. proposition, that <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> residual <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> is not one and the same with a binomiall line, and it hath also bene proued that the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> of a residuall line and of the fiue irrationall lines that follow it being applied to a rational line do make their breadthes one of the residuals of that order of which they were, whose square<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> were applied to the rationall line, likewise also the squares of a binomiall line, and of the fiue irrationall lines which follow it, being applied to a rationall line, do make the breadthes one
<pb n="305" facs="tcp:7062:334"/>
of the binomials of that order of which they were, whose squares were applied to the rationall line. Wherfore the irrationall lines which follow the binomiall line, and the irrationall lines which follow the residuall line, differ the one from the other, so that all the irrationall lynes are 13. in number, namely, these.</p>
                  <list>
                     <item>1 A mediall line.</item>
                     <item>2 A binomiall line.</item>
                     <item>3 A first bimediall line.</item>
                     <item>4 A second bimediall line.</item>
                     <item>5 A greater line.</item>
                     <item>6 A line containing in power a rationall superficies and a mediall superficies.</item>
                     <item>7 A line contayning in power two mediall superficieces.</item>
                     <item>8 A residuall line.</item>
                     <item>9 A first mediall residuall line.</item>
                     <item>10 A second mediall residuall line.</item>
                     <item>11 A lesse line.</item>
                     <item>12 A line making with a rationall superficies the whole superficies mediall.</item>
                     <item>13 A line making with a mediall superficies the whole superficies mediall.</item>
                  </list>
               </div>
            </div>
            <div type="part">
               <head>¶ The 88. Theoreme. The 112. Proposition. The square of a rationall line applyed vnto a binomiall line, maketh the breadth or other side a residuall line, whose names are commensurable to the names of the binomiall line, &amp; in the selfe same proportio<g ref="char:cmbAbbrStroke">̄</g>: &amp; moreouer that residuall line is in the selfe same order of residuall lines, that the bino<g ref="char:EOLhyphen"/>miall line is of binomiall lines.</head>
               <p>SVppose that <hi>A</hi> be a rationall line,<note place="margin">The determi<g ref="char:EOLhyphen"/>nation hath sundry partes orderly to be proued.</note> and <hi>BC</hi> a binomiall line whose greater name let be <hi>CD.</hi> And vnto the square of the line <hi>A</hi> let the parallelogramme contayned vnder the lines <hi>BC</hi> and <hi>EF</hi> (so that <hi>EF</hi> be the breadth) be equall. Then I say that <hi>EF</hi> is a residuall line, whose names are commensurable to the names of the binomiall line <hi>BC,</hi> which names let be <hi>CD</hi>
                  <figure/>
and <hi>DB,</hi> and are in the same proportion with them: and moreouer the line <hi>EF</hi> is in the selfe same order of residual lines, that the line <hi>BC</hi> is of binomiall lines. Vnto the square of the line <hi>A</hi> let the parallelogramme contayned vnder the lines <hi>BD</hi> and <hi>G</hi> be e<g ref="char:EOLhyphen"/>quall.<note place="margin">Construction.</note> Now forasmuch as that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines <hi>BC</hi> &amp; <hi>EF</hi> is equal to that which is contayned vnder the lines <hi>BD</hi> and <hi>G,</hi> therfore reciprocally (by the 14. of the sixth) as the line <hi>CB</hi> is to the <hi>BD,</hi> so is the line <hi>G</hi> to the line <hi>EF.</hi> But the line <hi>BC</hi> is greater then the line <hi>BD,</hi> wherefore the line <hi>G</hi> is greater then the line <hi>EF.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Vnto the line <hi>G</hi> let the line <hi>EH</hi> be equall. Wherefore (by the 11. of the fift) as the line <hi>CB</hi> is to the line <hi>BD,</hi> so is the line <hi>HE</hi> to the line <hi>FE.</hi> Wherefore by deuision (by the 17. of the fifth) as the line <hi>CD</hi> is to the line <hi>BD,</hi> so is the line <hi>HF</hi> to the line <hi>FE.</hi>
                  <note place="margin">This is an As<g ref="char:EOLhyphen"/>sumpt, proble<g ref="char:EOLhyphen"/>maticall, artifi<g ref="char:EOLhyphen"/>cially vsed and demonstrated.</note> As the line HF is to the FE, so let the line FK be to the line KE (how this is to be done we will declare at the end of this demonstration). Wherefore (by the 12. of the fift) the whole line <hi>HK</hi> 
                  <note n="*" place="margin">Therfore those three lines are in continuall proportion.</note> is to the whole line <hi>KF</hi> as the line <hi>FK</hi> is to the line <hi>KE.</hi> For as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the consequentes. But as the line <hi>FK</hi> is the line <hi>KE,</hi> so is the line <hi>CD</hi> to the line <hi>DB</hi> (for <hi>FK</hi> is to <hi>EK</hi> as <hi>HF</hi> is to <hi>FE,</hi> and <hi>HF</hi> is to <hi>FE</hi> as <hi>CD</hi> is <hi>DB</hi>). Wherfore (by the 11.
<pb facs="tcp:7062:335"/>
of the fift) as the line <hi>HK</hi> is to the line <hi>KF,</hi> so is the line <hi>CD</hi> to the line <hi>DB.</hi> But the square of the line <hi>CD</hi> is commensurable to the square of the line <hi>DB</hi>: wherefore (by the 10. of the tenth) the square of the line <hi>HK</hi> is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the square of the line <hi>FK.</hi> But these three lines <hi>HK, FK,</hi> and <hi>EK</hi> are proportionall in co<g ref="char:cmbAbbrStroke">̄</g>tinuall proportio<g ref="char:cmbAbbrStroke">̄</g> (as it hath already bene pro<g ref="char:EOLhyphen"/>ued). Wherefore (by the second corrollary of the 20. of the sixth) the square of the line <hi>HK</hi> is to the square of the line <hi>FK</hi> as the line <hi>HK</hi> is to the line <hi>EK</hi>: wherfore the line <hi>HK</hi> is com<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>EK.</hi> Wherefore (by the 15. of the tenth) the line <hi>HE</hi> is com<g ref="char:EOLhyphen"/>mensurable in length to the line <hi>FK.</hi> And forasmuch as the square of the line <hi>A</hi> is equall to that which is contayned vnder the lines <hi>EH</hi> and <hi>BD,</hi> but the square of the line <hi>A</hi> is ra<g ref="char:EOLhyphen"/>tionall, where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore that which is contayned vnder the lines <hi>EH</hi> and <hi>BD</hi> is rationall. And it is applyed vnto the rationall line <hi>BD.</hi> Wherefore (by the 20. of the tenth) the line <hi>EH</hi> is ra<g ref="char:EOLhyphen"/>tionall and commensurable in length to the line <hi>BD.</hi> Wherefore also the line <hi>EK</hi> which is commensurable in length to the line <hi>HE</hi> is
<figure/>
                  <milestone type="tcpmilestone" unit="unspecified" n="1"/> rational and commensurable in length to the line <hi>BD.</hi> Now for that as the line <hi>CD</hi> is to the line <hi>DB,</hi> so is t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e line <hi>FK</hi> to the line <hi>EK</hi> (for it was before proued, that as <hi>CD</hi> is to <hi>DB,</hi> so is <hi>HF</hi> to <hi>FE,</hi> and as <hi>HF</hi> is to <hi>FE,</hi> so is <hi>FK</hi> to <hi>EK</hi>) but the lines <hi>CD</hi> and <hi>DB</hi> are commensurable in power onely, wherefore (by the 10. of the tenth) the lines <hi>FK</hi> and <hi>KE</hi> are also commensurable in power onely. And for that as the line <hi>CD</hi> is to the line <hi>DB,</hi> so is the line <hi>FK</hi> to the line <hi>EK,</hi> therefore by contrary proportion as <hi>DB</hi> is to <hi>CD,</hi> so is <hi>EK</hi> 
                  <milestone type="tcpmilestone" unit="unspecified" n="2"/> to <hi>FK,</hi> and alternately as <hi>DB</hi> is to <hi>EK,</hi> so is <hi>CD</hi> to <hi>FK</hi>: but the lines <hi>BD</hi> and <hi>EK</hi> are commensurable in length (as it hath already bene proued). Wherfore also the lines <hi>CD</hi> and <hi>FK</hi> are commensurable in length. But the line <hi>CD</hi> is rationall: wherefore also the line <hi>FK</hi> is rationall. Wherefore the lines <hi>FK</hi> and <hi>EK</hi> are rationall commensurable in power onely. <milestone type="tcpmilestone" unit="unspecified" n="3"/> Wherefore the line <hi>FE</hi> is a residuall line: whose names <hi>FK</hi> and <hi>KE</hi> are commensurable to the names <hi>CD</hi> and <hi>BD</hi> of the binomiall line <hi>BC</hi> and in the same proportion as is proued.<note place="margin">FE concluded a residuall li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>, which is s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>what pre<g ref="char:EOLhyphen"/>p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>icro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sly, in respect o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>der propoun<g ref="char:EOLunhyphen"/>ded, both in the propositio<g ref="char:cmbAbbrStroke">̄</g>, and also in the determinatio<g ref="char:cmbAbbrStroke">̄</g>.</note> 
                  <milestone type="tcpmilestone" unit="unspecified" n="4"/> I say moreouer that it is a residuall line of the selfe same order that the binomiall line is. For the line <hi>CD</hi> is in power more then the line <hi>BD</hi> either by the square of a line commensura<g ref="char:EOLhyphen"/>ble in length to the line <hi>CD,</hi> or by the square of a line incommensurable in length. Now if the line <hi>CD</hi> be in power more then the line <hi>BD</hi> by the square of a line commensurable in length vnto the line <hi>CD,</hi> then (by the 13. of the tenth) the line <hi>FK</hi> is in power more then <milestone type="tcpmilestone" unit="unspecified" n="1"/> the line <hi>EK</hi> by the square of a line co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line <hi>FK.</hi> And so if the line <hi>CD</hi> be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rationall line put, the line <hi>FK</hi> also shalbe co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the same rationall line: wherfore then the line <hi>BC</hi> is a first binomiall line, &amp; the <milestone type="tcpmilestone" unit="unspecified" n="2"/> line <hi>FE</hi> is likewise a first residuall line. And if the line <hi>BD</hi> be commensurable in length to the rationall line, the line <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>K</hi> is also commensurable in length to the same, and then the line <hi>BC</hi> is a second binomiall line, and the line <hi>FE</hi> a second residuall line. And if neither of the <milestone type="tcpmilestone" unit="unspecified" n="3"/> lines <hi>CD</hi> nor <hi>DB</hi> be commensurable in length vnto the rationall line, neither of the lines <hi>FK</hi> nor <hi>EK</hi> are commensurable in length vnto the same, and then the line <hi>BC</hi> is a third binomiall line, &amp; the line <hi>FE</hi> is a third residuall line. And if the line <hi>CD</hi> be in power more then the line <hi>BD,</hi> by the square of a line incommensurable in length to the line <hi>CD,</hi> the line <hi>FK</hi> is also (by the 14. of the tenth) in power more then the line <hi>EK</hi> by the square of a line incommensurable in length to the line <hi>FK.</hi> And so if the line <hi>CD</hi> be commensurable in <milestone type="tcpmilestone" unit="unspecified" n="4"/> length to a rationall line put, the line <hi>FK</hi> also is commensurable in length to the same, where<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the line <hi>BC</hi> is a fourth binomiall line, and the line <hi>FE</hi> is a fourth residuall line. And if <milestone type="tcpmilestone" unit="unspecified" n="5"/> the line <hi>BD</hi> be commensurable in length to the rationall line, the line <hi>EK</hi> is likewise com<g ref="char:EOLhyphen"/>mensurable in length to the same, and then the line <hi>BC</hi> is a fifth binomiall line and the line
<pb n="306" facs="tcp:7062:335"/>
                  <hi>EF</hi> a fifth residuall line. And if neither of the lines <hi>CD</hi> nor <hi>DB</hi> be commensurable in length to the rationall line, neither also of the lines <hi>FK</hi> nor <hi>EK</hi> is commensurable in length to the same, and then the line <hi>BC</hi> is a sixth binomiall line, and the line <hi>FE</hi> a sixth residuall <milestone type="tcpmilestone" unit="unspecified" n="6"/> line. Wherfore the line <hi>FE</hi> is a residuall line, whose names, namely, <hi>FK</hi> and <hi>EK</hi> are comme<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>surable to the names of the binomiall line, namely, to the names <hi>CD</hi> and <hi>DB,</hi> and are in the selfe same proportion, and the residuall line <hi>EF</hi> is in the selfe same order of residuall lines, that the binomiall line <hi>BC</hi> is of binomiall lines. Wherefore the square of a rationall line applyed vnto a binomiall line maketh the breadth (or other side) a residuall line, whose names are commensurable to the names of the binomiall line, and in the selfe same proportion, and moreouer that residuall line is in the selfe same order of residuall lines, that the binomiall line is of binomiall lines: which was required to be demonstrated.</p>
               <div type="section">
                  <head>Here is the Assumpt (of the foregoing Proposition) confirmed.</head>
                  <p>Now let vs declare how as the line HF is to the line FE, so to make the line FK to the line EK. The line CD is greater then the line BD by supposition. Wherefore also the line HF is greater then the line FE (by alternate proportion, and the 14. of the fifth).<note place="margin">Construction.</note> From the line HF take away the line FL equall to the line FE. Wherefore the line remayning, name<g ref="char:EOLhyphen"/>ly, HL, is lesse then the line HF, for the line HF
<figure/>
is equall to the lines HL &amp; LF. As HL is to HF, so (by the 12. of the sixt) let FE be to FK.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore by contrary proportion (by the Corolla<g ref="char:EOLhyphen"/>ry of the 4. of the fifth) as HF is to HL, so is FK to FE. Wherefore by conuersion of pro<g ref="char:EOLhyphen"/>portion (by the Corollary of the 19. of the fifth) as HF is to LF, that is, to the line equall vnto it, namely, to FE, so is the line FK to the line EK.</p>
                  <div type="subsection">
                     <head>
                        <hi>M. Dee,</hi> of this Assumpt, maketh (<gap reason="foreign">
                           <desc>〈 in non-Latin alphabet 〉</desc>
                        </gap>, that is, <hi>Acquisiuely,</hi>) a Probleme vniuersall, thus:</head>
                     <p>Two vnequall right lines being propounded, to adioyne vnto the lesse, a right line, which take<g ref="char:cmbAbbrStroke">̄</g> with <milestone type="tcpmilestone" unit="unspecified" n="1"/> the lesse (as one right line) shall haue the same proportion, to the line adioyned, which, the greater of the two propounded, hath to the lesse.</p>
                     <p>
                        <hi>The construction and demonstration hereof, is worde for worde to be taken, as it standeth here before: after these wordes</hi>: The line HF is greater then the line FE.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ A Corollary also noted by <hi>I. Dee.</hi>
                     </head>
                     <p>
                        <q>It is therefore euident, that thus are three right lines (in our handling) in continuall proportion:<milestone type="tcpmilestone" unit="unspecified" n="2"/> it is to weete, the greater, the lesse and the adioyned, make the first, the lesse with the adioyned, make the second: and the adioyned line is the third.</q>
                     </p>
                     <p>This is proued in the beginning of the demonstration, after the Assumpt vsed.</p>
                  </div>
               </div>
               <div type="section">
                  <head>An other demonstration after <hi>Flussas.</hi>
                  </head>
                  <p>Take a rationall line A, and let GB be a binomiall line, whose greater line let be GD:<note place="margin">
                        <hi>Construction.</hi>
                     </note> and vpon the line GB apply (by the 45. of the first) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me BZ equall to the square of the line A, and making in breadth the line GZ. Likewise vpon the line DB (by the same) apply the parallelogramme BI equall also to the square of the line A, and making in breadth the line DI: and put the line GZT equall to the line DI. Then I say, that GZ is such a residuall line as is required in the Proposition.
<pb facs="tcp:7062:336"/>
Forasmuch as the paralle<g ref="char:EOLhyphen"/>logrammes
<figure/>
BZ &amp; BI are equall, therefore (by the 14. of the sixt) reciprocal<g ref="char:EOLhyphen"/>ly as the line GB is to the line BD, so is the line DI or the line GT, (which is equall vnto it) vnto the line GZ. Wherefore by diuision, as the line GD is to the line DB, so is the line TZ to the line ZG (by the 17. of the fifth). Wherefore the line TZ is greater then the line ZG. (For the line GD is the greater name of the binomiall line GB). Vnto the line ZG put the line ZC equall. And as the line TC is to the line TZ, so (by the 11. of the sixth) let the line ZG be to line ZK. Wherefore contrary wise (by the Corollary of the 4. of the fifth) the line TZ is to the line TC,<note place="margin">
                        <hi>Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> as the line ZK is to the line ZG. Wherefore by conuersion of proportion (by the 19. of the fifth) as the line TZ is to the line ZC (that is, to ZG, which is equall vnto it) so is the line ZK to the line <hi>K</hi>G. But the line TZ is to the line ZG, as the line GD is to the line DB. Wher<g ref="char:EOLhyphen"/>fore (by the 11. of the fifth) the line ZK is to the line KG, as the line GD is to the line DB. But the lines GD and DB are commensurable in power onely. Wherefore also the lines ZK and KG are com<g ref="char:EOLhyphen"/>mensurable in power onely, by the 10. of this booke. Farther, forasmuch as the line TZ is to the line ZG, as the line ZK is to the line KG, therefore by the 12. of the fifth, all the antecedentes, namely, the whole line TK are to all the consequentes, namely, to the line KZ, as one of the antecedentes, namely, the line ZK is to one of the consequentes, namely, to the line KG. Wherefore the line ZK is the meane proportionall betwene the lines TK and KG. And therefore (by the Corollary of the 20. of the sixth) as the first, namely, the line TK, is to the third, namely, to the line KG: so is the square of the line TK to the square of the second, namely, of the line KZ. And forasmuch as the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me BI (which is equall to the square of the rationall line A) is applied vpon the rationall line DB, it maketh the breadth DI rationall and commensurable in length vnto the line DB, by the 20. of the tenth. And ther<g ref="char:EOLhyphen"/>fore the line GT (which is equall vnto the line DI) is commensurable in length to the same line DB. And for that as the line GD is to the line DB, so is the line KZ to the line KG, but as the line KZ is to the line KG, so is the line TK to the line <hi>K</hi>Z, therefore (by the 11. of the fifth) as the line GD is to the line DB, so is the line T<hi>K</hi> to the line <hi>K</hi>Z. Wherefore (by the 22. of the sixth) as the square of the line GD is to the square of the line DB, so is the square of the line T<hi>K</hi> to the square of the line <hi>K</hi>Z. But the square of the line GD is commensurable to the square of the line DB (for the names GD and DB of the binomiall line GB are commensurable in power). Wherefore the square of the line T<hi>K</hi> shall be commensurable to the square of the line <hi>K</hi>Z, by the 10. of this booke. But as the square of the line T<hi>K</hi> is to the square of the line <hi>K</hi>Z, so is it proued, that the right line T<hi>K</hi> is to the right line <hi>K</hi>G. Wher<g ref="char:EOLunhyphen"/>fore the right line T<hi>K</hi> is commensurable in length to the right line <hi>K</hi>G. Wherefore it is also commen<g ref="char:EOLhyphen"/>surable in length to the line TG (by the 15. of the tenth). Which line TG is (as it hath bene proued a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ationall line, and equall to the line DI. Wherefore the lines T<hi>K</hi> and <hi>K</hi>G are rationall commensu<g ref="char:EOLhyphen"/>rable in length. And forasmuch as it hath bene proued, that the line Z<hi>K</hi> is commensurable in power <milestone type="tcpmilestone" unit="unspecified" n="1"/> onely vnto the rationall line <hi>K</hi>G, therefore the lines Z<hi>K</hi> and <hi>K</hi>G are rationall commensurable in power onely.<note place="margin">
                        <hi>Here are the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ower partes of the proposi<g ref="char:EOLhyphen"/>ti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> more order<g ref="char:EOLunhyphen"/>ly h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>dled the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> in the former de<g ref="char:EOLhyphen"/>möstration.</hi>
                     </note> Wherefore the line GZ is a residuall line. And forasmuch as the rationall line TG is commensurable in length to either of these lines DB and <hi>K</hi>G. Wherefore the lines DB &amp; <hi>K</hi>G shall be commensurable in length, by the 12. of the tenth. But the line Z<hi>K</hi> is to the line <hi>K</hi>G, as the line GD is to the line DB. Wherefore alternately, by the 16. of the fifth, the line <hi>K</hi>Z is to the line GD, as the line <hi>K</hi>G is to the line DB. Wherefore the line Z<hi>K</hi> is commensurable in length vnto the line GD. <milestone type="tcpmilestone" unit="unspecified" n="2"/> Wherefore the lines Z<hi>K</hi> and <hi>K</hi>G (the names of the residuall line GZ) are commensurable in length to the lines GD and DB, which are the names of the binomiall line GB: and the line Z<hi>K</hi> is to the line <hi>K</hi>G in the same proportion, that the line GD is to the line DB. <milestone type="tcpmilestone" unit="unspecified" n="3"/> Wherefore if the whole line Z<hi>K</hi> be in power more then the line conueniently ioyned <hi>K</hi>G, by the square of a line commensurable in length to the line Z<hi>K,</hi> then the greater name G<hi>D</hi> shall be in power more then the lesse name DB, by the square of a line commensurable in length to the line GD, by the 14. of the tenth. And if the line Z<hi>K</hi> be in power more then the line <hi>KG,</hi> by the square of a line incommensurable in length to the line Z<hi>K,</hi> the line also GD shall be in power more then the line DB, by the square of a line incommensurable in length vnto the line <hi>GD</hi> (by the same Propositi<g ref="char:EOLhyphen"/>on). And if the greater or lesse name of the one be commensurable in length to the rationall line put, the greater of l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sse name also of the other shall be commensurable in length to the same rationall line, (by the 12. of this booke). But if neither name of the one be commensurable in length to the rationall line put, neither name of the other also shall be commensurable in length to the same rationall line <milestone type="tcpmilestone" unit="unspecified" n="4"/> put (by the 13. of the same). Wherefore the residuall line <hi>G</hi>Z shall be in the selfe same order of residu<g ref="char:EOLhyphen"/>all lines, that the binomiall line GB is of binomiall lines (by the definitions of residuall and binomiall
<pb n="307" facs="tcp:7062:336"/>
lines. The square therefore of a rationall line applied to a binomiall line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> &amp;<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: which was required <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 89. Theoreme. The 113. Proposition. The square of a rational line applied vnto a residuall, maketh the breadth or other side a binomial line, whose names are commensurable to the names of the residuall line, and in the selfe same proportion: and moreouer that binomiall line is in the selfe same order of binomiall lynes, that the residual line is of residuall lynes.</head>
               <p>SVppose that A be a rationall line, and BD a residuall line. And vnto the square of the line A let that which is contained vnder the lines BD and KH be equal. Wherfore the square of the rationall line A applied vnto the residuall line BD maketh the breadth or other side KH. Then I say that the line KH is a binomi<g ref="char:EOLhyphen"/>all line, whose names are commensurable to
<figure/>
the names of the residuall line BD, and in the selfe same proportion, and that the line KH is in the selfe same order of binomiall lines, that the line BD is of residuall lines. Vnto the line BD let the line conueniently ioyned be DC.<note place="margin">Construction.</note> Wherfore the lines BC and DC are rationall commensurable in power onely. And vnto the square of the line A let the parallelogramme contained vnder the lines BC and G be equall. But the square of the line A is rationall.<note place="margin">Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the parallelogramme contained vnder the lines BC and G is also ra<g ref="char:EOLhyphen"/>tionall. Wherfore also the line G is rationall and commensurable in length to the line BC (by the 20. of the tenth). Now forasmuch as the parallelogramme contained vnder the lines BC and G is equall to that which is contained vnder the lines BD and KH, therfore (by the 16. of the sixt) as the line BC is to the line BD, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line KH to the line G. But the line BC is greater then the line BD. Wherfore also the line KH is greater then the line G. Vnto the line G l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t the line KE be equall. Wherfore the line KE is rationall and co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line BC, as also the line G was (by the 12. of the tenth) And for that as BC is to BD, so is KH to KE. Wherfore by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>duersion of proportio<g ref="char:cmbAbbrStroke">̄</g> (by the corollary of y<hi rend="sup">e</hi> 19. of the fift) as BC is to DC, so is KH, to EH,<note place="margin">An Assumpt.</note> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>KH into EH so let the line FH be to the line EF (how this is to be done, we will decare at the ende of this demonstration). Wherfore the residue KF is to the residue FH, as the whole KH is to the whole HE (by the 19. of the fift) that is, as the line BC is to the line CD. But the lines BC and CD are commensurable in power onely. Where<g ref="char:EOLhyphen"/>fore also the lines KF and FH are commensurable in power only. And for that as KH is to HE, so is KF to FH, but as KH is to HE, so is also HF to FE, therfore as KF is to FH, so is FH to FE. Wherfore (by the corollary of the 19. of the sixt) as the first is to the third, so is the square of the first, to the square of the second. Wherefore as KF is to FE, so is the square of the line KF to the square of the line FH, but these squares are commensurable, for the lines KF and FH are commensurable in power. Wherfore the lines KF and FE are commensurable in length. Wherfore (by the second part of the 15. of the tenth) the lines KE and EF are commensurable in length. Wherfore (by the same) the lines KF and FE are commensurable in length. But the line KE is rationall and commensurable in length to the line BC: wherefore the line KF is also rationall and commensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the line BC. And for that as the line BC is to the line CD, so it KF to EH, therfore alternately. (by the 16. of the fift) as BC is to KF, so is CD to FH. But the line BC is commensurable in length
<pb facs="tcp:7062:337"/>
to the line KF. Wherfore the line CD is commensurable in length to the line FH. But the line CD is rationall. Wherfore also the line FH is rationall. And the lines BC and CD are rationall commensurable in power onely. Wherfore the lines KF and FH are rationall com<g ref="char:EOLhyphen"/>mensurable <milestone type="tcpmilestone" unit="unspecified" n="1"/> in power onely. Wherfore the line KH is a binomiall line, whose names are com<g ref="char:EOLhyphen"/>mensurable <milestone type="tcpmilestone" unit="unspecified" n="2"/> to the names of the residuall line, and in the same proportion. I say moreouer that <milestone type="tcpmilestone" unit="unspecified" n="3"/> it is a binomiall of the selfe same order of binomial lines, that the line BD is of residual lines. <milestone type="tcpmilestone" unit="unspecified" n="4"/> For if the line BC be in power more then the line CD by the square of a line commensurable in length to the line BC, the line KF is also in power more then the line FH by the square of a line commensurable in length to the
<figure/>
line KF (by the 14. of the tenth). And if the line BC be commensurable in length to the rationall line put, the line KF is also (by the 12. of the tenth) commensurable in length to the rationall line, and so the <milestone type="tcpmilestone" unit="unspecified" n="1"/> lyne BD is a first residuall lyne, and the line KH is in like sort a first binomiall line. If the line CD be commensurable in length to the rational line, the line FH is also commen<g ref="char:EOLhyphen"/>surable <milestone type="tcpmilestone" unit="unspecified" n="2"/> in length to the same line, and so the line BD is a second residuall line, and the line KH a second binomiall line. And if neither of the lines BC nor CD be commensurable in length to the rationall line, neither also of the lines KF nor FH is commensurable in length <milestone type="tcpmilestone" unit="unspecified" n="3"/> to the same, and so the line BD is a third residuall line, and the line KH a third binomiall line. But if the line BC be in power more then the line CD by the square of a line incommen<g ref="char:EOLunhyphen"/>surable in length to the line BC, the line KF is in power more the<g ref="char:cmbAbbrStroke">̄</g> the line FH by the square of a line incommensurable in length to the line KF (by the 14. of the tenth) And if the line BC be commensurable in length to the rationall line put, the line KF is also commensura<g ref="char:EOLhyphen"/>ble in length to the same line, and so the line BD is a fourth residuall line, and the line KH <milestone type="tcpmilestone" unit="unspecified" n="4"/> a fourth binomiall line. And if the line CD be co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the rational line, the <milestone type="tcpmilestone" unit="unspecified" n="5"/> line FH is also co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the same, &amp; so the line BD is a fift residuall line, &amp; the line KH a fift binomiall line. And if neither of the lines BC nor CD be commensurable in length to the rationall line, neither also of the lines KF nor FH is commensurable in <milestone type="tcpmilestone" unit="unspecified" n="6"/> length to the same, and so the line BD is a sixt residuall line, and the line KH is a sixt bino<g ref="char:EOLhyphen"/>miall line. Wherfore KH is a binomiall line, whose names KF and FH are commensurable to the names of the residuall line BD, namely, to BC and CD, and in the selfe same propor<g ref="char:EOLhyphen"/>tion, and the binomiall line KH is in the selfe same order of binomiall lines, that the residu<g ref="char:EOLhyphen"/>all BD, is of residuall lines. Wherefore the square of a rationall line applied vnto a residuall line, maketh the breadth or other side a binomiall line, whose names are commensurable to the names of the residuall line, and in the selfe same proportion, and moreouer the binomiall line is in the selfe same order of binomiall lines, that the residuall line is of residuall lines: which was required to be demonstrated.</p>
               <div type="section">
                  <head>The Assumpt confirmed.</head>
                  <p>Now let vs
<figure/>
declare how, as the line KH is to the line EH, so to make the line HF to the line FE. Adde vnto the line KH directly a line equall to HE, and let the whole line be KL, and (by the tenth of the sixt) let the line HE be deuided as the whole line KL is deuided in the point H: let the line HE be so deuided in the point F. Wher<g ref="char:EOLhyphen"/>fore
<pb n="308" facs="tcp:7062:337"/>
as the line KH is to the line HL, that is, to the line HE, so is the line HF to the line FE.</p>
               </div>
               <div type="section">
                  <head>An other demonstration after <hi>Flussas.</hi>
                  </head>
                  <p>Suppose that <hi>A</hi> be a rationall line, and let <hi>BD</hi> be a residuall line. And vpon the line <hi>BD</hi> apply the parallelogramme <hi>DT</hi> equall to the square of the line <hi>A</hi> (by the 45. of the first) making in breadth the line <hi>BT.</hi>
                     <note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> af<g ref="char:EOLhyphen"/>ter Flussas.</hi>
                     </note> Then I say that <hi>BT</hi> is a binominall line such a one as is required in the proposition. Forasmuch as <hi>BD</hi> is a residuall line, let the line co<g ref="char:cmbAbbrStroke">̄</g>ueniently ioyned vnto it be <hi>GD.</hi> Wherfore the lines <hi>BG</hi> and <hi>GD</hi> are rationall commensurable in power onely.<note place="margin">
                        <hi>Construction.</hi>
                     </note> Vpon the rationall line <hi>BG</hi> apply the parallelogramme <hi>BI</hi> equall to the square of the line <hi>A</hi> and making in breadth the line <hi>BE.</hi> Wherefore the line <hi>BE</hi> is ra<g ref="char:EOLhyphen"/>tionall and commensurable in length to the line <hi>BG</hi> (by the 20. of the tenth). Now forasmuch as the pa<g ref="char:EOLhyphen"/>rallelogrammes <hi>BI</hi> and <hi>TD</hi> are equall (for
<figure/>
that they are eche equall to the square of the line <hi>A</hi>):<note place="margin">
                        <hi>Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> therfore reciprokally (by the 14. of the sixth) as the line <hi>BT</hi> is to the line <hi>BE,</hi> so is the line <hi>BG</hi> to the line <hi>BD.</hi> Wherefore by con<g ref="char:EOLhyphen"/>uersion of proportion (by the corrollary of the 19. of the fifth) as the line <hi>BT</hi> is to the line <hi>TE,</hi> so is the line <hi>BG</hi> to the line <hi>GD.</hi> As the line <hi>BG</hi> is to the line <hi>GD,</hi> so let the line <hi>TZ</hi> be to the line <hi>ZE</hi> by the corrollary of the 10. of the sixth. Wherefore by the 11. of the fifth the line <hi>BT</hi> is to the line <hi>TE,</hi> as the line <hi>TZ</hi> is to the line <hi>ZE.</hi> For either of them are as the line <hi>BG</hi> is to the line <hi>GD.</hi> Wherefore the residue <hi>BZ</hi> is to the residue <hi>ZT,</hi> as the whole <hi>BT</hi> is to the whole <hi>TE</hi> by the 19. of the fifth. Wherefore by the 11. of the fifth the line <hi>BZ</hi> is to the line <hi>ZT</hi> as the line <hi>ZT</hi> is to the line <hi>ZE.</hi> Wherfore the line <hi>TZ</hi> is the meane proportionall be<g ref="char:EOLhyphen"/>twene the lines <hi>BZ</hi> and <hi>ZE.</hi> Wherefore the square of the first, namely, of the line <hi>BZ,</hi> is to the square of the second, namely, of the line <hi>ZT,</hi> as the first, namely, the line <hi>BZ,</hi> is to the third, namely, to the line <hi>ZE</hi> (by the corollary of the 20. of the sixth). And for that as the line <hi>BG</hi> is to the line <hi>GD,</hi> so is the line <hi>TZ</hi> to the line <hi>ZE</hi>: but as the line <hi>TZ</hi> is to the line <hi>ZE,</hi> so is the line <hi>BZ</hi> to the line <hi>ZT.</hi> Wherefore as the line <hi>BG</hi> is to the line <hi>GD,</hi> so is the line <hi>BZ</hi> to the line <hi>ZT</hi> (by the 11. of the fifth). Wherfore the lines <hi>BZ</hi> and <hi>ZT</hi> are commensurable in pow<g ref="char:EOLhyphen"/>er onely, as also are the lines <hi>BG</hi> and <hi>GD</hi> (which are the names of the residuall line <hi>BD</hi>) by the 10. of this booke. Wherfore the right lines <hi>BZ</hi> and <hi>ZE</hi> are co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length, for we haue proued that they are in the same proportion that the squares of the lines <hi>BZ</hi> and <hi>ZT</hi> are. And therefore (by the co<g ref="char:EOLhyphen"/>rollary of the 15. of this booke) the residue <hi>BE</hi> (which is a rationall line) is commensurable in length vnto the same line <hi>BZ.</hi> Wherefore also the line <hi>BG</hi> (which is commensurable in length vnto the line <hi>BE</hi>) shall also be commensurable in length vnto the same line <hi>EZ</hi> (by the 12. of the tenth). And it is proued that the line <hi>RZ</hi> is to the line <hi>ZT</hi> commensurable in power onely. Wherefore the right lines <hi>BZ</hi> and <hi>ZT</hi> are rationall commensurable in power onely. Wherefore the whole line <hi>BT</hi> is a binomiall <milestone type="tcpmilestone" unit="unspecified" n="1"/> line (by the 36. of this booke). And for that as the line <hi>BG</hi> is to the line <hi>GD,</hi> so is the line <hi>BZ</hi> to the line <hi>ZT</hi>: therefore alternately (by the 16. of the fifth) the line <hi>BG</hi> is to the line <hi>BZ,</hi> as the line <hi>GD</hi> is to the line <hi>ZT.</hi> But the line <hi>BG</hi> is commensurable in length vnto the line <hi>BZ.</hi> Wherefore (by the 10. of this booke) the line <hi>GD</hi> is commensurable in length vnto the line <hi>ZT.</hi> Wherefore the names <hi>BG</hi> and <milestone type="tcpmilestone" unit="unspecified" n="2"/> 
                     <hi>GD</hi> of the residuall line <hi>BD</hi> are commensurable in length vnto the names <hi>BZ</hi> and <hi>ZT</hi> of the binomi<g ref="char:EOLhyphen"/>al line <hi>BT</hi>: and the line <hi>BZ</hi> is to the line <hi>ZT</hi> in the same proportion that the line <hi>BG</hi> is to the line <hi>GD</hi>
                     <milestone type="tcpmilestone" unit="unspecified" n="3"/> as before it was more manifest. And that they are of one and the selfe same order is thus proued. If the <milestone type="tcpmilestone" unit="unspecified" n="4"/> greater or lesse name of the residuall line, namely, the right lines <hi>BG</hi> or <hi>GD</hi> be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to any rationall line put: the greater name also or lesse, namely, <hi>BZ</hi> or <hi>ZT</hi> shalbe commensurable in length to the same rationall line put by the 12. of this booke. And if neither of the names of the residu<g ref="char:EOLhyphen"/>all line be commensurable in length vnto the rationall line put, neither of the names of the binomiall line shalbe commensurable in length vnto the same rationall line put (by the 13. of the tenth). And if the greater name <hi>BG</hi> be in power more then the lesse name by the square of a line commensurable in length vnto the line <hi>BG,</hi> the greater name also <hi>BZ</hi> shalbe in power more then the lesse by the square of a line commensurable in length vnto the line <hi>BZ.</hi> And if the one be in power more by the square of a line incommensurable in length, the other also shalbe in power more by the square of a line incommen<g ref="char:EOLhyphen"/>surable in length by the 14. of this booke. The square therefore of a rationall line. &amp;c. which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:338"/>
               <head>¶The 90. Theoreme. The 114. Proposition. <note place="margin">This is in a maner the con<g ref="char:EOLunhyphen"/>uerse of both the former pro<g ref="char:EOLhyphen"/>positions, ioynt<g ref="char:EOLunhyphen"/>ly.</note>If a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me be co<g ref="char:cmbAbbrStroke">̄</g>tained vnder a residuall line &amp; a binomiall lyne, whose names are commensurable to the names of the residuall line, and in the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e same proportion: the lyne which contayneth in power that super<g ref="char:EOLhyphen"/>ficies is rationall.</head>
               <p>SVppose that a parallelogramme be contained vnder a residuall line AB and a binomiall line CD, and let the greater name of the binomiall line be CE, and the lesse name be ED, and let the names of the binomiall line, namely, CE and ED be commensurable to the names of the residuall line, namely, to AF and F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and in the selfe same proportion. And let the
<figure/>
line which containeth in power that parallelo<g ref="char:EOLhyphen"/>gra<g ref="char:cmbAbbrStroke">̄</g>me be G. The<g ref="char:cmbAbbrStroke">̄</g> I say that the line G is ratio<g ref="char:EOLhyphen"/>nal. Take a rational line, namely, H. And vnto the line CD apply a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me equal to y<hi rend="sup">e</hi> square of the line H,<note place="margin">Construction.</note> and making in breadth the line KL. Wherefore (by the 112. of the tenth) KL is a residuall line,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> whose names let be KM and ML, which are (by the same) co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mensurable to the names of the binomiall line, that is to CE and ED, and are in the selfe same proportio<g ref="char:cmbAbbrStroke">̄</g>. But by position the lines CE and ED are co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the lines AF and FB, and are in the selfe same proportion. Wherfore (by the 12. of the tenth) as the line AF is to the line FB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the line KM to the line ML. Wherfore alternately (by the 16. of the fift) as the line AF is to the line KM, so is the line BF to the line LM. Wherfore the residue AB is to the residue KL, as the whole AF is to the whole KM. But the line AF is commensura<g ref="char:EOLhyphen"/>ble to the line KM, for either of the lines AF and KM is commensurable to the line CE. Wherfore also the line AB is commensurable to the line KL. And as the line AB is to the line KL, so (by the first of the sixt) is the parallelogramme contained vnder the lines CD and AB to the parallelogramme contained vnder the lines CD and KL. Wherfore the pa<g ref="char:EOLhyphen"/>rallelogramme contained vnder the lines CD and AB is commensurable to the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines CD and KL. But the parallelogramme contained vnder the lines CD and KL is equall to the square of the line H. Wherfore the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tained vnder the lines CD &amp; AB is co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the square of the line H. But the paral<g ref="char:EOLhyphen"/>lelogra<g ref="char:cmbAbbrStroke">̄</g>me contained vnder the lines CD and AB is equall to the square of the line G. Wher<g ref="char:EOLhyphen"/>fore the square of the line H is commensurable to the square of the line G. But the square of the line H is rationall. Wherfore the square of the line G is also rationall. Wherfore also the line G is rational, and it containeth in power the parallelogramme contained vnder the lines AB and CD. If therfore a parallelogramme be contained vnder a residuall line and a bino<g ref="char:EOLhyphen"/>miall line, whose names are commensurable to the names of the residuall line, and in the selfe same proportion, the line which containeth in power that superficies, is rationall: which was required to be proued.</p>
               <div type="section">
                  <head>¶Corollary.</head>
                  <p>Hereby it is manifest, that a rationall parallelogramme may be contained vnder irrationall lines.</p>
               </div>
               <div type="section">
                  <pb n="309" facs="tcp:7062:338"/>
                  <head>¶An ot<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>r <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> 
                     <hi>Flussas.</hi>
                  </head>
                  <p>
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D: whos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> names A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D let be commensurable in length vnto the names of the residuall line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, which let be AF and FB. And let the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e AE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be to the line ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in the same proportion that the line AF is to the line F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. And let the right line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> contayne in power the superficies D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Then I say, tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is a rationall lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ne,<note place="margin">
                        <hi>Construction.</hi>
                     </note> which l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: And vpon the line <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> describe (by the 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the first) a parallelogramme eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all to the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the line <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> and making in breadth the line DC. Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>12. of this booke) CD is a residu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> whose names (Which let be <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> and OD) shall be co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>mensurabl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gth vnto the names A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D, and the line C o shall be vnto the line OD,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> in the same pro<g ref="char:EOLhyphen"/>por<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion
<figure/>
that the line AE is to the line ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> But as the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is to the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D, so by supposition, is the line AF to the line FE. Wherfore as the line CO is to the line OD, so is the line AF to the line F<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> Where<g ref="char:EOLhyphen"/>fore the lines CO and OD are commensu<g ref="char:EOLhyphen"/>rable with the lines A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> (by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>. of this boke). Wherfore the residue, name<g ref="char:EOLhyphen"/>ly, the line CD is to the residue, namely, to the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, as the line CO is to the line AF (by the 19. of the fifth). But it is proued, that the line CO is co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto the line AF. Wherefore the line CD is commen<g ref="char:EOLhyphen"/>surable vnto the line AB. Wherefore (by the first of the sixth) the parallelogramme CA is commensurable to the parallelogramme D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. But the parallelogramme <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> (by construction) rationall (for it is equall to the square of the rationall line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>). Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>refore the parallelogramme <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s also rat<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which by supposition co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tayneth in power the superficies <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is also rationall. If therfore a parallelogra<g ref="char:cmbAbbrStroke">̄</g>me be contayned &amp;c: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 91. Theoreme. The 115. Proposition. Of a mediall line are produced infinite irrationall lines, of which none is of the selfe same kinde with any of those that were before.</head>
               <p>SVppose that A be a mediall line. Then I say, that of the line A may be produced infinite irrationall lines, of which none shall be of the selfe same kinde with any of those that were before. Take a rationall line B. And vnto that which is con<g ref="char:EOLhyphen"/>tained vnder the lines A and B let the square of the line C be equall (by the 14. of the second)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherefore the line C is irrationall. For a superficies
<figure/>
contained vnder a rationall line and an irrationall line, is (by the As<g ref="char:EOLhyphen"/>sumpt following the 38. of the tenth) irrationall: and the line which containeth in power an irrationall superficies, is (by the Assumpt going before the 21. of the tenth) irrationall. And it is not one and the selfe same with any of those thirtene that were before. For none of the lines that were before applied to a rationall line maketh the breadth mediall. Againe vnto that which is contained vnder the lines B and C, let the square of D be equall. Wherefore the square of D is irrationall. Wherefore also the line D is irrationall and not of the self same kinde with any of those that were before. For the square of none of the lines which were before, applied to a rationall line, maketh the breadth the line C. In like sort also shall it so followe, if a man proceede infinitely. Wherefore it is manifest, that of a mediall line are produced infinite irrationall lines, of which none is of the selfe same kinde with any of those that were before: which was required to be proued.</p>
               <div type="section">
                  <pb facs="tcp:7062:339"/>
                  <head>An other demonstratio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </head>
                  <p>
                     <note place="margin">An other de<g ref="char:EOLhyphen"/>monstration.</note>Suppose that AC be a mediall line. Then I say, that of the line AC may be produced in<g ref="char:EOLhyphen"/>finite irrationall lines, of which none shall be of the selfe same kinde with any of those irra<g ref="char:EOLhyphen"/>tionall lines before named. Vnto the line AC and from the point A, draw (by the 11. of the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>) a perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line AB, and let AB be a rationall line, and make perfect<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the parallelo<g ref="char:EOLhyphen"/>gramme BC. Wherefore BG is irrationall, by that which was declared and proued (in maner of an Assumpt) in the end of the demonstration of the 38: and the line that containeth it i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> power is also irrationall. Let the line CD con<g ref="char:EOLhyphen"/>taine
<figure/>
in power the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies BC. Wherefore CD is irrationall &amp; not of the selfe same kind with any of those that were before: for the square of the line CD applied to a rationall line, namely, AB, maketh the breadth a mediall line, namely, AC. But the square of none of the foresaid lines applied to a rationall line maketh the breadth a mediall line. Againe, make perfecte the parallelogramme ED. Wherefore the parallelogramme ED is also irrationall (by the sayd Assumpt in the end of the 98. his de<g ref="char:EOLhyphen"/>monstration brie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ly proued) and the line which containeth it in power is irrationall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> let the line which containeth it in power be DF. Wherefore DF is irrationall and not of the selfe same kinde with any of the foresaid irrationall lines. For the square of none of the foresayd irrationall lines applied vnto a rationall line, maketh the breadth the line CD. Wherefore of a mediall line are produced infinite irrationall lines, of which none is of the selfe same kinde with any of those that were before: which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 92. Theoreme. The 116. Proposition. Now let vs proue that in square figures, the diameter is incommensurable in length to the side.</head>
               <p>SVppose that ABCD be a square, and let the diameter therof be AC. Then I say that the diameter AC is incommensurable in length to the side AB. For if it be possible,<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> leading to an impossibili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ie.</note> let it be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth. I say that the<g ref="char:cmbAbbrStroke">̄</g> this will follow, that one and the selfe same nu<g ref="char:cmbAbbrStroke">̄</g>ber shall be both an euen number &amp; an odde number. It is manifest (by the 47. of the first) that the square of the line AC is double to the square of the line AB. And for that the line AC is commensurable in length to the line AB (by sup<g ref="char:EOLhyphen"/>position), therfore the lyne AC hath vnto the line AB
<figure/>
that proportion that a number hath to a number (by the 5. of the tenth). Let the lyne AC haue vnto the line AB that proportion that the number EF hath to the num<g ref="char:EOLhyphen"/>ber G. And let EF and G be the least numbers that haue one and the same proportion with them. Wherfore EF is not vnitie. For if EF be vnitie, and it hath to the num<g ref="char:EOLhyphen"/>ber G that proportion that the line AC hath to the lyne AB, and the line AC is greater then the lyne AB. Wher<g ref="char:EOLunhyphen"/>fore vnitie EF is greater then the number G, which is impossible. Wherfore FE is not vnitie, wherfore it is a number. And for that as the square of the line AC is to the square of the lyne AB, so is the square number of the number EF, to the square number
<pb n="310" facs="tcp:7062:339"/>
of the number G, for in eche is the proportion of their sides doubled (by the corollary of the 20 of the sixt and 11. of the eight): and the proportion of the line AC to the line AB doubled, is equal to the proportio<g ref="char:cmbAbbrStroke">̄</g> of the nu<g ref="char:cmbAbbrStroke">̄</g>ber EF to the number G, doubled, for as the line AC is to the line AB, so is the nu<g ref="char:cmbAbbrStroke">̄</g>ber EF to the number G. But the square of the line AC is double to the square of the line AB. Wherfore the square number produced of the number EF is double to the square number produced of the number G. Wherefore the square number produced of EF is an euen number. Wherfore EF is also an euen number. For if EF were an odde num<g ref="char:EOLhyphen"/>ber, the square number also produced of it, should (by the 23. and 29. of the ninth) be an odde number. For if odde numbers how many soeuer be added together, and if the multitude of the<g ref="char:cmbAbbrStroke">̄</g> be odde, the whole also shal be odde. Wherfore EF is an euen number. Deuide the number EF into two equall partes in H. And forasmuch as the numbers EF and G are the lest num<g ref="char:EOLhyphen"/>bers in that proportion, therfore (by the 24. of the seuenth) they are prime numbers the one to the other. And EF is an euen number. Wherfore G is an odde number. For if G were an e<g ref="char:EOLhyphen"/>uen number, the number two should measure both the number EF and the number G (for e<g ref="char:EOLhyphen"/>uery euen nu<g ref="char:cmbAbbrStroke">̄</g>ber hath an halfe part by the definition) but these numbers EF &amp; G are prime the one to the other. Wherfore it is impossible that they should be measured by two or by any other number besides vnitie. Wherfore G is an odde number. And forasmuch as the number EF is double to the number EH, therfore the square number produced of EF is quadruple to the square number produced of EH. And the square number produced of EF is double to the square number produced of G. Wherfore the square number produced of G is double to the square number produced of EH. Wherfore the square number produced of G is an euen number. Wherfore also by those thinges which haue bene before spoken, the number G is an euen number, but it is proued that it is an odde number, which is impossible. Wherefore the line AC is not commensurable in length to the line AB, wherfore it is incommensurable.</p>
               <div type="section">
                  <head>An other demonstration.</head>
                  <p>We may by an other demonstration proue, that the diameter of a square is incommensu<g ref="char:EOLhyphen"/>rable to the side thereof.<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration leading to an impossibili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ie.</note> Suppose that there be a square, whose diameter let be <hi>A</hi> and let the side thereof be <hi>B.</hi> Then I say that the line <hi>A</hi> is incommensurable in length to the line <hi>B.</hi> For if it be possible let it be commensurable in length. And agayne as the line <hi>A</hi> is to the line <hi>B</hi> so let the number <hi>EF</hi> be to the number <hi>G</hi>: and let them be the least that haue one and the same proportion with them: wherefore the numbers <hi>EF</hi> and
<figure/>
                     <hi>G,</hi> are prime the one to the other. First I say that <hi>G</hi> is not v<g ref="char:EOLhyphen"/>nitie. For if it be possible let it be vnitie. And for that the square of the line <hi>A</hi> is to the square of the line <hi>B</hi> as the square number produced of <hi>EF</hi> is to the square number produced of <hi>G</hi> (as it was proued in the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ormer demonstration) but the square of the line <hi>A</hi> is double to the square of the line <hi>B.</hi> Wher<g ref="char:EOLunhyphen"/>fore the square nu<g ref="char:cmbAbbrStroke">̄</g>ber produced of <hi>EF</hi> is double to the square number produced of <hi>G.</hi> And by your supposition <hi>G</hi> is vnitie. Wherefore the square number produced of <hi>EF</hi> is the number two which is impossible. Wherefore <hi>G</hi> is not vnitie. Wherefore it is a number. And for that as the square of the line <hi>A</hi> is to the square of the line <hi>B,</hi> so is the square number produced of <hi>EF</hi> to the square number produced of <hi>G.</hi> Wherefore the square number produced of <hi>EF</hi> is double to the square number produced of <hi>G.</hi> Wherefore the square number produced of <hi>G.</hi> measureth the square number produced of <hi>EF.</hi> Wherefore also (by the 14. of the eight) the number <hi>G</hi> measureth the number <hi>EF</hi>: and the number <hi>G</hi> also measureth it selfe. Wherefore the number <hi>G</hi> measureth these numbers <hi>EF</hi> and <hi>G,</hi> when yet they are prime the one to the
<pb facs="tcp:7062:340"/>
other: which is impossible. Wherefore the diameter <hi>A</hi> is not commensurable in length to the side <hi>B.</hi> Wherefore it is incommensurable: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>An other demonstration after <hi>Flussas.</hi>
                  </head>
                  <p>Suppose that vppon the line AB be described a square
<figure/>
whose diameter let be the line AC. Then I say that the side AB is incommensurable in length vnto the diameter AC. For<g ref="char:EOLhyphen"/>asmuch as the lines AB and BC are equall, therefore the square of the line AC is double to the square of the line AB by the 47. of the first. Take by the 2. of the eight nu<g ref="char:cmbAbbrStroke">̄</g>bers how many soeuer in continuall proportion fro<g ref="char:cmbAbbrStroke">̄</g> vnitie, and in the proporti<g ref="char:EOLhyphen"/>on of the squares of the lines AB and AC. Which let be the numbers D, E, F, G. And forasmuch as the first from vnitie name<g ref="char:EOLunhyphen"/>ly E is no square number, for that it is a prime number, neither is also any other of the sayd numbers a square number except the third from vnitie and so all the rest leuing one betwene, by the 10. of the ninth. Wherefore D is to E, or E to F, or F to G, in that proportion that a square number is to a number not square. Wherefore by the corrollary of the 25. of the eight, they are not in that proportion the one to the other that a square number is to a square number. Wherefore neither also haue the squares of the lines AB and AC (which are in the same pro<g ref="char:EOLhyphen"/>portion) that porportion that a square number hath to a square number. Wherefore by the 9. of this booke their sides, namely, the side AB and the diameter AC are incommensurable in length the one to the other which was required to be proued.</p>
                  <p>This demonstration I thought good to adde, for that the former demonstrations seme not so full, and they are thought of some to be none of <hi>Theons,</hi> as also the proposi<g ref="char:EOLhyphen"/>tion to be none of <hi>Euclides.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>Here followeth an instruction by some studious and skilfull Grecian (perchance <hi>Theon</hi>) which teacheth vs of farther vse and fruite of these irrationall lines.</head>
                  <p>Seing that there are founde out right lines incommensurable in length the one to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>ther, as the lines <hi>A</hi> and <hi>B,</hi> there may also be founde out many other magnitudes hauing le<g ref="char:cmbAbbrStroke">̄</g>gth and breadth (such as are playne superficieces) which shalbe incomme<g ref="char:cmbAbbrStroke">̄</g>surable the one to the o<g ref="char:EOLhyphen"/>ther. For if (by the 13. of the sixth) betwene the lines <hi>A</hi> and <hi>B</hi> there be taken the meane pro<g ref="char:EOLhyphen"/>portionall line, namely <hi>C,</hi> then (by the second corrollary of the 20. of the sixth) as the line <hi>A</hi> is to the line <hi>B,</hi> so is the figure described vpon the line <hi>A</hi> to the figure described vpon the line <hi>C,</hi> being both like and in like sort described, that is, whether they be squares (which are al<g ref="char:EOLhyphen"/>wayes like the one to the other), or whether they be any other like rectiline figures, or whether they be circles aboute the diameters <hi>A</hi> and <hi>C.</hi> For cir<g ref="char:EOLhyphen"/>cles
<figure/>
haue that proportion the one to the other, that the squares of their diameters haue (by the 2. of the twelfth). Wherfore (by the second part of the 10. of the tenth) the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igures described vpon the lines <hi>A</hi> and <hi>C</hi> being like and in like sort described are incommensurable the one to the other. Wherfore by this meanes there are founde out superficieces incommensurable the one to the other. In like sort there may be founde out figures co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable the one to the other, if ye put the lines <hi>A</hi> and <hi>B</hi> to be co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth the one to the other. And seing that it is so, now let vs also proue that euen in soli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es also or bodyes there are some commensurable the one to the other, and other some incommensurable the one to the other. For if from eche of the squares of the lines <hi>A</hi> and <hi>B,</hi> or from any other rectiline figures equal to these squares
<pb n="311" facs="tcp:7062:340"/>
be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rected solides of equall alti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ude, whether those solides be comp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>sed of equidistant supersi<g ref="char:EOLhyphen"/>cieces, or whether they be p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ramids or prismes, thos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> solides s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> er<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed shalbe in that proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other that theyr bases are (by the 32. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the eleuenth and 5. and 6. of the twelfth) Howbeit there is no such proposition concerning prismes, And so if the bases of the solides b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> commensurable the one to the other, the solides also shall be commensurable the one to the other, and if the bases be incommensurable the one to the other, the solides also shall be incom<g ref="char:EOLhyphen"/>mensurable the one to the other (by the 10. of the tenth). And if there be two circles <hi>A</hi> and <hi>B</hi>: and vpon ech of the circles be erected Cones or Cilinders of equal altitude, those Cones &amp; Ci<g ref="char:EOLhyphen"/>linders s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all be in that proportion the one to the other that the circles are, which are their ba<g ref="char:EOLhyphen"/>ses (by the 11. of the twelfth): and so if the circles be commensurable the one to the other, the Cones and Cilinders also shall be commensurable the one to the other. But if the circles be in<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>mensurable the one to the other, the Cones also and Cilinders shalbe inco<g ref="char:cmbAbbrStroke">̄</g>mensurable the one to the other, (by the 10. of the tenth). Wherefore it is manifest that not onely in lines and super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icieces, but also in solides or bodyes is found commensurabilitie or incommensurability.</p>
               </div>
               <div type="section">
                  <head>An aduertisement by <hi>Iohn Dee.</hi>
                  </head>
                  <p>Although this proposition were by <hi>Euclide</hi> to this booke alotted, (as by the auncient grecian pub<g ref="char:EOLhyphen"/>lished vnder the name of <hi>Aristoteles</hi> 
                     <gap reason="foreign">
                        <desc>〈 in non-Latin alphabet 〉</desc>
                     </gap>, it would seme to be, and also the property of the same, agreable to the matter of this booke, and the proposition it selfe, so famous in Philosophy and Logicke, as it was, would in maner craue his eleme<g ref="char:cmbAbbrStroke">̄</g>tal place, in this te<g ref="char:cmbAbbrStroke">̄</g>th boke) yet the dignitie &amp; perfec<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of Mathematicall Method can not allow it here: as in due order following: But most aptly after the 9. propositio<g ref="char:cmbAbbrStroke">̄</g> of this booke, as a Corrollary of the last part thereof. And vndoubtedly the propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>itio<g ref="char:cmbAbbrStroke">̄</g> hath for this 2000. yeares bene notably regarded among the greke Philosophers: and before <hi>Aristotles</hi> time was concluded with the very same inconuenience to the gaynesayer, that the first demonstration here induceth, namely, <hi>Odde number to be equall to euen</hi>: as may appear<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in <hi>Aristotles</hi> worke, named <hi>Analitica prima,</hi> the first booke and 40. chapter. But els in very many places of his workes he maketh mention of the proposition. Euident also it is that <hi>Euclide</hi> was about <hi>Aristotles</hi> time, and in that age the most excel<g ref="char:EOLhyphen"/>lent Geometrician among the Grekes. Wherefore, seing it was so publike in his time, so famous, and so appertayning to the property of this booke: it is most likely, both to be knowne to <hi>Euclide,</hi> and also to haue bene by him in apt order placed. But of the disordring of it, can remayne no doubt, if ye consider in <hi>Zamberts</hi> translation, two other propositions going next before it, so farre misplaced, that where they are, word for word, before du<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ly placed, being the 105. and 106. yet here (after the booke ended), they are repeated with the numbers of 116. and 117. proposition. <hi>Zambert</hi> therein was more faythfull to fol<g ref="char:EOLhyphen"/>low as he found in his greke example, than he was skilfull or carefull to doe what was necessary. Yea and some greke written auncient copyes haue them not so: Though in deede they be well demonstrated, yet truth disorded, is halfe disgraced<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> especially where the patterne of good order, by profession is auouched to be. But through ignoraunce, arrogan<g ref="char:EOLhyphen"/>cy and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>emerltie of vnskilfull Methode Masters, many thinges remayne yet, in these Geometricall Elementes, vnduely tumbled in: though true, yet with disgrace: which by helpe of so many wittes and habilitie of such, as now may haue good cause to be skilfull herein, will (I hope) ere long be taken a<g ref="char:EOLhyphen"/>way: and thinges of impor<g ref="char:EOLhyphen"/>tance (wanting) sup<g ref="char:EOLhyphen"/>plied.</p>
               </div>
            </div>
            <trailer>The end of the tenth booke of Euclides Elementes</trailer>
         </div>
         <div n="11" type="book">
            <pb facs="tcp:7062:341"/>
            <head>¶The eleuenth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>
               <note place="margin">
                  <hi>The argument of the eleuenth booke.</hi>
               </note>HITHERTO HATH <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>VCLID<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> IN TH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>S<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> former bookes with a wonderfull Methode and order entreated of such kindes of figures superficial, which are or may be described in a superficies or plaine. And hath taught and set forth their properties, natures, generati<g ref="char:EOLhyphen"/>ons, and productions euen from the first roote, ground, and beginning of them: namely, from a point, which al<g ref="char:EOLhyphen"/>though it be indiuisible, yet is it the beginning of all quantitie,<note place="margin">
                  <hi>A point the be<g ref="char:EOLhyphen"/>ginning of all quantitie con<g ref="char:EOLhyphen"/>tinuall.</hi>
               </note> and of it and of the motion and slowing ther<g ref="char:EOLhyphen"/>of is produced a line, and consequently all quantitie co<g ref="char:cmbAbbrStroke">̄</g>
               <g ref="char:EOLhyphen"/>tinuall, as all figures playne and solide what so euer. <hi>Eu<g ref="char:EOLhyphen"/>clide</hi> therefore in his first booke began with it,<note place="margin">
                  <hi>The methode vsed by Euclide in the ten <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>mer boo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es.</hi>
               </note> and from thence went he to a line, as to a thing most simple next vnto a point, then to a superficies, and to angles, and so through the whole first booke,<note place="margin">
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst bo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e.</hi>
               </note> he intreated of these most simple and plaine groundes. In the second booke he entrea<g ref="char:EOLhyphen"/>ted further,<note place="margin">
                  <hi>Second <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> and went vnto more harder matter, and taught of diuisions of lines, and of the multiplication of lines, and of their partes, and of their passions and properties. And for that rightlined <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>igures are far distant in nature and propertie from round and circular figures, in the third booke he instructeth the reader of the nature and conditio<g ref="char:cmbAbbrStroke">̄</g> of circles.<note place="margin">
                  <hi>Third boo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> In the fourth booke he compareth figures of right lines and circles together,<note place="margin">
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ourth b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> and teacheth how to describe a figure of right lines with in or about a circle: and con<g ref="char:EOLhyphen"/>tra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>iwi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e a circle with in or about a rectiline figure. In the fifth booke he searcheth out the nature of proportion (a matter of wonderfull vse and deepe consideration),<note place="margin">
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iueth bo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e.</hi>
               </note> for that otherwise he could not compare <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>igure with figure, or the sides of figures together. For whatsoeuer is compared to any other thing, is compared vnto it vndoubtedly vn<g ref="char:EOLhyphen"/>der some kinde of proportion. Wherefore in the sixth booke he compareth figures to<g ref="char:EOLhyphen"/>gether,<note place="margin">
                  <hi>Sixth boo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> one to an other, likewise their sides. And for that the nature of proportion, can not be fully and clearely sene without the knowledge of number, wherein it is first and chiefely found: in the seuenth,<note place="margin">
                  <hi>Seuenth book<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi>
               </note> eight,<note place="margin">
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ight boo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</hi>
               </note> and ninth bookes,<note place="margin">
                  <hi>Ninth booke.</hi>
               </note> he entreat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>th of number, &amp; of the kindes and properties thereof. And because that the sides of solide bodyes, for the most part are of such sort, that compared together, they haue such proportion the one to the other,<note place="margin">
                  <hi>Tenth boo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</hi>
               </note> which can not be expres<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ed by any number certayne, and therefore are cal<g ref="char:EOLhyphen"/>led irrational lines, he in the te<g ref="char:cmbAbbrStroke">̄</g>th boke hath writte<g ref="char:cmbAbbrStroke">̄</g> &amp; taught which line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> are co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>sura<g ref="char:EOLhyphen"/>ble or inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable the one to the other, and of the diuersitie of kindes of irrationall lines, with all the conditions &amp; proprieties of them. And thus hath <hi>Euclide</hi> in these ten foresayd bokes, fully &amp; most ple<g ref="char:cmbAbbrStroke">̄</g>teously in a meruelous order taught, whatsoeuer semed necessary, and requisite to the knowledge of all superficiall figures, of what sort &amp; forme so euer they be. Now in these bookes following he entreateth of figures of an other kinde, namely, of bodely figures:<note place="margin">
                  <hi>What is entrea<g ref="char:EOLunhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ea of in the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e boo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es fol<g ref="char:EOLhyphen"/>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>wi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</hi>
               </note> as of Cubes, Piramids, Cones, Columnes, Cilinders, Parallelipipedons. Spheres and such others<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and sheweth the diuersitie of the<g ref="char:cmbAbbrStroke">̄</g>, the gene<g ref="char:EOLhyphen"/>ration, and production of them, and demonstrateth with great and wonderfull art, their proprieties and passions, with all their natures and conditions. He also compareth one o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> them to an other, whereby to know the reason and proportion of the one to the o<g ref="char:EOLhyphen"/>ther, chiefely of the fiue bodyes which are called regular bodyes.<note place="margin">
                  <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>ular bo<g ref="char:EOLhyphen"/>dies<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>all ende <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> I u<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>es <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eo<g ref="char:EOLhyphen"/>me<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>all <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e<g ref="char:EOLhyphen"/>mentes.</hi>
               </note> And these are the thinges of all other entreated of in Geometrie, most worthy and of greatest dignitie, and as it were the end and finall entent of the whole are of Geometrie, and for whose cause hath bene written, and spoken whatsoeuer hath hitherto in the former bookes bene sayd or written. As the first booke was a ground, and a necessary entrye to all the r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>st <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ollowing, so is this eleuenth booke a necessary entrie and ground to the rest which follow.<note place="margin">
                  <hi>Co<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>is<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e and <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> booke <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>.</hi>
               </note> And as that contayned the declaration of wordes, and definitions of thinge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>
               <pb n="312" facs="tcp:7062:341"/>
requisite to the knowledge of superficiall figures, and entreated of lines (and of their di<g ref="char:EOLunhyphen"/>uisions and sections) which are the termes and limites of superficiall figures: so in this booke is set forth the declaration of wordes and definitions of thinges pertayning to solide and corporall figures: and also of superficieces which are the termes &amp; limites of solides: moreouer of the diuision and intersection of them, and diuers other thinges, without which the knowledge of bodely and solide formes can not be attayned vnto. And first is set the definitions as follow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>th.</p>
            <div type="part">
               <head>Definitions. A solide or body is that which hath length, breadth, and thicknes,<note place="margin">
                     <hi>First dif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ini<g ref="char:EOLhyphen"/>tion</hi>
                  </note> and the terme or limite of a solide is a superficies.</head>
               <p>There are three kindes of continuall quantitie, a line, a superficies, and a solide or body: the begin<g ref="char:EOLhyphen"/>ning of all which (as before hath bene sayd) is a poynt, which is indiuisible. Two of these quantities, namely, a line and a superficies, were defined of <hi>Euclide</hi> before in his first booke. But the third kinde, namely, a solide or body he there defined not, as a thing which pertayned not then to his purpose: but here in this place he setteth the definitio<g ref="char:cmbAbbrStroke">̄</g> therof, as that which chiefely now pertayneth to his purpose, and without which nothing in these thinges can profitably be taught. <hi>A solide</hi> (sayth he) <hi>is that which hath le<g ref="char:cmbAbbrStroke">̄</g>gth, breadth, and thicknes, or depth.</hi> There are (as before hath bene taught) three reasons or meanes of measuring, which are called co<g ref="char:cmbAbbrStroke">̄</g>monly dimensions, namely, l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ngth, breadth, and thicknes. These di<g ref="char:EOLhyphen"/>mensions are ascribed vnto quantities onely. By these are all kindes of quantitie de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ined, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> are counted perfect or imperfect, according as they are pertaker of fewer or more of them. As <hi>Euclide</hi> defined a line, ascribing vnto it onely one of these dimensions, namely, length: Wherefore a line is the imperfectest kinde of quantitie. In defining of a superficies, he ascribed vnto it two dimensions, namely, length, and breadth: whereby a superficies is a quantitie of greater perfection then is a line, but here in the defini<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> of a solide or body. <hi>Euclide</hi> attributeth vnto it all the three dimensio<g ref="char:cmbAbbrStroke">̄</g>s, le<g ref="char:cmbAbbrStroke">̄</g>gth, breadth, and thicknes. Wherfore a solide is the most perfectest quantitie,<note place="margin">
                     <hi>A solide the most perfec<g ref="char:EOLhyphen"/>test quantitie.</hi>
                  </note> which wanteth no dimension at all, passing a lyne by two dimensions, and passing a super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies by one. This definition of a solide is without any designation of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orme or figure easily vnderstanded, onely conceiuing in minde, or beholding with the eye a piece of timber or stone, or what matter so euer els, whose dimension<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> let be equall or vnequall. For example let the length therof be 5. inches, the breadth 4. and the thicknes 2. if the dimensions were equall, the reason is like, and all one, as it is in a Sphere and in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cube. For in that respect and consideration onely, that it is long, broade, and thicke, it beareth the name of a solide or body, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd hath the nature and pro<g ref="char:EOLhyphen"/>perties therof. There is added to the end<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the definition of a solide, that the terme and limite of a so<g ref="char:EOLhyphen"/>lide <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s a superficies. Of thinges infinitie there i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> no Arte or Scien<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e. All quantities therfore in this Arte entreated of, are imagined to be finite,<note place="margin">
                     <hi>No science of thinges infi<g ref="char:EOLhyphen"/>nite.</hi>
                  </note> and to haue their endes and borders as hath bene shewed in the first booke, that the limites and endes of a line are pointes, and the limites or borders of a superficies are lines, so now he saith tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the endes, limites, or borders of a solide<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are superficieces. As the side of any <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quare piece of timber, or of a table, or die, or any other lik<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, are the termes and limites of them.</p>
            </div>
            <div type="part">
               <head>2 A right line is then erected perpendicularly to a pl<gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>erficies, whe<g ref="char:cmbAbbrStroke">̄</g> the right line maketh right angles with all the lines <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> it,<note place="margin">
                     <hi>Second diffi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> and are drawen vpon the ground plaine superficies.</head>
               <p>Suppose that vpon the grounde playne superfi<g ref="char:EOLhyphen"/>cies,
<figure/>
CDEF from the pointe B be erected a right line, namely, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>A, so that let the point A be a lo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e in the ayre. Drawe also from the poynte <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the playne superficies C<g ref="char:EOLhyphen"/>DBF, as many right lines as ye list, as the lines BC, BD, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>, BF, BG, HK, BH, and BL. If the erected line BA with all these lines drawen in the superficies CDEF make a right angle, so that all those <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ngles A<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>, A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D, A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E, ABF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G, A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>K, ABH, ABL, and so of others, be right angles, then by this definition, the line AB, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a line <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>cted vpon the superficies CDEF: it is also called com<g ref="char:EOLhyphen"/>monly a perpendicular line or a plumb line, vnto or vpon a superficies.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:342"/>
               <head>
                  <note place="margin">
                     <hi>Third diffini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>3 A plaine superficies is then vpright or erected perpendicularly to a plaine superficies, when all the right lines drawen in one of the plaine su<g ref="char:EOLhyphen"/>perficieces vnto the common section of those two plaine superficieces, ma<g ref="char:EOLhyphen"/>king therwith right angles, do also make right angles to the other plaine superficies. Inclination or leaning of a right line, to a plaine superficies, is an acute angle, contained vnder a right line falling from a point aboue to the plaine superficies, and vnder an other right line, from the lower end of the sayd line (let downe) drawen in the same plaine superficies, by a certaine point assigned, where a right line from the first point aboue, to the same plaine superficies falling perpendicularly, toucheth.</head>
               <p>In this third definition are included two definitions: the first is of a plaine superficies erected per<g ref="char:EOLhyphen"/>pendicularly vpon a plaine superficies.<note place="margin">
                     <hi>Two dif<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ini<g ref="char:EOLhyphen"/>tions included in this di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>i<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> The second is of the inclination or leaning of a right line vnto a superficies: of the first take this example. Suppose ye haue two super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces ABCD and CDEF. Of which let the superficies CDEF be a ground plaine superficies, and let the superficies ABCD be e<g ref="char:EOLhyphen"/>rected vnto it, and let the line CD be a common terme or in<g ref="char:EOLhyphen"/>tersection
<figure/>
to them both, that is, let it be the end or bound of either of them,<note place="margin">
                     <hi>Declaratio<g ref="char:cmbAbbrStroke">̄</g> of the first part.</hi>
                  </note> &amp; be drawen in either of them: in which line note at pleasure certaine pointes, as the point G, H. From which pointes vnto the line CD, draw perpendicular lines in the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies ABCD, which let be GL and HK, which fal<g ref="char:EOLhyphen"/>ling vpon the superficies CDEF, if they cause right angles with it, that is, with lines drawen in it from the same pointes G and H, as if the angle LGM or the angle LGN contayned vnder the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G drawen in the superficies erected, and vnder the GM or GN drawen in the ground superficies CDEF ly<g ref="char:EOLhyphen"/>ing flat, be a right angle, then by this definition, the superficies ABCD is vpright or erected vpon the superficies CDEF. It is also commonly called a superficies perpendicular vpon or vnto a superficies.</p>
               <p>
                  <note place="margin">
                     <hi>Declaration of the second part.</hi>
                  </note>For the second part of this definition, which is of the inclination of a right line vnto a plaine su<g ref="char:EOLhyphen"/>perficies, take this example. Let ABCD be a ground plaine superficies, vpon which from a point being a loft, namely, the point E, suppose a right line to fall, which let be the line EG, touching the plaine superficies ABCD at the poynt G. Againe, from the point E, being the toppe or higher limite and end of the inclining line EG, let a perpendicular line fall vnto the plaine superficies ABCD, which let be the line EF, and let F be the point where EF toucheth the plaine superficies ABCD. Then from the point of the fall of the line inclining vpon the superficies vnto
<figure/>
the point of the falling of the perpendicular line vpon the same super<g ref="char:EOLhyphen"/>ficies, that is, from the point G to the point F, draw a right line GF. Now by this definition, the acute angle EGF is the inclination of the line EG vnto the superficies ABCD. Because it is contayned of the inclining line, and of the right line drawen in the superficies, from the point of the fall of the line inclining to the point of the fall of the per<g ref="char:EOLhyphen"/>pendicular line: which angle must of necessitie be an acute angle. For the angle EFG is by construction a right angle, and three angles in a triangle are equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ight angles. Wherefore the other two angles, namely, the angles EGF, and GEF, are equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> right angle. Wherfore either of them is lesse then a right angle. Wherfore the angle EGF is an <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>gle.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Fourth diffi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note>4 Inclination of a plaine superficies to a plaine superficies, is an acute an<g ref="char:EOLhyphen"/>gle contayned vnder the right lines, which being drawen in either of the plaine superficieces to one &amp; the self same point of the co<g ref="char:cmbAbbrStroke">̄</g>mon section, make with the section right angles.</head>
               <p>
                  <pb n="313" facs="tcp:7062:342"/>Suppose that there be two superficieces ABCD &amp; EFGH,
<figure/>
and let the superficies ABCD be supposed to be erected not perpendicularly, but somewhat leaning and inclining vnto the plaine superficies EFGH, as much or as litle as ye will: the co<g ref="char:cmbAbbrStroke">̄</g>mon terme or section of which two superficieces let be the line CD. From some one point, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> from the point M assigned in the common section of the two superficieces, namely, in the line CD, draw a perpendicular line in either superficies. In the ground superficies EFGH draw the line MK, and in the su<g ref="char:EOLhyphen"/>perficies ABCD draw the line ML. Now if the angle LMK be an acute angle, then is that angle the inclination of the su<g ref="char:EOLhyphen"/>perficies ABCD vnto the superficies EFGH, by this defini<g ref="char:EOLhyphen"/>tion, because it is contained of perpendicular lines drawen in either of the superficieces to one and the self same point being the common section of them both.</p>
            </div>
            <div type="part">
               <head>5 Plaine superficieces are in like sort inclined the on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>her,<note place="margin">
                     <hi>Fifth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> when the sayd angles of inclination are equall the one to the o<gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>
               </head>
               <p>This definition needeth no declaration at all, but is most manifest by the definition last going before. For in considering the inclinations of diuers superficieces to others, if the acute angles contayned vn<g ref="char:EOLhyphen"/>der the perpendicular lines drawen in them from one point assigned in ech of their common sections be equall, as if to the angle LMK in the former example be geuen an other angle in the inclination of two other superficieces equall, then is the inclination of these superficieces like, and are by this defini<g ref="char:EOLhyphen"/>tion sayd in like sort to incline the one to the other.</p>
               <p>
                  <hi>Theodosius</hi> geueth an other definition of like inclination of plaine superficieces the one to the other, after this maner.
<q>
                     <hi>One plaine superficies is like inclined to an other, as an other superficies is to an other, when in either of the plaine superficieces right lines being drawen, and making right angles with their common section, containe in the same pointes equall angles.</hi> This definition is in substance the same with that geuen of <hi>Euclide,</hi> and is an elucidation of it.</q>
For example let ABCD be a ground plaine superficies, vnto which let the superficies EFIK incline and leane. And let the common section of these two superficieces be the line EF. Then drawe in eche of thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> superficieces right lines to some one point of the common section EF, which let be the point G: with which section let them make right angles. As in the super<g ref="char:EOLhyphen"/>ficies
<figure/>
ABCD draw the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G, which in the point G let it make with the common sectio<g ref="char:cmbAbbrStroke">̄</g> a right angle HGF or HGE. Also in the superficies EFIK draw the line LG, which in the point G together with the common section EF let make also a right angle LGF, or the right angle LG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.</p>
               <p>Now also let there be an other ground plaine superficies, namely, the superficies MNOP, vnto whom also let leane and incline the superficies Q<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>T, and let the common section or segment of them be the line QR. And draw in the superficies MNOP to some one point of the co<g ref="char:cmbAbbrStroke">̄</g>mon section as to the point X the line VX, making with the common section right angles, namely, the angle VXR, or the angle VXQ: also in the superficies STQR draw the right line YX to the same point X in the common secti<g ref="char:EOLhyphen"/>on, making therwith right angles, as the angle YX<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, or the angle YXQ. Now (as sayth the definition) if the angles contayned vnder the right lines drawen in these superficieces &amp; making right angles with the common section, be in the pointes, that is, in the pointes of their meting in the common section, e<g ref="char:EOLhyphen"/>quall: then is the inclination of the superficieces equall. As in this example, if the angle LGH con<g ref="char:EOLhyphen"/>tayned vnder the line LG being in the inclining superficies <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>KEF and vnder the line HG being in the ground superficies ABCD, be<g ref="char:cmbAbbrStroke">̄</g> equall, to the angle YXV contayned vnder the line VX being in the
<pb facs="tcp:7062:343"/>
ground superficies MNOP and vnder the line YX being in the inclining superficies STQR: then is the inclination of the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies IKEF vnto the superficies ABCD, like vnto the inclination of the superficies STQR vnto the superficies MNOP. And so by this definition these two superficieces are sayd to be in like sort inclined.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Sixth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>6 Parallell plaine superficieces are those, which being produced or exten<g ref="char:EOLhyphen"/>ded any way neuer touch or concurre together.</head>
               <p>Neither needeth this definition any declaration, but is very easie to be vnderstanded by the defini<g ref="char:EOLhyphen"/>tion of parallell lines: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or as they being drawen on any part, neuer touch or come together: so parallel plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces are such, which admitte no touch, that is, being produced any way infinitely neuer meete or come together.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Seuenth def<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<g ref="char:EOLhyphen"/>nition.</hi>
                  </note>7 Like solide or bodily figures are such, which are contained vnder like plaine superficieces, and equall in multitude.</head>
               <p>What plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces are called like, hath in the beginning of the sixth booke, bene sufficiently declared. Now when solide figures or bodies be contained vnder such like plaine superficieces as there are defined, and equall in number, that is, that the one solide haue as many in number as the other, in their sides and limites: they are called like solide figures, or like bodies.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Eighth di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>fi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note>8 Equall and like solide (or bodely) figures are those which are contained vnder like superficieces, and equall both in multitude and in magnitude.</head>
               <p>In like solide figures it is su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ficient, that the superficieces which containe them be like and equall in number onely, but in like solide figures and equall, it is necessary that the like superficieces contaynyng them, be also equal in magnitude. So that besides the likenes betwene them, they be (eche being com<g ref="char:EOLhyphen"/>pared to his correspondent super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies) o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> one greatnes, and that their areas or fieldes be equal. When such super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces contayne bodies or solides, then are such bodies equall and like solides or bodies.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Ninth di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>i<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ition.</hi>
                  </note>9 A solide or bodily angle, is an inclination of moe then two lines to all the lines which touch themselues mutually, and are not in one and the selfe same super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies.</head>
               <head type="sub">
                  <hi>Or els thus:</hi> A solide or bodily angle is that which is contayned vnder mo then two playne angles, not being in one and the selfe same plaine superfi<g ref="char:EOLhyphen"/>cies, but consisting all at one point.</head>
               <p>Of a solide angle doth Euclide here geue two seue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all definitio<g ref="char:cmbAbbrStroke">̄</g>s. The first is geuen by the concurse and touch of many lines. The second by the touch &amp; concurse of many superficiall angles. And both these definitions tende to one, and are not much different, for that lynes are the limittes and termes of superficieces. But the second geuen by super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iciall angles is the more naturall definition, because that supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ficieces a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the next and immediate limites of bodies, and so are not lines. An example of a solide angle cannot wel and at <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ully be geue<g ref="char:cmbAbbrStroke">̄</g> or described in a pla<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e superficies. But touchyng this first defini<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>, lay before you a cube or a die, and co<g ref="char:cmbAbbrStroke">̄</g>sider any of the corners or angles therof so shal ye see that at eue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y angle there concurre thre lines (for two lines co<g ref="char:cmbAbbrStroke">̄</g>curring cannot make a solide angle) namely, the line or edge of his breadth, of his le<g ref="char:cmbAbbrStroke">̄</g>gth, and of his thicknes, which their so inclining &amp; co<g ref="char:cmbAbbrStroke">̄</g>curring toue<g ref="char:EOLhyphen"/>ther, make a solide angle, and so of others. And now co<g ref="char:cmbAbbrStroke">̄</g>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rning the second definitio<g ref="char:cmbAbbrStroke">̄</g>, what super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icial or plaine angles be hath bene taught before in the first bok<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, namely, that it is the touch of two right lines. And as a super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iciall or playne angle is caused &amp; co<g ref="char:cmbAbbrStroke">̄</g>tained of right lines, so si a solide angle caused &amp; co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tayned of plaine superficiall angles. Two right lines touching together, make a plaine angle, but two plaine angles ioyned together can not make a solide angle, but according to the definitio<g ref="char:cmbAbbrStroke">̄</g>, they must be moe the<g ref="char:cmbAbbrStroke">̄</g> two, as three, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oure, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue, or mo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>: which also must not be in one &amp; the selfe same superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, but must be in diuers superficieces, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eeting at one point. This definition is not hard, but may easily be co<g ref="char:cmbAbbrStroke">̄</g>ceiued in a cube or a die, where ye see three angles of any three superficieces or sides of the die con<g ref="char:EOLhyphen"/>curre and meete together in one point, which three playne angles so ioyned together, make a solide angle. Likewise in a Pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or a spi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e of a steple or any other such thing, all the sides therof te<g ref="char:cmbAbbrStroke">̄</g>ding vp<g ref="char:EOLhyphen"/>ward
<pb n="314" facs="tcp:7062:343"/>
narower and narower, at length ende their angles (at the heig<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <figure/>
or toppe therof) in one point. So all their angles there ioyned toge<g ref="char:EOLhyphen"/>ther, make a solide angle. And for the better <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t thereof, I haue set here a figure wherby ye shall more easily conceiue <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the base of the figure is a triangle, namely, ABC, if on euery side of the triangle ABC, ye rayse vp a triangle, as vpon the side AB, ye raise vp the triangle AFB, and vpon the side AC the triangle AFC, and vpon the side BC, the triangle BFC, and so bowing the triangles raised vp, that their toppes, namely, the pointes F meete and ioyne together in one point, ye shal easily and plainly see how these three superficiall angles AFBBFC, CFA, ioyne and close together, touching the one the other in the point F, and so make a solide angle.</p>
            </div>
            <div type="part">
               <head>10 A Pyramis is a solide figure contained vnder many playne superficieces set vpon one playne superficies, and gathered together to one point.<note place="margin">
                     <hi>Tenth diffini<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>
               </head>
               <p>Two superficieces raysed vpon any ground can not make a Pyramis, for that two superficiall angles ioyned together in the toppe, cannot (as before is sayd) make a solide angle. Wherfore whe<g ref="char:cmbAbbrStroke">̄</g> thre, foure, fiue, or moe (how many soeuer) superficieces are raised vp fro<g ref="char:cmbAbbrStroke">̄</g> one superficies being the ground, or base, and euer asce<g ref="char:cmbAbbrStroke">̄</g>ding diminish their breadth, till at the le<g ref="char:cmbAbbrStroke">̄</g>gth all their angles co<g ref="char:cmbAbbrStroke">̄</g>curre in one point, making there a solide angle: the solide inclosed, bounded, and terminated by these superficieces is called a Py<g ref="char:EOLhyphen"/>ramis, as ye see in a taper of foure sides, and in a spire of a towre which containeth many sides, either of which is a Pyramis.</p>
               <p>And because that all the superficieces of euery Pyramis ascend from one playne superficies as from the base, and tende to one poynt, it must of necessitie come to passe; that all the superficieces of a Pyra<g ref="char:EOLhyphen"/>mis are trianguler, except the base, which may be of any forme or figure except a circle. For if the base be a circle, then it ascendeth not with sides, or diuers superficieces, but with one round superficies, and hath not the name of a Pyramis, but is called (as hereafter shall appeare) a Cone.</p>
               <p>Of Pyramid, there are diuers kindes. For according to the varietie of the base is brought forth the varietie and diuersitie of kindes of Pyramids. If the base of a Pyramis be a triangle, then is it called a triangled Pyramis. If the base be a figure of fower angles, it is called a quadrangled Pyramis. If the base be a Pentagon, then is it a Pentagonall or fiue angled Pyramis. And so forth according to the increase of the angles of the base infinitely. Although the fi<g ref="char:EOLhyphen"/>gure
<figure/>
of a Pyramis can not be well expressed in a playne superficies, yet may ye sufficiently conceaue of it both by the figure before set in the de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition of a solide angle, and by the figure here set, if ye ima<g ref="char:EOLhyphen"/>gine the point A together with the lines AB, AC, and AD, to be eleuated on high. And yet that the reader may more clerely see the forme of a Pyramis, I haue h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re set two sundry Pyramids which will appeare bodilike, if ye erecte the papers wherin are drawen the trian<g ref="char:EOLhyphen"/>gular sides of eche Pyramis, in such sort that the pointes of the angles F of ech triangle may in euery Pyramis concurre in one point, and make a solide angle: one of which hath to his base a fower sided fi<g ref="char:EOLhyphen"/>gure, and the other a fiue sided figure. T<gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>me of a triangled Pyra<g ref="char:EOLhyphen"/>mis ye may before beholde in the examp<gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> solide angle. And by these may ye conceaue of all other kindes o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>yramids.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:344"/>
               <head>
                  <note place="margin">
                     <hi>Eleuenth diffi<g ref="char:EOLunhyphen"/>nition.</hi>
                  </note>11 A prisme is a solide or a bodily figure contained vnder many plaine superficieces, of which the two superficieces which are opposite, are equall and like, and parallells, &amp; all the other superficieces are parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes.</head>
               <p>Although you may in a plaing superfi<g ref="char:EOLhyphen"/>cies
<figure/>
by this figure here set,<note place="margin">
                     <hi>This diffiniti<g ref="char:EOLhyphen"/>on agreeth to the generall name of a prisme, which also may be applied to Pa<g ref="char:EOLhyphen"/>rallel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>pedons (a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> may ap<g ref="char:EOLhyphen"/>peare in the</hi> 12. <hi>booke fol<g ref="char:EOLhyphen"/>lowing &amp;</hi> 10. <hi>propo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tio<g ref="char:cmbAbbrStroke">̄</g>, his demonstratio<g ref="char:cmbAbbrStroke">̄</g>) and to some o<g ref="char:EOLhyphen"/>ther bodie likewise.</hi>
                  </note> without any hardne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> conceaue what a prisme is, name<g ref="char:EOLhyphen"/>ly, if ye imagine the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies ABDC to be the ground &amp; base of the solid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and the two sup<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ficieces, namely, the superficies AEFB, and the superficies CEFD to be erected vpon the sides of the base, the one on the one side, namely; on the line AB, and the other on the other side, namely, on the line DC, not perpendicularly, but in<g ref="char:EOLhyphen"/>clining and bending the one to the other, till they meete in the toppe, namely, on the line EF. For so ye see that this solide figure is contained vnder many plaine superficieces, of which two, namely, the superficies AEC, and the su<g ref="char:EOLhyphen"/>perficies BFD, which are the endes of the solide, and opposite the one to the other, are equall like and parallels, and all the other superficieces, namely, the base ABCD, &amp; the two erected superficieces, that is, the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies AEFB, and the superficies CEED are parallelogrammes. Yet notwithstan<g ref="char:EOLhyphen"/>ding, to make the thing more clere vnto the reader, I haue here set a Prisme which will appeare bodi<g ref="char:EOLhyphen"/>like,
<figure/>
if you erecte bending wise the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
papers wherein are drawen the pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>mes ABEF, &amp; CDE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that they may concurr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> EF in the toppe, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> pers wherein are drawen th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> gles AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and BDE, that the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> AE of the one triangle may exactly agree with the side AE of the one parallelogramme ABEF, and the side CE of the same triangle, with the side CE of the parallelo<g ref="char:EOLhyphen"/>gramme CDEF: and moreouer, the side BF of the other triangle, with the side DF of the parallelogramme CDEF: and finally, the side BF of the same triangle, with the side BF of the parallelo<g ref="char:EOLhyphen"/>gramme ABEF. And so shall you most easilie see the forme of a Prisme: that it consisteth of two equall, like, and parallell triangular superficieces, and of three parallelogrammes: wherof the one is the base, and the other two are erected bending wise. Here also be<g ref="char:EOLhyphen"/>holde the forme thereof as it is by arte described in a plaine to ap<g ref="char:EOLhyphen"/>peare bodilike.</p>
               <p>
                  <hi>Flussas</hi> here noteth that <hi>Theon</hi> and <hi>Campane</hi> disagree in defi<g ref="char:EOLhyphen"/>ning a Prisme, and he preferreth the definition geuen of <hi>Campane</hi> before the de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition geuen of <hi>Euclide</hi> (which because he may seme with out lesse offence to reiect, he calleth it <hi>Theons</hi> definition) and following <hi>Campane</hi> he geueth an other definition, which is this.</p>
               <p>
                  <note place="margin">
                     <hi>An other dif<g ref="char:EOLhyphen"/>finition of a prisme, which is a speciall diffinition of a prisme: as it is commonly called and vsed.</hi>
                  </note>
                  <hi>A Prisme is a solide figure, which is contayned vnder fiue playne superficieces, of which two are triangles, like, equall, and parallels, and the rest are parallelogrammes.</hi>
               </p>
               <p>The example before set agreeth likewise with this definition, and manifestly declareth the same. For in it were <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue superficieces, the base, the two erected superficieces, and the two endes: of which the two endes are triangles like, equall and parallels, and all the other are parallelogrammes as this definiti<g ref="char:EOLhyphen"/>on requireth. The cause why he preferreth the difinion of <hi>Campane</hi> before the difinition of <hi>Theon</hi> (as he calleth it, but in very deede it is <hi>Euclides</hi> definition, as certainely, as are all those which are geuen of him in the former bookes, neither is there any cause at all, why it should be doubted in this one defini<g ref="char:EOLhyphen"/>tion more then in any of the other) as he him selfe alledgeth, is, for that it is (as he sayth) to large, and comprehendeth many mo kindes of solide figures besides Prismes, as Columnes hauing sides, and all Parallelipipedons, which a definition should not doo: but should be conuertible with the thing defi<g ref="char:EOLhyphen"/>ned, and declare the nature of it onely, and stretch no farther.</p>
               <p>Me <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hinketh <hi>Flussas</hi> ought not to haue made so much a doo in this matter, nor to haue bene so sharpe in sight and so quicke as to see and espy out such faultes, which can of no man that will see right<g ref="char:EOLhyphen"/>ly withou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> affection be espyed for such great faultes. For it may well be aunswered that these faultes which he noteth (if yet they be faultes) are not to be found in this definion. It may be sayd that it exten<g ref="char:EOLhyphen"/>deth it selfe not <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>arther then it should, but declareth onely the thing defined, namely, a Prisme. Neither
<pb n="315" facs="tcp:7062:344"/>
doth it agree (as <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>lussas</hi> cauilleth) with all Parallelipipedons and Columnes hauing sides. All Paralleli<g ref="char:EOLhyphen"/>pipedons what so euer right angled, or not right angled which are described of equidistant sides or su<g ref="char:EOLhyphen"/>perficieces, haue their sides opposit. So that in any of them there is no one side, but it hath a side oppo<g ref="char:EOLhyphen"/>sit vnto it. So likewise is it of eue<g ref="char:cmbAbbrStroke">̄</g> sided Columnes, eche hath his opposite side directly agaynst it, which agreeth not with this definition of <hi>Euclide.</hi> Here it is euidently sayd, that of all the superficieces, the two which are opposite are equall, like, and parallels, meaning vndoubtedly onely two &amp; no moe. Which is manifest by that which followeth. The other (sayth he) are parallelogrammes, signifiing most euidently that none of the rest besides the two aforesayd, which are equall, like, and parallels, are opposite: but two of necessitie are raysed vp, and concurre in one common line, and the other is the base. So that it contayneth not vnder it the figures aforesayd, that is sided Columnes, &amp; al Parallelipipedons, as <hi>Flussas</hi> hath not so aduisedly noted.</p>
               <p>Agayne where <hi>Flussas</hi> setteth in his definition, as an essentiall part thereof, that of the fiue superfi<g ref="char:EOLhyphen"/>cieces, of which a Prisme is contayned, two of them must be triangles, that vndoubtedly is not of ne<g ref="char:EOLhyphen"/>cessitie, they may be of some other figure. Suppose that in the figure before geuen that in the place of the two opposite figures, which there were two triangl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, were placed two pentago<g ref="char:cmbAbbrStroke">̄</g>s: yet should the fi<g ref="char:EOLhyphen"/>gure remayne a Prisme still, and agree with the definition of <hi>Euclide,</hi> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>alleth not vnder the definiti<g ref="char:EOLhyphen"/>on of <hi>Flussas.</hi> So that his definitio<g ref="char:cmbAbbrStroke">̄</g> semeth to be to narrow and stretcheth not so farre as it ought to do, nor declareth the whole nature of the thing defined. Wherefore it is not to be preferrd before <hi>Euclides</hi> definition, as he woulde haue it. This figure of <hi>Euclide</hi> called a Prisme, is called of <hi>Campane</hi> and certayne others <hi>Figura Serr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tilis,</hi> for that it repres<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>teth in some maner the forme of a Sawe.<note place="margin">
                     <hi>This bodie called Figura Serratilis.</hi>
                  </note> And of some others it is called <hi>Cuneus,</hi> that is, a Wedge, because it beareth the figure of a wedge.</p>
               <p>Moreouer although it were so, that the definitio<g ref="char:cmbAbbrStroke">̄</g> of a Prisme should be so large, that it should co<g ref="char:cmbAbbrStroke">̄</g>taine all these figures noted of <hi>Flussas</hi> as sided Columnes, &amp; all Parallelipipedons: yet should not <hi>Flussas</hi> haue so great a cause to finde so notably a fault, so vtterly to reiect it. It is no rare thing in all learninges, chiefely in the Mathematicalls, to haue one thing more generall then an other. Is it not true that euery Isosceles is a triangle, but not euery triangle is an Isosceles? And why may not likewise a Prisme be more generall, then a Parallelepipedon, or a Columne hauing sides (and contayne them vnder it as a triangle co<g ref="char:cmbAbbrStroke">̄</g>tayneth vnder it an Isosceles and other kinds of triangles). So that euery Prallelipipedon, or euery si<g ref="char:EOLhyphen"/>ded Columne be a Prisme, but not euery Prisme a Parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> or a sided Columne. This ought not to be so much offensiue. And indeede it semeth manifestly of many, yea &amp; of the learned so to be take<g ref="char:cmbAbbrStroke">̄</g>, as clearely appeareth by the wordes of <hi>Psellus</hi> in his Epitome of Geometrie, where he entreateth of the production and constitution of these bodyes. His wordes are these.<note place="margin">
                     <hi>Psellus.</hi>
                  </note> 
                  <hi>All r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ctili<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e figures being erected vpon their playnes or bases by right angles, make Prismes.</hi> Who perceaueth not but that a Pentagon erected vpo<g ref="char:cmbAbbrStroke">̄</g> his base of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue sides maketh by his motion a sided Columne of fiue sides? Likewise an Hexagon erected at right angles produceth a Columne hauing sixe sides: and so of all other rectillne figures. All which solides or bodyes so produced, whether they be sided Columnes or Parallelipipedons, be here in most plaine words (of this excelle<g ref="char:cmbAbbrStroke">̄</g>t and auncient Greke author <hi>Psellus</hi>) called Prismes. Wherfore if the defini<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> of a Prisme geue<g ref="char:cmbAbbrStroke">̄</g> of <hi>Euclide</hi> should extend it selfe so largely as <hi>Flussas</hi> imagineth, and should enclude such figures or bodyes, as he noted: he ought not yet for all that so much to be offended, and so na<g ref="char:EOLhyphen"/>rowly to haue sought faultes. For <hi>Euclide</hi> in so defining mought haue that meaning &amp; sense of a Prisme which <hi>Psellus</hi> had. So ye see that <hi>Euclide</hi> may be defended either of these two wayes, either by that that the definition extendeth not to these figures, and so not to be ouer generall nor stretch farther then it ought: or ells by that that if it should stretch so far it is not so haynous. For that as ye se many haue tak<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> it in that sense. In deede co<g ref="char:cmbAbbrStroke">̄</g>monly a Prisme is taken in that significatio<g ref="char:cmbAbbrStroke">̄</g> and meaning in which <hi>Campa<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> Flussas</hi> and others take it. In which sense it semeth also that in diuers propositions in these bookes fol<g ref="char:EOLhyphen"/>lowing it ought of necessitie to be taken.</p>
            </div>
            <div type="part">
               <head>12 A Sphere is a figure which is made,<note place="margin">
                     <hi>Twelueth dif<g ref="char:EOLhyphen"/>finition.</hi>
                  </note> when the diameter of a semicircle abiding fixed, the semicircle is turned round about, vntill it returne vnto the selfe same place from whence it began to be moued.</head>
               <p>To the end we may fully and perfectly vnderstand this definiti<g ref="char:EOLhyphen"/>on,
<figure/>
how a Sphere is produced of the motion of a semicircle, it shall be expedient to co<g ref="char:cmbAbbrStroke">̄</g>sider how quantities Mathematically are by ima<g ref="char:EOLhyphen"/>gination conceaued to be produced, by flowing and motion, as was somewhat touched in the beginning of the first booke. Euer the lesse quantitie by his motion bringeth for<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h the qua<g ref="char:cmbAbbrStroke">̄</g>titie next aboue it. As a point mouing, flowing, or gliding, bringeth forth a line, which is the first quantitie, and next to a point. A line mouing pro<g ref="char:EOLhyphen"/>duceth a superficies, which is the second quantitie, and next vnto a line. And last of all, a superficies mouing bringeth forth a solide or body, which is the third &amp; last quantitie. These thinges well mar<g ref="char:EOLhyphen"/>ked, it shall not be very hard to attaine to the right vnderstanding of this definition. Vpon the line AB being the diameter, describe a
<pb facs="tcp:7062:345"/>
                  <figure/>
semicircle ACB, whose
<figure/>
centre let be D: the dia<g ref="char:EOLhyphen"/>meter AB being sixed on his endes or<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> pointes, imagine the whole su<g ref="char:EOLhyphen"/>perficies of the semicir<g ref="char:EOLhyphen"/>cle to moue round from some one point assigned, till it returne to the same point againe. So shall it produce a perfect Sphere or Globe, the forme whereof you see in a ball or bowle. And it is fully round and solide, for that it is described of a semicircle which is per<g ref="char:EOLhyphen"/>fectly round, as our countrey man <hi>Iohannes de Sacro Busco</hi> in his booke of the Sphere,<note place="margin">
                     <hi>What is to be ta<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>n heede of in the diffini<g ref="char:EOLhyphen"/>tion of a sphere geuen by Io<g ref="char:EOLhyphen"/>hannes de Sa<g ref="char:EOLhyphen"/>cro Busco.</hi>
                  </note> of this definition which he taketh out of <hi>Euclide,</hi> doth well collecte. But it is to be noted and taken heede of, that none be deceaued by the definition of a Sphere geuen by <hi>Iohannes de Sacro Busco: A Sphere</hi> (sayth he) <hi>is the passage or mouing of the circumference of a semicircle, till it returne vnto the place where it beganne,</hi> which agreeth not with <hi>Euclide. Euclide</hi> plain<g ref="char:EOLhyphen"/>ly sayth, that a Sphere is the passage or motion of a semicircle, and not the passage or motion of the cir<g ref="char:EOLhyphen"/>cumference of a semicircle: neither can it be true that the circumference of a semicircle, which is a line, should describe a body. It was before noted that euery quantitie moued, describeth and produceth the quantitie next vnto it. Wherefore a line moued can not bring forth a body, but a superficies onely. As if ye imagine a right line fastened at one of his endes to moue about from some one point till it returne to the same againe, it shall describe a plaine superficies, namely, a circle. So also if ye likewise conceaue of a crooked line, such as is the circumference of a semicircle, that his diameter fastened on both the endes it should moue from a point assigned till it returne to the same againe, it should describe &amp; pro<g ref="char:EOLhyphen"/>duce a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ound superficies onely, which is the superficies and limite of the Sphere, and should not pro<g ref="char:EOLhyphen"/>duce the body and soliditie of the Sphere. But the whole semicircle, which is a superficies, by his moti<g ref="char:EOLhyphen"/>on, as is before said, produceth a body, that is, a perfect Sphere. So see you the errour of this definiti<g ref="char:EOLhyphen"/>on of the author of the Sphere: which whether it happened by the author him selfe, which I thinke not: or that that particle was thrust in by some one after him, which is more likely, it it not certaine. But it is certaine, that it is vnaptly put in, and maketh an vntrue definition: which thing is not here spoken, any thing to derogate the author of the booke, which assuredly was a man of excellent know<g ref="char:EOLhyphen"/>ledge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> neither to the hindrance or diminishing of the worthines of the booke, which vndoubtedly is a very necessary booke, then which I know none more meere to be taught and red in scholes touching the groundes and principles of Astronomie and Geographie: but onely to admonishe the young and vnskil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ull reader of not falling into errour.<note place="margin">
                     <hi>Theodosiu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>inition of a sphere.</hi>
                  </note> 
                  <hi>Theodosius</hi> in his booke <hi>De Sphericis</hi> (a booke very necessa<g ref="char:EOLhyphen"/>ry for all those which will see the groundes and principles of Geometrie and Astronomie, which also I haue translated into our vulgare tounge, ready to the presse) defineth a Sphere after thys maner: <hi>A Sphere is a solide or body contained vnder one superficies, in the midle wherof there is a point, fro<g ref="char:cmbAbbrStroke">̄</g> which all lines drawen to the circumference are equall.</hi> This definition of <hi>Theodosius</hi> is more essentiall and naturall, then is the other geuen by <hi>Euclide.</hi> The other did not so much declare the inward nature and substance of a Sphere, as it shewed the industry and knowledge of the producing of a Sphere, and therfore is a causall definition geuen by the cause efficient, or rather a description then a definition. But this definition is very es<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>entiall, declaring the natu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e and substance of a Sphere. As if a circle should be thus defined, as it well may: <hi>A circle is the passage or mouing of a line from a point till it returne to the same point againe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> it is a causall definition, shewing the efficient cause wherof a circle is produced, namely, of the motion of a line. And it is a very good description fully shewing what a circle is. Such like description is the de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition of a Sphere geuen o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>Euclide</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the motion of a semicircle. But when a circle is defined to be a plaine superficies, in the middest wherof is a point, from which all lines drawen to the circumfe<g ref="char:EOLhyphen"/>rence therof, are equall: this definition is essentiall and formall, and declareth the very nature of a cir<g ref="char:EOLhyphen"/>cle. And vnto this definition of a circle, is correspondent the de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition of a Sphere geue<g ref="char:cmbAbbrStroke">̄</g> by <hi>Theodosius,</hi> saying: that it is a solide o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> body, in the middest, whereof there is a point, from which all the lines drawen to the circumference are equall. So see you the affinitie betwene a circle and a Sphere. For what a circle is in a plaine, that is a Sphere in a Solide. The fulnes and content of a circle is described by the motion of a line moued about: but the circumference therof, which is the limite and border thereof, is described of the end and point of the same line moued about. So the fulnes, content, and body of a Sphere or Globe is described of a semicircle moued about. But the Sphericall superficies, which is the limite and border of a Sphere,<note place="margin">
                     <hi>The circumfe<g ref="char:EOLhyphen"/>rence of a sphere.</hi>
                  </note> is described of the circumference of the same semicircle moued about. And this is the superficies ment in the definition, when it is sayd, that it is contained vnder one super<g ref="char:EOLhyphen"/>ficies, which superficies is called of <hi>Iohannes de <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>acro Busco</hi> &amp; others, the circumference of the Sphere.</p>
               <p>
                  <note place="margin">
                     <hi>Galens diffi<g ref="char:EOLhyphen"/>nition <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> a sph<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note>Galene in his booke <hi>de diffinitionibus medici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> geueth yet an other definitio<g ref="char:cmbAbbrStroke">̄</g> of a Sphere, by his proper<g ref="char:EOLhyphen"/>tie or co<g ref="char:cmbAbbrStroke">̄</g>mon accide<g ref="char:cmbAbbrStroke">̄</g>ce of mouing, which is thus. <hi>A Sphere is a figure most apt to all motion, as hauing no base whereon th stay.</hi> This is a very plaine and witty de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition, declaring the dignitie thereof aboue all figures generally.<note place="margin">
                     <hi>The dig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>itie of a s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>here.</hi>
                  </note> All other bodyes or solides, as Cubes, Pyramids, and others haue sides, bases, and angles, all which are stayes to rest vpon, or impedimentes and lets to motion. But the Sphere hauing no side or
<pb n="316" facs="tcp:7062:345"/>
base to stay one, nor angle to let the course thereof, but onely in a poynt touching the playne wherein <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> standeth, moueth freely and fully with out let. And for the dignity and worthines thereof, this circular and Sphericall motion is attributed to the heauens, which are the most worthy bodyes. Wherefore there is ascribed vnto them this chiefe kinde of motion. This solide or bodely figure is also commonly called a Globe.<note place="margin">
                     <hi>A sphere cal<g ref="char:EOLhyphen"/>led a Globe.</hi>
                  </note>
               </p>
            </div>
            <div type="part">
               <head>13 The axe of a Sphere is that right line which abideth fixed, about which the semicircle was moued.<note place="margin">
                     <hi>Thirtenth dif<g ref="char:EOLhyphen"/>finition.</hi>
                  </note>
               </head>
               <p>As in the example before geuen in the definition of a Sphere, the line AB, about which his endes being fixed, the semicircle was moued (which line also yet remayneth after the motion ended) is the axe of the Sphere described of that semicircle. <hi>Theodosius</hi> defineth the axe of a Sphere after this maner.<note place="margin">
                     <hi>Theodosius diffinition of the axe of a sphere.</hi>
                  </note> 
                  <hi>The axe of a Sphere is a certayne right line drawen by the centre, ending on either side in the superficies of the Sphere, about which being fixed the Sphere is turned.</hi> As the line AB in the former example. There nedeth to this definition no other declaration, but onely to consider, that the whole Sphere turneth vpon that line AB, which passeth by the centre D, and is extended one either side to the superficies of the Sphere, wherefore by this definition of <hi>Theodosius</hi> it is the axe of the Sphere.</p>
            </div>
            <div type="part">
               <head>14 The centre of a Sphere is that poynt which is also the centre of the se<g ref="char:EOLhyphen"/>micircle.<note place="margin">
                     <hi>Fourtenth diffinition.</hi>
                  </note>
               </head>
               <p>This definition of the centre of a Sphere is geuen as was the other definition of the axe, namely, hauing a relation to the definition of a Sphere here geuen of <hi>Euclide</hi>: where it was sayd that a Sphere is made by the reuolution of a semicircle, whose diameter abideth fixed. The diameter of a circle and of a semicrcle is all one. And in the diameter either of a circle or of a semicircle is contayned the center of either of them, for that they diameter of eche euer passeth by the centre. Now (sayth <hi>Euclide</hi>) the poynt which is the center of the semicircle, by whose motion the Sphere was described, is also the centre of the Sphere. As in the example there geuen, the poynt D is the centre both of the semicircle &amp; also of the Sphere. <hi>Theodosius</hi> geueth as other definition of the centre of a Sphere which is thus.<note place="margin">
                     <hi>Theodosius diffinition of the center of a sphere.</hi>
                  </note> 
                  <hi>The centre of a Sphere is a poynt with in the Sphere, from which all lines drawen to the superficies of the Sphere are equall.</hi> As in a circle being a playne figure there is a poynt in the middest, from which all lines drawen to the circum<g ref="char:EOLhyphen"/>frence are equall, which is the centre of the circle: so in like maner with in a Sphere which is a solide and bodely figure, there must be conceaued a poynt in the middest thereof, from which all lines drawen to the superficies thereof are equall. And this poynt is the centre of the Sphere by this definition of <hi>Theodosius. Flussas</hi> in defining the centre of a Sphere comprehendeth both those definitions in one, after this sort. <hi>The centre of a Sphere is a poynt assigned in a Sphere, from which all the lines drawen to the superfi<g ref="char:EOLhyphen"/>cies are equall, and it is the same which was also the centre of the semicircle which described the Sphere.</hi>
                  <note place="margin">
                     <hi>Flussas diffi<g ref="char:EOLhyphen"/>nition of the center of a sphere.</hi>
                  </note> This defi<g ref="char:EOLhyphen"/>nition is superfluous and contayneth more the<g ref="char:cmbAbbrStroke">̄</g> nedeth. For either part thereof is a full and sufficient dif<g ref="char:EOLhyphen"/>finition, as before hath bene shewed. Or ells had <hi>Euclide</hi> bene insufficient for leauing out the one part, or <hi>Theodosius</hi> for leauing out the other. Paraduenture <hi>Flussas</hi> did it for the more explication of either, that the one part might open the other.</p>
            </div>
            <div type="part">
               <head>15 The diameter of a Sphere is a certayne right line drawen by the ce<g ref="char:cmbAbbrStroke">̄</g>tre, and one eche side ending at the superficies of the same Sphere.<note place="margin">
                     <hi>Fiuetenth dif<g ref="char:EOLhyphen"/>finition</hi>
                  </note>
               </head>
               <p>This definitio<g ref="char:cmbAbbrStroke">̄</g> also is not hard, but may easely be couceaued by the definitio<g ref="char:cmbAbbrStroke">̄</g> of the diameter of a cir<g ref="char:EOLhyphen"/>cle. For as the diameter of a circle is a right line drawne fro<g ref="char:cmbAbbrStroke">̄</g> one side of the circu<g ref="char:cmbAbbrStroke">̄</g>frence of a circle to the other, passing by the centre of the circle: so imagine you a right line to be drawen from one side of the superficies of a Sphere to the other, passing by the center of the Sphere,<note place="margin">
                     <hi>Difference be<g ref="char:EOLhyphen"/>twene the di<g ref="char:EOLhyphen"/>ameter &amp; axe of a sphere.</hi>
                  </note> and that line is the diameter of the Sphere. So it is not all one to say, the axe of a Sphere, and the diameter of a Sphere. Any line in a Sphere drawen from side to side by the centre is a diameter. But not euery line so drawen by the centre is the axe of the Sphere, but onely one right line about which the Sphere is imagined to be moued<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> So that the name of a diameter of a Sphere is more general, then is the name of an axe. For euery axe in a Sphere is a diameter of the same: but not euery diameter of a Sphere is an axe of the same. And there<g ref="char:EOLhyphen"/>fore <hi>Flussas</hi> setteth a diameter in the definition of an axe as a more generall word <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n this maner. <hi>The axe
<pb facs="tcp:7062:346"/>
of a Sphere, is that fixed diameter aboue which the Sphere is moued.</hi> A Sphere (as also a circle) may haue infi<g ref="char:EOLhyphen"/>nite diameters, but it can haue but onely one axe.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Seuententh diffinition.</hi>
                  </note>16 A cone is a solide or bodely figure which is made, when one of the sides of a rectangle triangle, namely, one of the sides which contayne the right angle, abiding fixed, the triangle is moued about, vntill it returne vnto the selfe same place from whence it began first to be moued. Now if the right line which abideth fixed be equall to the other side which is moued about and containeth the right angle: then the cone is a rectangle cone. But if it be lesse, then is it an obtuse angle cone. And if it be greater, the<g ref="char:cmbAbbrStroke">̄</g> is it an a cute<g ref="char:EOLhyphen"/>angle cone,</head>
               <p>This definition of a Cone is of the nature and condition that the definition of a Sphere was, for either is geuen by the motion of a superficies. There, as to the production of a Sphere was imagined a semicircle to moue round, from some one point till it returned to the same point againe: so here must ye imagine a rectangle triangle to moue about till it come againe to the place where it beganne. Let ABC be a rectangle triangle, hauing
<figure/>
the angle ABC a right angle, which let be contained vnder the lines AB and BC.
<figure/>
Now suppose the side AB, namely, one of the lines which co<g ref="char:cmbAbbrStroke">̄</g>taine the right angle ABC to be fastened, and about it suppose the triangle ABC to be moued from some one poynt assigned till it re<g ref="char:EOLhyphen"/>turne to the same agayne (as vppon the diameter in the definition of a Sphere ye imagined a se<g ref="char:EOLhyphen"/>micircle to moue about): so shall the solide or body thus described be a perfect Cone. As you may imagine by this figure here set. And the forme of a Cone you may sufficiently conceaue by the figure set in the margent. There are of Cone three kindes, namely, a rectangle Cone, an obtuseangle Cone, and an acute angle Cone, all which were before in the former definitio<g ref="char:cmbAbbrStroke">̄</g> defined: Name<g ref="char:EOLhyphen"/>ly, the first kinde after this maner.</p>
               <p>
                  <hi>If the right line which abideth fixed, be equall to the other side which moueth ro<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>d about, and containeth the right angle, then the Cone is a rectangle Cone.</hi>
               </p>
               <p>
                  <note place="margin">
                     <hi>First kinde of Cones.</hi>
                  </note>As suppose in the former example, that the line AB which is fixed, and about which the triangle was moued, and after the motion yet remayneth, be equall to the line BC, which is the other line con<g ref="char:EOLhyphen"/>tayning the right angle, which also is moued about together with the whole triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> then is the Cone described, as the Cone ADC in this example, a right angled Cone: so called for that the angle at the toppe of the Cone is a right angle. For forasmuch as the lines AB and BC of the triangle ABC are e<g ref="char:EOLhyphen"/>quall, the angle BAC is equall to the angle BCA (by the 5. of the first). And eche of them is the halfe of the right angle ABC (by the 32. of the first). In like sort may it be shewed in the triangle ABD, that the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DA is equall to the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>AD, and that eche of them is the halfe of a right angle. Wherefore the whole angle CAD, which is composed of the two halfe right angles, namely, DA<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and CA<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is a right angle. And so haue ye what is a right angled Cone.</p>
               <p>
                  <hi>But if it be lesse, then is it an obtuseangle Cone.</hi> As in this ex<g ref="char:EOLhyphen"/>ample,
<figure/>
the line AB fixed is lesse then the line BC moued a<g ref="char:EOLhyphen"/>bout. Wherefore the Cone described of the circumuoluti<g ref="char:EOLhyphen"/>on of the triangle ABC about the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is an obtusean<g ref="char:EOLhyphen"/>gle Cone, for that the angle at the toppe DAC is greater then a right angle. Wherefore it is an obtuseangle. And therefore the Cone is called an obtuse angle Cone.</p>
               <p>
                  <pb n="317" facs="tcp:7062:346"/>
                  <hi>And if it be greater, then i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> an acuteangle Cone.</hi> As in
<figure/>
this figure, the line AB fastened, is greater then the line BC moued about. Wherefore the Cone de<g ref="char:EOLhyphen"/>scribed by the motion and turning of the triangle ABC about AB is an acuteangle Cone, hauing the angle at the toppe BAC an acute angle. Of whome the Cone is called an acuteangle Cone. For the ea<g ref="char:EOLhyphen"/>sier sight &amp; co<g ref="char:cmbAbbrStroke">̄</g>sideration of all these kindes of Cones, and also for the plainer demonstration of the varie<g ref="char:EOLhyphen"/>ties of their angles in their toppes, I haue described them all three in one playne figure, of which the Cone ACB is a right angled Cone, hauyng his fix<g ref="char:EOLhyphen"/>ed side CF equall to the line FB, and hys angle ACB a right angle: the Cone AEB is an obtuse angle Cone, and ADB an acuteangle Cone. By which figure ye may easily demonstrate (by the 21. of the first) that the angle ADB of the Cone ADB, whose fixed line DF is greater then the side FB, is lesse then the right angle ACB, and so is an acute angle. And also (by the same 21. of the first) ye shall with like facilitie perceaue how the angle AEB of the Cone AEB whose fix<g ref="char:EOLhyphen"/>ed line EF is lesse then the side FB, is grea<g ref="char:EOLhyphen"/>ter then the right angle ACB: and there<g ref="char:EOLhyphen"/>fore is an obtuse angle.</p>
               <p>This figure of a Cone is of <hi>Campane,</hi> of <hi>Vitellio,</hi> and of others which haue written in these latter times, called a round Pyramis,<note place="margin">
                     <hi>A Cone called of Campane a ro<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>de Pira<g ref="char:EOLhyphen"/>mis.</hi>
                  </note> which is not so aptly. For a Pyramis, and a Cone, are farre distant, &amp; of sundry natures. A Cone is a regular body produced of one circumuolution of a rectangle triangle, and limited and bordered with one onely round super<g ref="char:EOLhyphen"/>ficies. But a Pyramis is terminated and bordered with diuers superficieces. Therefore can not a Cone by any iust reason beare the name of a Pyramis. This solide of many is called <hi>Turbo,</hi> which to our purpose may be Englished a <hi>Top</hi> or <hi>Ghyg</hi>: and moreouer, peculiarly <hi>Campane</hi> calleth a Cone the Py<g ref="char:EOLhyphen"/>ramis of a round Columne, namely, of that Columne which is produced of the motion of a parallelo<g ref="char:EOLhyphen"/>gramme (contained of the lines AB and BC) moued about, the line AB being fixed. Of which Co<g ref="char:EOLhyphen"/>lumnes shall be shewed hereafter.</p>
            </div>
            <div type="part">
               <head>17 The axe of a Cone is that line, which abideth fixed, about which the triangle is moued.<note place="margin">
                     <hi>Seuententh diffinition.</hi>
                  </note> And the base of the Cone is the circle which is described by the right line which is moued about.</head>
               <p>As in the example the line <hi>AB</hi> is sup<g ref="char:EOLhyphen"/>posed
<figure/>
to be the line about which the right angled triangle <hi>ABC</hi> (to the pro<g ref="char:EOLhyphen"/>duction of the Cone) was moued: and that line is here of <hi>Euclide</hi> called the axe of the Cone described. <hi>The base of the Cone is the circle which is described by the right line which is moued about.</hi> As the line <hi>AB</hi> was fixed and slayed, so was the line <hi>BC</hi> (together with the whole triangle <hi>ABC</hi>) moued and turned a<g ref="char:EOLhyphen"/>bout. A line moued, as hath bene sayd before, produceth a superficies: and be<g ref="char:EOLhyphen"/>cause the line <hi>BC</hi> is moued about a point, namely, the point <hi>B,</hi> being the end of the axe of the Cone <hi>AB,</hi> it produceth by his motion, and reuolution a circle, which circle is the base of the Cone: as in this example, the circle <hi>CDE.</hi>
               </p>
               <p>
                  <pb facs="tcp:7062:347"/>The line which produceth the base of the Cone, is the line of the triangle which together with the axe of the Cone contayneth the right angle. The other side also of the triangle, namely, the line <hi>AC,</hi> is moued about also with the motion of the triangle, which with his reuolution describeth also a super<g ref="char:EOLhyphen"/>ficies,<note place="margin">
                     <hi>A conicall su<g ref="char:EOLhyphen"/>perficies.</hi>
                  </note> which is a round superficies, &amp; is erected vpon the base of the Cone, &amp; endeth in a point, name<g ref="char:EOLhyphen"/>ly, in the higher part or toppe of the Cone. And it is commonly called a Conicall superficies.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Eightenth diffinition.</hi>
                  </note>18 A cylinder is a solide or bodely figure which is made, when one of the sides of a rectangle parallelogramme, abiding fixed, the parallelogramme is moued about, vntill it returne to the selfe same place from whence it be<g ref="char:EOLhyphen"/>gan to be moued.</head>
               <p>This definition also is of the same sort and condition, that the two definition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> before geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> were, namely, the definition of a Sphere and the definition of a Cone. For all are geuen by mouing of a su<g ref="char:EOLhyphen"/>perficies about a right line fixed, the one of a semicircle about his diameter, the other of a rectangle tri<g ref="char:EOLhyphen"/>angle about one of his sides And this solide or body here de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ined is caused of the motion of a rectangle
<figure/>
parallelogra<g ref="char:cmbAbbrStroke">̄</g>me hauing one of his sides contayning
<figure/>
the right angle fixed from some one poynt till it re<g ref="char:EOLhyphen"/>turne to the same agayne where it began. As suppose ABCD to be a rectangle parallelogramme, hauing his side AB fastned, about which imagine the whole parallelogramme to be turned, till it returne to the poynt where it began, then is that solide or body, by this motion described, a Cylinder: which because of his roundnes can not at full be described in a playne superficies, yet haue you for an example thereof a suf<g ref="char:EOLhyphen"/>ficient designation therof in the margent <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> as in a plaine may be. If you wil perfectly behold the forme of a cilinder. Consider a round piller that is perfect<g ref="char:EOLhyphen"/>ly round.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Ninetenth diffinition.</hi>
                  </note>19 The axe of a cilinder is that right line which abydeth fixed, about which the parallelogramme is moued. And the bases of the cilinder are the circles described of the two opposite sides which are moued about.</head>
               <p>Euen as in the description of a Sphere the line fastened was the axe of the Sphere pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uced: and in the description of a cone, the line fastened was the axe of the cone brought forth: so in this descripti<g ref="char:EOLhyphen"/>on of a cilinder the line abiding, which was fixed, about which the rectangle parallelogramme was mo<g ref="char:EOLhyphen"/>ued is the axe of that cilinder. As in this example is the line <hi>AB. The bases of the cilinder <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>c.</hi> In the reuo<g ref="char:EOLhyphen"/>lution of a parallelogramme onely one side is fixed, therefore the three other sides are moued about: of which the two sides which with the axe make right angles, and which also are opposite sides, in their motion describe eche of them a circle, which two circles are called the bases of the cilinder. As ye see in the figure before put two circles described of the motio<g ref="char:cmbAbbrStroke">̄</g> of the two opposit lines AD and BC, which are the bases of the Cilinder.</p>
               <p>
                  <note place="margin">
                     <hi>A cillindricall superficies.</hi>
                  </note>The other line of the rectangle parallelogramme moued, by his motion describeth the round su<g ref="char:EOLhyphen"/>perficies about the Cilinder. As the third line or side of a rectangle triangle by his motion described the round Conical superficies about the Cone. And as the circu<g ref="char:cmbAbbrStroke">̄</g>fere<g ref="char:cmbAbbrStroke">̄</g>ce of the semicircle described the round sphericall superficies about the Sphere. In this example it is the superficies described of the line DC.</p>
               <p>
                  <note place="margin">
                     <hi>Corollary.</hi>
                  </note>By this definition it is playne that the two circles, or bases of a cilinder are euer equall and paral<g ref="char:EOLhyphen"/>lels: for that the lines moued which produced them remayned alwayes equall and parallels. Also the axe of a cilinder is euer an erected line vnto either of the bases. For with all the lines described in the bases, and touching it, it maketh right angles,</p>
               <p>
                  <note place="margin">
                     <hi>A round<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> Co<g ref="char:EOLhyphen"/>lumne or sphere.</hi>
                  </note>
                  <hi>Campane, Vitell<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o,</hi> with other later writers, call this solide or body a round Column<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or piller. And <hi>Campane</hi> addeth vnto this definition this, as a corrollary. That of a round Columne, of a Sphere, and
<pb n="318" facs="tcp:7062:347"/>
of a circle the ce<g ref="char:cmbAbbrStroke">̄</g>tre is one and the selfe same.<note place="margin">
                     <hi>A Corollary added by Cam<g ref="char:EOLunhyphen"/>pane.</hi>
                  </note> That is (as he him selfe declareth it &amp; proueth the same) where the Columne, the Sphere, and the circle haue one diameter.</p>
            </div>
            <div type="part">
               <head>20 Like cones and cilinders are those,<note place="margin">
                     <hi>Twenty diffi<g ref="char:EOLhyphen"/>nition.</hi>
                  </note> whose axes and diameters of their bases are proportionall</head>
               <p>The similitude of cones and cilin<g ref="char:EOLhyphen"/>ders
<figure/>
standeth in the proportion of those right lines, of which they haue their ori<g ref="char:EOLhyphen"/>ginall and spring. For by the diameters of their bases is had their length and breadth, and by their axe is had their heigth or deepenes. Wherefore to see whether they be like or vnlike, ye must compare their axes together, which is their depth, and also their diameters to<g ref="char:EOLhyphen"/>gether, which is thier length &amp; breadth. As if the axe <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>G</hi> of the cone <hi>ABC</hi> be to to the axe <hi>EI</hi> of the cone <hi>DEF,</hi> as the diameter <hi>AC</hi> of the cone <hi>ABC</hi> is to the diameter <hi>DF</hi> of the cone <hi>DEF,</hi> then a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the cones <hi>ABC</hi> and <hi>DEF</hi> like cones. Likewise in the cilinders. If the axe <hi>LN</hi> of the cilinder <hi>LHMN</hi> haue that pro<g ref="char:EOLhyphen"/>portion to the axe <hi>OQ</hi> of the cilinder <hi>ROPQ,</hi> which the diameter <hi>HM</hi> hath to the diameter <hi>RP</hi>: then are the cilinders <hi>HLMN</hi> and <hi>ROPQ</hi> like cilinders, and so of all others.</p>
            </div>
            <div type="part">
               <head>21 A Cube is a solide or bodely figure contayned vnder sixe e<g ref="char:EOLhyphen"/>quall squares.<note place="margin">
                     <hi>Twenty one diffinitio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note>
               </head>
               <p>As is a dye which hath sixe sides, and eche of
<figure/>
them is a full and perfect square, as limites or bor<g ref="char:EOLhyphen"/>ders vnder which it is contayned. And as ye may conceiue in a piece of timber contayning a foote square euery way, or in any such like. So that a Cube is such a solide whose three dimensions are equall, the length is equall to the breadth thereof, and eche of them equall to the depth. Here is as it may be in a playne superficies set an image therof, in these two figures wherof the first is as it is com<g ref="char:EOLhyphen"/>monly described in a playne, the second (which is in the beginning of the other side of this leafe) is drawn as it is described by arte vpo<g ref="char:cmbAbbrStroke">̄</g> a playne su<g ref="char:EOLhyphen"/>perficies to shew somwhat bodilike. And in deede the latter descriptio<g ref="char:cmbAbbrStroke">̄</g> is for the sight better the<g ref="char:cmbAbbrStroke">̄</g> the first. But the first for the demo<g ref="char:cmbAbbrStroke">̄</g>strations of Euclides propositions in the fiue bookes following is of more vse, for that in it may be considered and sene
<pb facs="tcp:7062:348"/>
all the fixe sides of the Cube. And so any lines or sections drawen
<figure/>
in any one of the sixe sides. Which can not be so wel sene in the o<g ref="char:EOLhyphen"/>ther figure described vpon a playnd. And as touching the first figure (which is set at the ende of the other side of this leafe) ye see that there are sixe parallelogrammes which ye must conceyue to be both equilater and rectangle, although in dede there can be in this description onely two of them rectangle, they may in dede be described al equilater. Now if ye imagine one of the sixe paral<g ref="char:EOLhyphen"/>lelogrammes, as in this example, the parallelogramme ABCD to be the base lieng vpon a ground playne superfices. And so con<g ref="char:EOLhyphen"/>ceiue the parallelogramme EFGH to be in the toppe ouer it, in such sort, that the lines AE, CG, DH, &amp; BF may be erected per<g ref="char:EOLhyphen"/>pendicularly from the pointes A, C, B, D, to the ground playne su<g ref="char:EOLhyphen"/>perficies or square ABCD. For by this imagination this figure wil shew vnto you bodilike. And this imagination perfectly had, wil make many of the propositions in these fiue bookes following, in which are required to be descri<g ref="char:EOLhyphen"/>bed such like solides (although not all cubes) to be more plainly and easily conceiued.</p>
               <p>In many examples of the Greeke and also of the Latin, there is in this place set the diffinition of a Te<g ref="char:EOLhyphen"/>trahedron, which is thus.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Twenty two diffinition.</hi>
                  </note>22 A Tetrahedron is a solide which is contained vnder fower triangles equall and equilater.</head>
               <p>A forme of this solide ye may see in these two examples here set,
<figure/>
whereof one is as it is commonly described in a playne. Neither is it hard to conceaue. For (as we before taught in a Pyramis) if ye imagine the triangle BCD to lie vpon a ground plaine superficies, and the point A to be pulled vp together with the lines AB, AC, and AD, ye shall perceaue the forme of the Tetrahedron to be contayned vnder 4. triangles, which ye must imagine to be al fower equilater and equiangle, though they can not so be drawen in a plaine. And a Te<g ref="char:EOLhyphen"/>trahedron thus described, is of more vse in these fiue bookes follow<g ref="char:EOLhyphen"/>ing, then is the other, although the other appeare in forme to the eye more bodilike.<note place="margin">
                     <hi>A Tetrahe<g ref="char:EOLhyphen"/>dron one of the fiue regu<g ref="char:EOLhyphen"/>lar bodyes.</hi>
                  </note>
               </p>
               <p>Why this definition is here left out both of <hi>Campane</hi> and of <hi>Flussas,</hi>
                  <figure/>
I can not but maruell, considering that a Tetrahedron, is of all Philo<g ref="char:EOLhyphen"/>sophers counted one of the fiue chiefe solides which are here de<g ref="char:EOLhyphen"/>fined of <hi>Euclide,</hi> which are called co<g ref="char:cmbAbbrStroke">̄</g>monly regular bodies, with<g ref="char:EOLhyphen"/>out mencion of which, the entreatie of these should seeme much maimed: vnlesse they thought it sufficiently defined vnder the definition of a Pyramis,<note place="margin">
                     <hi>Di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>erence betwene a Tetrahedron and a Pira<g ref="char:EOLhyphen"/>mis.</hi>
                  </note> which plainly and generally taken, inclu<g ref="char:EOLhyphen"/>deth in deede a Tetrahedron, although a Tetrahedron properly much differe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h from a Pyramis, as a thing speciall or a particular, from a more generall. For so taking it, euery Tetrahedron is a Py<g ref="char:EOLhyphen"/>ramis, but not euery Pyramis is a Tetrahedron. By the generall definition of a Pyramis, the superficieces of the sides may be as many in number as ye list, as 3.4. 5.6. or moe, according to the forme of the base, whereon it is set, whereof before in the defi<g ref="char:EOLhyphen"/>nition of a Pyramis were examples geuen. But in a Tetrahedron the superficieces erected can be but three in number according to the base therof, which is euer a triangle. Againe, by the generall definition of a Pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, the superfi<g ref="char:EOLhyphen"/>cieces erected may ascend as high as ye list, but in a Tetrahedron they must all be equall to the base. Wherefore a Pyramis may seeme to be more generall then a Tetrahedron, as before a Prisme seemed to be more generall then a Parallelipipedon, or a sided Columne: so that euery Parallelipipedon is a Prisme, but not euery Prisme is a Parallelipipedon. And euery axe in a Sphere is a diameter: but not euery diameter of a Sphere is the axe therof. So also noting well the definition of a Pyramis, euery Te<g ref="char:EOLhyphen"/>trahedron may be called a Pyramis, but not euery Pyramis a Tetrahedron. And in dede <hi>Psellus</hi> in num<g ref="char:EOLhyphen"/>bring of these fiue solides or bodies, calleth a Tetrahedron a Pyramis in manifest wordes.<note place="margin">
                     <hi>Psellus calleth a Tetrahedron a Piramis.</hi>
                  </note> This I say might make <hi>Flussas</hi> &amp; others (as I thinke it did) to omitte the definition of a Tetrahedron in this place, as sufficiently comprehended within the definition of a Pyramis geuen before. But why then did he not count that de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition of a Pyramis faultie, for that it extendeth it selfe to large, and comprehendeth vnder it a Tetrahedron (which differeth from a Pyramis by that it is contayned of equall triangles) as he not so aduisedly did before the definition of a Prisme.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Twenty three definition.</hi>
                  </note>23 An Octohedron is a solide or bodily figure co<g ref="char:cmbAbbrStroke">̄</g>tained vnder eight equall and equilater triangles.</head>
               <p>
                  <pb n="319" facs="tcp:7062:348"/>As a Cube is a solide figure contayned vnder sixe superficiall fi<g ref="char:EOLhyphen"/>gures
<figure/>
of foure sides or squares which are equilater, equiangle, and equall the one to the other: so is an Octohedron a solide figure contained vnder eight triangles which are equilater and equall the one to the other. As ye may in these two figures here set beholde. Whereof the first is drawen according as this solide is commonly described vpon a plaine superficies. The second is drawen as it is described by arte vpon a plaine, to shewe bodilike. And in deede although the second appeare to the eye more bodilike, yet as I be<g ref="char:EOLhyphen"/>fore noted in a Cube, for the vnderstanding of diuers Propositions in these fiue bookes following, is the first description of more vse yea &amp; of necessitie. For without it, ye can not co<g ref="char:cmbAbbrStroke">̄</g>ceaue the draught of lines and sections in any one of the eight sides which are some<g ref="char:EOLhyphen"/>times in the descriptions of some of those Propositions required. Wherefore to the consideration of this first description, imagine first that vppon the vpper face of the superficies of the parallelo<g ref="char:EOLhyphen"/>gramme <hi>ABCD,</hi> be described a Pyramis, hauing his fower trian<g ref="char:EOLhyphen"/>gles <hi>AFB, AFC, CFD,</hi> and <hi>DFB,</hi> equilater and equiangle, and con<g ref="char:EOLhyphen"/>curring in the point <hi>F.</hi> The<g ref="char:cmbAbbrStroke">̄</g> co<g ref="char:cmbAbbrStroke">̄</g>ceaue that on the lower face of the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies of the former parallelogramme be described an other Pyramis, hauing his fower triangles <hi>AEB, AEC, CED,</hi> &amp; <hi>DEB,</hi> equilater and equiangle, and concurring in the point <hi>E.</hi> For so al<g ref="char:EOLhyphen"/>though somewhat grosly by reason the triangles can not be de<g ref="char:EOLhyphen"/>scribed equilater, you may in a plaine perceaue the forme of this solide, and by that meanes conceaue any lines or sections requi<g ref="char:EOLhyphen"/>red to be drawen in any of the sayd eight triangles which are the sides of that body.</p>
            </div>
            <div type="part">
               <head>24 A Dodecahedron is a solide or bodily fi<g ref="char:EOLhyphen"/>gure co<g ref="char:cmbAbbrStroke">̄</g>tained vnder twelue equall, equilater,<note place="margin">
                     <hi>Twe<g ref="char:cmbAbbrStroke">̄</g>ty <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>er definition.</hi>
                  </note> and equiangle Pentagons.</head>
               <p>As a Cube, a Tetrahedron, and an Octohe<g ref="char:EOLhyphen"/>dron,
<figure/>
are contayned vnder equall plaine figures, a Cube vnder squares, the other two vnder tri<g ref="char:EOLhyphen"/>angles: so is this solide figure contained vnder twelue equilater, equiangle, and equall Penta<g ref="char:EOLhyphen"/>gons, or figures of fiue sides. As in these two fi<g ref="char:EOLhyphen"/>gures here set you may perceaue. Of which the first (which thinge also was before noted of a Cube, a Tetrahedron, and an Octohedron) is the common description of it in a plaine, the other is the description of it by arte vppon a plaine to make it to appeare somwhat bodilike. The first description in deede is very obscure to conceaue, but yet of necessitie it must so, ney<g ref="char:EOLhyphen"/>ther can it otherwise, be in a plaine described to vndersta<g ref="char:cmbAbbrStroke">̄</g>d those Propositions of <hi>Euclide</hi> in these fiue bokes a following which concerne the same. For in it although rudely, may you see all the twelue Pentagons, which should in deede be all equall, equilater, and equiangle. And now how you may somewhat conceaue the first figure de<g ref="char:EOLhyphen"/>scribed in the plaine to be a body. Imagine first the Pentagon ABCDE <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o be vpon a ground plaine superficies, then imagine the Pentagon FGHKL to be on high opposite vnto the Penta<g ref="char:EOLhyphen"/>gon ABCDE. And betwene those two Pentagons there will be ten Pentagons pulled vp, fiue fro<g ref="char:cmbAbbrStroke">̄</g> the fiue sides of the ground Pen<g ref="char:EOLhyphen"/>tagon, namely, from the side AB the Pentagon ABONM, from the side BC the Pentagon BCQPO, from the side CD the Pentagon CDSRQ, from the side DE, the Pentagon DEVTS, from the side EA the Pentagon EAMXV, the other fiue Penta<g ref="char:EOLhyphen"/>gons haue eche one of their sides common with one of the sides of the Pentagon FGHKL, which is opposite vnto the Pentagon in the ground superficies: namely, these are the other fiue Penta<g ref="char:EOLhyphen"/>gons FGNMX, GHPON, HKRQP, KLRST, LFXVT.
<pb facs="tcp:7062:349"/>
So here you may behold twelue Pentagons, which if you imagine to be equall, equilater, &amp; equiangle, and to be lifted vp, ye shall (although somewhat rudely) conceaue the bodily forme of a Pentagon. And some light it will geue to the vnderstanding of certaine Propositions of the fiue bookes following con<g ref="char:EOLhyphen"/>cerning the same.</p>
            </div>
            <div type="part">
               <head>
                  <note place="margin">
                     <hi>Twenty fiue diffinition.</hi>
                  </note>25 An Icosahedron is a solide or bodily figure contained vnder twentie equall and equilater triangles.</head>
               <p>As the solides before last mentioned are all
<figure/>
described by the number and forme of the su<g ref="char:EOLhyphen"/>perficieces which containe them: so this body likewise is de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ined by that that it is contayned of twentie triangles equall, equilater, and e<g ref="char:EOLhyphen"/>quiangle. And although this solide also be ve<g ref="char:EOLhyphen"/>ry hard to conceaue, as it is commonly descri<g ref="char:EOLhyphen"/>bed vpon a plaine (an example wherof you haue in the first figure here set): yet is it of necessitie that in that forme it be described, if we will vnderstand such descriptions as are set forth of <hi>Euclide</hi> touching that body in the fiue bookes following. Howbeit you may by it (although somewhat rudely) see the 20. triangles, which are imagined to be equall, equilater, and equian<g ref="char:EOLhyphen"/>gle, if you consider <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iu: angles of fiue triangles to concurre together at a point. And forasmuch as there are in this solide 20. triangles, and euery t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iangle hath three angles, the concurse of the said triangles will be in twelue pointes. As in this example the pointes of the concurse are A, B, C, D, E, F, G, H, K, L, M, &amp; N. Where note that in this plaine the two poyntes M and N are but one point, yet must ye imagine one of those pointes to be erected vpward, and the other down<g ref="char:EOLhyphen"/>ward. Now the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue triangles which concurre in the point M, are these, BMD, DMF, FMH, HML, and LMB: the fiue triangles which concurre in the point N, and are imagined to be erected downward, are these, ANC, CNE, ENG, GNK, and KNA: the other ten triangles which include this body, are these, ABC, BCD, CDE, DEF, EFG, FGH, GHK, HKL, KLA, LAB. The second figure here appeareth more bodilike vnto the eye.</p>
               <p>These <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue solides now last defined, namely, a Cube, a Tetrahedro<g ref="char:cmbAbbrStroke">̄</g>, an Octohedron, a Dodecahedron and an Icosahedro<g ref="char:cmbAbbrStroke">̄</g> are called regular bodies.<note place="margin">
                     <hi>Fiue regular bodies.</hi>
                  </note> As in plaine superficieces, those are called regular figures, whose sides and angles are equal, as are equilater triangles, equilater pentagons, hexagons, &amp; such lyke, so in solides such only are counted and called regular, which are co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder equal playne super<g ref="char:EOLhyphen"/>ficieces, which haue equal sides and equal angles, as all these fiue foresayd haue, as manifestly appeareth by their definitions, which were all geuen by this proprietie of equalitie of their superficieces, which haue also their sides and angles equall. And in all the course of nature there are no other bodies of this condition and perfection, but onely these fiue. Wherfore they haue euer of the auncient Philosophers bene had in great estimation and admiration, and haue bene thought worthy of much contemplacion, about which they haue bestowed most diligent study and endeuour to searche out the natures &amp; pro<g ref="char:EOLhyphen"/>perties of them. They are as it were the ende and perfection of all Geometry, for whose sake is written whatsoeuer is written in Geometry. They were (as men say) first inuented by the most witty <hi>Pithago<g ref="char:EOLhyphen"/>ras</hi> then afterward set forth by the diuine <hi>Plato,</hi> and last of all meruelously taught and declared by the most excellent Philosopher <hi>Euclide</hi> in these bookes following, and euer since wonderfully embraced of all learned Philosophers.<note place="margin">
                     <hi>The dignity of these bodies.</hi>
                  </note> The knowledge of them containeth infinite secretes of nature. <hi>Pithag<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ras, Ti<g ref="char:EOLhyphen"/>meus</hi> and <hi>Plato,</hi> by them searched out the co<g ref="char:cmbAbbrStroke">̄</g>position of the world, with the harmony and preseruation therof, and applied these <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iue solides to the simple partes therof, the Pyramis, or Tetrahedro<g ref="char:cmbAbbrStroke">̄</g> they ascri<g ref="char:EOLhyphen"/>bed to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ire,<note place="margin">
                     <hi>A Tetrahe<g ref="char:EOLhyphen"/>dron ascribed vnto the fire.</hi>
                  </note> for that it ascendeth vpward according to the figure of the Pyramis. To the ayre they ascribed the Octohedron,<note place="margin">
                     <hi>An octohe<g ref="char:EOLhyphen"/>dron ascribed vnto the ayre.</hi>
                  </note> for that through the subtle moisture which it hath, it extendeth it selfe euery way to the one side, and to the other, accordyng as that figure doth. Vnto the water they assigned the
<pb n="340" facs="tcp:7062:349"/>
Ikosahedron, for that it is continually flowing and mouing,<note place="margin">
                     <hi>An Ikosahedron assigned vnto the water.</hi>
                  </note> and as it were makyng angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ide according to that figure. And to the earth they attributed a Cube,<note place="margin">
                     <hi>A cube assigned vnto the earth.</hi>
                  </note> as to a thing stable<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> and sure as the figure <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ignifieth. Last of all a Dodecahedron,<note place="margin">
                     <hi>A dodecahe<g ref="char:EOLhyphen"/>dron assigned to heauen.</hi>
                  </note> for that it is made of P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ntago<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, whose angles are more ample and large then the angles of the other bodies, and by that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ea<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> draw more <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> roun<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nes, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> &amp; to the forme and nature of a sphere, they assigned to a sphere, namely, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>. Who so will <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> in his <hi>Tineus,</hi> shall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ead of these figures, and of their mutuall proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>raunge ma<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, which h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re are not to be entreated of, this which is sayd, shall be sufficient for the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of them and for th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> declaration of their diffinitions.</p>
               <p>After all these diffinitions here set of <hi>Euclide, Flussas</hi> hath added an other diffinition, which <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of a Parallelipipedon, which bicause it hath not hitherto of <hi>Euclide</hi> in any place bene defined, and because it is very good and necessary to be had, I thought good not to omitte it, thus it is.</p>
            </div>
            <div type="part">
               <head>A parallelipipedon is a solide figure comprehended vnder foure playne qua<g ref="char:EOLhyphen"/>drangle figures,<note place="margin">
                     <hi>Diffinition of a parallelipipe<g ref="char:EOLhyphen"/>don.</hi>
                  </note> of which those which are opposite are parallels.</head>
               <p>As in playne superficieces a parallelogramme is that which is contained
<figure/>
vnder foure sides beyng lines, and whose opposite sides are equidistant and parallel lynes, so in solide figures a Parallelipipedon is that solide which is contayned vnder foure quadrangle superficieces, whose opposite sides are al<g ref="char:EOLhyphen"/>so parallels, as it is easily to be sene and conceaued in a cube or die, all whose opposite sides are parallel superficieces, &amp; so of others like, ye may also some<g ref="char:EOLhyphen"/>what conceiue therof by the example in the margent.</p>
               <p>There is also in these bookes following, mencion made of solides, whose two bases are Poligonon figures, lyke, equall, equilater, and parallels, and the sides set vpon the bases are parallelogrammes: which kynde of solides Cam<g ref="char:EOLhyphen"/>pane calleth sided Columnes (and which as was before noted, may be co<g ref="char:cmbAbbrStroke">̄</g>pre<g ref="char:EOLhyphen"/>ded vnder the definition of a Prisme) a forme wherof although grosely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">
                     <hi>A sided Co<g ref="char:EOLhyphen"/>lumne.</hi>
                  </note> be<g ref="char:EOLhyphen"/>hold in this example, whose bases are two like equall, equilater, equiangle, and parallel hexagons, and the sides set vppon those bases are sixe parallelo<g ref="char:EOLhyphen"/>gra<g ref="char:cmbAbbrStroke">̄</g>mes: ye may better co<g ref="char:cmbAbbrStroke">̄</g>ceiue the forme therof by
<figure/>
the figure put vnder the figure of the parallelipipe<g ref="char:EOLhyphen"/>don, which apeareth more bodilike. There may of these be infinite formes according to the diuersitie of their bases.</p>
               <p>Because these fiue regular bodies here defined are not by these figures here set, so fully and liuely expressed, that the studious beholder can through<g ref="char:EOLhyphen"/>ly according to their definitions conceyue them. I haue here geuen of them other descriptions drawn in a playne, by which ye may easily attayne to the knowledge of them. For if ye draw the like formes in matter that wil bow and geue place, as most apt<g ref="char:EOLhyphen"/>ly ye may do in fine pasted paper, such as pastwiues make wome<g ref="char:cmbAbbrStroke">̄</g>s pastes of, &amp; the<g ref="char:cmbAbbrStroke">̄</g> with a knife cut eue<g ref="char:EOLhyphen"/>ry line finely, not through, but halfe way only, if the<g ref="char:cmbAbbrStroke">̄</g> ye bow and bende them accordingly, ye shall most plainly and manifestly see the formes and shapes of these bodies, euen as their definitions shew. And it shall be very necessary for you to had<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tore of that pasted paper by you, for so shal yo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vpon it <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the formes of other bodies, as Prismes and Parallelipopedons, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> set forth in these fiue bookes following, and see the very <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se bodies there me<g ref="char:cmbAbbrStroke">̄</g>cioned: which will make these bokes concerning bodies, as easy vnto you as were the other bookes, whose figures you might plainly see vpon a playne superficies.</p>
               <p>
                  <pb facs="tcp:7062:350"/>If ye draw this figure con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ing a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
ye s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ower <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quila<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er and equian<g ref="char:EOLhyphen"/>gle triangles vpo<g ref="char:cmbAbbrStroke">̄</g> pasted paper, or vp<g ref="char:EOLhyphen"/>pon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ny other such like matter that will bowe and geue place,<note place="margin">
                     <hi>A Tetrahe<g ref="char:EOLhyphen"/>dron.</hi>
                  </note> and then cut not through the paper, but as it were halfe the thicknes of the p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>per, the three lines contained within the figure, and bowe &amp; folde in the fower triangles accordingly<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> they will close together in such sort, that they will make the perfecte forme of a T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tra<g ref="char:EOLhyphen"/>hedron.</p>
               <p>This figure (consisting of
<figure/>
sixe equall squares) drawen vp<g ref="char:EOLhyphen"/>on pasted paper,<note place="margin">
                     <hi>A c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>be.</hi>
                  </note> and the fiue lines contained within the fi<g ref="char:EOLhyphen"/>gure being cut finely halfe the thicknes of the paper, or not through, if their ye bowe and folde accordingly the sixe e<g ref="char:EOLhyphen"/>quall squares, they will so close together, that they will caus<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the perfecte forme of a Cube.</p>
               <p>This figure (which consisteth of eight <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>quilater
<figure/>
and equiangle triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) drawen vp<g ref="char:EOLhyphen"/>on the foresayd matter, and the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n lin<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> contained within the figure being <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> as b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>fore was taught,<note place="margin">
                     <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> octo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<g ref="char:EOLhyphen"/>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi>
                  </note> and the triangles bowed and folded accordingly, they will clos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ether in such sort, that they will mak<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> per<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> forme of an Octohedron.</p>
               <p>
                  <pb n="341" facs="tcp:7062:350"/>Describe thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> figur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, which consist<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th of tw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quil<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quiangl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nt<gap reason="illegible" resp="#PDCC" extent="5 letters">
                     <desc>•••••</desc>
                  </gap>, vpo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the fore<g ref="char:EOLhyphen"/>said matt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r, and finely cut as before was <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ught t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n lines contain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d within th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> figur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and bow and folde the Pen<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>gon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> accordingly. And they will so close to<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>eth<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>, tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y will <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>k<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> very forme of a Dodecahedron.<note place="margin">
                     <hi>A D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>d<gap reason="illegible" resp="#PDCC" extent="4 letters">
                           <desc>••••</desc>
                        </gap>e<g ref="char:EOLhyphen"/>dron.</hi>
                  </note>
               </p>
               <figure/>
               <p>
                  <pb facs="tcp:7062:351"/>
                  <note place="margin">
                     <hi>An Icosa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<g ref="char:EOLhyphen"/>dron.</hi>
                  </note>If ye describe this figure which consisteth of twentie equilater and equiangle triangles vpon the foresaid matter, and finely cut as before was shewed the nin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne lines which are contayned within the figure, and then bowe and folde them accordingly, they will in such sort close together, that ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> will be made a perfecte forme of an Icosahedron.</p>
               <figure/>
               <p>Because in these fiue bookes there are sometimes required other bodies besides the foresaid fiue regular bodies, as Pyramises of diuers formes, Prismes, and others, I haue here set forth three figures of three sundry Pyramises, one hauing to his base a triangle, an other a quadrangle figure, the other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Pentagon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which if ye describe vpon the foresaid matter &amp; finely cut as it was before taught the lines contained within ech figure, namely, in the first, three lines, in the second, fower lines, and in the third, fiue lines, and so bend and folde them accordingly, they will so close together at the toppes, that they will <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ake Pyramids of that forme that their bases are of. And if ye conceaue well the describing of these, ye may most easily describe the body of a Pyramis of what forme so euer ye will.</p>
               <p>
                  <note place="margin">
                     <hi>A triangled Pyramis.</hi>
                  </note>
                  <figure/>
               </p>
               <p>
                  <pb n="327" facs="tcp:7062:351"/>
                  <figure/>
                  <note place="margin">
                     <hi>The forme of a quadrang<g ref="char:EOLhyphen"/>led Pyramis.</hi>
                  </note>
                  <figure/>
                  <note place="margin">
                     <hi>The forme of a fiue angled Pyramis.</hi>
                  </note>
               </p>
               <p>
                  <pb facs="tcp:7062:352"/>Likewise if ye describe this figure
<figure/>
vpon the foresaid matter, and finely cutte the fower lines co<g ref="char:cmbAbbrStroke">̄</g>tained within the figure, and bowe and folde them together accordingly, the three paral<g ref="char:EOLhyphen"/>lelogrammes and the two triangles will so close together, that they will cause the perfecte forme of a Prisme co<g ref="char:cmbAbbrStroke">̄</g>tained vnder three parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes and two equedistant triangles.<note place="margin">
                     <hi>The forme of a prisme.</hi>
                  </note> And conceauing this description well, it shall not be hard to describe any o<g ref="char:EOLhyphen"/>ther Prisme of any other forme.</p>
               <p>Touching the descrip<g ref="char:EOLhyphen"/>tion
<figure/>
of Parallelipipedons I shall not neede to speake. For if ye consider well the description of a Cube, it shall not be hard to de<g ref="char:EOLhyphen"/>scribe a Parallelipipedon of what forme ye will.<note place="margin">
                     <hi>The forme of a parallelip<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>pedon.</hi>
                  </note> Onely where as in a Cube all the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes in the description of that fi<g ref="char:EOLhyphen"/>gure are squares, in the de<g ref="char:EOLhyphen"/>scribing of a Parallelipipe<g ref="char:EOLhyphen"/>don, the sayd parallelo<g ref="char:EOLhyphen"/>gramme may be of what forme ye will. So that ye take heede that the oppo<g ref="char:EOLhyphen"/>site parallelogrammes be equal &amp; equiangle. Which opposite parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes in the figure as it lieth in a plaine, is any two paralle<g ref="char:EOLhyphen"/>logrames leauing one pa<g ref="char:EOLhyphen"/>rallelogramme betwene them. An example wher<g ref="char:EOLhyphen"/>of beholde in this figure.</p>
               <p>Because these fiue bookes following are somewhat hard for young beginners, by reason they must in the figures described in a plaine imagine lines and superficieces to be eleuated and erected, the one to the other, and also conceaue solides or bodies, which, for that they haue not hitherto bene acquain<g ref="char:EOLhyphen"/>ted with, will at the first sight be somwhat s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>raunge vnto the<g ref="char:cmbAbbrStroke">̄</g>, I haue for their more <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ase, in this eleuenth booke, at the end of the demonstration of euery Proposition either set new figures, if they concerne the eleuating or erecting of lines or superficieces, or els if they concerne bodies, I haue shewed how they shall describe bodies to be compared with the constructions and demonstrations of the Proposi<g ref="char:EOLhyphen"/>tions to them belonging. And if they diligently weigh the maner obserued in this eleuenth booke tou<g ref="char:EOLhyphen"/>ching the description of new figures agreing with the figures described in the plaine, it shall not be hard for them of them selues to do the like in the other bookes following, when they come to a Propositi<g ref="char:EOLhyphen"/>on which concerneth either the eleuating or erecting of lines and superficieces, or any kindes of bodies to be imagined.</p>
            </div>
            <div type="part">
               <pb n="323" facs="tcp:7062:352"/>
               <head>¶The 1. Theoreme. The 1. Proposition. That part of a right line should be in a ground playne superficies, &amp; part eleuated vpward is impossible.</head>
               <p>FOr if it be possible, let part of the right line ABC, name<g ref="char:EOLhyphen"/>ly, the part AB be in a ground playne superficies, and the other part therof, namely, BC be eleuated vpwarde. And produce directly
<figure/>
vpo<g ref="char:cmbAbbrStroke">̄</g> the ground playne superfi<g ref="char:EOLhyphen"/>cies the right line AB beyond the point B vn<g ref="char:EOLhyphen"/>to the point D. Wherfore vnto two right lines geuen ABC, and ABD, the line AB is a common section or part, which is impossi<g ref="char:EOLhyphen"/>ble.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For a right line can not touche a right line in <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> pointes then one, v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lesse those right be exactly agreing and laid the one vpon the other. Wherfore that part of a right line should be in a ground plaine superficies, and part eleuated vpward is impossible: which was required to be proued.</p>
               <p>
                  <hi>This figure more plainly setteth forth the foresaid de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g>,
<figure/>
if ye eleuate the superficies wheri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the line BC.</hi>
               </p>
               <div type="section">
                  <head>An other demonstration after <hi>Fl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>s.</hi>
                  </head>
                  <p>If it be possible let there be a right line ABG,
<figure/>
whose part AB let be in the ground playne super<g ref="char:EOLhyphen"/>ficies AED;<note place="margin">
                        <hi>An other de<g ref="char:EOLhyphen"/>monstration after Flussas.</hi>
                     </note> and let the rest therof BG be eleuated on high, that is, without the playne AED. Then I say that ABG is not one right line. For forasmuch as AED is a plaine superficies, produce directly &amp; equally vpon the sayd playne AED the right lyne AB towardes D, which by the 4. definition of the first shall be a right line. And from some one point of the right line ABD, namely, from C, dra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> vnto the point G a right lyne CG. Wherefore in the triangle <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the outward ang<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is eq<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ll to the two inward and opposite angles (by the 32. of the first) and therfore it is lesse then two right angles (by the 17. of the same) Wherfore the lyne ABG forasmuch as it maketh an angle, is not <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> right line. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>refore that part of a right line should be in a ground playne superficies, and part eleuated vpward is impossible.</p>
                  <p>If ye marke well the figure before added for the play<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er declaration of Euclides demonstration, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> will not be hard for you to co<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>e this figure which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>luss<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s putteth for his demonst<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>tion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wherein is no difference but onely the draught of the lyne GC.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 2. Theoreme. The 2. Proposition. If two right line cut the ou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the other, they are <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne and the selfe same playne superficies: &amp; euery triangle is in one &amp; the selfe same superficie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.</head>
               <p>
                  <pb facs="tcp:7062:353"/>SVppose that these two right lines <hi>AB</hi> and <hi>CD</hi> doo
<figure/>
cutte the one the other in the point <hi>E.</hi> Then I say that these lines <hi>AB</hi> and <hi>CD</hi> are in one and the selfe same superficies, and that euery triangle is in one &amp; selfe same playne superficies.<note place="margin">Construction.</note> Take in the lines <hi>EC</hi> and <hi>EB</hi> points at all auentures, and let the same be <hi>F</hi> and <hi>G,</hi> and draw a right line from the poynt <hi>B</hi> to the point <hi>C,</hi> and an other from the point <hi>F</hi> to the point <hi>G.</hi> And draw the lines <hi>FH</hi> and <hi>GK.</hi> First I say that the triangle <hi>EBC</hi> is in one and the same ground superficies.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if part of the triangle <hi>EBC,</hi> namely the triangle <hi>FCH,</hi> or the triangle <hi>GBK</hi> be in the ground superficies, and the residu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be in an other, then also part of one of the right lines <hi>EC</hi> or <hi>EB</hi> shall be in the ground superficies, and part in an other. So also if part of the triangle <hi>EBC,</hi> namely, the part <hi>EFG</hi> be in the ground superficies and the residue be in an other, then also one part of eche of the right lines <hi>EC</hi> and <hi>EB</hi> shall be in the ground superficies, &amp; an other part in an other superficies, which (by the first of the eleuenth) is proued to be impossible. Wherfore the triangle <hi>EBC</hi> is in one and the selfe same playne su<g ref="char:EOLhyphen"/>perficies. For in what superficies the triangle <hi>BCE</hi> is, in the same also is either of the lines <hi>EC</hi> and <hi>EB,</hi> and in what superficies either of the lines <hi>EC</hi> and <hi>EB</hi> is, in the selfe same al<g ref="char:EOLhyphen"/>so are the lines <hi>AB</hi> and <hi>CD.</hi> Wherfore the right lines lines <hi>AB</hi> and <hi>CD</hi> are in one &amp; the selfe same playne superficies, and euery triangle is in one &amp; the selfe same playne superficies: which was required to be proued.</p>
               <p>
                  <hi>In this figure here set may ye more playnely conceaue the demon<g ref="char:EOLhyphen"/>stration
<figure/>
of the former proposition where <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> may ele<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap> what part of the triangle ECB ye will, namely the part FCH or the part GBK, or finally the part FCGB as is required in the demonstration.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 3. Theoreme. The 3. Proposition. If two playne superficieces cutte the one the other: their common section is a right line.</head>
               <p>SVppose that these two superficieces <hi>AB</hi> &amp; <hi>BC</hi> do
<figure/>
cutte the one the other, and let their common secti<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e the line <hi>DB.</hi> Then I say that <hi>DB</hi> is a right line. For if not, draw from the poynt <hi>D</hi> to the point <hi>B</hi> a right line <hi>DFB</hi> in the playne superficies <hi>AB,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> and likewise from the same poyntes draw an other right line <hi>DEB</hi> in the playne superficies <hi>BC.</hi> Now therfore two right lines <hi>DEB</hi> and <hi>DFB</hi> shall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>aue the selfe sa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and therefore doo include a superficies which (by the last common sentence) is impossible<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore the lines <hi>DEB</hi> and <hi>DFB</hi> are not right lines. In
<pb n="324" facs="tcp:7062:353"/>
like sort also may we proue that no other right line can be drawne from the poynt <hi>D</hi> to the point <hi>B</hi> besides the line <hi>DB</hi> which is the common section of the two superficieces <hi>AB</hi> and <hi>BC.</hi> If therefore two playne superficieces cutte the one the other, their common section is a right line: which was required to be demonstrated.</p>
               <p>
                  <figure/>
               </p>
               <p>
                  <hi>This figure here set, sheweth most playnely not onely this third proposition, but also the demonstra<g ref="char:EOLhyphen"/>tion thereof, if ye eleuate the superficies AB, and so compare it with the demonstration.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 4. Theoreme. The 4. Proposition. If from two right lines, cutting the one the other, at their common section, a right line be perpendicularly erected: the same shall also be perpendicular<g ref="char:EOLhyphen"/>ly erected from the playne superficies by the sayd two lines passing.</head>
               <p>SVppose that there be two right lines <hi>AB</hi>
                  <figure/>
and <hi>CD</hi> cutting the one the other in the poynt <hi>E.</hi> And from the poynt <hi>E</hi> let there be erected a right line <hi>EF</hi> perpendicularly to the sayd two right lines <hi>AB</hi> and <hi>CD</hi>: then I say that the right line <hi>EF,</hi>
                  <note place="margin">Construction.</note> is also erected perpendicular to the plaine superficies which passeth by the lines A<hi>B</hi> and <hi>CD.</hi> Let these lines <hi>AE, EB, EC,</hi> and <hi>ED</hi> be put equall the one to the other. And by the poynt <hi>E</hi> extend a right line at all auentures, and let the same be <hi>GEH.</hi> And drawe these right lines <hi>AD, CB, FA, FG, FD, FC, FH,</hi> and <hi>FB.</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as these two right lines <hi>AE</hi> &amp; <hi>ED</hi> are equall to these two lines <hi>CE</hi> and <hi>EB,</hi> and they com<g ref="char:EOLhyphen"/>prehend equall angles (by the 15. of the first): therefore (by the 4. of the first) the base <hi>AD</hi> is equall to the base <hi>CB,</hi> and the triangle <hi>AED</hi> is equall to the triangle <hi>CEB.</hi> Wherefore also the angle <hi>DAE,</hi> is equall to the angle <hi>EBC.</hi> But the angle <hi>AEG</hi> is equall to the angle <hi>BEH</hi> (by the 15 of the first). Wherefore there are two triangles <hi>AGE,</hi> and <hi>BEH</hi> hauing two angles of the one equall to two angles of the other, eche to his correspondent angle, and one side of the one equall to one side of the other, namely one of the sides which lye betwene the equall angles, namely, the side <hi>AE</hi> is equall to the side <hi>EB.</hi> Wherefore (by the 26. of the first) the sides remayning are equall to the sides remayning. Wherefore the side <hi>GE</hi> is equall to the side <hi>EH,</hi> and the side <hi>AG</hi> to the side <hi>BH.</hi> And for<g ref="char:EOLhyphen"/>asmuch as the line <hi>AE</hi> is equall to the line <hi>EB,</hi> and the line <hi>FE</hi> is common to them both,
<pb facs="tcp:7062:354"/>
and maketh with them right angles, wherefore (by the fourth of the first) the base <hi>FA</hi> it e<g ref="char:EOLhyphen"/>quall to the base <hi>FB.</hi> And (by the same reason) the base <hi>FC</hi> is equall to the base <hi>FD.</hi> And forasmuch as the line <hi>AD</hi> is equall to the line <hi>BC,</hi> and the line <hi>FA</hi> is equall to the line <hi>FB</hi> as it hath bene proued. Therefore these two lines <hi>FA</hi> and <hi>AD</hi> are equall to these two lines <hi>FB</hi> &amp; <hi>BC,</hi> the one to the other, &amp; the base <hi>FD</hi> is equall to the base <hi>FC.</hi> Wherfore also the angle <hi>FAD</hi> is equall to the angle <hi>FBC.</hi> And againe forasmuch as it hath bene proued, that the line <hi>AG</hi> is equall to the line <hi>BH,</hi> but the line <hi>FA</hi> is equall to the line <hi>FB.</hi> Where<g ref="char:EOLhyphen"/>fore there are two lines <hi>FA</hi> and <hi>AG</hi> equall to two lines <hi>FB</hi> and <hi>BH</hi> and it is proued that the angle <hi>FAG</hi> is equall to the angle <hi>FBH</hi>: wherefore (by the 4. of the first) the base <hi>FG</hi> is equal to the base <hi>FH.</hi> Agayne forasmuch as it hath bene proued that the line <hi>GE</hi> is equal to the line <hi>EH,</hi> and the line <hi>EF</hi> is common to them both: wherefore these two lines <hi>GE</hi> and <hi>EF</hi> are equall to these two lines <hi>HE</hi> and <hi>EF,</hi> and the base <hi>FH</hi> is equall to the base <hi>FG</hi>: wherefore the angle <hi>GEF</hi> is equall to the angle <hi>HEF.</hi> Wherefore either of the angles <hi>GEF,</hi> and <hi>HEF</hi> is a right angle. Wherefore the line <hi>EF</hi> is erected, from the point <hi>E,</hi> perpendi<g ref="char:EOLhyphen"/>cularly to the line <hi>GH.</hi> In like sort may we proue, that the same line <hi>FE</hi> maketh right angles with all the right lines which are drawne vpon the ground playne superficies and touch the point <hi>B.</hi> But a right line is then erected perpendicularly to a plaine superficies, when it maketh right angles with all the lines which touch it, and are drawne vpon the ground playne super<g ref="char:EOLhyphen"/>ficies (by the 2. definition of the eleuenth). Wherefore the right line <hi>FE</hi> is erected perpendi<g ref="char:EOLhyphen"/>cularly to the ground playne superficies. And the ground plaine superficies is that which pas<g ref="char:EOLhyphen"/>seth by these right lines <hi>AB</hi> and <hi>CD.</hi> Wherefore the right line <hi>FE</hi> is erected perpendicu<g ref="char:EOLhyphen"/>larly to the playne superficies which passeth by the right lines <hi>AB</hi> and <hi>CD.</hi> If therefore from two right lines cutting the one the other and at their common section a right line be perpendicularly erected: it shall also be erected perpendicularly to the plaine superficies by the sayd two lines passing: which was required to be proued.</p>
               <p>
                  <hi>In this figure you may most euidently conceaue the former
<figure/>
proposition and demonstration, if ye erect perpendicularly vnto the ground playne superficies ACBD the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> triangle AFB: and eleuate the triangles AFD, &amp; CFB in such sort, that the line AF of the tri<g ref="char:EOLhyphen"/>angle AFB may ioyne &amp; make one line with the line AF of the tri<g ref="char:EOLhyphen"/>angle AFD: and likewise that the line BF of the triangle AFB may ioyne &amp; make one right line with the line BF of the triangle BFC.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶The 5. Theoreme. The 5. Proposition. If vnto three right lines which touch the one the other, be erected a per<g ref="char:EOLhyphen"/>pendicular line from the common point where those three lines touch: those three right lines are in one and the selfe same plaine superficies.</head>
               <p>SVppose that vnto these three right lines <hi>BC, BD,</hi> and <hi>BE,</hi> touching the one the other in the poynt <hi>B,</hi> be erected perpendicularly from the poynt <hi>B,</hi> the line <hi>AB.</hi> Then I say, that those thre right lines <hi>BC, BD</hi> and <hi>BE,</hi> are in one &amp; the selfe same plaine superficies. For if not, then if it be possible, let the lines <hi>BD</hi> &amp;
<pb n="325" facs="tcp:7062:354"/>
                  <hi>BE</hi> be in the ground superficies, and let the line <hi>BC</hi> be e<g ref="char:EOLhyphen"/>rected
<figure/>
vpward (now the lines <hi>AB</hi> and <hi>BC</hi> are in one and the same playne superficies (by the 2. of the eleuenth) for they touch the one the other in the point <hi>B</hi>). Extend the plaine superficies wherein the lines <hi>AB</hi> and <hi>BC</hi> are,<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> and it shall make at the length a common section with the ground superficies, which common section shall be a right line (by the 3. of the eleuenth): let that common section be the line <hi>BF.</hi> Wherefore the three right lines <hi>AB, BC,</hi> and <hi>BF</hi> are in one and the selfe same su<g ref="char:EOLhyphen"/>perficies, namely, in the superficies wherein the lines <hi>AB</hi> and <hi>BC</hi> are. And forasmuch as the right line <hi>AB</hi> is erected per<g ref="char:EOLhyphen"/>pendicularly to either of these lines <hi>BD</hi> and <hi>BE,</hi> therefore the line <hi>AB</hi> is also (by the 4. of the eleuenth) erected perpendicu<g ref="char:EOLhyphen"/>larly to the plaine superficies, wherein the lines <hi>BD</hi> and <hi>BE</hi> are. But the superficies wherein the lines <hi>BD</hi> and <hi>BE</hi> are is the ground superficies. Wherefore the line <hi>AB</hi> is erected per<g ref="char:EOLhyphen"/>pendicularly to the ground plaine superficies. Wherefore (by the 2. definition of the eleuenth) the line <hi>AB</hi> maketh right angles with all the lines which are drawne vpon the ground super<g ref="char:EOLhyphen"/>ficies and touch it. But the line <hi>BF</hi> which is in the ground superficies doth touch it. Wherfore the angle <hi>ABF</hi> is a right angle. And it is supposed that the angle <hi>ABC</hi> is a right angle. Wherefore the angle <hi>ABF</hi> is equall to the angle <hi>ABC,</hi> and they are in one and the selfe same plaine superficies which is impossible. Wherefore the right line <hi>BC</hi> is not in an higher superficies. Wherefore the right lines <hi>BC, BD, BE</hi> are in one and the selfe same plaine su<g ref="char:EOLhyphen"/>perficies. If therefore vnto three right lines touching the one the one the other, be erected a perpendicular line from the common point where those three lines touch: those three right lines are in one and the selfe same plaine superficies: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
               <p>
                  <hi>This figure here set more playnely
<figure/>
declareth the demonstration of the for<g ref="char:EOLhyphen"/>mer proposition, if ye erect perpendicu<g ref="char:EOLhyphen"/>larly vnto the ground superficies, the s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>perficies wherein is drawne the line <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> and so compare it with the sayd de<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>stration.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 6. Theoreme. The 6. Proposition. If two right lines be erected perpendicularly to one &amp; the selfe same plaine superficies: those right lines are parallels the one to the other.</head>
               <p>
                  <pb facs="tcp:7062:355"/>SVppose that these two right lines AB and CD be erected perpendicularly to a ground plaine superficies. Then I say that the line AB is a parallel to the line CD. Let the pointes which those two right lines touch in the plaine superficies be B and D.<note place="margin">Construction.</note> And draw a right line from the point B to the point D. And (by the 11. of the first) from the point D, draw vnto the line BD in the ground superficies a perpen<g ref="char:EOLhyphen"/>dicular line DE. And (by the <hi>2.</hi> of the first)<note n="*" place="margin">An Assumpt as M. Dee pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ueth it. Demonstra<g ref="char:EOLhyphen"/>tion.</note> put the line DE equall to the line AB. And draw these right lines BE, AE, and AD. And forasmuch as the line
<figure/>
AB is erected perpendicularly to the ground superficies, ther<g ref="char:EOLhyphen"/>fore (by the 2. definition of the eleuenth) the line AB maketh right angles with all the lines which are drawne vpon the ground playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies and touch it. But either of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ese lines BD and BE which are in the ground superficies, touch the line AB, wherefore either of these angles ABD and ABE is a right angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and by the same reason also either of the angles CDB, &amp; CDE is a right angle. And forasmuch as the line AB is equall to the line DE, and the line BD is common to them both, therfore these two lines AB and BD, are equall to these two lines ED and DB, and they contayne right angles: wherefore (by the 4. of the first) the base AD is equall to the base BE. And forasmuch as the line AB is equall to the line DE, and the line AD to the line BE, therefore these two lines AB and BE are equall to these two lines ED and DA, and the line AE is a common base to them both. Wherefore the angle ABE is (by the 8. of the first) equal to the angle EDA. But the angle ABE is a right angle: whefore also the angle EDA is a right angle: wherfore the line ED is erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the line DA: and it is also erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to either of these lines BD and DC, wherefore the line ED is vnto these three right lines BD, DA, and DC erected perpendicularly from the poynt where these three right lines touch the one the other: wherefore (by the 5. of the eleuenth) these three right lines BD, DA, and DC are in one and the selfe same superficies. And in what superficies the lines BD and DA are, in the selfe same also is the line BA: for euery triangle is (by the 2. of the eleuenth) in one and the selfe same superficies. Wherefore these right lines AB, BD, and DC are in one and the selfe same superficies, and either of these angles ABD and BDC is a right angle (by sup<g ref="char:EOLhyphen"/>position). Wherefore (by the 28. of the first) the line AB is a parallel to the line CD. If there<g ref="char:EOLhyphen"/>fore two right lines be erected perpendicularly to one and the selfe same playne superficies, those right lines are parallels the one to the other: which was required to be proued.</p>
               <p>
                  <hi>Here for the better vnderstanding of this 6. proposition I
<figure/>
haue described an other figure: as touching which if ye erect the superficies</hi> ABD <hi>perpendicularly to the superficies</hi> BDE, <hi>and imagine only a line to be drawne from the poynt</hi> A <hi>to the poynt</hi> E (<hi>if ye will ye may extend a thred from the saide poynt</hi> A <hi>to the poynt</hi> E) <hi>and so compare it with the demonstration, it will make both the proposition, and also the demonstration most cleare vnto you.</hi>
               </p>
               <div type="section">
                  <head>¶An other demonstration of the sixth proposition by <hi>M. Dee.</hi>
                  </head>
                  <p>Suppose that the two right lines <hi>AB</hi> &amp; <hi>CD</hi> be perpendicularly erected to one &amp; the same playne su<g ref="char:EOLhyphen"/>perficies,
<pb n="326" facs="tcp:7062:355"/>
namely the playne superficies OP. Then I say that <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> and CD are parallels. Let the end points of the right lines AB and CD, which touch the plaine sup<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>cies O<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, be the poyntes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and D, fro<g ref="char:cmbAbbrStroke">̄</g> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to D let a straight line be drawne (by the first petition) and (by the second petition) let the straight line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D be exte<g ref="char:cmbAbbrStroke">̄</g>ded, as to the poynts M &amp; N. Now forasmuch
<figure/>
as the right line AB, from the poynt <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> produced, doth cutte the line MN (by construction). There<g ref="char:EOLhyphen"/>fore (by the second proposition of this eleuenth booke) the right lines AB &amp; MN are in one plain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> superficies. Which let be QR, cutting the superfi<g ref="char:EOLhyphen"/>cies OP in the right line MN. By the same meanes may we conclude the right line CD to be in one playne superficies with the right line MN. But the right line MN (by supposition) is in the plaine su<g ref="char:EOLhyphen"/>perficies QR: wherefore CD is in the plaine su<g ref="char:EOLhyphen"/>perficies QR. And A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the right line was proued to be in the same plaine superficies QR. Therfore AB and CD are in one playne superficie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, namely QR. And forasmuch as the lines A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and CD (by supposition) are perpendicular vpon the playne superficies OP, therefore (by the second definition of this booke) with all the right lines drawne in the superficies OP and touching AB and CD, the same perpe<g ref="char:cmbAbbrStroke">̄</g>diculars A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and CD, do make right angles. But (by construction) MN, being drawne in the plaine superficies OP toucheth the perpendiculars AB and CD at the poyntes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and D. Therefore the perpen<g ref="char:EOLhyphen"/>diculars A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and CD, make with the right line MN two right angles namely ABN, and CDM: and MN the right line is proued to be in the one and the same playne superficies, with the right lines AB &amp; CD: namely in the playne superficies QR. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>refore by the second part of the 28. proposition of the first booke, the right line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> AB and CD are parallel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. If therefore two right lines be erected perpendicularly to one and the selfe same playne superficies those right lines are parallels the one to the other: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>A Corollary added by <hi>M. Dee.</hi>
                  </head>
                  <p>
                     <hi>Hereby it is euident that any two right lines perpendicularly erected to one and the selfe same <milestone type="tcpmilestone" unit="unspecified" n="1"/> playne superficies, are also them selues in one and the same playne superficies, which is likewis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> perpen<g ref="char:EOLhyphen"/>dicularly <milestone type="tcpmilestone" unit="unspecified" n="2"/> erected to the same playne superficies, vnto which the two right lines are perpendicular.</hi>
                  </p>
                  <p>The first part hereof is proued by the former construction and demonstration, that the right lines <milestone type="tcpmilestone" unit="unspecified" n="1"/> AB and CD are in one and the same playne superficies Q<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. The second part is also manifest (that is, that <milestone type="tcpmilestone" unit="unspecified" n="2"/> the playne superficies QR is perpendicularly erected vpon the playne superficies OP) for that A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and CD being in the playne superficies QR, are by supposition perpendicular to the playne superficies OP: wherefore by the third definition of this booke QR is perpendicularly erected to, or vpon OP: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>
                     <hi>Io. Dee</hi> his aduise vpon the Assumpt of the 6.</head>
                  <p>As concerning the making of the line DE, equall to the right line AB, verely the second of the first, without some farther consideration, is not properly enough alledged. And no wonder it is, for that in the former booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, whatsoe<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h of lines
<figure/>
bene spoken, the same hath alway<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s bene imagined to be in one onely playne superficies considered or execu<g ref="char:EOLhyphen"/>ted. But here the perpendicular line AB, is not in the same playn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> superficies, that the right line DB is. Ther<g ref="char:EOLhyphen"/>fore some other helpe must be put into the handes of young beginners, how to bring this probleme to exe<g ref="char:EOLhyphen"/>cution: which is this, most playne and briefe. Vnder<g ref="char:EOLhyphen"/>stand that BD the right line, is the common section of the playne superficies, wherein the perpendiculars AB and CD are, &amp; of the other playne superficies, to which they are perpendiculars. The first of these (in my former demonstration of the 6)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> I noted by the playne superfi<g ref="char:EOLhyphen"/>cies QR: and the other, I noted by the plaine superficies OP. Wherfore BD being a right line common to both the playne sup<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rficieces QR &amp; OP, therby the ponits B and D are co<g ref="char:cmbAbbrStroke">̄</g>mon to the playnes QR and OP. Now
<pb facs="tcp:7062:356"/>
from BD (sufficiently extended) cutte a right line equall to AB, (which suppose to be BF) by the third of the first, and orderly to BF make DE equall, by the 3. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the first, if DE be greater then BF. (Which alwayes you may cause so to be, by producing of DE sufficiently). Now forasmuch as BF by constructi<g ref="char:EOLhyphen"/>on is cutte equall to AB, and DE also, by construction, put equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll to BF, therefore by the 1. common sentence, DE is put equall to AB: which was required to be done.</p>
                  <p>In like sort, if DE were a line geuen to whome AB
<figure/>
were to be cutte and made equall, first out of the line DB (su<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>iciently produced) cutting of DG, equall to DE by the third of the first: and by the same 3. cutting from BA (sufficiently produced) BA, equall to DG: then is it euide<g ref="char:cmbAbbrStroke">̄</g>t, that to the right line DE, the perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line AB is put equall. And though this be easy to conceaue, yet I haue designed the figure accordingly, wherby you may instruct your imagination. Many such helpes are in this booke requisite, as well to enforme the young stu<g ref="char:EOLhyphen"/>dentes therewith, as also to master the froward gayne<g ref="char:EOLhyphen"/>sayer of our conclusion, or interrupter of our demon<g ref="char:EOLhyphen"/>strations course.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 7. Theoreme. The 7. Proposition. If there be two parallel right lines, and in either of them be taken a point at all aduentures: a right line drawen by the said pointes is in the self same superficies with the parallel right lines.</head>
               <p>SVppose that these two right lines AB and CD be parallels, and in either of the<g ref="char:cmbAbbrStroke">̄</g> take a point at all aduentures, namely, E and F. Then I say, that a right line drawen from the point E to the point F, is in the selfe same plaine superficies that the pa<g ref="char:EOLhyphen"/>rallel lines are. For if not, then if it be possible,
<figure/>
let it be in an higher superficies,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> as the line EGF is, and draw the superficies wherin the line EGF is, &amp; extend it, and it shall make a common secti<g ref="char:EOLhyphen"/>on with the ground superficies, which section shall (by the 3. of the eleuenth) be a right line: let that section be the right line EF. Wherefore two right lines EGF and EF include a superficies: which (by the last common sentence) is impossible. Wherfore a right line drawen from the point E to the point F, is not in an higher superficies. Wherfore a right line drawen from the point E to the point F, is in the selfe same superficies wherein are the parallel right lines AB and CD. If therefore there be two parallel right lines, and in either of them be taken a point at all aduentures, a right line drawen by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se pointes is in the selfe same plaine superficies with the parallel right lines: which was requi<g ref="char:EOLhyphen"/>red to be demonstrated.</p>
               <p>
                  <hi>By this figure it is easie to see
<figure/>
the former demonstration, if ye eleuate the superficies wherin is drawen the line EGF.</hi>
               </p>
            </div>
            <div type="part">
               <pb n="327" facs="tcp:7062:356"/>
               <head>The 8. Theoreme. The 8. Proposition. If there be two parallel right lines, of which one is erected perpendicularly to a round playne superficies: the other also is erected perpendicularly to the selfe same ground playne superficies.</head>
               <p>SVppose that there be two parallel right lines AB and CD, and let one of them, namely, AB be erected perpendiculerly to a ground superficies.<note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is as it were the con<g ref="char:EOLhyphen"/>uerse of the sixth Construction.</note> Then I say that the line CD is also erected perpendiculerly, to the selfe same ground superficies. Let the lines AB and CD fall vpon the ground superficies in t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e pointes B and D, and (by the first peticion) draw a righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line from the point B to the point D. And drawe (by the 11. of the first) in the ground superficies from the point D vnto the line BD a per<g ref="char:EOLhyphen"/>pendiculer line DE, and (by the 2. of the first) put the line
<figure/>
DE equall to the line AB, and draw a right line from the point B to the point E, and an other from the point A to the point E, and an other from the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oint A to the point D.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AB is erested perpendicularly to the ground superficieces, therfore (by the 2. definition of the e<g ref="char:EOLhyphen"/>leuenth) the line AB is erected perpendicularly to all the right lines that are in the ground superficies and touche it. Wherfore either of these angles ABD &amp; ABE is a right angle. And forasmuch as vpon these parallel lines AB and CD falleth a certaine right line BD, therefore (by the 29. of the first) the angles ABD and CDB are equal to two right angles. But the angle ABD is a right angle, wherfore also the angle CDB is a right angle. Wherfore the line CD is e<g ref="char:EOLhyphen"/>rected perpendic<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>larly to the line BD. And forasmuch as the line AB is equall to the line DE, and the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D is common to them both, therfore these two lines AB and BD are equal to these two lines ED and DB, and the angle ABD is equall to the angle EDB for either of them is a right angle. Wherfore (by the 4. of the first) the base AD is equall to the base BE. And forasmuch as the line AB is equall to the line DE, and the line BE to the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> AD, therfore thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two lines AB and BE are equall to these two lines AD &amp; DE, the on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the other, and the line AE is a common base to them both. Wherfore (by the 8. of the first) the angle ABE is equall to the angle ADE: but the angle A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E is a right angle, wherfore th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ngle EDA, is also a right angle. Wherefore the line ED is erected perpendicularly to the line AD, and it is also erected perpendicularly to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line DB. Wherfore the line ED is erect<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d perpendicularly to the plaine superficies wherin th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s BD and BA are (by the 4. of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>his booke) Wherfore (by the 2. definition of the eleuenth) the line ED is erected perpen<g ref="char:EOLhyphen"/>dicularly to all the right lines that touche it and are in the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>perficies wherein the lines BD and AD are. But in what superficies the lines BD and DA are, in the selfe same superficies is the line DC. For the line AD being drawen from two pointes taken in the parallel lines AB and CD is by the former proposition in the selfe same superficies with them. Now f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ras<g ref="char:EOLhyphen"/>much as the lines AB and BD ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the superficies wherin the lines BD and DA are, but in what superficies the lines AB &amp; BD are, in the same is the line DC. Wherfore the line ED is erected perpendicularly to the line DC. Wherfore also the line CD is erected perpendi<g ref="char:EOLhyphen"/>cularly to the line DE. And the line CD is erected perpendicularly to the line DB. For by the 29. of the first, the angle CDB being equall to the angle ABD is a right angle. Where<g ref="char:EOLhyphen"/>fore the line CD is from the point D erected perpendicularly to two right lines DE and DB cutting the one the other in the point D. Wherfore by the 4. of the eleuenth, the line CD is erected perpendiculaaly to the plaine superficies, wherein are the lines DE and DB. But
<pb facs="tcp:7062:357"/>
the ground plaine superficies is that wherin are the lines DE and DB, to which superficies also the line AB is supposed to be erected perpendiculerly. Wherefore the line CD is erected perpendicularly to the ground plaine superficies, wherunto the line AB is erected perpendicu<g ref="char:EOLhyphen"/>larly. If therfore there be two parallel right lines, of which one is erected perpendicularly to a ground plaine superficies, the other also is erected perpendicularly to the selfe same ground plaine superficies: which was required to be demonstrated.</p>
               <p>
                  <hi>This figure will more clearely set forth the former de<g ref="char:EOLhyphen"/>monstration,
<figure/>
if ye erect perpendicularly the superficies ABD to the superficies BDE, and imagine a lyne to be drawen from the point A to the point D, in stede wher<g ref="char:EOLhyphen"/>of, as in the 6. proposition ye may extende a threede.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 9. Pro<gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> Right lines which are parallels to one and the selfe same right line, and are not in the selfe same superficies that it is in: are also parallels the one to the other.</head>
               <p>SVppose that either of these right lines AB and CD be a parallel to the line EF not being in the selfe same superficies with it. Then I say that the line AB is a parallel to the line CD. Take in the line EF a point at all aduentures, and let the same be G.<note place="margin">Construction.</note> And from the point G raise vp in the superficies wherin are the lines EF and AB, vnto the line EF a perpendiculer line GH, and againe in the superficies wherin are the lines EF and CD, raise vp from the same point G to the line EF a perpen<g ref="char:EOLhyphen"/>diculer line GK.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line
<figure/>
EF is erected perpendiculerly to either of the lines GH and GK, therfore (by the 4. of the eleuenth) the line EF is erected perpendicu<g ref="char:EOLhyphen"/>larly to the superficies wherein the lines GH and GK are, but the line EF is a parallel line to the line AB. Wherfore (by t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e 8. of the eleuenth) the line AB is erected perpendicu<g ref="char:EOLhyphen"/>larly to the plaine superficies, wherin are the lines GH and GK. And by the same reason al<g ref="char:EOLhyphen"/>so the line CD is erected perpendicularly to the plaine superficies wherin are the lines GH &amp; GK. Wherefore either of these lines AB and CD is erected perpendicularly to the plaine superficies, wherin the lines GH and GK are. But if two right lines be erected perpendicu<g ref="char:EOLhyphen"/>larly to one and the selfe same plaine superficies, those right lines are parallels the one to the other (by the 6. of the eleuenth) Wherfore the line AB is a parallel to the line CD. Wherfore right lines which are parallels to one &amp; the selfe same right line, and are not in the self same superficies with it are also parallels the one to the other: which was required to be proued.</p>
               <p>
                  <pb n="328" facs="tcp:7062:357"/>
                  <hi>This figure more clearely manifesteth the former propo<g ref="char:EOLhyphen"/>sition
<figure/>
and demonstration, if ye eleuate the superficieces ABEF and CDEF that they may incline and concurre in the lyne EF.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 10. Theoreme. The 1<gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> If two right lines touching the one the othe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>her right lines touching the one the other, and no<gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>lfe same superficies with the two first: those right lines co<g ref="char:cmbAbbrStroke">̄</g>taine equall angles.</head>
               <p>SVppose that these two right lines AB and BC touching the one the other, be parallells to these two lines DE and EF touching also the one the other, and not being in the selfe same superficies that the lines AB and BC are. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the angle ABC is equall to the angle DEF.<note place="margin">Construction.</note> For let the lines BA, BC, ED, EF, be put equall the one to the other: and draw these
<figure/>
right lines AD, CF, BE, AC, and DF.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line BA is equall to the line ED, and also parallell vnto it, therefore (by the 33. of the first) the line AD is e<g ref="char:EOLhyphen"/>quall and parallell to the line BE: and by the same reason also the line CF is equall &amp; parallell to the line BE. Wher<g ref="char:EOLhyphen"/>fore either of these lines AD and CF is equall &amp; parallell to the line EB. But right lines which are parallells to one and the selfe same right line, and are not in the selfe same su<g ref="char:EOLhyphen"/>perficies with it, are also (by the 9. of the eleuenth) parallells the one to the other. Wherefore the line AD is a parallell line to the line CF. And the lines AC and DF ioyne them together. Wherefore (by the 33. of the first) the line AC is equall and parallell to the line DF. And forasmuch as these two right lines AB &amp; BC are equall to these two right lines DE and EF, and the base AC also is equall to the base DF: therefore (by the 8. of the first) the angle ABC is equall to the angle DEF. If therfore two right lines touching the one the other be parallells to two other right lines touching the one the other, and not being in one and the selfe same superficies with the two first: those righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> lines containe equall angles: which was required to be demonstrated.</p>
               <p>
                  <hi>This figure here set more
<figure/>
plainly declareth the former Pro<g ref="char:EOLhyphen"/>position and demonstration, if ye eleuate the superficieces DABE, and FCBE, till they concurre in the line FE.</hi>
               </p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:358"/>
               <head>¶ The 1. Probleme. The 11. Proposition. From a point geuen on high, to drawe vnto a ground plaine superficies a perpendicular right line.</head>
               <p>
                  <note place="margin">
                     <p>Construction.</p>
                     <p>Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</p>
                     <p>The first case.</p>
                  </note>SVppose that the point geuen on high be A, and suppose a ground plaine superficies, namely, BCGH. It is required from the point A to draw vnto the ground super<g ref="char:EOLhyphen"/>ficies a perpendicular line. Drawe in the ground superficies a right line at aduen<g ref="char:EOLhyphen"/>tures, and let the same be BC.<note place="margin">Iohn Dee.</note> And (by the 12. of the first) from the point A draw vnto the line BC a perpendicular line AD. <note n="*" place="margin">
                     <hi>This requireth the imagination of a plaine superficies passing by the pointe A, and the straight line BC. And so helpe your selfe in the lyke cases either Mathematically imagi<g ref="char:EOLhyphen"/>ning, or Me<g ref="char:EOLhyphen"/>chanically practising.</hi>
                  </note> Now if AD be a perpendicular line to the ground superficies, then is that done which was sought for. But if not, then (by the 11. of the first) from the point D raise vp in the ground super<g ref="char:EOLhyphen"/>ficies vnto the line BC a perpendicular line DE. And (by the 12. of the first) from the point A draw vnto the line DE a perpendicular line AF. And by the point F draw (by the 31. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst) vnto the
<figure/>
line BC a parallell line FH: And extend the line FH from the point F to the point G.<note place="margin">
                     <p>Second cas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</p>
                     <p>Demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>a<g ref="char:EOLhyphen"/>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on.</p>
                  </note> And forasmuch as the line BC is erected perpendicularly to either of these lines DE and DA, therefore (by the 4. of the eleuenth) the line BC is erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the superficies wher<g ref="char:EOLhyphen"/>in the lines ED and AD are: and to the line BC the line GH is a parallell. But i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> there be two parallell right lines, of which one is erected perpendicularly to a certaine plaine superficies, the other also (by the 8. of the eleuenth) is erected perpendicularly to the selfe same superficies. Wherefore the line GH is erected perpendicularly to the plain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> superficies wherein the lines ED and DA are. Wherfore also (by the 2. definition of the eleuenth) the line GH is erected perpendicularly to all the right lines which touch it, and are in the plaine superficies wherein the lines ED and AD are. But the line AF toucheth it being in the superficies wherein the lines ED and AD are (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of this booke). Wherefore the line GH is erected perpen<g ref="char:EOLhyphen"/>dicularly to the line FA. Wherefore also the line FA is erected perpendicularly to the line GH: and the line AF is also erected perpendicularly to the line DE. Wherefore AF is e<g ref="char:EOLhyphen"/>rected perpendicularly to either of these lines HG and DE. But if a right line be erected per<g ref="char:EOLhyphen"/>pendicularly from the common section of two right lines cutting the one the other, it shall also be erected perpendicularly to the plaine superficies of the said two lines (by the 4. of the ele<g ref="char:EOLhyphen"/>uenth). Wherefore the line AF is erected perpendicularly to that superficies wherin the lines ED and GH are. But the superficies wherein the lines ED and GH are, is the ground su<g ref="char:EOLhyphen"/>perficies. Wherefore the line AF is erected perpendicularly to the ground superficies. Wher<g ref="char:EOLhyphen"/>fore from a point geuen on high, namely, fro<g ref="char:cmbAbbrStroke">̄</g> the point A, is drawen to the ground superficies a perpendicular line: which was required to be done.</p>
               <p>
                  <hi>In this figure shall ye much more plainely see both
<figure/>
the cases of this former demonstratio<g ref="char:cmbAbbrStroke">̄</g>. For as touching the first case, ye must erecte perpendicularly to the ground superficies, the superficies wherein is drawen the line AD, and compare it with the demonstration, and it will be clere vnto you. For the second case ye must erecte perpendicularly vnto the ground superfi<g ref="char:EOLhyphen"/>cies the superficies wherein is drawen the line AF, and vnto it let the other superficies wherein is drawen the line AD, incline, so that the point A of the one may concurre with the point A of the other: and so with your figure thus ordered, compare it with the demon<g ref="char:EOLhyphen"/>stration, and there will be in it no hardnes at all.</hi>
               </p>
            </div>
            <div type="part">
               <pb n="329" facs="tcp:7062:358"/>
               <head>¶The 2. Probleme. The 12. Proposition. Vnto a playne superficies geuen, and from a poynt in it geuen, to rayse vp a perpendicular line.</head>
               <p>SVppose that there be a ground playne superficies geuen,
<figure/>
and let the poynt in it geuen be <hi>A.</hi> It is required from the point <hi>A</hi> to raise vp vnto the ground plaine superficies a perpendicular line. Vnderstand some certayne poynt on high, and let the same be <hi>B.</hi>
                  <note place="margin">Construction.</note> And from the poynt <hi>B</hi> draw (by the 11. of the eleuenth) a perpendicular line to the ground superficies, and let the same be <hi>BC.</hi> And (by the 31. of the first) by the poynt <hi>A</hi> drawe vnto the line <hi>BC</hi> a parallel line <hi>DA.</hi> Now forasmuch as there are two parallel right lines <hi>AD</hi> and <hi>CB,</hi> the one of them, namely,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> 
                  <hi>CB</hi> is erected perpendicularly to the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> superficies: wherefore the other line also, namely, <hi>AD,</hi> is <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> perpendicularly to the same ground superficies (by the eight of <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>leuenth). Wherefore vnto a playne superficies geuen, and <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> poynt in it geuen, namely, <hi>A,</hi> is raysed vp a perpendicular lyn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>required to be doone.</p>
               <p>
                  <hi>In this second figure ye may consider playnely the demonstration of the former proposition if ye erect per<g ref="char:EOLhyphen"/>pendicularly the superficies wherein are drawne the lines AD and CB.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 11. Theoreme. The 13. Pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>position. From one and the selfe poynt, and to one and the selfe same playne superfi<g ref="char:EOLhyphen"/>cies, can not be erected two perpendicular right lines on one and the selfe same side.</head>
               <p>FOr if it be possible from the poynt <hi>A</hi> let there be erected perpendicularly to one and the selfe same playne superficies two righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> lines <hi>AB</hi> and <hi>AC</hi> on one and the selfe same side.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> And extende
<figure/>
the superficies wherein are the lines <hi>AB</hi> and <hi>AC</hi>:<note place="margin">Note this maner of i<g ref="char:EOLhyphen"/>magination Mathemati<g ref="char:EOLhyphen"/>call.</note> and it shall make at length a common section in the ground super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies which common section shall be a right line, and shall passe by the poynt <hi>A</hi>: let that common section be the line <hi>DAE.</hi> Wherefore (by the 3. of the ele<g ref="char:EOLhyphen"/>uenth) the lines <hi>AB, AC,</hi> and <hi>DAE</hi> are in one and the selfe same playne superficies. And foras<g ref="char:EOLhyphen"/>much as the line <hi>CA</hi> is erected perpendicularly to the ground superficies, therfore (by the
<pb facs="tcp:7062:359"/>
2. definition of the eleuenth) it maketh right angles with all the right lines that touch it, and are in the ground superficies. But the line <hi>DAE</hi> toucheth it, being in the ground superficies. Wherefore the angle <hi>CAE</hi> is a right angle, and by the same reason also the angle <hi>BAE</hi> is a right angle. Wherefore (by the 4 petition) the angle <hi>CAE</hi> is equall to the angle <hi>BAE</hi> the lesse to the more, both angles being in one &amp; the selfe same playne superficies: which is im<g ref="char:EOLhyphen"/>possi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>le. Wherefore from one and the selfe same poynt, and to one and the selfe same playne su<g ref="char:EOLhyphen"/>perficies can not be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rected two perpendicular right lines on one &amp; the selfe same side: which was required to be demonstrated.</p>
               <p>
                  <hi>In this figure if ye erect perpendicularly the su<g ref="char:EOLhyphen"/>perficies
<figure/>
wherein are drawne the lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>A and CA to the ground super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies wherein is drawn the line DAE, and so compare it with the the demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the for<g ref="char:EOLunhyphen"/>mer proposition it will be cleare vnto you.</hi>
               </p>
               <div type="section">
                  <head>
                     <hi>M. Dee</hi> his annotation.</head>
                  <p>
                     <hi>Euclides</hi> wordes in this 13. proposition admit two cases: one, if th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> in the playne superficies, (as co<g ref="char:cmbAbbrStroke">̄</g>monly the demonstrations suppose) the other, if the poynt assigned be any where with<g ref="char:EOLunhyphen"/>out the sayd playne superficies, to which, the perpendiculars fall, is considered. Contrary to either of which, if the aduersarie affirme, admitting from one poynt two right lines, perpendiculars to one and the selfe same playne superficies, and on one and the same side thereof, by the 6. of the eleuenth he may be bridled: which will <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ore him to confesse his two perpendiculars to be also parallels. But by supposi<g ref="char:EOLhyphen"/>tion agreed one, they concurre at one and the same poyn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, which (by the definition of parallels) i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> im<g ref="char:EOLhyphen"/>possible. Therefore our aduersary must recant a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d yelde to out proposition.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 14. Proposition. To whatsoeuer plaine superficieces one and the selfe same right line is e<g ref="char:EOLhyphen"/>rected perpendicularly: those superficieces are parallels the one to the other.</head>
               <p>SVppose that a right line AB be
<figure/>
erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to either of these plaine superficieces CD and EF. Then I say, that these superficieces CD and EF are paral<g ref="char:EOLhyphen"/>lels the one to the other.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if not, then if they be extended they will at the length meete. Let them meete, if it be possible. Now then their common section shall (by the 3. of the eleuenth) be a right line. Let that common section be GH. And in the line GH take a point at all aduen<g ref="char:EOLhyphen"/>tures and let the same be K. And drawe a right line from the point A to the point K, and an other from the point B to the point K. And forasmuch as the line AB is e<g ref="char:EOLhyphen"/>rected
<pb n="330" facs="tcp:7062:359"/>
perpendicularly to the plaine superficies EF, therefore the line AB is also erected perpendicularly to the line BK which is in the extended superficies EF. Wherfore the angle ABK is a right angle. And by the same reason also the angle BAK is a right angle. Wher<g ref="char:EOLhyphen"/>fore in the triangle ABK, these two angles ABK &amp; BAK, are equall to two right angles: which (by the 17. of the first) is impossible. Wherefore these superficieces CD and EF being extended meete not together. Wherefore the superficieces CD and EF are parallells. Wher<g ref="char:EOLhyphen"/>fore to what soeuer plaine superficieces one and the selfe same right line i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> erected perpendicu<g ref="char:EOLhyphen"/>larly: those superficies are parallells the one to the other: which was required to be proued.</p>
               <p>
                  <hi>In this figure may ye plainly see the former demonstra<g ref="char:EOLhyphen"/>tion
<figure/>
if ye erecte the three superficieces, GD, GE, and KLM perpe<g ref="char:cmbAbbrStroke">̄</g>diculary to the ground plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies: but yet in such sort that the two superfici<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> may concurre in the common line G <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the demonstration.</hi>
               </p>
               <div type="section">
                  <head>A corollary added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <hi>If a right line be erected perpendicularly to one of those superficies, it <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> erected perpen<g ref="char:EOLhyphen"/>dicularly to the other.</hi>
                  </p>
                  <p>For if it should not be erected perpendicularly to the other, then it falling vpon that other shall make with some one line thereof an angle lesse then a right angle: which line should (by the 5. petition of the first) concurre with some one line of that super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies whereunto it is perpendicular. So that those superficieces should not be parallels: which is contrary to the supposition. For they are suppsed to be parallels.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 13. Theoreme. The 15. Proposition. If two right lines touching the one the other be parallels to two other right lines touching also the one the other and not being in the selfe same plaine superficies with the two first: the plaine superficieces extended by those right lines, are also parallells the one to the other.</head>
               <p>SVppose that these two right lines AB and BC touching the one the other be pa<g ref="char:EOLhyphen"/>rallells to these two right lines DE &amp; EF touching also the one the other, and not being in the selfe same plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies with the right lines AB and BC. Then I say, that the plaine superficieces by the lines AB and BC, and the lines DE and EF being extended, shall not meete together, that
<figure/>
is, they are equedistant and parallels:<note place="margin">Construction.</note> From the point B draw (by the 11. of the eleuenth) a perpendicular line to the su<g ref="char:EOLhyphen"/>per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies wherein are the lines DE and EF, and let that per<g ref="char:EOLhyphen"/>pendicular line be BG. And by the point G in the plaine su<g ref="char:EOLhyphen"/>perficies passing by DE, and EF, draw (by the 31. of the first) vnto the line ED a parallell line GH: and likewise by that point G drawe in the same superficies vnto the line EF a pa<g ref="char:EOLhyphen"/>rallell line GK.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line BG is erected per<g ref="char:EOLhyphen"/>pendicularly to the superficies wherein are the lines DE and EF, there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore (by the 2. definition of the eleuenth) it is also erected perpendicularly to all the right lines which touch it, and are in the selfe same superficies wherein are the lines DE
<pb facs="tcp:7062:360"/>
and EF. But either of these lines GH and GK touch it, and
<figure/>
are also in the superficies wherein are the lines DE and EF, therefore either of these angles BGH, and BGK, is a right angle. And forasmuch as the line BA is a parallell to the line GH (that the lines GH and GK are parallells vnto the lines AB and BC it is manifest by the 9. of this booke): there<g ref="char:EOLhyphen"/>fore (by the 29. of the first) the angles GBA and BGH are equall to two right angles. But the angle BGH is (by con<g ref="char:EOLhyphen"/>structio<g ref="char:cmbAbbrStroke">̄</g>) a right angle, therfore also the angle GBA is a right angle: therefore the line GB is erected perpendicularly to the line BA. And by the same reason also may it be proued, that the line BG is erected perpendicularly to the line BC. Now forasmuch as the right line BG is erected perpendicularly to these two right lines BA and BC touching the one the other, therefore (by the 4. of the eleuenth) the line BG is erected perpendicularly to the superficies wherein are the lines BA and BC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And it is also erected perpendicularly to the superficies wherein are the lines GH and GK. But the superficies wherein are the lines GH and GK, is that superficies wherein are the lines DE and EF: wherefore the line BG is erected perpendicularly to the superficies wherein are the lines DE and EF. Wherefore the line BG is erected perpendicularly to the superficies wherein are the lines DE and EF, and to the superficies wherein are the lines AB and BC. But if one and the selfe same right line be erected perpendicularly to plaine superficieces, those superfi<g ref="char:EOLhyphen"/>cieces are (by the 14. of the eleuenth) parallels the one to the other. Wherefore the superficies wherin are the lines AB and BC is a par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>llel to the superficies wherin are the lines DE and EF. If therefore two right lines touching the one the other be parallels to two other right lines touching also the one the other, and not being in the selfe same plaine superficies with the two first, the plaine superficieces extended by those right lines are also parallels the one to the other which was required to be demonstrated.</p>
               <p>
                  <hi>By this figure here put, ye may more clerely see both
<figure/>
the former 15. Proposition and also the demonstration therof: if ye erecte perpendicularly vnto the ground superficies, the three superficieces ABC, KHE, and LHBM, and so compare it with the demonstration.</hi>
               </p>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Vnto a plaine superficies being geuen, to drawe by a point geuen without it, a parallel plaine superfi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> Suppose as in the former description that the superficies geue<g ref="char:cmbAbbrStroke">̄</g> be ABC, &amp; let the point geue<g ref="char:cmbAbbrStroke">̄</g> without it be G. Now then by the point G drawe (by the 31. of the first) vnto the lines AB and BC parallel lines GH and HK. And the superficies extended by the lines GH and GK shall be parallel vnto the superficies ABC, by this 15. Proposition.</p>
               </div>
            </div>
            <div type="part">
               <head>The 14. Theoreme. The 16. Proposition. If two parallel playne superficieces be cut by some one playne superficies: their common sections are parallel lines.</head>
               <p>SVppose that these two plaine superficieces AB and CD be cut by this plaine su<g ref="char:EOLhyphen"/>perficies EFGH, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd let their common sections be the right lines EF and GH. Then I say that the line EF is a parallel to the line GH. For if not, then the lines EF and GH being produced, shall at the length meete together either on
<pb n="331" facs="tcp:7062:360"/>
the side that the pointes FH, are, or on
<figure/>
the side that the pointes E, G are. First let them be produced on that side that the pointes F, H are, and let them mete in the point K. And forasmuch as the line EFK is in the superficies AB, therfore all the points which are in the line EF are in the superficies AB (by the first of this booke) But one of the pointes which are in the right line EFK is the point K,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an absurditie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> therfore the point K is in the superficies AB. And by the same reason also the point K is in the superficies CD. Wherfore the two su<g ref="char:EOLhyphen"/>perficieces AB and CD being produ<g ref="char:EOLhyphen"/>ced do mete together, but by sup<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ositio<g ref="char:cmbAbbrStroke">̄</g> they mete not together, for they are sup<g ref="char:EOLunhyphen"/>posed to be parallels. Wherfore the right lines EF and GH produced shall not meete together on that side that the pointes F, H are. In like sort also may we proue that the right lines EF and GH produced meete not together on that side that the pointes E, G are. But right lines which being produced on no side mete together, are parallels (by the last definicion of the first). Wherfore the line EF is a parallel to the line GH. If therfore two parallel plaine su<g ref="char:EOLhyphen"/>perficieces be cut by some one plaine superficies their common sections are parallel lines: which was required to be proued.</p>
               <p>
                  <hi>This figure here set more plainl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                     <figure/>
demonstration, if ye erect perpendicu<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> superficies the three superficieces A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> and so compare it with the demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                  </hi>
               </p>
               <div type="section">
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>If two plaine superficieces be parallels to one and the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>lfe same playne <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> also be parallels the one to the other, or they shall make one and the selfe same plaine sup<gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                     </hi>
                  </p>
                  <p>For if the plaine superficieces DG and GH
<figure/>
being parallels to one and the selfe same superfi<g ref="char:EOLhyphen"/>cies, namely, to AB be not also parallels the one to the other, then being produced they shall con<g ref="char:EOLhyphen"/>curre (by the conuerse of the sixt definition of the eleuenth) Let them concurre in the right line GE. Then I say that the superficieces GD and GH are in one and the selfe same playne superficies. Draw in the playne superficies AB a right line at all aduentures AC. And by that right lyne &amp; the point E extende a playne superficies, cutting the two superficieces DG and GH by the right lines ED and EI. Wherfore the right lines AC and DE, also AC and EI are parallels by this propositi<g ref="char:EOLhyphen"/>on. But the lines DE and EI forasmuch as they concurre in the point E are not parallels the one to the other. Wherefore the right lines DE and EI make directly one right line (by that which is added after the 30. propositon of the first.) And therfore the plaine superficieces DG and GH are in one and the selfe same playne superficies. For
<pb facs="tcp:7062:361"/>
i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hey be not, then part of the right line DI, namely, the part DE is in the playne superficies DG, and an other part therof, namely, EI is on high in an other superficies GH, which by the first of the eleue<g ref="char:cmbAbbrStroke">̄</g>th i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> impossible. Wherfore the superficieces DG and GH are in one and the selfe same playne superficies. But i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the superficieces DG and GH neuer concurre, then are they parallels by the 6. definition of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>leuenth.</p>
                  <p>
                     <hi>In this figure here set, ye may more plainely see the
<figure/>
form<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r demonstration, if ye eleuate to the ground super<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>icieces ACDI, the three super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>icieces AB, DG, &amp; GI, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o compare it with the demonstration.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The 15. Theoreme. The 17. Proposition. I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> two right lines be cut by playne superficieces being parallels: the partes o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lines deuided shall be proportionall.</head>
               <p>
                  <note place="margin">
                     <p>In t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rono<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>o<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> must vnd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>and the prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rtio<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ll <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>artes or s<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>ions to be th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>se which are c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ntai<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ed <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> the parallel super<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>cies.</p>
                     <p>Construction.</p>
                     <p>Dem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nstra<g ref="char:EOLhyphen"/>tion.</p>
                  </note>S<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ppose that these two right lines AB and CD be deuided by these plaine super<g ref="char:EOLhyphen"/>fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ces being parallels, namely, GH, KL, MN in the points A, E, B, C, F, D. The<g ref="char:cmbAbbrStroke">̄</g> I say that as the right line AE is to the right line EB, so is the right line CF to the right line FD. Draw these right lines AC, BD and AD. And let the line AD and the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies KL concurre in the point X. And
<figure/>
draw a right line from the point E to the point X and an o<g ref="char:EOLhyphen"/>ther from the point X to the point F. And forasmuch as these two parallel superficieces KL and MN are cut by the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies EBDX, ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore their common sections which are the lines EX and BD, are (by the 16. of the eleuenth) parallels the one to the other. And by the same reason also <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as the two parallel superficies GH and KL be cut by the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies AXFC, their common sections AC and XF are (by the 16. of the eleuenth) parallels. And <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as to one of the sides of the triangle ABD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> namely, to the side BD is drawne a parallel line EX, therfore (by the 2. of the sixt) proportionally as the line AE is to the line EB, so is the line AX to the line XD. Againe forasmuch as to one of the sides of the triangle ADC, namely, to the side AC is drawen a parallel line XF, therfore by the 2. of the sixt, pro<g ref="char:EOLhyphen"/>portionally as the line AX is to the line XD, so is the line CF to the line FD. And it was proued that as the line AX is to the line XD, so is the line AE to the line EB, therefore also (by the 11. of the fift) as the line AE is to the line EB, so is the line CF to the line FD. If therfore two right lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e deuided by plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces being parallels, the parts of the lines deuided shal be propor<g ref="char:EOLhyphen"/>tionall: which was required to be demonstrated.</p>
               <p>
                  <pb facs="tcp:7062:361"/>
                  <hi>In this figure it is more easy to see the former demonstration, if ye e<g ref="char:EOLhyphen"/>rect
<figure/>
perpendicularly vnto the ground superficies ACBD, the thre su<g ref="char:EOLhyphen"/>perficieces, GH, KL, and MN, or if ye so <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ct them that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y be equedi<g ref="char:EOLhyphen"/>stant one to the other.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 16. Theoreme. The 18. Proposition. If a right line be erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to a plaine superficies: all the superficieces extended by that right line, are erected perpendicularly to the selfe same plaine superficies.</head>
               <p>SVppose that a right line AB be erected perpendicularly to a ground superficies. The<g ref="char:cmbAbbrStroke">̄</g> I say, that all the superficieces passing by the line AB, are erected perpendicularly to the ground superficies. Extend a superficies by the line AB, and let the same be ED, &amp; let the co<g ref="char:cmbAbbrStroke">̄</g>mon section of the plaine
<figure/>
superficies and of the ground superficies be the right line CE. And take in the line CE a point at all aduentures,<note place="margin">Construction.</note> and let the same be F: and (by the 11. of the first) from the point F drawe vnto the line CE a perpendicular line in the superficies DE, and let the same be FG. And forasmuch as the line AB is erected perpendicularly to the ground super<g ref="char:EOLhyphen"/>ficies, therefore (by the 2. definition of the e<g ref="char:EOLhyphen"/>leuenth) the line AB is erected perpendicu<g ref="char:EOLhyphen"/>larly to all the right lines that are in the ground plaine superficies,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and which touch it. Wher<g ref="char:EOLhyphen"/>fore it is erected perpendicularly to the line CE. Wherefore the angle ABE is a right angle. And the angle GFB is also a right angle (by construction). Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>8. of the first) the line AB is a parallel to the line FG. But the line AB is erected perpendicularly to the ground superficies: wherefore (by the 8. of the eleuenth) the line FG is also erected per<g ref="char:EOLhyphen"/>pendicularly to the ground superficies. And forasmuch as (by the 3. definition of the eleuenth) a plaine superficies is then erected perpendicularly to a plaine superficies, when all the right lines drawen in one of the plaine superficieces vnto the common section of those two plaine su<g ref="char:EOLhyphen"/>perficieces making therwith right angles, do also make right angles with the other plaine su<g ref="char:EOLhyphen"/>perficies and it is proued that the line FG drawen in one of the plaine superficieces, namely, in DE, perpendicularly to the common section of the plaine superficieces, namely, to the line CE, is erected perpendicularly to the ground superficies: wherefore the plaine superficies DE is erected perpendicularly to the ground superficies. In like sort also may we proue, that all the plaine superficieces which passe by the line AB, are erected perpendicularly to the ground su<g ref="char:EOLhyphen"/>perficies. If therefore a right line be erected perpendicularly to a plaine superficies all the su<g ref="char:EOLhyphen"/>perficieces passing by the right line, are erected perpendicularly to the selfe same plaine super<g ref="char:EOLhyphen"/>ficies: which was required to be demonstrated.</p>
               <p>
                  <pb facs="tcp:7062:362"/>
                  <hi>In this figure here set ye may erect per<g ref="char:EOLhyphen"/>pe<g ref="char:cmbAbbrStroke">̄</g>dicularly
<figure/>
at your pleasure the superficies wherin are drawen the lines DC, GF, AB, and HE, to the ground superficies wherin is drawen the line CFBE, and so plainly compare it with the demonstration before put.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 17. Theoreme. The 19. Proposition.
If two plaine superficieces cutting the one the other be erected perpendicu<g ref="char:EOLhyphen"/>larly to any plaine superficies: their common section is also erected perpen<g ref="char:EOLhyphen"/>dicularly to the selfe same plaine superficies.</head>
               <p>SVppose that these two plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces AB &amp; BC cutting the one the other be e<g ref="char:EOLhyphen"/>rected p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ndicularly to a ground superficies, and let their common section be the line BD. Then I say, that the line BD is erected perpendicularly to the ground super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> For if not, then (by the 11. of the first)
<figure/>
from the point D draw in the superficies AB vnto the right line DA a perpendicular line DE. And in the superficies CB draw vnto the line DC a per<g ref="char:EOLhyphen"/>pendicular line DF. And forasmuch as the super<g ref="char:EOLhyphen"/>ficies AB is erected perpendicularly to the ground superficies, and in the plaine superficies AB vnto the common section of the plaine superficies and of the ground superficies, namely, to the line DA is erected a perpendicular line DE, therefore (by the conuerse of the 3. de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>inition of this booke) the line DE is e<g ref="char:EOLhyphen"/>rected perpendicularly to the ground super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. And in like sort may we proue, that the line DF is erected perpendicularly to the ground superficies. Wherefore from one and the selfe same point, namely, from D, are erected perpendicularly to the ground superficies two right lines both on one and the self same side: which is (by the 15. of the eleuenth) impossible. Wher<g ref="char:EOLhyphen"/>fore from the point D can not be erected perpendicularly to the ground superficies any other right lines besides BD, which is the common section of the two superficieces AB and BC. If therefore two plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces cutting the one the other be erected perpendicularly to any plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies, their common section is also erected perpendicularly to the selfe same plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies: which was required to be proued.</p>
               <p>
                  <hi>Here haue I set an other figure which
<figure/>
will more plainly shewe vnto you the for<g ref="char:EOLhyphen"/>mer demonstration, if ye erecte perpendi<g ref="char:EOLhyphen"/>cularly to the ground superficies AC the two superficieces AB and BC which cut the one the other in the line BD.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 18. Theoreme. The 20. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>roposition. If a solide angle be contayned vnder three playne superficiall angles: euery
<pb n="333" facs="tcp:7062:362"/>
two of those three angles, which two so euer be taken, are greater then the third.</head>
               <p>SVppose that the solide angle <hi>A</hi> be contayned vnder three playne superficiall an<g ref="char:EOLhyphen"/>gles, that is, vnder <hi>BAC, CAD,</hi> and <hi>DAB.</hi> Then I say that two of these superficiall angles how so euer they be taken, are greater then the third. If the angles <hi>BAC, CAD,</hi> &amp; <hi>DAB</hi>
                  <figure/>
be equall the one to the other, then is it manifest that two of them which two so euer be taken are greater then the third. But if not, let the angle <hi>BAC</hi> be the greater of the three angles. And vnto the right line <hi>AB</hi> and from the poynt <hi>A</hi> make in the playne superficies <hi>BAC</hi> vnto the angle <hi>DAB</hi> an equall angle <hi>BAE.</hi> And (by the 2. of the first) make the line <hi>AE</hi> equall to the line <hi>AD.</hi> Now a right line <hi>BEC</hi> drawne by the poynt <hi>E,</hi> shall cut the right lines <hi>AB</hi> and <hi>AC</hi> in the poyntes <hi>B</hi> and <hi>C</hi>: draw a right line from <hi>D</hi> to <hi>B,</hi> and an other from <hi>D</hi> to <hi>C.</hi> And forasmuch as the line <hi>DA</hi> is equall to the line <hi>AE,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and the line <hi>AB</hi> is common to the<g ref="char:cmbAbbrStroke">̄</g> both, therefore these two lines <hi>DA</hi> and <hi>AB</hi> are equall to these two lines <hi>AB</hi> and <hi>AE</hi> and the angle <hi>DAB</hi> is equall to the angle <hi>BAE.</hi> Wherefore (by the 4. of the first) the base <hi>DB</hi> is equall to the base <hi>BE.</hi> And forasmuch as these two lines <hi>DB</hi> and <hi>DC</hi> are greater then the line <hi>BC,</hi> of which the line <hi>DB</hi> is proued to be equall to the line <hi>BE.</hi> Wherefore the re<g ref="char:EOLhyphen"/>sidue, namely, the line <hi>DC</hi> is greater then the residue, namely, then the line <hi>EC.</hi> And foras<g ref="char:EOLhyphen"/>much as the line <hi>DA</hi> is equall to the line <hi>AE,</hi> and the line <hi>AC</hi> is common to them both, and the base <hi>DC</hi> is greater then the base <hi>EC,</hi> therefore the angle <hi>DAC</hi> is greater then the angle <hi>EAC.</hi> And it is proued that the angle <hi>DAB</hi> is equall to the angle <hi>BAE</hi>: wher<g ref="char:EOLunhyphen"/>fore the angles <hi>DAB</hi> and <hi>DAC</hi> are greater then the angle <hi>BAC.</hi> If therefore a solide angle be contayned vnder three playne superficiall angles euery two of those three angles, which two so euer be taken are greater then the third: which was required to be proued.</p>
               <p>
                  <hi>In this figure ye may playnely behold the
<figure/>
former demonstration, if ye eleuate the three triangles ABD, A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>C and ACD in such <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>that they may all meete together in the poynt A.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 19. Theoreme. The 21. Proposition. Euery solide angle is comprehended vnder playne angles lesse then fower right angles.</head>
               <p>
                  <hi>SVppose that</hi> A <hi>be a solide angle contayned vnder these superficiall angles</hi> BAC, DAC <hi>and</hi> DAB. <hi>Then I say that the angles</hi> BAC, DAC <hi>and</hi> DAB <hi>are lesse then fower right angles.</hi>
                  <note place="margin">
                     <hi>Construction</hi>
                  </note> 
                  <hi>Take in euery one of these right lines</hi> ACAB <hi>and</hi>
                  <pb facs="tcp:7062:363"/>
AD <hi>a poynt at all aduentures and let the same be</hi> B, C, D. <hi>And draw these right lines</hi> BC, CD <hi>and</hi> DB.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion.</hi>
                  </note> 
                  <hi>And forasmuch as the angle</hi> B <hi>is a solide angle, for it is contayned vnder three superficiall angles, that is, vnder</hi> CBA, ABD <hi>and</hi> CBD, <hi>therefore (by the 20. of the eleuenth) two of them which two so euer be taken are greater then the third. Wherefore the angles</hi> CBA <hi>and</hi> ABD <hi>are greater
<figure/>
then the angle</hi> CBD: <hi>and by the same rea<g ref="char:EOLhyphen"/>son the angles</hi> BCA <hi>and</hi> ACD <hi>are grea<g ref="char:EOLhyphen"/>ter then the angle</hi> BCD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <hi>and moreouer the angles</hi> CDA <hi>and</hi> ADB <hi>are greater then the angle</hi> CDB. <hi>Wherefore these sixe angles</hi> CBA, ABD, BCA, ACD, CDA, <hi>and</hi> ADB <hi>are greater the<g ref="char:cmbAbbrStroke">̄</g> these thre angles, namely,</hi> CBD, BCD, <hi>&amp;</hi> CDB. <hi>But the three angles</hi> CBD, BDC, <hi>and</hi> BCD <hi>are equall to two right angles. Wherefore the sixe angles</hi> CBA, ABD, BCA, ACD, CDA, <hi>and</hi> ADB <hi>are greater the<g ref="char:cmbAbbrStroke">̄</g> two right an<g ref="char:EOLhyphen"/>gles. And forasmuch as in euery one of these triangles</hi> ABC, <hi>and</hi> ABD <hi>and</hi> ACD <hi>three angles are equall two right angles (by the 32. of the first). Wherefore the nine angles of the thre triangles, that is, the angles</hi> CBA, ACB, BAC, ACD, DAC, CDA, ADB, DBA <hi>and</hi> BAD <hi>are equall to sixe right angles. Of which angles the sixe angles</hi> ABC, BCA, ACD, CDA, ADB <hi>and</hi> DBA <hi>are greater then two right angles. Wherefore the angles remayning, namely, the angles</hi> BAC, CAD <hi>and</hi> DAB <hi>which contayne the solide angle are lesse then sower right angles. Wherefore euery solide angle is comprehended vnder playne angles lesse then fower right angles: which was required to be proued.</hi>
               </p>
               <p>If ye will more fully see this demonstration compare it with the figure which I put for the better sight of the demonstration of the proposition next going before. Onely here is not required the draught of the line AE.</p>
               <p>Although this demonstration of <hi>Euclide</hi> be here put for solide angles contayned vnder three super<g ref="char:EOLhyphen"/>ficiall angles, yet after the like maner may you proceede if the solide angle be contayned vnder superfi<g ref="char:EOLhyphen"/>ciall angles how many so euer. As for example if it be contayned vnder fower superficiall angles, if ye follow the former construction, the base will be a quadrangled figure, whose fower angles are equall to fower right angles: but the 8. angles at the bases of the 4. triangles set vpon this quadrangled figure may by the 20. proposition of this booke be proued to be greater then those 4. angles of the quadrangled fi<g ref="char:EOLhyphen"/>gure: As we sawe by the discourse of the former demonstration. Wherefore those 8. angles are greater then fower right angles: but the 12. angles of those fower triangles are equall to 8. right angles. Where<g ref="char:EOLhyphen"/>fore the fower angles remayning at the toppe which make the solide angle are lesse then fower right angles. And obseruing this course ye may proceede infinitely.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Theoreme. The 22. Proposition. If there be three superficiall plaine angles of which two how soeuer they be taken, be greater then the third, and if the right lines also which con<g ref="char:EOLhyphen"/>tayne those angles be equall: then of the lines coupling those equall right lines together, it is possible to make a triangle.</head>
               <p>SVppose that there be thre superficial angles ABC, DEF, and GHK, of which let two, which two soeuer be taken, be greater then the third, that is, let the an<g ref="char:EOLhyphen"/>gles ABC, and DEF be greater then the angle GHK, and let the angles DEF and GHK be greater then the angle ABC: and moreouer let the angles GHK and ABC be greater then the angle DEF. And let the right lines AB, BC, DE, EF GH, and HK be equall the one to the other, and draw a right line from the point A to the
<pb n="334" facs="tcp:7062:363"/>
point C, and an other from the point D to the point F, and moreouer an other from the point G to the point K.<note place="margin">Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</note> Then I say that it is possible of three right lines equall to the lines AC, DF
<figure/>
and GK, to make a triangle, that is, that two of the right lynes AC, DF, and GK, which two soeuer be taken, are greater then the third. Now if the angles ABC, DEF,<note place="margin">The first case.</note> and GHK be equall the one to the other, it is manifest that these right lines AC, DF, and GK being also (by the 4. of the first) equall the one to the other, it is possible of three right lines equall to the lines AC, DF, and GK to make a triangle.<note place="margin">Second case.</note> But if they be not equall, let them be vnequall. And (by the 23. of the first) vnto the right line HK,<note place="margin">Constructi<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>
                  </note> and at the point in it H, make vnto the angle ABC an equall angle KHL. And by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first) to one of the lines AB, BG, DE, EF, GH,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> or HK make the line HL equal, &amp; draw these right lines KL and GL. And forasmuch as these two lines AB and DC, are equall to these two lines KH and HL, and the angle B is equall to the angle KHL, ther<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> (by the 4. of the first) the base AC is equall to the base KL. And forasmuch as the angles ABC, and GHK are greater then the angle DEF, but the angle GHL is equall to the angles ABC, &amp; GHK<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore the angle GHL is greater then the angle DEF. And forasmuch as these two lines GH and HL are equall to these two lines DE and EF, and the angle GHL is grea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er then the angle DEF, therfore (by the 25. of the first) the base GD is greater the<g ref="char:cmbAbbrStroke">̄</g> the base DF. But the lines GK and KL are greater then the line GL. Wherfore then lines GK &amp; KL are much greater then the line DF. But the line KL is equall to the line AC. Wher<g ref="char:EOLhyphen"/>fore the lines AC and GK are greater then the line DF. In like sort also may we pro<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lines AC and DF are greater then the line GK, and that the lines GK and DF are greater then the lyne AC. Wherfore it is possible to make a triangle of three lynes equall to the lines AC, DF, and GK: which was required to be demonstrated.</p>
               <div type="section">
                  <head>An other demonstration.</head>
                  <p>Suppose that the three superficiall angles be ABC, DEF, and GHK, of which angles,<note place="margin">An other de<g ref="char:EOLhyphen"/>monstration.</note> two howsoeuer they be taken, are greater then the third. And let them be contained vnder these equall right lines AB, BC, DE, EF, GH, HK, which equall right lines let th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se lines AC, DF, and GK
<figure/>
ioyne together. Then I say that it is possible of three right lines equal to the lines AC, DF, and GK to make a tri<g ref="char:EOLhyphen"/>angle, whi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h againe is as much to say, as that two of those lines which two soeuer be ta<g ref="char:EOLhyphen"/>ken, are greater then the third. Now againe if the angles B, E, H, be equall, the lines also AC
<pb facs="tcp:7062:364"/>
DF, and GK are equall, and so two of them shall be greater then the third. But if not, l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the angles B, E, H, be vnequall, and let the angle B be greater then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ither of the angles E and H. Therfore (by the 24. of the first) the right line AC is greater then either of the lines DF &amp; GK. And it is manifest that the line AC with either of the lines DF or GK is greater then the third. I say also that the lines DF and GK are greater then the line AC. Vnto the right line AB,<note place="margin">Construction.</note> and to the point in it B, make (by the 22. of the first) vnto the angle GHK an equall angle AB
<figure/>
L, and vnto one of the lines AB, BC, DE, EF, GH or HK, make by the 2. of the first) an equall line BL. And draw a right line from the point A to y<hi rend="sup">e</hi> point L, and an other fro<g ref="char:cmbAbbrStroke">̄</g> the point L to the point C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>n.</note> And forasmuch as these two lines AB &amp; BL are equal to these two lines GH &amp; HK the one to the other, and they containe equal an<g ref="char:EOLhyphen"/>gles: therfore (by the 4. of the first) the base AL is equall to the base GK. And forasmuch as the angles E and H be greater then the angle ABC, of which the angle GHK is equal to the angle ABL, therfore the angle remayning, namely, the angle E is greater then the angle LBC. And forasmuch as these two lines LB and BC, are equall to these two lines DE and EF the one to the other, and the angle DEF is greater then the angle LBC, therfore (by the 25. of the first) the base DF is greater then the base LC: and it is proued that the line GK is e<g ref="char:EOLhyphen"/>quall to the line AL. Wherfore the lines DF &amp; GK are greater then the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s AL &amp; LC. B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es AL and LC are greater then the line AC. Wherfore the lines DF &amp; GK are much greater the<g ref="char:cmbAbbrStroke">̄</g> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> line AC. Wherfore two of these right lines AC, DF &amp; GK which two soeuer be taken are greater then th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> third. Wherfore it is possible of three right lines equall to the lines AC, DF and GK to make a triangle: which was required to be demonstrated.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 3. Probleme. The 23. Proposition. Of three plaine superficiall angles, two of which how soeuer they be taken, are greater then the third, to make a solide angle: Now it is necessary that those three superficiall angles be lesse then fower right angles.</head>
               <p>SVppose that the superficiall angles geuen be ABC, DEF, GHK: of which let two how soeuer they be taken, be greater then the third: and moreouer, let those three angles be l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>sse then fower right angles. It is required of three super<g ref="char:EOLhyphen"/>ficiall angles equall to the angles ABC, DEF, and GHK, to make a solide
<figure/>
                  <pb n="335" facs="tcp:7062:364"/>
or bodily angle. Let the lines AB, BC, DE, EF, GH, and HK, be made equall:<note place="margin">Construction.</note> and drawe a right line from the point A to the point C, &amp; an other from the point D to the point F, and an other from the point G to the point K. Now (by the 22. of the eleuenth) it is pos<g ref="char:EOLhyphen"/>sible of three right lines equall to the right lines AC, DF, and GK, to make a triangle. Make such a triangle, and let the same be LMN,
<figure/>
so that let the line AC be equall to the line LM, and the line DF to the line MN, and the line GK to the line LN. And (by the 5. of the fourth) about the triangle LMN describe a circle LMN, and take (by the 1. of the third) the centre of the same circle,<note place="margin">
                     <p>Three cases in this propo<g ref="char:EOLhyphen"/>sition.</p>
                     <p>The first case.</p>
                  </note> which centre shall either be within the tri<g ref="char:EOLhyphen"/>angle LMN, or in one of the sides therof, or with<g ref="char:EOLhyphen"/>out it. First let it be within the triangle, and let the same be the point X, &amp; drawe these right lines LX, MX, and NX. Now I say, that the line AB is greater then the line LX.<note place="margin">A necessary thing to be proued before he p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oceede any <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>arther in the con<g ref="char:EOLhyphen"/>struction of the Problem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> For if not, then the line AB is either equall to the line LX, or els it is lesse then it. First let it be equall. And forasmuch as the line AB is equall to the line LX, but the line AB is equall to the line BC, therefore the line LX is equall to the line BC: and vnto the line LX the line XM is (by the 15. definition of the first) equall: wherefore these two lines AB and BC, are equall to these two lines LX and XM the one to the other: and the base AC is supposed to be equall to the base LM. Wherefore (by the 8. of the first) the angle ABC is equall to the angle LXM. And by the same reason also the angle DEF is equall to the angle MXN, and moreouer the angle GHK to the angle NXL. Wherefore these three angles ABC, DEF, and GHK, are e<g ref="char:EOLhyphen"/>quall to these three angles LXM, MXN, &amp; NXL. But the three angles LXM, MXN, and NXL, are equall to fower right angles (as it is manifest to see by the 13. of the first, if a<g ref="char:EOLhyphen"/>ny one of these lines MX, LX, or NX, be extended on the side that the point X is). Wher<g ref="char:EOLhyphen"/>fore the three angles ABC, DEF, and GHK, are also equall to fower right angles. But they are supposed to be lesse then fower right angles: which is impossible. Wherefore the line AB is not equall to the line LX. I say also that the line AB is not lesse then the line LX. For if it be possible, let it be lesse: and (by the 2. of the first) vnto the line AB put an equall line XO: and to the line BC put an equall line XP, and draw a right line from the point O to the point P. And forasmuch as the line AB is equall to the line BC, therefore also the line XO is equall to the line XP. Wherefore the residue OL is equall to the residue MP. Wher<g ref="char:EOLhyphen"/>fore (by the 2. of the sixt) the line LM is a parallel to the line OP<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the triangle LMX is equiangle to the triangle OPX. Wherefore as the line XL is to the line LM, so is the line XO to the line OP. Wherefore alternately (by the 16. of the fift) as the line LX is to the line XO, so is the line LM to the line OP. But the line LX is greater then the line XO. Wher<g ref="char:EOLhyphen"/>fore also the line LM is greater then the line OP. But the line LM is put to be equall to the line AC: wherefore also the line AC is greater then the line OP. Now forasmuch as these two right lines AB and BC are equall to these two right lines OX and XP, and the base AC is greater then the base OP, therefore (by the 25. of the first) the angle ABC is greater then the angle OXP. In like sort also may we proue, that the angle DEF is greater then the an<g ref="char:EOLhyphen"/>gle MXN, and that the angle GHK is greater then the angle NXL. Wherefore the three angles ABC, DEF, and GHK, are greater then the three angles LXM, MXN, and NXL. But the angles ABC, DEF, and GHK, are supposed to be lesse then fower right angles: wherefore much more are the angles LXM, MXN, &amp; NXL lesse then fower right angles. But they are also equall to fower right angles: which is impossible. Wherefore the line AB is not lesse then the line LX: and it is also proued that it is not equall vnto it. Wherfore
<pb facs="tcp:7062:365"/>
the line AB is greater then the line LX.</p>
               <p>Now from the point X raise vp vnto the plaine superficies of the circle LMN a perpendi<g ref="char:EOLhyphen"/>cular line XR (by the 12. of the eleuenth). And vnto that which the square of the line AB excedeth the square of the line XL <note n="*" place="margin">
                     <hi>Which how to finde out is taught at the end of this de<g ref="char:EOLunhyphen"/>monstration, and also was taught in the as<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>umpt put before the 14. proposition of the te<g ref="char:cmbAbbrStroke">̄</g>th boke.</hi>
                  </note> let the
<figure/>
square of the line XR be equall. And draw a right line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rom the point R to the point L, and an other from the point R to the point M, and an other from the point R to the point N. And forasmuch as the line RX is erected perpendicularly to the plaine su<g ref="char:EOLhyphen"/>perficies of the circle LMN, therefore (by conuer<g ref="char:EOLhyphen"/>sion of the second definition of the eleuenth) the line RX is erected perpendicularly to euery one of these lines LX, MX, and NX.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the first case.</note> And forasmuch as the line LX is equall to the line XM, &amp; the line XR i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> common to them both, and is also erected per<g ref="char:EOLhyphen"/>pendicularly to them both, therefore (by the 4. of the first) the base RL is equall to the base RM. And by the same reason also the line RN is equall to either of these lines RL and RM. Where<g ref="char:EOLhyphen"/>fore these three lines RL, RM, and RN, are equall the one to the other. And forasmuch as vnto that which the square of the line AB excedeth the square of the line LX, the square of the line RX is supposed to be equall, therefore the square of the line AB is equall to the squares of the lines LX and RX. But vnto the squares of the lines LX and XR, the square of the line LR is (by the 47. of the first) equall, for the angle LXR is a right angle. Where<g ref="char:EOLhyphen"/>fore the square of the line AB is equall to the square of the line RL. Wherefore also the line AB is equall to the line RL. But vnto the line AB is equall euery one of these lines BC, DE, EF, GH, and HK, and vnto the line RL is equall either of these lines RM and RN. Wher<g ref="char:EOLhyphen"/>fore euery one of these lines AB, BC, DE, EF, GH, and HK, is equall to euery one of these lines RL, RM, and RN. And forasmuch as these two lines RL and RM are equall to these two lines AB and BC, and the base LM is supposed to be equall to the base AC, ther<g ref="char:EOLhyphen"/>fore (by the 8. of the first) the angle LRM is equall to the angle ABC. And by the same reason also the MRN is equall to the angle DEF, and the angle LRN to the angle GHK. Wherefore of three superficiall angles LRM, MRN, and LRN, which are e<g ref="char:EOLhyphen"/>quall to three superficiall angles geue<g ref="char:cmbAbbrStroke">̄</g>, namely, to the angles ABC, DEF, &amp; GHK, is made a solide angle R, comprehended vnder the superficiall angles LRM, MRN, and LRN: which was required to be done.</p>
               <p>
                  <note place="margin">Second case.</note>But now let the centre of the circle be in one of the sides of the triangle, let it be in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> side MN, and let the centre be X. And draw a right line from the point L to the point X. I say againe, that the line AB is greater then the line LX. For if not, then AB is either e<g ref="char:EOLhyphen"/>quall to
<figure/>
LX, or els it is lesse then it. First let it be equall. Now the<g ref="char:cmbAbbrStroke">̄</g> these two lines AB and
<pb n="336" facs="tcp:7062:365"/>
BC, that is, DE &amp; EF are equall to these two lines
<figure/>
MX and XL, that is, to the line MN. But the line MN is supposed to be equall to the line DF. Where<g ref="char:EOLhyphen"/>fore also the lines DE and EF are equall to the line DF: which (by the 20. of the first) is impossible. Wher<g ref="char:EOLhyphen"/>fore the line AB is not equall to the line LX. In like sort also may we proue, that it is not lesse. Wherefore the line AB is greater then the line LX. And now if as before vnto the plaine superficies of the circle be e<g ref="char:EOLhyphen"/>rected fro<g ref="char:cmbAbbrStroke">̄</g> the point X a perpendicular line RX whose square let be equall vnto that which the square of the line AB excedeth the square of the line LX, and if the rest of the construction and demonstration be ob<g ref="char:EOLhyphen"/>serued in this that was in the forme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> case, then shall the Probleme be finished.</p>
               <p>But now let the centre of the circle be without the triangle LMN, and let it be in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> point X. And draw these right lines LX, MX, and NX.<note place="margin">Third case.</note> I say that in this case also the line AB is greater then the line LX. For if not, then is it either equall or lesse. First let it be e<g ref="char:EOLhyphen"/>quall.
<figure/>
Wherefore these two lines AB and BC are
<figure/>
equall to these two lines MX and XL the one to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> other, and the base AC is equall to the base ML. Wherefore (by the 8. of the first) the angle ABC is equall to the angle MXL: and by the same reason also the angle GHK is equall to the angle LXN. Wherefore the whole angle MXN is equall to these two angles ABC and GHK. But the angles ABC and GHK are greater then the angle DEF. Wherefore the angle MXN is greater then the an<g ref="char:EOLhyphen"/>gle DEF. And forasmuch as these two lines DE and EF are equall to these two lines MX and XN, and the base DF is equall to the base MN, there<g ref="char:EOLhyphen"/>fore (by the 8. of the first) the angle MXN is equall to the angle DEF. And it is proued that it is also greater: which is impossible. Wherefore the line AB is not equall to the line LX. In like sort also may we proue, that it is not lesse. Wherfore the line AB is greater then the line LX. And againe if vnto the plaine superficies of the circle be erected perpendicularly from the point X a line XR, whose square is equall to that which the square of the line AB exceedeth the square of the line LX, and the rest of the construction be done in this that was in the for<g ref="char:EOLhyphen"/>mer cases, then shall the Probleme be finished.</p>
               <p>I say moreouer, that the line AB is not lesse then the line LX.<note place="margin">An other demo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>stratio<g ref="char:cmbAbbrStroke">̄</g> to proue that th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> line AB is not lesse the<g ref="char:cmbAbbrStroke">̄</g> the line LX.</note> For if it be possibl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, let it be lesse. And vnto the line AB, put (by the 2. of the first) the line XO equall: and vnto the line BC put the line XP equall: And draw a right line from the point O to the point P.
<pb facs="tcp:7062:366"/>
And forasmuch as the line AB is equall to the line BC, therefore the line XO is equall to the line XP. Wherefore the residue OL is equall to the residue MP. Wherefore (by the 2. of the sixt) the line LM, is a parallel to the line PO. And the triangle LXM is equiangle to the triangle PXO. Wherefore (by the 6. of the sixt) as
<figure/>
the line LX is to the line LM, so is the line XO to the line PO. Wherefore alternately (by the 16. of the fift) as the line LX is to the line XO, so is the line LM to the line OP. But the line LX is greater then the line XO. Wherefore also the line LM is greater then the line OP. But the line LM is equall to the line AC. Wherefore also the line AC is greater then the line OP. Now forasmuch as these two lines AB and BC are equall to these two lines OX &amp; XP the one to the other, and the base AC is greater then the base OP, therefore (by the 25. of the first) the angle ABC is greater then the angle OXP. And in like sort if we put the line XR equall to either of these lines XO or XP, and draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point O to the point R, we may proue that the angle GHK is greater then the angle OXR. Vnto the right line LX, and vn<g ref="char:EOLhyphen"/>to the point in it X, make (by the 23. of the first) vnto the angle ABC an equall angle LXS: and vnto the angle GHK make an equall angle LXT. And (by the second of the first) let either of these lines SX, and XT, be equall to the line OX. And drawe these lines OS, OT, and ST. And forasmuch as these two lines AB and BC are e<g ref="char:EOLhyphen"/>quall to these two lines OX and XS, and the angle ABC is equall to the angle OXS, there<g ref="char:EOLhyphen"/>fore (by the 4. of the first) the base AC, that is, LM, is equall to the base OS. And by the same reason also the line LN is equall to the line OT. And forasmuch as these two lines LM and LN are equall to these two lines SO and OT, and the angle MLN is greater then the angle SOT, therefore (by the 25. of the first) the base MN is greater then the base ST. But the line MN is equall to the line DF. Wherefore the line DF is greater then the line ST. Now forasmuch as these two lines DE and EF are equall to these two lines SX and XT, and the base DF is greater then the base ST, therefore (by the 25. of the first) the angle DEF is greater then the angle SXT. But the angle SXT is equall to the angles ABC and GHK. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore also the angle DEF is greater then the angles ABC and GHK. But it is also lesse: which is impossible.</p>
               <p>
                  <note place="margin">This was before ta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ght in the tenth booke in the assumpt put before the <hi>14.</hi> proposition.</note>But now let vs declare how to finde out the line <hi>XR,</hi> whose square shall be equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o that which the square of the line <hi>AB</hi> exceedeth the square of the line <hi>LX.</hi> Take the two right lines <hi>AB</hi> and <hi>LX,</hi> and let <hi>AB</hi> be the greater, and vpon <hi>AB</hi>
                  <figure/>
describe a semicircle <hi>ACB,</hi> and from the poynt <hi>A</hi> apply into the semicircle a right line <hi>AC</hi> equall to the right line <hi>LX</hi>: and draw a right line from the poynt <hi>C</hi> to the poynt <hi>B.</hi> And forasmuch as in the semicircle <hi>ACB</hi> is an angle <hi>ACB,</hi> therefore (by the 31. of the third) the angle <hi>ACB</hi> is a right angle. Wherefore (by the 47. of the first) the square of the line <hi>AB</hi> is equall to the squares of the lines <hi>AC</hi> and <hi>CB</hi>: wherefore the square of the line <hi>AB</hi> is in power more then the square of the line <hi>AC</hi> by the square of the line <hi>CB</hi>: but the line <hi>AC</hi> is equall to the line <hi>LX,</hi> wherefore the square of the line <hi>AB</hi> is in power more then the square of the line <hi>LX</hi> by the square of the line <hi>CB.</hi> If therefore vnto the line <hi>CB</hi> we make the line <hi>XR</hi> equall, then is the square of the line <hi>AB</hi> greater then the square of the line <hi>LX</hi> by the square of the line <hi>XR</hi>: which was required to be doone.</p>
               <p>
                  <pb n="337" facs="tcp:7062:366"/>
                  <hi>In this figure may ye more fully s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> the
<figure/>
construction and demonstration of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rst case of the former 23. Propositio<g ref="char:cmbAbbrStroke">̄</g>, if ye erect perpendicularly the triangle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> RN, and vn<g ref="char:EOLhyphen"/>to it bend the triangle LMR, that the an<g ref="char:EOLhyphen"/>gles R of eche may ioyne together in the point R. And so fully vnderstanding this case, the other cases will not be hard to conceaue.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 21. Theoreme. The 24. Proposition. If a solide or body be contayned vnder<note n="*" place="margin">
                     <p>
                        <hi>M. Dee (to auoide cauillation) addeth to Euclides proposition this worde sixe: whome I haue follow<g ref="char:EOLhyphen"/>ed accordingly, and not Zamberts, in this.</hi>
                     </p>
                     <p>
                        <hi>This kinde of body mencioned in the proposition is called a Paral<g ref="char:EOLhyphen"/>lelipipedo<g ref="char:cmbAbbrStroke">̄</g> according to the di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>fini<g ref="char:EOLhyphen"/>tion before geuen thereof.</hi>
                     </p>
                  </note> sixe parallel playne superficieces, the opposite plaine superficieces of the same bo<g ref="char:EOLhyphen"/>dy are equall and parallelogrammes.</head>
               <p>SVppose that this solide body CDHG be contained vnder these 6. parallel plaine superficieces, namely, AC, GF, BG, CE, FB, and AE. Then I say that the opposite superficieces of the same body, are equal and parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes, it is to wete, the two opposites AC and GF, and the two opposites BG
<figure/>
and CE, and the two opposites FB and AE to be equall, and al to be parallelogrammes.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion that the opposite sides are parallelo<g ref="char:EOLhyphen"/>grammes.</note> For forasmuch as two pa<g ref="char:EOLhyphen"/>rallel plaine superficieces, that is, BG, and CE are deui<g ref="char:EOLhyphen"/>ded by the plaine superficies AC, their common sections are (by the 16. of the eleuenth) parallels. Wherfore the line AB is a parallel to the line CD. Again<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> forasmuch as two parallel plaine superficieces FB and AE are deui<g ref="char:EOLhyphen"/>ded by the plaine superficies AC their common sections are by the same proposition, parallels. Wherfore the lyne AD is a parallel to the line BC. And it is also proued, that the line AB is a parallel to the line DC. Wherfore the superficies AC is a parallelogramme. In like sort also may we proue, that euery one of these superficices CE, GF, BG, FB, and AE are parallelo<g ref="char:EOLhyphen"/>grammes. Draw a right line from the point A, to the point H, and an other from the point D to the point F.<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> that the oppo<g ref="char:EOLhyphen"/>site superficies are equall.</note> Aud forasmuch as the line AB is proued a parallel to the line CD, and the lyne BH to the line CF, therfore these two right lines AB and BH touching the one the other, are parallels to these two right lines DC and CF touching also the one the other, and not being in one and the selfe same plaine superficies. Wherfore (by the 10. of the eleuenth) they compre<g ref="char:EOLhyphen"/>hend equall angles. Wherfore the angle ABH is equall to the angle DCF. And forasmuch as these two lines AB and BH are<note n="*" place="margin">
                     <hi>AB is equall to DC, because the superficies AC is proued a pa<g ref="char:EOLhyphen"/>rallelogra<g ref="char:cmbAbbrStroke">̄</g>me, and by the same rea<g ref="char:EOLhyphen"/>son, is BH equall to CF, because the superficies FB is proued a pa<g ref="char:EOLhyphen"/>rallelogramme: therefore the 34. of the first is our proofe.</hi>
                  </note> equall to these two lines DC and CF, and the angle ABH is proued equall to the angle DCF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore (by the 4. of the first) the base AH is equall to the base DF, and the triangle ABH is e<g ref="char:EOLhyphen"/>quall to the triangle DCF. And forasmuch as (by the 41. of the first) the parallelogramme BG is double to the triangle ABH, and the parallelo<g ref="char:EOLhyphen"/>gramme CE is also double to the triangle DCF, therfore the parallelo<g ref="char:EOLhyphen"/>gramme
<pb facs="tcp:7062:367"/>
BG is equall to the parallelogramme CE. In like sort
<figure/>
also may we proue that the parallelogramme AC is equall to the parallelogramme GF, and the parallelograme AE to the parallelogramme FB. If therfore a solide or body be contained vnder sixe parallel plaine superficieces, the opposite plaine super<g ref="char:EOLunhyphen"/>ficieces of the same body are equal &amp; parallelogrammes which was required to be demonstrated.</p>
               <p>
                  <hi>I haue for the better helpe of young beginners, described here an other figure whose forme if it be described vpon pa<g ref="char:EOLhyphen"/>sted paper with the letters placed in the same order that it is here, and then if ye cut finely these lines AG, DE and CF not through the paper, and folde it accordingly, compare it with the demonstration, and it will shake of all hardenes from it.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 22. Theoreme. The 25. Proposition. If a Parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> be cutte of a playne superfi<g ref="char:EOLhyphen"/>cies beyng a parallel to the two opposite playne superficieces of the same body: then, as the base is to the base, so is the one solide to the other solide.</head>
               <p>SVppose that this solide ABCD being contained
<figure/>
vnder parallel plaine superficieces (and therfore called a parallelipipedo<g ref="char:cmbAbbrStroke">̄</g>) be cut of the plaine super<g ref="char:EOLhyphen"/>ficies VE, being a parallel to the two opposite superficieces of the same body, namely, to the superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ces AR &amp; DH. The<g ref="char:cmbAbbrStroke">̄</g> I say that as the base AEFW is to the base EHCF, so is the solide ABFV to the solide EGGD.<note place="margin">Construction.</note> Exte<g ref="char:cmbAbbrStroke">̄</g>d the line AH on either side, &amp; put vnto y<hi rend="sup">e</hi> line EH as many equal lines as you wil, namely, HM, &amp; MN: &amp; likewise vnto the line AE, put as many equal lines as you will, namely, AK &amp; KL, &amp; make perfect these parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes LO, KW, HZ, &amp; MS, and likewise make perfect these solides or bodies LP, KR, DM, and MT.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as these right lines, LK, KA, and AE are equall the one to the other, therfore these parallelogrammes LO, KW, and AF, are also (by the first of the sixt) equall the one to the other: and so also (by the same) are these parallelogrammes KX, KB, and AG equall the one to the other. And likewise (by the 24. of the eleuenth) are the parallelogrammes LY, KP, and AR, e<g ref="char:EOLhyphen"/>qual, for they are opposite the one to the other. And by the same reason also the parallelogrammes EC, HZ, and MS are equall the one to the other, and the parallelogrammes HG, HI, and IN are equal the one to the other. And more<g ref="char:EOLunhyphen"/>uer the parallelogrammes DH, MQ, and NT are (by the 24. of the eleuenth) equall the one to the other, for they
<pb n="338" facs="tcp:7062:367"/>
are opposite: wherefore three plaine superficies of the solides LP, RK, and AV are equall to three plaine superficies: but vnto eche of these three superficieces are equall the three opposite superficieces (by the 24. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) Wherfore these three solides or bodies LP, KR, and AV, are equal the one to the other, by the 8. definition of the eleue<g ref="char:cmbAbbrStroke">̄</g>th. And by the same reason also the three solides, ED, DM &amp; MT are equal the one to the other. Wherfore how multiple x the base LF is, to the base AF, so multiplex is the solide LV to the solide AV. And by the same reason also how multiplex the base NF is to the base FH, so multiplex is the solide NV to the solide HV: so that if the base LF be equall to the base NF, the solide also LV shall be equall to the solide VN, and if the base LF exceede the base NF, the solide also LV shall exceede the solide VN, and if the base LF be lesse then the base NF, the solide also LV shall be lesse then the solide VN (by the 1. and 14. of the fift.) Now then there are foure magnitudes, namely, the two bases AF and FH, and the two solides or bodies AV and VH, of which are take<g ref="char:cmbAbbrStroke">̄</g> their equemultiplices, namely, the equemultiplices of the base AF, &amp; of the solide AV, or the base LF, &amp; the solide LV, &amp; the equemultiplices of the base HF, &amp; of the solide HV are the base NF, and the solide NV. And it is proued that if the base LF excede the base NF, the solide also LV excedeth the solide NV, &amp; if it be equal, it is equal, and if it be lesse, it is lesse. Wherfore (by the 6. definitio<g ref="char:cmbAbbrStroke">̄</g> of the fift) as the base AF is to the base FH, so is the solide AV to the solide HV. If therfore a parallelipipedon be cut of a playne su<g ref="char:EOLhyphen"/>perficies being a parallel to the two opposite playne superficieces of the same body, then, as the base is to the base, so is the one solide to the other solide: which was required to be proued.</p>
               <p>
                  <hi>I haue for the better
<figure/>
sight of the co<g ref="char:cmbAbbrStroke">̄</g>structio<g ref="char:cmbAbbrStroke">̄</g> &amp; demo<g ref="char:cmbAbbrStroke">̄</g>stration of the for<g ref="char:EOLhyphen"/>mer 25. propositio<g ref="char:cmbAbbrStroke">̄</g>, here set another figure, whose forme if ye describe vp<g ref="char:EOLhyphen"/>pon pasted paper, and finely cut the three lines XI, BS, and TY, not through the paper but halfe way, and then fold it accordingly, and compare it with the construction and demo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>stration, you shall see that it will geue great light therunto.</hi>
               </p>
               <p>
                  <hi>Here</hi> Flussas <hi>addeth three Corollaries.</hi>
               </p>
               <div type="section">
                  <head>First Corollary.</head>
                  <p>
                     <hi>If a Prisme be cutte of a playne superficies parallel to the opposite superficieces,</hi>
                     <note place="margin">
                        <hi>F<gap reason="illegible" resp="#PDCC" extent="4 letters">
                              <desc>••••</desc>
                           </gap> Corolla<g ref="char:EOLhyphen"/>ry.</hi>
                     </note> 
                     <hi>the se<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ions of the Prisme shall be the one to the other in that proportion, that the sections of the base are the one to the other.</hi>
                  </p>
                  <p>
                     <pb facs="tcp:7062:368"/>For the sections of the bases, which bases (by the 11. definitio<g ref="char:cmbAbbrStroke">̄</g> of this booke) are parallelogrammes, shall likewise be parallelogrammes, by the 16. of this booke (when as the superficies which cutteth is parallelel to the opposite super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icieces) and shal also be equiangle. Wherfore if vnto the bases (by produ<g ref="char:EOLhyphen"/>cing the right lines) be added like and equall bases, as was before shewed in a parallelipipedon, of those sections shalbe made as many like Prismes as ye will. And so by the same reason, namely, by the commo<g ref="char:cmbAbbrStroke">̄</g> excesse, equalitie, or want of the multiplices of the bases &amp; of the sections by the 5. definitio<g ref="char:cmbAbbrStroke">̄</g> of the fifth may be proued, that euery section of the Prisme multiplyed by any multiplycation whatsoeuer, shall haue to any other section that proportion that the sections of the bases haue.</p>
               </div>
               <div type="section">
                  <head>Second Corollary.</head>
                  <p>
                     <hi>Solides whose two opposite superficie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>es are poligonon figures like equall and parallels, the other su<g ref="char:EOLhyphen"/>perficies, which of necessitie are parallelogrammes,</hi>
                     <note place="margin">
                        <p>
                           <hi>Second Corol<g ref="char:EOLhyphen"/>ry.</hi>
                        </p>
                        <p>
                           <hi>These solides which he speaketh of in this Corollary are of some called sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>d column<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>s.</hi>
                        </p>
                     </note> 
                     <hi>being cutte of a playne superficies parallel to the two opposite superficies: the sections of the base are the one to the other, as the sections of the solide are th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> one to the other.</hi>
                  </p>
                  <p>Which thing is manifest, for such solides are deuided into Prismes, which haue one co<g ref="char:cmbAbbrStroke">̄</g>mon side, namely, the axe or right line, which is drawne by the centers of the opposite bases. Wherefore how many pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>allelogra<g ref="char:cmbAbbrStroke">̄</g>mes or bases are set vpon the opposite poligonon figures, of so many Prismes shal the whole solide be co<g ref="char:cmbAbbrStroke">̄</g>posed. For those poligonon figures are deuided into so many like triangles by the 20. of the sixth, which describe Prismes. Which Prismes being cut by a superficies parallel to the opposite superficieces, the sectio<g ref="char:cmbAbbrStroke">̄</g>s of the bases shal (by the former Corollary) be proportional with the sectio<g ref="char:cmbAbbrStroke">̄</g>s of the Prismes. Wherefore by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of the fifth, as the sections of the one are the one to the other, so are the sections of the whole the one to the other.</p>
                  <p>Of these solides there are infinite kindes, according to the varietie of the opposite and parallel po<g ref="char:EOLhyphen"/>ligonon figures, which poligonon figures doo alter the angles of the parallelogrammes set vpon them according to the diuersitie o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> their situation. But this is no let at all to this corollary, for that which we haue proued will alwayes follow. When as the superficieces which are set vpo<g ref="char:cmbAbbrStroke">̄</g> the opposite like &amp; equal poligonon and parallel superficieces are alwayes parallelogrammes.</p>
               </div>
               <div type="section">
                  <head>Third Corollary.</head>
                  <p>
                     <note place="margin">
                        <hi>Third Corol<g ref="char:EOLhyphen"/>lary.</hi>
                     </note>
                     <hi>T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e foresayd solides <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>omposed of Prismes, being cutte by a playne superficies parallel to the opposit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> superficieces, are the one to the other as the heades or higher parts cutte are.</hi>
                  </p>
                  <p>For it is proued that they are the one to the other as the bases are. Which bases forasmuch as they are par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>llelogrammes, are the one to the other as the right lines are vpon which they are set by, the first of the sixth, which right lines are the heddes or higher parts of the Prismes.</p>
               </div>
            </div>
            <div type="part">
               <head>The 4. Probleme. The 26. Proposition. Vpon a right lyne geuen, and at a point in it geuen, to make a solide angle equall to a solide angle geuen.</head>
               <p>SVppose that the right line geuen be AB, and let the point in it geuen be A, and let the solide or bodily angle geuen be D being contained vnder these superfici<g ref="char:EOLhyphen"/>all angles EDC, EDF and FDC. It is required vpon the right line AB, &amp; at the point in it geuen A to make a solide angle equall to the solide angle D. Take in the line DF a point at all aduentures, and let the same be F.<note place="margin">Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>struc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> And (by the 11. of the eleuenth) fro<g ref="char:cmbAbbrStroke">̄</g> the point F. Draw vnto the superficies wherin are the lines ED &amp; DC a per<g ref="char:EOLhyphen"/>pendicular line FG, and let it fall vpon the plaine superficies in the point G, &amp; draw a right line from the point D to the point G. And (by the 23. of the first) vnto the line AB, and at the point A make vnto the angle EDC an equall angle BAL, and vnto the angle EDG put the angle BAK equall: and by the 2. of the first, put the line AK equall to the line DG, and (by the 12. of the eleuenth) from the point K raise vp vnto the plaine superficies BAL a perpendicular line KH, and put the line KH equall to the line GF, and draw a right lyne
<pb n="339" facs="tcp:7062:368"/>
from the point H to the point A. Now I say that the so<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ide angle A contained vnder the su<g ref="char:EOLhyphen"/>perficiall angles BAL, BAH, and HAL is equall to the solide angle D contained vnder the superficiall angles EDCEDF, and FD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es AB and DE be put e<g ref="char:EOLhyphen"/>quall, and draw these right lines. HB, BK, FE, and EG. And forasmuch as the line FG is erected perpendicularly to the ground superficies,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> therfore by the 2. definition of the eleuenth, the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> FG is also erected perpendicularly to all the right lines that are in the ground super<g ref="char:EOLhyphen"/>ficies and touche it. Wherfore either of these angles FGD and FGE is a right angle, and by the same reason also either of the angles HKA and HKB is a right angle. And foras<g ref="char:EOLhyphen"/>much as these two lines KA &amp; AB are equall to these two lines GD &amp; DE, the one to the other, and they containe equall
<figure/>
angles (by construction). Wher<g ref="char:EOLunhyphen"/>fore (by the 4. of the first) the base KB is equall to the base EG, and the line KH is equall to the line GF, and they co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>d right angles. Wherfore the line BH is equall to the line FE. Agayne, forasmuch as these two lines AK and KH are equal to these two lines DG and GF, and they containe right angles. Wherfore y<hi rend="sup">e</hi> base AH is (by the 4. of the first) equall to the base DF. And the line AB is equall to the line DE. Wherfore these two lines AB and AH are equall to these two lines FD and DE, and the base BH is equall to the base FE. Wherfore (by the 8. of the first) the angle BAH is equall to the angle EDF. And by the same reason also the angle HKL is equall to the angle FGC. Wherfore if we put these lines AL and DC equall, and draw these right lines KL, HL, GC, and FC: for<g ref="char:EOLhyphen"/>asmuch as the whole angle BAL is equall to the whole angle EDC, of which the angle BAK is supposed to be equall to the angle EDG, therfore the angle remayning, namely, KAL is equall to the angle remayning GDC. And forasmuch as these two lines KA and AL are equall to these two lines GD and DC, and they containe equall angles, therefore by the 4. of the first, the base KL is equall to the base GC, and the line KH is equall to the line GF, wherfore thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wo lines LK and KH are equall to these two lines CG and GF, and they co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>taine right angles. Wherfore the base HL is (by the 4. of the first) equal to the base FC. And forasmuch as these two lines HA and AL are equall to these two FD and DC, and the base HL is equall to the base FC, therfore (by the 8. of the first) the angle HAL is equall to the angle FDC, and by construction, the angle BAL is equall to the angle EDC. Wherefore vnto the right line geuen, and at the point in it geuen, namely, A, is made a solide angle equal to the solide angle geuen D: which was required to be done.</p>
               <p>
                  <hi>In thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> two <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                     <figure/>
here put, you may in <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> clearely concerne the <gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>mer construction and d<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>monstratio<g ref="char:cmbAbbrStroke">̄</g>, if ye erect pe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>pendicularly vnto the ground superficies the tri<g ref="char:EOLhyphen"/>angles ALB and DCE, &amp; eleuate the triangles ALH and DCF that the lynes
<pb facs="tcp:7062:369"/>
LA and CD of them may exactly agree with the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> LA and CD of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>riangles erec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For so or<g ref="char:EOLhyphen"/>dering them, if ye compare the former construction and demonstration with them, they will be playn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> vnto you.</hi>
               </p>
               <p>
                  <hi>Although Euclides demo<g ref="char:cmbAbbrStroke">̄</g>stration be touching solide angles which are contained only vnder three superficiall angles, that is, whose bases are triangles: yet by it may ye performe the Probleme touching solide angles contained vnder superficiall angles how many soeuer, that is, hauing to their bases any kinde of Poligonon figures. For euery Poligonon figure may by the 20. of the sixt, be resolued into like tringles. And so also shall the solide angle be deuided into so many solide angles as there be triangles in the base. Vnto euery one of which solide angles you may by this proposition describe <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> equall solide angle vpon a line geuen, and at a point in it geuen. And so at the length the whole solide angle after this maner described shall be equall to the solide angle geuen.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 5. Theoreme. The 27. Proposition. Vpon a right line geuen to describe a parallelipipedon like and in like sort si<g ref="char:EOLhyphen"/>tuate to a parallelipipedon geuen.</head>
               <p>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>Vppose that the right line geuen be <hi>AB,</hi> and let the parallelipipedon geuen be <hi>CD.</hi> It is required vpon the right line geuen AB to describe a parallelipipe<g ref="char:EOLhyphen"/>don like and in like sort situate to the parallelipipedon geuen, namely, to <hi>CD.</hi>
                  <note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ction.</note> Vnto the right line <hi>AB</hi> and at the poynt in it <hi>A</hi> describe (by the 26. of the ele<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth
<figure/>
(a solide angle equall to the solide angle <hi>C,</hi> and let it be contayned vnder these superfi<g ref="char:EOLhyphen"/>ciall angles <hi>BAH, HAK,</hi> and <hi>KAB,</hi> so that let the angle <hi>BAH</hi> be equall to the angle <hi>ECF,</hi> and the angle <hi>BAK</hi> to the angle <hi>ECG,</hi> and moreouer the angle <hi>KAH</hi> to the an<g ref="char:EOLhyphen"/>gle <hi>GCF.</hi> And as the line <hi>EC</hi> is to the line <hi>CG,</hi> so let the <hi>AB</hi> be to the line <hi>AK</hi> (by the 12. of the sixth) and as the line <hi>GC</hi> is to the line <hi>CF,</hi> so let the <hi>KA</hi> be <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the line <hi>AH.</hi> Wherefore of equalitie (by the 22. of the fift) as the line <hi>EC</hi> is to the line <hi>CF,</hi> so is the line <hi>BA</hi> to the line <hi>AH.</hi>
                  <note place="margin">Demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>tion.</note> Make perfect the parallelogramme <hi>BH,</hi> and also the solide <hi>AL.</hi> Now for that as the line <hi>EC</hi> is to the line <hi>CG,</hi> so is the line <hi>BA</hi> to the line <hi>AK,</hi> therefore the sides which contayne the equall angles, namely, the angles <hi>ECG</hi> and <hi>BAK</hi> are proportio<g ref="char:EOLhyphen"/>nall: wherefore (by the first definition of the sixth) the parallelogramme <hi>GE</hi> is like to the pa<g ref="char:EOLhyphen"/>rallelogramme <hi>KB.</hi> And by the same reason the parallelogramme <hi>KH</hi> is like to the paralle<g ref="char:EOLhyphen"/>logramme <hi>GF,</hi> and moreouer the parallelogramme <hi>FE</hi> to the parallelogramme <hi>HB</hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wher<g ref="char:EOLhyphen"/>fore there are three parallelogrammes of the solide <hi>CD</hi> like to the three parallelogrammes of the solide <hi>AL.</hi> But the three other sides in eche of these solides are equall and like to their opposite sides. Wherefore the whole solide <hi>CD</hi> is like to the whole solide <hi>AL.</hi> Wherfore vpon the right line geuen <hi>AB</hi> is described the solide <hi>AL</hi> contayned vnder parallel playne superfi<g ref="char:EOLhyphen"/>cieces like and in like sort situate to the solide geue<g ref="char:cmbAbbrStroke">̄</g> 
                  <hi>CD</hi> contayned also vnder parall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>l playne superficieces: which was required to be doone.</p>
               <p>
                  <hi>This demonstration is not hard to conceaue by the former figure as it is described in a playne, if ye
<pb n="340" facs="tcp:7062:369"/>
that imagination of parallelipipedons described in a playne which we before taught in the diffinition of a cube. Howbeit I haue here for the more ease of such as are not yet acquainted with solides, de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>cribed
<figure/>
two figures, whose formes first describe vpon pasted paper with the like letters noted in them, and then finely cutte the three midle lines of eche figure, and so fold them accordingly, and that doone compare them with the construction and demonstration of this 27. proposition, and they will be very easy to conceaue.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 23. Theoreme. The 28. Proposition. If a parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> be cutte by a plaine superficies drawne by the diagonall lines of those playne superficieces which are opposite: that solide is by this playne superficies cutte into two equall partes.</head>
               <p>SVppose that the parallelipipedon
<figure/>
                  <hi>AB</hi> be cutte by the playne super<g ref="char:EOLhyphen"/>ficies <hi>CDEF</hi> drawne by the dia<g ref="char:EOLhyphen"/>gonal lines of y<hi rend="sup">e</hi> plaine superficieces which are opposite, namely, of the superficieces <hi>CF</hi> and <hi>DE.</hi> Then I say that the parallelipi<g ref="char:EOLhyphen"/>pedon <hi>AB</hi> is cutte into two equall partes by the superficies <hi>CDEF.</hi> For forasmuch as (by the 34. of the first) the triangle <hi>CGF</hi> is equall to the triangle <hi>CBF,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and the triangle <hi>ADE</hi> to the triangle <hi>DEH,</hi> and the parallelograme <hi>CA</hi> is equall to the parallelogramme <hi>BE,</hi> for they are opposite, and the parallelogramme <hi>GE</hi> is also equall to the parallelogramme <hi>CH,</hi> and the parallelogramme <hi>CE,</hi> is the common section by supposition: Wherfore the prisme contai<g ref="char:EOLhyphen"/>ned vnder the two triangles <hi>CGF,</hi> and <hi>DAE,</hi> and vnder the three parallelogrammes <hi>GE, AC,</hi> and <hi>CE</hi> is (by the 8. definition of the eleuenth) equall to the prisme contayned vnder the two triangles <hi>CFB</hi> and <hi>DEH</hi> and vnder the three parallelogrammes <hi>CH, BE,</hi> and <hi>CE.</hi> For they are co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder playne superficieces equall both in multitude and in magnitude. Wherefore the whole parallelipipedon <hi>AB</hi> is cutte into two equall partes by the playne superficies <hi>CDEF</hi>: which was required to be demonstrated.</p>
               <p>
                  <pb facs="tcp:7062:370"/>A diagonall line is a right line which in angular figures is drawne from one angle and extended to his contrary angle as you see in the figure AB.</p>
               <p>
                  <hi>Describe for the better sight of this demonstration a figure vpon pasted
<figure/>
paper like vnto that which you described for the 24. proposition onely altering the letters: namely, in steade of the letter A put the letter F, and in steade of the letter H the letter C: moreouer in steade of the letter C put the letter H, and the letter E for the letter D, and the letter A for the letter E, and finally put the letter D for the letter F. And your figure thus ordered compare it with the de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g>, only imagining a superficies to passe through the body by the dia<g ref="char:EOLhyphen"/>gonall lines CF and DE.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 24. Theoreme. The 29. Proposition. Parallelipipedons consisting vpon one and the selfe same base, and vn<g ref="char:EOLhyphen"/>der one and the selfe same altitude, whose <note n="*" place="margin">
                     <hi>Looke at the end of the de<g ref="char:EOLhyphen"/>monstratio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> what is vn<g ref="char:EOLhyphen"/>derstanded by sta<g ref="char:cmbAbbrStroke">̄</g>ding lines.</hi>
                  </note> standing lines are in the selfe same right lines, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that that these parallelipipedons</hi> CM <hi>and</hi> CN <hi>doo consist vpon one and the selfe same base, namely</hi> AB, <hi>and let them be vnder one and the selfe same altitude, whose standing lines, that is, the fower sides of eche solide which fall vpon the base, as the lines</hi> AF, CD, BH, LM <hi>of the solide</hi> CM, <hi>and the lines</hi> CE, BK, AG, <hi>and</hi> LN
<figure/>
                  <hi>of the solide</hi> CN, <hi>let be in the selfe same right lines or paral<g ref="char:EOLhyphen"/>lel lines</hi> FN, DK.<note place="margin">
                     <p>
                        <hi>Iohn Dee his figure.</hi>
                     </p>
                     <p>
                        <hi>By this figure it app<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>th why <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>uch Pris<g ref="char:EOLunhyphen"/>mes were cal<g ref="char:EOLhyphen"/>led <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>edges: of <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ry shape of a wedge, as is the solide DEFGA<g ref="char:EOLhyphen"/>C. &amp;c.</hi>
                     </p>
                  </note> 
                  <hi>Then I say that the solide</hi> CM <hi>is equall to the solide</hi> CN. <hi>For forasmuch as either of these superficieces</hi> CBDH, CBEK <hi>is a paralle<g ref="char:EOLhyphen"/>logramme, therefore (by the 34. of the first) the line</hi> CB <hi>is e<g ref="char:EOLhyphen"/>quall to either of these lines</hi> DH <hi>and</hi> EK. <hi>Wherefore also the line</hi> DH <hi>is equall to the line</hi> EK. <hi>Take away</hi> EH <hi>which is common to them both, where<g ref="char:EOLhyphen"/>fore the residue namely</hi> DE <hi>is equall to the residue</hi> HK. <hi>Wherfore also the triangle</hi> DCE <hi>is equall to the triangle</hi> HKB. <hi>And the parallelogramme</hi> DG <hi>is equall to the parallelo<g ref="char:EOLhyphen"/>gramme</hi> HN. <hi>And by the same reason the triangle</hi> AGF <hi>is equall to the triangle</hi> MLN, <hi>and the parallelogramme</hi> CF <hi>is equall to the parallelogramme</hi> BM. <hi>But the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me</hi> CG <hi>is equall to the parallelogramme</hi> BN, <hi>by the 24. of the tenth for they are opposite the one to the other. Wherefore the prisme contayned vnder the two triangles</hi> FAG <hi>and</hi> DCE <hi>and vnder the three parrallelogra<g ref="char:cmbAbbrStroke">̄</g>mes</hi> AD, DG, <hi>and</hi> CG <hi>is equall to the prisme co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the two triangles</hi> MLN <hi>and</hi> HBK, <hi>and vnder the three parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes, that is,</hi> BM, NH, <hi>and</hi> BN. <hi>Put that solide common to them both, whose base is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me</hi> AB, <hi>and the opposite side vnto the base is the superficies</hi> GEHM. <hi>Wherefore the whole pa<g ref="char:EOLhyphen"/>rallelipipedon</hi> CM <hi>is equall to the whole parallelipipedon</hi> CN: <hi>Wherfore parallelipipedons consisting vpon one and the selfe same base, and vnder one and the selfe same altitude, whose standing lines are in the selfe same right lines, are equall the one to the other: which was re<g ref="char:EOLhyphen"/>qui<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed to be demonstrated.</hi>
               </p>
               <p>
                  <pb n="341" facs="tcp:7062:370"/>Although this demonstration
<figure/>
be not hard to a good imaginati<g ref="char:EOLhyphen"/>on to conceaue by the former fi<g ref="char:EOLhyphen"/>gure (which yet by <hi>M. Dee<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> refor<g ref="char:EOLhyphen"/>ming is much better then the figure of this proposition commonly des<g ref="char:EOLhyphen"/>cribed in other copyes both greake and lattin): yet for the ease of those which are young beginners in thys matter of solides, I haue here set an other figure whose forme if it be described vpon pasted paper, with the like letters to euery line as they be here put, and then if ye finely cut not thorough but as it were halfe way the three lines LA, NMGF, and KHED, &amp; so folde it accordingly, &amp; compare it with the demonstratio<g ref="char:cmbAbbrStroke">̄</g>, it will geue great light thereunto.</p>
               <p>Sta<g ref="char:cmbAbbrStroke">̄</g>ding lines are called those fower right lines of euery parallelipipedon which ioyne together the angles of the vpper and nether bases of the same body.<note place="margin">
                     <hi>Sta<g ref="char:cmbAbbrStroke">̄</g>ding lines.</hi>
                  </note> Which according to the diuersitie of the angles of the solides, may either be perpendicular vpon the base, or fall obliquely. And forasmuch as in thys proposition and in the next proposition following, the solides compared together are supposed to haue one and the selfe same base, it is manifest that the standing lines are in respect of the lower base in the selfe same parallel lines, namely, in the two sides of the lower base. But because there are put two solides vpon one and the selfe same base, and vnder one and the selfe same altitude, the two vpper bases of the solides may be diuersly placed. For forasmuch as they are equall and like (by the 24. of this booke) either they may be placed betwene the selfe same parallel lines: and the<g ref="char:cmbAbbrStroke">̄</g> the standing lines are in either solide sayd to be in the selfe same parallel lines, or right lines: namely, when the two sides of eche of the vpper bases are contayned in the selfe same parallel lines: but contrariwise if those two sides of the vp<g ref="char:EOLhyphen"/>per bases be not contayned in the selfe same parallel or right lines, neither shal the standing lines which are ioyned to those sides be sayd to be in the selfe same parallel or right lines. And therefore the stan<g ref="char:EOLhyphen"/>ding lines are sayd to be in the selfe same right lines, when the sides of the vpper bases, at the least two of the sides are contayned in the selfe same right lines: which thing is required in the supposition of this 29, proposition. But the standing lines are sayd not to be in the selfe same right lines, when none of the two sides of the vpper bases are contayned in the selfe same right lines, which thing the next propositi<g ref="char:EOLhyphen"/>on following supposeth.</p>
            </div>
            <div type="part">
               <head>¶The 25. Theoreme. The 30. Proposition. Parallelipipedons consisting vpon one and the selfe same base, and vnder the selfe same altitude, whose standing lines are not in the selfe same right lines, are equall the one to the other.</head>
               <p>SVppose that these Parallelipipedons <hi>CM</hi> and <hi>CN,</hi> do consist vpon one and the selfe same base, namely, <hi>AB,</hi> and vnder one and the selfe same altitude, whose standing lines, namely, the lines <hi>AF, CD, BH,</hi> and <hi>LM,</hi> of the Parallelipipedon <hi>CM,</hi> and the standing lines <hi>AG, CH, BK,</hi> and <hi>LN,</hi> of the Parallelipipedon <hi>CN,</hi> let not be in the selfe same right lines. Then I say, that the Parallelipipedon <hi>CM,</hi> is equall to the Parallelipipedon <hi>CN.</hi> Forasmuch as the vpper superficieces <hi>FH</hi> &amp; <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>K,</hi> of the former Parallelipipedons, are in one and the selfe same superficies (by reason they are supposed to be of one and the selfe same alti<g ref="char:EOLhyphen"/>tude):<note place="margin">Construction.</note> Extend the lines <hi>FD</hi> and <hi>MH,</hi> till they concurre with the lines <hi>N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </hi> and <hi>KE</hi> (suffi<g ref="char:EOLhyphen"/>ciently both waies extended: for in diuers cases their concurse is diuers). Let <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>D</hi> extended, meete with <hi>NG,</hi> and cut it in the point <hi>X</hi>: and with <hi>KE</hi> in the point <hi>P.</hi> Let likewise <hi>MH</hi> extended, meete with <hi>NG</hi> (sufficiently produced) in the point <hi>O,</hi> and with <hi>KE</hi> in the point <hi>R.</hi> And drawe these right lines <hi>AX, LO, CP,</hi> and <hi>BR.</hi> Now (by the <hi>29.</hi> of the eleuenth) the solide <hi>CM,</hi> whose base is the parallelogramme <hi>ACBL,</hi> and the parallelogramme opposite vn<g ref="char:EOLhyphen"/>to
<pb facs="tcp:7062:371"/>
it is <hi>FDHM,</hi> is equall to the solide <hi>CO,</hi> whose base is <hi>ACBL</hi> and the opposite side the parallelogramme <hi>XPRO,</hi>
                  <note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> for they consiste vpon one and the selfe same base, namely, vpon the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me <hi>ACBL,</hi> whose standing lines <hi>AF, AX, LM, LO, CD, CP, BH,</hi> and <hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>K,</hi>
                  <figure/>
                  <note place="margin">I. Dees figure.</note>
are in the selfe same right lines <hi>FP</hi> and <hi>MR.</hi> But the solide <hi>CO,</hi> whose base is the paralle<g ref="char:EOLhyphen"/>logramme <hi>ACBL,</hi> and the opposite superficies vnto it is <hi>XPRO,</hi> is equall to the solide <hi>CN,</hi> whose base is the parallelogramme <hi>ACBL,</hi> and the opposite superficies vnto it is the superfi<g ref="char:EOLhyphen"/>cies <hi>G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>KN,</hi> for they are vpon one and the selfe same base, namely, <hi>ACBL,</hi> and their stan<g ref="char:EOLhyphen"/>ding lines <hi>AG, AX, CF, CP, LN, LO, BK,</hi> and <hi>BR,</hi> are in the selfe same right lines <hi>NX,</hi> and <hi>PK.</hi> Wherefore also the solide <hi>CM,</hi> is equall to the solide <hi>CN.</hi> Wherefore Parallelipi<g ref="char:EOLhyphen"/>pedons consisting vpon one and the selfe same base, and vnder the selfe same altitude, whose standing lines are not in the selfe same right lines, are equall the one to the other: which was required to be proued.</p>
               <p>
                  <hi>This demonstra<g ref="char:EOLhyphen"/>tion
<figure/>
is somwhat har<g ref="char:EOLhyphen"/>der then the former to conceaue by the fi<g ref="char:EOLhyphen"/>gure described in the plaine (and yet before</hi> M. Dee <hi>inuented that figure which is placed for it, it was much harder) by reason one solide is contained in an other. And there<g ref="char:EOLhyphen"/>fore for the clerer light therof, describe vpo<g ref="char:cmbAbbrStroke">̄</g> pasted paper this figure here put with the like letters and finely cut the lines AC, CB, EG, BL, EPK, ROHM, and folde it accordingly that euery line may exactly agree with his correspondent lyne (which obseruing the letters of euery line
<pb n="342" facs="tcp:7062:371"/>
ye shall easily do) and so co<g ref="char:cmbAbbrStroke">̄</g>pare your figure with the demonstration, and it will make it very plaine vnto you.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 26. Theoreme. The 31. Proposit. Parallelipipedons consisting vp<g ref="char:EOLhyphen"/>on equall bases, and being vnder one and the selfe same altitude, are equall the one to the other.</head>
               <p>
                  <hi>SVppose that vppon these equall
<figure/>
bases AB and CD do consiste these parallelipipedons AE and CF, being vnder one &amp; the self same altitude.</hi>
                  <note place="margin">
                     <p>
                        <hi>Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</hi>
                     </p>
                     <p>
                        <hi>Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> first case.</hi>
                     </p>
                     <p>
                        <hi>Construction.</hi>
                     </p>
                  </note> 
                  <hi>The<g ref="char:cmbAbbrStroke">̄</g> I say, that the solide AE is equall to the solide CF. First let the stan<g ref="char:EOLhyphen"/>ding lines, namely, HK, BE, AG, LM, OP, DF, CX, and RS, be erected perpendicular<g ref="char:EOLhyphen"/>ly to the bases AB and CD, and let the an<g ref="char:EOLhyphen"/>gle ALB not be equall to the angle CRD. Extend the line CR to the point T. And (by the 23. of the first) vpon the right line RT, and at the point in it R, describe vn<g ref="char:EOLhyphen"/>to</hi>
                  <note place="margin">
                     <hi>We are behol<g ref="char:EOLhyphen"/>ding to M. Dee for inuenting this figure, with other, which till his reforming were as much mishappen as this was, and so both in the Greeke and La<g ref="char:EOLhyphen"/>tine copies re<g ref="char:EOLhyphen"/>maine.</hi>
                  </note>
                  <pb facs="tcp:7062:372"/>
                  <hi>the angle ALB an equall angle TRV.
<figure/>
And (by the third of the first) put the line RT equall to the line LB, and the line RV equall to the line AL. And (by the 31. of the first) by the point V draw vnto the line RT a parallel line VW:</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>and make perfecte the base RW, and the solide YV. Now forasmuch as these two lines TR and RV are equall to these two lines BL and LA, and they containe equall angles, ther<g ref="char:EOLhyphen"/>fore the parallelogramme RW is equall and like to the parallelogramme AB. Againe, forasmuch as the line LB is equall to the line RT, and the line LM to the line RS (for the lines LM and RS are the altitudes of the Parallelipipedons AE and CF, which altitudes are supposed to be equall) and they containe right angles by supposition, there<g ref="char:EOLhyphen"/>fore the parallelogramme RY is equall and like to the parallelogramme BM. And by the same reason also the parallelogramme LG is equall and like to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me SV. Wherefore three parallelogrammes of
<pb n="343" facs="tcp:7062:372"/>
the solide AE are equall and like to the three parallelogrammes of the solide YV. But these three parallelogrammes are equall and like to the three opposite sides. Wherefore the whole Parallelipipedon AE is equall and like to the whole Parallelipipedon YV. Extend (by the second petition) the lines DR and WV, vntill they concurre, and let them concurre in the point Q. And (by the 31. of the first) by the point T drawe vnto the line RQ a parallel line T 4, and extend duely the lines Ta and DO vntill they concurre, and let them concurre in the point <g ref="char:cross">✚</g>. And make perfecte the solides QY and RI. Now the solide QY, whose</hi> 
                  <note n="*" place="margin">
                     <hi>Note now, how the base respec<g ref="char:EOLhyphen"/>tiuely is taken<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> for so may alte<g ref="char:EOLhyphen"/>ratio<g ref="char:cmbAbbrStroke">̄</g> of respects alter the name of the bowndes eyther of solides or playnes.</hi>
                  </note> 
                  <hi>base is the parallelogramme RY, and the opposite side vnto the base the parallelogramme Qb, is equall to the solide YV, whose base is the paral<g ref="char:EOLhyphen"/>lelogramme RY, and the opposite side vnto the base the parallelogramme VZ. For they con<g ref="char:EOLhyphen"/>siste vpon one and the selfe fame base, namely, RY, and are vnder one and the selfe same al<g ref="char:EOLhyphen"/>titude, and their standing lines, namely, RQ, RV, Ta, TW, SN, Sd, Yb, and YZ, are in the selfe same right lines, namely, QW, and NZ. But the solide YV is proued e<g ref="char:EOLhyphen"/>quall to the solide AE. Wherefore also the solide YQ is equall to the solide AE. Now forasmuch as the parallelogramme RVWT is equall to the parallelogramme QT (by the 35. of the first) and the parallelogramme AB is equall to the parallelogramme RW: there<g ref="char:EOLhyphen"/>fore the parallelogramme QT is equall to the parallelogramme AB, and the parallelo<g ref="char:EOLhyphen"/>gramme CD is equall to the parallelogramme AB (by supposition). Wherefore the paralle<g ref="char:EOLhyphen"/>logramme CD is equall to the parallelogramme QT. And there is a certaine other super<g ref="char:EOLhyphen"/>ficies, namely, DT. Wherefore (by the 7. of the fift) as the base CD is to the base DT, so is the base QT to the base DT. And forasmuch as the whole Parallelipipedon CI is cut by the plaine superfi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>es RF, which is a parallel to either of the opposite plaine superficieces, ther<g ref="char:EOLhyphen"/>fore as the base CD is to the base DT, so is the solide CF to the solide RI (by the 25. of the eleuenth). And by the same reason also, forasmuch as the whole Parallelipipedon QI is cut by the plaine superficies RY, which is a parallel to either of the opposite plaine superficieces, therefore as the base QT is to the base DT, so is the solide QY to the solide RI. But as the base CD is to the base DT, so is the base QT to the base TD. Wherefore (by the 11. of the fift) as the solide CF is to the solide RI, so is the solide QY to the solide RI. Wherefore either of these solides CF and QY, haue to the solide RI one and the same proportion. Wher<g ref="char:EOLhyphen"/>fore the solide CF is equall to the solide QY. But it is proued that the solide QY is equall to the solide AE. Wherefore also the solide CF is equall to the solide AE.</hi>
               </p>
               <p>
                  <hi>But now suppose that the
<figure/>
sta<g ref="char:cmbAbbrStroke">̄</g>ding lines, namely, AG, HK,</hi>
                  <note place="margin">
                     <hi>Second case.</hi>
                  </note> 
                  <hi>BE, LM, CX, OP, DF, and RS, be not erected perpendicu<g ref="char:EOLhyphen"/>larly to the bases AB and CD. Then also I say, that the solide AE is equall to the solide CF. Draw (by the 11. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) vnto the ground plaine superfi<g ref="char:EOLhyphen"/>cieces AB and CD from these pointes K, E, C, M, P, F, X, S, these perpendicular lines KN, ET, GV, MZ, PW, FY, XQ, and SI. And draw these right lines NT, NV, ZV, ZT, WY, WQ, IQ, and IY. Now (by that which hath before bene pro<g ref="char:EOLhyphen"/>ued in this 31. Proposition) the
<pb facs="tcp:7062:373"/>
solide KZ is equall to the solide
<figure/>
PI, for they consist vpon equall</hi> 
                  <note n="*" place="margin">
                     <hi>There you perceaue how the base is di<g ref="char:EOLhyphen"/>uersly consi<g ref="char:EOLhyphen"/>dered &amp; cho<g ref="char:EOLhyphen"/>sen: as before we aduertised you.</hi>
                  </note> 
                  <hi>bases, namely, KM, and PS, and are vnder one and the selfe same altitude, whose standing lines also are erected perpendicu<g ref="char:EOLhyphen"/>larly to the bases. But the solide KZ is equall to the solide AE (by the 29. of the eleuenth): and the solide PI is (by the same) e<g ref="char:EOLhyphen"/>quall to the solide CF, for they consist vppon one and the selfe same base, and are vnder one &amp; the selfe same altitude, whose standing lines are vpon the selfe same right lines. Wherefore also the solide AE is equall to the solide CE. Wherefore Parallelipipedons consisting vpon equall bases and being vnder one and the selfe same altitude, are equall the one to the other: which was required to be demonstrated.</hi>
               </p>
               <p>The demonstration of the first case of this proposition is very hard to conceaue by the figure des<g ref="char:EOLhyphen"/>cribed for it in a playne. And yet before <hi>M. Dee</hi> inuented that figure which we haue there placed for it, it was much harder. For both in the Greke and Lattin <hi>Euclide,</hi> it is very ill made, and it is in a maner im<g ref="char:EOLhyphen"/>possible to conceaue by it the construction and demonstration thereto appertayning. Wherefore I haue here described other figures, which first describe vpon pasted paper, or such like matter and then order them in maner following.</p>
               <figure/>
               <p>As touching the solide AE in the first case, I neede not to make any new description. For it is playne inough to conceaue as it is there drawne. Although you may for your more ease of imagination describe of pasted paper a parallelipipedo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> hauing his sides equall with the sides of the parallelipipedon AE before described, and hauing also the sixe parallelogrammes thereof (contayned vnder those sides)
<pb n="344" facs="tcp:7062:373"/>
equiangle with the sixe parallelogrammes of that figure, ech side and eche angle equall to his correspon<g ref="char:EOLunhyphen"/>dent side, and to his correspondent angle.</p>
               <p>But concerning the other solide. When ye haue described these three figures vpon pasted pa<g ref="char:EOLhyphen"/>per: Where note for the proportion of eche line, to make your figure of pasted paper equall with the fi<g ref="char:EOLhyphen"/>gure before described vpon the playne, let your lines OP, CX, RS, DF, &amp;c. namely, the rest of the stan<g ref="char:EOLhyphen"/>ding lines, of these figures, be equall to the standing lines OP, CX, RS, DF, &amp;c. of that figure. Likewise let the lines OC, CR, RD, DO, &amp;c. namely, the sides which co<g ref="char:cmbAbbrStroke">̄</g>tayne the bases of these figures be equal to the lines OC, CR, RD, DO, &amp;c. namely, to the sids which co<g ref="char:cmbAbbrStroke">̄</g>tayne the bases of that figure. Moreouer let the lines PX, X<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, SF, FP, &amp;c. namely, the rest of the lines which co<g ref="char:cmbAbbrStroke">̄</g>taine the vpper superficieces of these figures, be equal to the lines PX, XS, SF, F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, &amp;c. namely, to the rest of the lines which co<g ref="char:cmbAbbrStroke">̄</g>taine the vpper superficieces of that figure (to haue described all those foresaid lines of these figures equal to all the lines of that figure, would haue required much more space then here can be spared: I haue made them equall onely to the halues of those lines, but by the example of these ye may, if ye will describe the like figures hauing their lines equall to the whole lines of the figure in the playne, eche line to his correspondent line). When I say ye haue as before is taught described these three figures, cut finely the lines XC, SR, FD of the first figure, and the lines SR, YT, and I <g ref="char:cross">✚</g> of the second figure: likewise the lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>R, NQ, ZVV, and YT, of the third figure, and fold these figures accordingly, which ye can not chuse but doo if ye marke well the letters of euery line.</p>
               <p>The three former figures being after this sort described, set them vpon this figure here described vpon a playne, as vpo<g ref="char:cmbAbbrStroke">̄</g> their bases, namely, the lines OC, CR, RD, DO: RT, T <g ref="char:cross">✚</g>, <g ref="char:cross">✚</g> D, DR: VR, RT, TW, Wa, VQ, and QR of these three figures vpon the lines correspondent vnto them in this figure. And they so standing compare them with the
<figure/>
construction and demonstration of the first case, and they will geue great light vnto it. This also ye must note,<note place="margin">
                     <hi>Construction.</hi>
                  </note> that if ye make the lines of the fore sayd three figures equall to the lines of the figure of the plaine described before in the demo<g ref="char:cmbAbbrStroke">̄</g>strati<g ref="char:EOLhyphen"/>on of the first case: then must ye make a new base for them like vnto this, which is easy to doo, if ye draw a pallelogramme equall and like to the parallelogramme OCT <g ref="char:cross">✚</g>, and the<g ref="char:cmbAbbrStroke">̄</g> cut of from the same a parallelogramme DRT <g ref="char:cross">✚</g> equall and in like sort situate to the parallelogramme DRT <g ref="char:cross">✚</g> of that figure: &amp; vpon the line RT describe two parallelogrammes, the one equall like, and in like sort situate to the parallelogramme RTQa of that figure, and the other equall, like and in like sort situate to the parallelogramme RTVW of the same figure. The lines of this base which I haue here put are equall onely to the halues of the lines of that figure in the demonstration.</p>
               <p>As touching the second case ye neede no new figures, for it is playne to see by the figures (now reformed by <hi>M. Dee</hi>) described for it in the playne, especially if ye remember the forme of the figure of the 29. proposition of this booke. Only that which there ye conceaue to be the base, imagine here in both the figures of this second case to be the vpper superficies opposite to the base, and that which was there supposed to be the vpper superficies conceaue here to be the base. Ye may describe them vpon pasted paper for your better sight, taking hede ye note the letters rightly according as the construction requireth.</p>
               <div type="section">
                  <head>
                     <hi>Flussas</hi> demonstrateth this proposition an otherway taking onely the bases of the solides, and that after this maner.</head>
                  <p>Take equall bases (which
<figure/>
yet for the surer vnderstan<g ref="char:EOLhyphen"/>ding let be vtterly vnlike) namely, AEBF and ADCH, and let one of the sides of eche concurre in one &amp; the same right line AED, &amp; the bases being vpon one and the selfe same playne let there be supposed to be set vpon the<g ref="char:cmbAbbrStroke">̄</g> parallelipipedons vnder one &amp; the selfe same altitude. Then I say that the solide set vpo<g ref="char:cmbAbbrStroke">̄</g> the base AB is equal to the solide set vpon the base AH. By the poynt E draw vnto the line AC a
<pb facs="tcp:7062:374"/>
parallel line EG, which if it
<figure/>
fall without the base AB, produce the right line HC to the poynt I. Now foras<g ref="char:EOLhyphen"/>much as AB and AH are pa<g ref="char:EOLhyphen"/>rallelogrmaes, therefore by the 24. of this booke, the triangles ACI and EGL shall be equaliter the one to the other: and by the 4. of the first, they shal be equian<g ref="char:EOLhyphen"/>gle and equall: and by the first definition of the sixth, and fourth Proposition of the same, they shall be like. Wherfore Prismes erected vppon those triangles and vnder the same altitude that the solides AB and AH a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e, shall be equall and like, by the 8. definition of this booke. For they are contayned vnder like playne superficieces equall both in multitude and mag<g ref="char:EOLunhyphen"/>nitude. Adde the solide set vpon the base ACLE common to them both. Wherefore the solide set vp<g ref="char:EOLhyphen"/>pon the base AEGC, is equall to the solide set vpon the base AELI. And forasmuch as the superficie<g ref="char:EOLhyphen"/>ces AEBF, and ADHC are equall (by supposition): and the part taken away AG is equall to the part taken away AL: therefore the residue BI shall be equall to the residue GD. Wherefore as AG is to GD as AL is to BI (namely, equalls to equalls). But as AG is to GD, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the solide set vpon AG to the solide set vpon GD by the 25. of this booke, for it is cut by a playne superficies set vpon the line GE, which superficies is parallel to the opposite superficieces. Wherefore as AL is to BI, so is the solide set vpon AL to the solide set vpon BI. Wherefore (by the 11. of the fifth) as the solide set vpon AG (or vpon AL which is equall vnto it) is to the solide set vpon GD, so is the same solide set vpon AG or AL to the solide set vpon BI. Wherefore (by the 2. part of the 9. of the fifth) the solides set vpon GD and BI shall be equall. Vnto which solides if ye adde equall solides, namely, the solide set vpon AG to the solide set vpon GD, and the solide set vpon AL to the solide set vpon BI: the whole solides set vpon the base AH and vpon the base AB <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hall be equall. Wherefore Parallelipedons consisting vpon equall bases and being vnder one and the selfe same altitude are equall the one to the other: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 27. Theoreme. The 32. Proposition. Parallelipipedons being vnder one and the selfe same altitude, are in that proportion the one to the other that their bases are.</head>
               <p>SVppose that these parallelipipedons AB and CD be vnder one &amp; the selfe same altitude. Then I say that those parallelipipedons AB and CD are in that pro<g ref="char:EOLhyphen"/>portion the one to the other, that their bases are, that is, that as the base AE is to the base CF, so is the parallelipipedon AB to the parallelipipedon CD.<note place="margin">Construction.</note> Vpon the line FG describe (by the 45.
<figure/>
of the first) the parallelo<g ref="char:EOLhyphen"/>gramme FH equall to the parallelogramme AE and e<g ref="char:EOLhyphen"/>quiangle with the parallelo<g ref="char:EOLhyphen"/>gramme CF. And vpon the base FH describe a paralleli<g ref="char:EOLunhyphen"/>pipedo<g ref="char:cmbAbbrStroke">̄</g> of the selfe same alti<g ref="char:EOLhyphen"/>tude that the parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> CD is, &amp; let y<hi rend="sup">e</hi> same be GK.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now (by the 31. of the ele<g ref="char:EOLhyphen"/>uenth) the parallelipipedon AB is equall to the parallelipipedon GK, for they consist vpon e<g ref="char:EOLhyphen"/>quall
<pb n="345" facs="tcp:7062:374"/>
bases, namely, AE and FH, and are vnder one and the selfe same altitude. And foras<g ref="char:EOLhyphen"/>much as the parallelipipedon CK is cut by a plaine superficies DG, being parallel to either of the opposite plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces, therfore (by the 25. of the eleuenth) as the base HF is to the base FC, so is the parallelipipedon GK to parallelipipedon CD: but the base HF is equal to the base AE, and the parallelipipedon GK is proued equall to the parallelipipedon AB. Wherfore as the base A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E is to the base CF, so is the parallelipedon AB, to the parallelipipe<g ref="char:EOLhyphen"/>don CD. Wherfore parallelipipedons being vnder one and the selfe same altitude, are in that proportion the one to the other that their bases are: which was required to be demonstrated.</p>
               <p>
                  <hi>I neede not to put any other figure for the declaration of this demonstration, for it is easie to see by the figure there described. Howbeit ye may for the more full sight therof, describe solides of pasted paper according to the construction there set forth, which will not be hard for you to do, if ye remem<g ref="char:EOLhyphen"/>ber the descriptions of such bodies before taught.</hi>
               </p>
               <div type="section">
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Equall parallelipipedons co<g ref="char:cmbAbbrStroke">̄</g>tained vnder one and the selfe same altitude, haue also their bases equal.</hi> For if the bases should be vnequall, the parallelipipedons also should be vnequal by this 32 propositio<g ref="char:cmbAbbrStroke">̄</g>. <hi>And equall parallelipipedons hauing equall bases, haue also one and the selfe same altitude.</hi> For if they should haue a greater altitude, they should exceede the equall parallelipipedons which haue the selfe same altitude: But if they should haue a lesse they should want so much of those selfe same equal parallelipipedons.</p>
               </div>
            </div>
            <div type="part">
               <head>The 28. Theoreme. The 33. Proposition. Like parallelipipedons are in treble proportion the one to the other of that in which their sides of like proportion are.</head>
               <p>SVppose that these parallelipipedons AB and CD be like, &amp; let the sides AE and CF be sides of like proportion. Then I say, the parallelipipedon AB is vnto the parallelipipedon CD in treble proportion of that in which the side AE is to the side CF. Extend the right lines AE, GE and HE to the pointes K, L, M.<note place="margin">Construction.</note> And (by the 2. of the first) vnto the
<figure/>
line CF put the line EK equal, and vnto the line FN put the line EL equall, and moreouer vnto the line FR put the line EM equall, and make perfect the parallelogramme KL, and the parallelipipedon KO.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now for<g ref="char:EOLhyphen"/>asmuch as these two lines EK and EL are equall to these two lines CF and FN, but the angle KEL is equall to the angle CFN (for the angle AEG is equall to the angle CFM by reason that the solides AB and CD are like). Wher<g ref="char:EOLhyphen"/>fore the parallelogramme KL is equall and like to the pa<g ref="char:EOLhyphen"/>rallelogramme CN, and by the same reason also the parallelogramme KM is equall and
<pb facs="tcp:7062:375"/>
like to the parallelogramme C<g ref="char:EOLhyphen"/>R,
<figure/>
and moreouer the parallelo<g ref="char:EOLhyphen"/>gramme OE to the parallelo<g ref="char:EOLhyphen"/>gramme FD. Wherefore three parallelogrammes of the paralle<g ref="char:EOLhyphen"/>lipipedon KO are like and equall to three parallelogrammes of the parallelipipedon CD: but those three sides are equall and like to the three opposite sides: wherfore the whole parallelipipedon KO is equal and like to the whole pa<g ref="char:EOLhyphen"/>rallelipipedon CD. Make perfect the parallelogramme GK. And vpon the bases GK and KL make perfect to the altitude of the parallelipipedon AB, the pa<g ref="char:EOLhyphen"/>rallelipipedons EX &amp; LP. And forasmuch as by reason that the parallelipipedons AB &amp; CD are like, as the line AE is to the line CF, so is the line EG, to the line FN, and the line EH to the line FR. But the line CF is equall to the line EK, and the line FN to the line EL, and the line FR to the line EM, therefore as the line AE is to the line EK, so is the line GE to the line EL, and the line HE to the line EM (by construc<g ref="char:EOLhyphen"/>tion). But as the line AE is to the line EK, so is the parallelogramme AG to the parallelo<g ref="char:EOLhyphen"/>gramme GK (by the first of the sixt). And as the line GE is to the line EL, so is the paralle<g ref="char:EOLhyphen"/>logramme GK to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me KL. And moreouer as the line HE is to the line EM, so is the parallelogramm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> PE to the parallelogramme KM. Wherefore (by the 11. of the fift) as the parallelogramme AG is to the parallelogramme GK, so is the parallelogramme GK to the parallelogramme KL, and the parallelograme PE to the parallelogramme KM. But as the parallelogramme AG is to the parallelogramme GK, so is the parallelipipedon AB to the parallellpipedon EX, by the former proposition, and as the parallelogramme GK is to the parallelogramme KL, so by the same is the parallelipipedon XE to the parallelipipe<g ref="char:EOLhyphen"/>don PL: and agayne by the same, as the parallelogramme PE is to the parallelogramme K<g ref="char:EOLhyphen"/>M, so is the parallelipipedon PL to the parallelipipedon KO. Wherfore as the parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> AB is to the parallelipipedon EX, so is the parallelipipedon EX to the parallelipipedon PL, and the parallelipipedon PL to the parallelipipedon KO. But if there be fower magnitudes in* continuall proportion, the first shalbe vnto the fourth in treble proportion that it is to the second (by the 10. definition of the fift). Wherefore the parallelipipedon AB is vnto the pa<g ref="char:EOLhyphen"/>rallelipedon KO in treble proportion that the parallelipipedon AB is to the parallelipipedon EX. But as the parallelipipedon AB is to the parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> EX, so is the parallelogramme AG to the parallelogramme GK, and the right line AE to the right line EK. Wherefore also the parallelipipedon AB is to the parallelipipedon KO in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that which the line AE hath to the line EK. But the parallelipipedon KO is equall to the parallelipi<g ref="char:EOLhyphen"/>pedon CD, and the right line EK to the right line CF. Wherefore the parallelipipedon AB is to the parallelipipedon CD in treble proportion that the side of like proportion, namely, A<g ref="char:EOLhyphen"/>E is to the side of like proportion, namely, to CF. Wherefore like parallelipipedons are in tre<g ref="char:EOLhyphen"/>ble proportion the one to the other of that in which their sides of like proportion are: which was required to be demonstrated.</p>
               <div type="section">
                  <pb n="346" facs="tcp:7062:375"/>
                  <head>¶ Corellary.</head>
                  <p>
                     <hi>Hereby it is manifest, that if there be fower right lines in</hi> 
                     <note place="margin">*</note> 
                     <hi>continuall pro<g ref="char:EOLhyphen"/>portion, as the first is to the fourth, so shall the Parallelipipedon described of the first line, be to the Parallelipipedon described of the second, both the Parallelipipedons being like and in like sort described.</hi> For the first line is to the fourth in t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eble proportion that it is to the second: and it hath before bene
<figure/>
proued that the Parallelipipedon described of the first, is to the Parallelipipedon described of the second, in the same proportion that the first line is to the fourth.</p>
                  <p>Because the one of the figures before, described in a plaine, pertayning to the demonstration of this 33. Proposition, is not altogether so easie to a younge begin<g ref="char:EOLhyphen"/>ner to conceaue, I haue here for the same described an other figure, which if ye first drawe vpon pasted paper, and afterward cut the lines &amp; folde the sides accordingly, will agree with the construction &amp; demon<g ref="char:EOLhyphen"/>stration of the sayd Proposition. Howbeit this ye must note that ye must cut the lines OQ &amp; MR on the contrary side <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o that which ye cut the other lines. For the so<g ref="char:EOLhyphen"/>lides which haue to their base the paralle<g ref="char:EOLhyphen"/>logramme LK are set on vpward and the other downward: Ye may if ye thinke good describe after the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>me maner of pasted pa<g ref="char:EOLhyphen"/>per a body equall to the solide CD: though that be easie inough to conceaue by the fi<g ref="char:EOLhyphen"/>gure thereof described in the plaine.</p>
               </div>
               <div type="section">
                  <head>¶ Certaine most profitable Corollaries, Annotations, Theo<g ref="char:EOLhyphen"/>remes, and Problemes, with other practises, Logisticall, and Mechanicall, partly vpon this 33. and partly vpon the 34. 36. and other following, added by Master Iohn Dee. ¶ A Corollary. 1.</head>
                  <p n="1">1. <hi>Hereby it is manifest, that two right lines may be found, which shall haue that proportion, the one to the other, that any two like Parallelipi<g ref="char:EOLhyphen"/>pedons, and in like sort described, geuen, haue the one to the other.</hi>
                  </p>
                  <p>Suppose Q and X to be two like Parallelipipedons, and in like sort described. Of Q take any of the three lines, of which it is produced: as namely, RG. Of X, take that right line of his production, which line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s aunswerable to R G in proportion (which most aptly, after the Greke name, may be
<pb facs="tcp:7062:376"/>
called Omologall to RG) &amp;
<figure/>
let that be TV. By the 11. of the sixth, to RG and TV, let the third line in propor<g ref="char:EOLhyphen"/>tion with them be founde, and let that be Y. By the same 11. of the sixth, to TV and Y, let the thirde right line be fou<g ref="char:cmbAbbrStroke">̄</g>d, in the sayd pro<g ref="char:EOLunhyphen"/>portion of TV to Y: &amp; let that be Z. I say now that RG hath that proportion to Z, which Q hath to X. For by construction, we haue fower right lines in continuall pro<g ref="char:EOLhyphen"/>piotion, namely, RG, TV, Y, and Z. Wherfore by <hi>Euclides</hi> Corollary, here before, RG is to Z, as Q is to X. Where<g ref="char:EOLhyphen"/>fore we haue fou<g ref="char:cmbAbbrStroke">̄</g>d two right lines hauing that proportion the one to the other, which any two like Parallelipipedons of like descrip<g ref="char:EOLhyphen"/>tion, geuen, haue the one to the other: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <hi>As a Conuerse, of my <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ormer Corollary, doth it followe: To finde two like Parallelipipedons of like description, which shall haue that proportion the one to the other, that any two right lines, geuen, haue the one to the other.</hi>
                  </p>
                  <p>Suppose the two right lines geuen to be A and B: Imagine of foure right lines in continuall pro<g ref="char:EOLhyphen"/>portion, A. to be the first, and B to be the fourth: (or contrariwise, B to be first, &amp; A to be the fourth). The second and third are to be found, which may, betwene A &amp; B, be two meanes in continuall pro<g ref="char:EOLhyphen"/>po<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>tion: as now,<note n="*" place="margin">
                        <hi>Note this famous L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>m<g ref="char:EOLhyphen"/>ma.</hi>
                     </note> suppose such two lines, found: and let them be C and D. Wherefore by <hi>Euclides</hi> Corollary, as A is to B (if A were taken as first) so shall the Parallelipipedon described of A, be to the like Parallelipipedon and in like sort described of C: being the second of the fower lines in continuall proportion: it is to we<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e, A, C, D, and B. Or, if B shall be taken as first, (and that thus they are orderly in continuall proportion, B, D, C, A,) then, by the sayd Corollary, as B is to A, so shall the Parallelipi<g ref="char:EOLhyphen"/>pedon described of B, be vnto the like Parallelipipedon and in like sort described o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> D. And vnto a Pa<g ref="char:EOLhyphen"/>rallelipipedon of A or B, at pleasure described, may an other of C or D be made like, and in like sort si<g ref="char:EOLhyphen"/>tuated or described, by the 27. of this eleuenth booke. Wherefore any two right lines being geue<g ref="char:cmbAbbrStroke">̄</g>, &amp;c: which was required to be done.</p>
                  <p>Thus haue I most briefly brought to your vnderstanding, if (first) B were double to A, then what Parallelipipedon soeuer, were described of A, the like Parallelipipedon and in like sort described of C, shall be double to the Parallelipipedon described of A. And so likewise (secondly) if A were double to B, the Parallelipipedon of D, shoulde be double to the like, of B described, both being like situated. Wherefore if of A or B, were Cubes made, the Cubes of C and D are proued double to them:<note place="margin">
                        <hi>The doubling o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>he Cube.</hi>
                     </note> as that of C, to the Cube of A: and the Cube of D to the Cube of B: in the second case. <note n="*" place="margin">
                        <hi>Note.</hi>
                     </note> And so of any proportion els betwene A and B.</p>
                  <p>Now also do you most clerely perceaue the Mathematicall occasion, whereby (first of all men) <hi>Hippocrates,</hi> to double any Cube geuen, was led to the former Lemma: <hi>Betwene any two right lines geuen, to finde two other right lines, which shall be with the two first lines, in conti<g ref="char:EOLhyphen"/>nuall proportion:</hi> After whose time (many yeares) diuine <hi>Plato, Heron, Philo, Appollonius, Di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>, Pappus, Sporus, Menech<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>us, Archytas Tarentinus</hi> (who made the wodden doue to slye) <hi>Erato<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>hene, Nicomedes,</hi> with many other (to their immortall fame and renowme)
published, <note n="*" place="margin">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap> Lemma.</hi>
                     </note> diuers their witty deuises, methods, and engines (which yet are extant) whereby to execute thys Problematicall Lemma. But not withstanding all the trauailes of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>oresayd Philosophers and Mathematiciens, yea and all others doinges and con<g ref="char:EOLhyphen"/>triuinges (vnto this day) about the sayd Lemma, yet there remaineth sufficient matter, <hi>Mathematically so to demonstrate the same, that most exactly &amp; readily, it may also be Mechanically practis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>d:</hi> that who soeuer shall achieue that feate, shall not be counted a second <hi>Archi<g ref="char:EOLhyphen"/>medes,</hi>
                     <pb n="347" facs="tcp:7062:376"/>
but rather a per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>les Mathematicien, and <hi>Mathematicorum Princeps.</hi> I will sundry wayes (in my briefe additions and annotations vpon <hi>Euclide</hi>) excite you thereto,<note place="margin">
                        <hi>Note what i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> yet lacking re<g ref="char:EOLhyphen"/>quisite to the doubling of the Cube.</hi>
                     </note> yea and bring before your eyes sundry new wayes, by meinuented: and in this booke so placed, as matter thereof, to my inuentions appertayning, may geue occasion: Leauing the farther, full, &amp; absolute my concluding of the Lemma, to an other place and time: which will, now, more co<g ref="char:cmbAbbrStroke">̄</g>pendiously be done: so great a part therof, being before hand in thys booke published.</p>
               </div>
               <div type="section">
                  <head>¶A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Parallelipipedons consisting vpon equall bases, are in proportion the one to the other as their alti<g ref="char:EOLhyphen"/>tudes are.</hi> For if those altitudes be cut by a plaine superficies parallel to the bases: the sections shall be in proportion the one to the other as the sections of the bases cut, by the 25. of this booke. Which sections of the bases are the one to the other in that proportion that their sides or the altitudes of the solides are, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst of the sixt. Wherefore the solides are the one to the other as their altitudes are. But if the bases be vnlike, the selfe same thing may be proued by the Corollary of the 25. of this booke, which by the 25. Proposition was proued in like bases.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 29. Theoreme. The 34. Proposition. In equall Parallelipipedons the bases are reciprokall to their altitudes. And Parallelipipedons whose bases are reciprokall to their altitudes, are equall the one to the other.</head>
               <p>SVppose that these Parallelipipedons AB &amp; CD be equall the one to the other.<note place="margin">Two cases in the first part of this propo<g ref="char:EOLhyphen"/>tion.</note> Then I say, that the bases of the Parallelipipedons AB and CD are reciprokall to their altitudes, that is, as the base EH is to the base NP, so is the altitude of the solide CD to the altitude of the solide AB. First let the standing lines AG, EF, LB, HK, of the solide AB, &amp; the sta<g ref="char:cmbAbbrStroke">̄</g>ding lines CM, NX, OD, and PR, of the solide CD, be erectedperpe<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>dicularly to the bases EH &amp; NP.<note place="margin">First case, which also may be two wayes.</note> The<g ref="char:cmbAbbrStroke">̄</g> I say, that as the base EH is to the base NP, so is the line CM to the line AG.<note place="margin">First way.</note> Now if
<figure/>
the base EH be equal to the base NP, and the solide AB is equall to the solide CD, wherefore the line CM is equall to the line AG<note n="*" place="margin">This follow<g ref="char:EOLhyphen"/>eth also of the Corollary ad<g ref="char:EOLhyphen"/>ded of F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>as after the <hi>3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> proposition of this booke.</note>. For if the bases EH and NP being equall, the altitudes AG and CM be not equall, nei<g ref="char:EOLhyphen"/>ther also shall the solide AB be equall to the solide CD, but they are supposed to be equall. Wherefore the altitude CM is not vnequall to the altitude AG. Wherefore it is equall. And therefore as the base EH is to the base PN, so is the altitude CM to the altitude AG. Wher<g ref="char:EOLhyphen"/>fore it is manifest, that the bases of the Parallelipipedons AB and CD are reciprokall to their altitudes.</p>
               <p>But now suppose that the base EH be not equall to the base NP.<note place="margin">Second way.</note> But let the base EH be the greater. Now the solide AB is equall to the solide CD. Wherefore also the altitude CM is greater then the altitude AG <note n="*" place="margin">This also followeth of the former Corollary.</note>. For if not, then againe are not the solides AB and CD equall: but they are by supposition equall. Wherefore (by the 2. of the first) put vnto the line AG an equall line CT. And vpon the base NP and the altitude being CT, make perfecte a solide contained vnder parallel plaine superficieces, and let the same be CZ. And forasmuch
<pb facs="tcp:7062:377"/>
as the solide AB is equall to the solide CD, and there is a certaine other solide, namely, CZ, but vnto one and the selfe same magnitude equall magnitudes haue one and the selfe same proportion (by the 7. of the fift). Wherefore as the solide AB is to the solide CZ, so is the so<g ref="char:EOLhyphen"/>lide CD to the solide CZ. But
<figure/>
as the solide AB is to the solide CZ, so is the base EH to the base NP (by the 32. of the ele<g ref="char:EOLhyphen"/>uenth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for the solides AB and CZ are vnder equall altitudes. And as the solide CD is to the solide CZ, so is the base MP to the base PT, and the line MC to the line CT. Wherefore (by the 11. of the fift) as the base EH is to the base NP, so is the line CM to the line CT. But the line CT is equall to the line AG. Wherefore (by the 7. of the fift) as the base EH is to the base NP, so is the altitude CM to the altitude AG. Wherfore in these Parallelipipedons AB and CD the bases are reciprokall to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ir altitudes.</p>
               <p>But now againe suppose that the bases of the Parallelipipedons AB and CD be recipro<g ref="char:EOLhyphen"/>kall to their altitudes,<note place="margin">The conuerse of both the partes of the first case.</note> that is, as the base EH is to the base NP, so let the altitude of the so<g ref="char:EOLhyphen"/>lide CD be to the altitude of the solide AB. Then I say, that the solide AB is equall to the solide CD. For againe let the standing lines be erected perpendicularly to their bases.</p>
               <p>And now if the base EH be equall to the base NP: but as the base EH is to the base <milestone type="tcpmilestone" unit="unspecified" n="1"/> NP, so is the altitude of the solide CD to the altitude of the solide AB. Wherefore the al<g ref="char:EOLhyphen"/>titude o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the solide CD is equall to the altitude of the solide AB. But Parallelipipedons con<g ref="char:EOLhyphen"/>sisting vpon equall bases and vnder one and the selfe same altitude, are (by the 31. of the eleuenth) equall the one to the other. Wherefore the solide AB is equall to the solide CD.</p>
               <p>
                  <milestone type="tcpmilestone" unit="unspecified" n="2"/> But now suppose that the base EH be not equall to the base NP: but let the base EH be the greater. Wherefore also the altitude of the solide CD, that is, the line CM is greater then the altitude of the solide AB, that is, then the line AG. Put againe (by the 3. of the first) the line CT equall to the line AG, and make perfecte the solide CZ. Now for that as the base EH is to the base NP, so is the line MC to the line AG. But the line AG is e<g ref="char:EOLhyphen"/>quall to the line CT. Wherefore as the base EH is to the base NP, so is the line CM to the line CT. But as the base EH is to the base NP, so (by the 32. of the eleuenth) is the solide AB to the solide CZ, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or the solides AB and CZ are vnder equall altitudes. And as the line CM is to the line CT, so (by the 1. of the sixt) is the base MP to the base P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>T, and (by the 32. of the eleuenth) the solide CD to the solide CZ. Wherefore also (by the 11. and 9. of the fift) as the solide AB is to the solide CZ, so is the solide CD to the solide CZ. Wher<g ref="char:EOLhyphen"/>fore either of these solides AB and CD haue to the solide CZ one and the same proportion. Wherefore (by the 7. of the fift) the solide AB is equall to the solide CD: which was requi<g ref="char:EOLhyphen"/>red to be demonstrated.</p>
               <p>
                  <note place="margin">Second case.</note>But now suppose that the standing lines, namely, FE, BL, GA, KH: XN, DO, MC, and RP, be not erected perpendicularly to their bases. And (by the 11. of the eleuenth) from the pointes F, G, B, K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> X, M, D, R,<note place="margin">Construction.</note> draw vnto the plaine superficies of the bases EH and NP perpendicular lines, and let those perpendicular lines light vpo<g ref="char:cmbAbbrStroke">̄</g> the pointes S, T, V, Z: W, Y, d, and Q, and make perfecte the Parallelipipedons FZ, and XQI say that euen in this case also, if the solides AB and CD be equall, their bases are reciprokall to their altitudes, that is, as the base EH is to the base NP, so is the altitude of the solide CD to the altitude of the solide AB. For forasmuch as the solide AB is equall to the solide
<pb n="348" facs="tcp:7062:377"/>
CD, but the solide AB is equall to
<figure/>
the solide BT (by the 20. of the e<g ref="char:EOLhyphen"/>leuenth) for they are vpon one and the selfe same base,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> namely, the pa<g ref="char:EOLhyphen"/>rallelogramme KF, and vnder one and the selfe same altitude, whose standing lines are in the selfe same right lines, namely, HZAT, and LVES: and the solide CD is by the same reason equall to the solide DY, for they both consist vpon one and the selfe same base, namely, the parallelogramme XR, are vn<g ref="char:EOLhyphen"/>der one and the selfe same altitude, whose standing lines are in the selfe same right lines, namely, PQCY, and OhNW. Wherefore the solide
<figure/>
BT is equall to the solide DY. But in equall Parallelipipedons, whose altitudes are erected perpendicularly to their bases, their bases are re<g ref="char:EOLhyphen"/>ciprokall to their altitudes (by the first part of this Proposition). Wherefore as the base FK is to the base XR, so is the altitude of the solide DY to the altitude of the solide BT. But the base FK is equall to the base EH, and the base XR to the base NP. Wherefore as the base EH is to the base NP, so is the altitude of the solide DY to the altitude of the solide BT. But the altitudes of the solides DY &amp; BT, and of the solides DC &amp; BA are one and the selfe same. Wherefore as the base EH is to the base NP, so is the altitude of the solide CD to the altitude of the solide AB. Wherfore the bases of the Parallelipipedons AB and CD are reciprokall to their altitudes.</p>
               <p>Againe suppose that the bases of the Paralleli<g ref="char:EOLhyphen"/>pipedons AB and CD be reciprokall to their al<g ref="char:EOLhyphen"/>titudes,<note place="margin">The conuerse of the second case.</note> that is, as the base EH is to the base NP, so let the altitude of the solide CD be to the altitude of the solide AB. Then I say, that the solide AB is equall to the solide CD. For the same order of construction remayning, for that as the base EH is to the base NP, so is the altitude of the solide CD to the altitude of the solide AB: but the base EH is equall to the parallelogramme FK, and the base NP to the parallelogramme XR: wherefore as the base FK is to the base XR, so is the altitude of the solide CD to the altitude of the solide AB But the altitudes of the solides AB and BT are equall, and so also are the altitudes of the solides DC and DY. Wherefore as the base FK is to the base XR, so is the altitude of the solide DY to the altitude of the solide BT. Wherefore the bases of the Parallelipipedons BT and DY are reciprokall to their altitudes. But Parallelipipedons whose altitudes are erected per<g ref="char:EOLhyphen"/>pendicularly to their bases, and whose bases are reciprokall to their altitudes, are equall the one to the other (by this Proposition). Wherefore the solide BT is equall to the solide DY. But the solide BT is equall to the solide BA (by the 29. of the eleuenth) for they consist vpon one and the selfe same base, namely, FK, and are vnder one and the self same altitude, whose
<pb facs="tcp:7062:378"/>
standing lines are in the selfe same right lines. And the solide DY is equall to the solide DC, for they consiste vpon one and the selfe same base, namely, XR, and are vnder one and the selfe same altitude, whose standing lines are in the selfe same right lines. Wherefore also the solide AB is equall to the solide CD.<note place="margin">The generall conclusion.</note> Wherefore in equall Parallelipipedons the bases are reciprokall to their altitudes. And Parallelipipedons whose bases are reciprokall with their altitudes, are equall the one to the other: which was required to be proued.</p>
               <p>
                  <hi>The demonstration of the first case of this Proposition is easie to conceaue by the figure as it is descri<g ref="char:EOLhyphen"/>bed in the plaine. But ye may for your more full sight describe Parallelipipedons of pasted paper, ac<g ref="char:EOLhyphen"/>cording as the construction teacheth you.</hi>
               </p>
               <p>
                  <hi>And for the second case, if ye reme<g ref="char:cmbAbbrStroke">̄</g>ber well the forme of the figure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which you made for the second case of the 31. Proposition: and describe the like for this, taking <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eede to the letters that ye place them like as the co<g ref="char:cmbAbbrStroke">̄</g>struction in this case requireth, ye shall most easily by them come to the full vnderstanding of the construction and demonstration of the said case.</hi>
               </p>
               <div type="section">
                  <head>M. Iohn Dee, his sundry Inuentions and Annotacions, very necessary, here to be added and considered.</head>
                  <div type="subsection">
                     <head>A The<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>reme.</head>
                     <p>
                        <hi>If fower right lines be in continuall proportion, and vpon the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of the first, as a base, be erec<g ref="char:EOLhyphen"/>ted a rectangle parallelipipedon, whose heith is the fourth line: that rectangle parallelipipedon is equal to the Cube made of the second line. And if vpon the square of the fourth line, as a base, be erected a rectangle parallelipipedon, whose heith is the first line, that parallelipipedon is equall to the Cube made of the third line.</hi>
                     </p>
                     <p>Suppose AB,
<figure/>
CD, E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>H to be fower right lines in co<g ref="char:cmbAbbrStroke">̄</g>tinuall proportion:<note place="margin">
                           <hi>Construction.</hi>
                        </note> and vpo<g ref="char:cmbAbbrStroke">̄</g> the square of AB (which let be AI) as a base, l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t be erected a rec<g ref="char:EOLhyphen"/>ta<g ref="char:cmbAbbrStroke">̄</g>gle parallelipi<g ref="char:EOLhyphen"/>pedo<g ref="char:cmbAbbrStroke">̄</g>, hauing his heith IK, equall to GH, the fourth line. And let that parallelipipedon be AK. Of the se<g ref="char:EOLhyphen"/>cond line CD, let a Cube be made: whose square base, let be noted wi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h CQ: and let his heith b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> no<g ref="char:EOLhyphen"/>ted by QL: &amp; let the whole Cube be signified by C<g ref="char:EOLhyphen"/>L. I say that AK is equall to CL. Let the like co<g ref="char:cmbAbbrStroke">̄</g>structi<g ref="char:EOLhyphen"/>on be for the cube of the third line: that is, vpon the square of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>H (which suppose to be GN) let a rectangle paralle<g ref="char:EOLhyphen"/>lipipedon be e<g ref="char:EOLhyphen"/>rected,
<pb n="349" facs="tcp:7062:378"/>
hauyng his heith NO, equall to A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, the first line: which parallelipipedon let be noted with GO. And suppose the cube of the third line (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>F) to be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> whose square base, let be noted by <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> and hys heith by RM. I say now (secondly) that GO is equal to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>M. For the first part consider,<note place="margin">
                           <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the first part.</hi>
                        </note> that AI (the square base of AK) is to CQ, the square base of CL, as A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to the third line EF, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. Corollary of the 20. of the sixth. But as A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, is to EF, so (by alternate proportion) is CD to GH, to CD. The cubes roote, is QL, the same cubes heith equall: and to GH is IK (by construction) equall: wherefore, as AI is to CQ, so is QL to IK. The bases therefore and heithes of AK and CL, are reciprocally in proporti<g ref="char:EOLhyphen"/>on: wherefore by the second part of this 34. proposition, AK and CL are equall.<note place="margin">
                           <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the second part.</hi>
                        </note> For proofe of the se<g ref="char:EOLhyphen"/>cond part of my theoreme, I say, that as AB, CD, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>F, and GH, are in continuall proportion forward, so are they backward in continuall proportion, as by the fourth of the fift may be proued. Wherefore now considering GH to be as first, and so A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to be the fourth: the square base GN, is to the square base <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>K, as GH is to CD, by the 2. corollary of the 20. of the sixth: But as GH is to CD, so is <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> to A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, by alternate proportion: to the Cubik roote <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>F, is RM (the heith of the same Cube <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>M) equall. And to AB, is the heith NO equall, by construction: wherfore as GN is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>R, so is RM to NO. Therfore by the second part of this 34. proposition, GO is equall to EN. If fowre right lines (therefore) be in continu<g ref="char:EOLhyphen"/>all proportion &amp;c. as in the proposition: which was required to be demonstrated.</p>
                  </div>
               </div>
               <div type="section">
                  <head>A Corollary logisticall.</head>
                  <p>Of my former Theoreme it followeth: Any two numbers being geuen, betwene which two we would haue two other numbers, middle, in continuall proportion:<note place="margin">To finde two midd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e propor<g ref="char:EOLunhyphen"/>tionals be<g ref="char:EOLhyphen"/>twene two numbers geue<g ref="char:cmbAbbrStroke">̄</g>.</note> That if we multiply the square of the first number geuen, by the other geuen number (as if it were the fourth): the roote Cubik of that of come or product, shall be the second number sought: And farther, if we multiply the square of the other number geuen, by the first geuen number, the roote Cubike of that of come shall be the thirde number sought.</p>
                  <p>For (by my Theoreme) those rectangle parallelipipedo<g ref="char:cmbAbbrStroke">̄</g>s made of the squares of the first &amp; fourth, multiplied by the fourth &amp; the first, accordingly, are equall to the Cubes made of the second &amp; third numbers: which we make our two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iddl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportionals<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wherefore of those parallelipipedo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s (as Cubes) the Cubik rootes, by good and vsuall arte sought and found, geue the very two middle num<g ref="char:EOLhyphen"/>bers desired. And where those numbers, are not by logisticall consideration accounted Cubik num<g ref="char:EOLhyphen"/>bers, ye may vse the logistical secret of approching nere to the precise verytye:<note place="margin">Note the prac<g ref="char:EOLunhyphen"/>tise of app<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o<g ref="char:EOLhyphen"/>ching to pre<g ref="char:EOLhyphen"/>cie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>es in Cu<g ref="char:EOLhyphen"/>bik rootes.</note> so that therof most easily you shall p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>aue, that your fayle is of the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nce neuer to be perceaued: it is to wete, as in a lyne of an inch long, not to want or exceede the thousand thousand part: or farther you may (infinitely approche at pleasure. O Mechanicall frend, be of good comfort, put to thy hand: <hi>Labor improbus, om<g ref="char:EOLhyphen"/>nia vincit.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>A Probleme. 1.</head>
                  <p>Vppon a right lined playne superficies geuen, to apply a rectangle parallelipipedon geuen.</p>
               </div>
               <div type="section">
                  <head>Or we may thus expresse the same thing.</head>
                  <p>
                     <hi>Vppon a right lined playne superficies geuen, to erect a rectangle parallelipipedon, equall to a rec<g ref="char:EOLhyphen"/>tangle parallelipipedon geuen.</hi>
                  </p>
                  <p>Suppose the right lined playne superficies geuen to be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: and the rectangle parallelipipedon ge<g ref="char:EOLhyphen"/>uen to be AM. Vppon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, as a base must AM be applyed: that is, a rectangle par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>llipipedon must be ere<g ref="char:EOLhyphen"/>cted vppon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>,
<figure/>
as a base, whi<g ref="char:EOLhyphen"/>che shall be e<g ref="char:EOLhyphen"/>quall to AM. By the laste of the second, to the right ly<g ref="char:EOLhyphen"/>ned figure <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, let an equall square be made: which suppose to be FRX.<note n="*" place="margin">
                        <hi>This is the way to apply any square geuen, to a line also geue<g ref="char:cmbAbbrStroke">̄</g>, sufficienty ex<g ref="char:EOLhyphen"/>tended.</hi>
                     </note> produce one side of the base of the pa<g ref="char:EOLhyphen"/>rallelipipedo<g ref="char:cmbAbbrStroke">̄</g>
                     <pb facs="tcp:7062:379"/>
AM, which
<figure/>
let be AC, extended to the point P. Let the o<g ref="char:EOLhyphen"/>ther side of the sayde base, con<g ref="char:EOLhyphen"/>curring with AC, be CG. As CG is to FR (the side of the square FRX) so let the same FR be to a line cut of from CP sufficiently extended: by the 11. of the sixth: and let that third proportionall line be CP. Let the rectangle parallelogramme be made perfect, as CD. It is euident, that CD, is equall to the square FRX by the 17. of the sixth: and by construction FRX, is equall to B. Wherfore CD, is equall to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. By the 12. of the sixth, as CP, is to AC, so let AN (the heith of AM) be to the right line O. I say that a solide perpen<g ref="char:EOLhyphen"/>dicularly erected vppon the base <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, hauinge the heith of the line O, is equall to the parallelipipedon AM. For CD is to AG, as CP is to A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> by the firste of the sixth, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is proued equall to CD: Wherfore by the 7. of the fifth, B is to AG as CP is to AC: But as CP is to AC, so is AN to O, by construction: Wherefore B is to AG as AN is to O. So than the bases <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and AG are reciprocally in proportion with the heithes AN and O. By this 34 therefore, a solide erected perpendicularly vppon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> as a base, hauing the height O, is equall to AM. Wherefore vppon a right lyned playn superficies geuen, we haue applied a rectangle parallelipipedon geuen: Which was requisite to be donne.</p>
               </div>
               <div type="section">
                  <head>A Probleme 2.</head>
                  <p>
                     <hi>A rectangle parallelipipedon being geuen to make an other equall to it of any heith assigned.</hi>
                  </p>
                  <p>Suppose the rectangle parallelipipedon geuen to be A, and the heith assigned to be the right line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: Now must we make a rectangle parallelipipedon, equal to A: Whose heith must be equall to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Ac<g ref="char:EOLhyphen"/>cording to the manner before vsed, we must frame our co<g ref="char:cmbAbbrStroke">̄</g>struction to a reciprokall proportio<g ref="char:cmbAbbrStroke">̄</g> betwene the bases and heithes. Which will be done if, as the heith assigned beareth it selfe in proportion to the heith of the parallelipipedon giuen: so, one of the sides of the base of the parallelipipedon giuen, be to a fourth line, by the 12. of the sixth found. For that line founde, and the other side of the base of the ge<g ref="char:EOLhyphen"/>uen parallelipipedon, contayne a parallelogramme, which doth serue for the base, (which onely, we wanted) to vse with our giuen heith: and so is the Probleme to be executed.</p>
               </div>
               <div type="section">
                  <head>Note.</head>
                  <p>
                     <hi>Euclide</hi> in the 27. of this eleuenth hath taught, how, of a right line geue<g ref="char:cmbAbbrStroke">̄</g>, to describe a parallepipedo<g ref="char:cmbAbbrStroke">̄</g>, like, &amp; likewise situated, to a parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> geue<g ref="char:cmbAbbrStroke">̄</g>: I haue also added, How, to a parallepipedon geuen, an other may be made equall, vppon any right lined base geuen, or of any heith assigned: But if either Euclide, or any other before our time (answerably to the 25. of the sixth, in playns) had among solids inuented this proposi<g ref="char:EOLhyphen"/>tion:<note place="margin">
                        <hi>A probleme worth the searching for.</hi>
                     </note>
                     <q>
                        <hi>Two vnequall and vnlike parallelipipedons being geuen, to describe a parallelipipedon equall to the one, and like to the other,</hi> we would haue geuen them their deserued praise: and I would also haue ben right glad to haue ben eased of my great trauayles and discourses about the inuenting thereof.</q>
                  </p>
                  <trailer>
                     <hi>Here ende I. Dee his additions vppon this</hi> 34. <hi>Proposition.</hi>
                  </trailer>
               </div>
            </div>
            <div type="part">
               <head>The 30. Theoreme. The 35. Proposition. If there be two superficiall angles equall, and from the pointes of those an<g ref="char:EOLhyphen"/>gles
<pb n="350" facs="tcp:7062:379"/>
be eleuated on high right lines, comprehending together with those right lines which containe the superficiall angles, equall angles, eche to his corespo<g ref="char:cmbAbbrStroke">̄</g>dent angle, and if in eche of the eleuated lines be take<g ref="char:cmbAbbrStroke">̄</g> a point at all auentures, and from those pointes be drawen perpendicular lines to the ground playne superficieces in which are the angles geuen at the begin<g ref="char:EOLhyphen"/>ning, and from the pointes which are by those perpendicular lines made in the two playne superficieces be ioyned to those angles which were put at the beginning right lines: those right lines together with the lines eleua<g ref="char:EOLhyphen"/>ted on high shall contayne equall angles.</head>
               <p>SVppose that these two rectiline superficiall angles BAC, and EDF be equall the one to the other:<note place="margin">Construction.</note> and from the pointes A and D let there be eleuated vpward these right lines AG and DM, comprehendinge together with the lines put at the beginninge equall angles, ech to his correspondent angle, that is, the angle MDE to the angle GAB, and the angle MDF to the angle GAC, and take in the lines AG and DM pointes at all aue<g ref="char:cmbAbbrStroke">̄</g>tures and let the same be G and M. And (by the 11. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) from the pointes G and M draw vnto the ground playne superficieces wherein are the an<g ref="char:EOLhyphen"/>gles BAC and E
<figure/>
DF perpendicular lines GL and MN and let them fall in the sayd playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces in the pointes N and L, and drawe a right line from the point L to the point A and an other from the pointe N to the pointe D. Then I say that the angle GAL is equall to the angle MDN. Fro<g ref="char:cmbAbbrStroke">̄</g> the greater of the two lines AG and DM, (which let be AG) cut of by the 3. of the first the line AH equall vnto the line DM. And (by the 31. of the first) by the point H, drawe vnto the line GL a parallel line, and let the same be HK. Now the line GL is erected perpendicularly to the grounde playne superficies BAL: Wherfore also (by the 8. of the eleuenth) the line HK is erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the same grounde plaine superficies BAC. Drawe (by the 12. of the first) fro<g ref="char:cmbAbbrStroke">̄</g> the pointes K and N vnto the right lines AB, AC, DF, &amp; DE perpe<g ref="char:cmbAbbrStroke">̄</g>dicular right lines, and let the same be KC, NF, KB, NE. And drawe these right lines HC, CB, MF, FE. Now forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (by the 47. of the first) the square of the line HA is equall to the squares of the lines HK and KA,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> but vnto the square of the line KA are equall the squares of the lines KC and CA: Wherefore the square of the line HA is equall to the squares of the lines HK, KC and CA. But by the same vnto the squares of the lines HK and KC is equall the square of the line HC: Wherefore the square of the line HA is equall to the squares of the lines HC and CA: wherfore the angle HCA is (by the 48. of the first) a right angle. And by the same reason also the angle MFD is a right angle. Wherefore the angle HCA is equall to the angle MFD. But the angle HAC is (by suppositio<g ref="char:cmbAbbrStroke">̄</g>) equal to the angle MDF. Wherfore there are two triangles MDF and HAC hauing two an<g ref="char:EOLhyphen"/>gles of the one equall to twoo angles of the other, eche to his correspondent angle, and one side of the one equall to one side of the other, namely, that side which subtendeth one of the equall
<pb facs="tcp:7062:380"/>
angles, that is, the side HA is equall to the side DM by construction. Wherefore the sides remayning are (by the 26. of the first) equall to the sides remayning. Wherefore the side AC is equall to the side DF. In like sort may we proue that the side AB is equall to the side DE, if ye drawe a right line from the point H to the point B, and an other from the point M to the point E. For forasmuch as the square of the line AH is (by the 47. of the firste) equall to the squares of the lines AK and KH, and (by the same) vnto the square of the line AK are equall the squares of the lines AB and BK. Wherefore the squares of the lines AB, BK, and KH are equall to the square of the line AH. But vnto the squares of the lines BK and KH is equall the square of the line BH (by the 47. of the first) for the angle HKB is a right angle, for that the line HK is erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the ground playne superficies: Wherefore the square of the line AH is equall to the squares of the lines AB and BH. Wherefore (by the 48. of the first) the angle ABH is a right angle. And by the same rea<g ref="char:EOLhyphen"/>son the angle DEM is a right angle. Now the angle BAH is equall to the angle EDM, for it is so supposed, and the line AH is equall to the line DM. Wherefore (by the 26. of the firste) the line AB is equall to the line DE. Now forasmuch as the line AC is equall to the line DF, and the line AB to the line DE, therefore these two lines AC and AB are equall to these two lines FD and DE. But the angle also CAB is by suppositi<g ref="char:EOLhyphen"/>on equall to the angle FDE. Wherefore (by the 4. of the firste) the base BC is equall to the base EF, and the triangle to the triangle, and the rest of the angles to the reste of the angles. Wherefore the angle ACB is equall to the angle DFE. And the right angle ACK is equal to the right angle DFN. Wherfore the angle remayning, namely, BCK, is equall to the an<g ref="char:EOLhyphen"/>gle remayning, namely, to EFN. And by the same reaso<g ref="char:cmbAbbrStroke">̄</g> also the angle CBK is equal to the angle FEN. Wher<g ref="char:EOLunhyphen"/>fore
<figure/>
there are two triangles BCK, &amp; EFN, hauing two angles of the one equal to two angles of the other, eche to his correspondent angle, and one side of the one equall to one side of the o<g ref="char:EOLhyphen"/>ther, namely, that side that lieth betwene the equall angles, that is the side BC is equall to the side EF: Where<g ref="char:EOLhyphen"/>fore (by the 26. of the first) the sides remaininge are equall to the sides remayning. Wherfore the side CK is equall to the side FN: but the side AC is equall to the side DF. Wherefore these two sides AC and CK are equall to these two sides DF and FN, and they contayne equall angles: Wherefore (by the 4. of the first) the base AK is equall to the base DN. And forasmuch as the line AH is equall to the line DM, therefore the square of the line AH is equall to the square of the line DM. But vnto the square of the line AH are equall the squares of the lines AK and KH (by the 47. of the first) for the angle AKH is a right angle. And to the square of the line DM are equall the squares of the lines DN and NM, for the angle DNM is a right angle. Wherefore the squares of the lines AK and KH are equall to the squares of the lines DN and NM: of which two, the square of the line AK is equall to the square of the line DN (for the line AK is proued equall to the line AN). Wherefore the residue, namely, the square of the line KH is equal to the residue, namely, to the square of the line NM. Wherefore the line HK is equall to the line MN. And forasmuch as these two lines HA and AK are equall to these two lines MD and DN, the one to the other, and the
<pb n="351" facs="tcp:7062:380"/>
base HK is equall to the base MN: therfore (by the 8. of the first) the angle HAK is equall to the angle MDN. If therefore there be two superficiall angles equall, and fro<g ref="char:cmbAbbrStroke">̄</g> the pointes of those angles be eleuated on high right lines, comprehending together with those right lines which were put at the beginning, equall angles, ech to his corespondent angle, and if in ech of the erected lines be taken a point at all aduentures, and from those pointes be drawen perpen<g ref="char:EOLhyphen"/>dicular lines to the plaine superficieces in which are the angles geuen at the beginning, and fr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the pointes which are by the perpendicular lines made in the two plaine superficieces be ioyned right lines to those angles which were put at the beginning, those right lines shall to<g ref="char:EOLhyphen"/>gether with the lines eleuated on high make equall angles which was required to be proued.</p>
               <p>
                  <hi>Because the figures of the former demonstration are somewhat hard to conceaue as they are there drawen in a plaine, by reason of the lines that are imagined to be eleuated on high, I haue here set o<g ref="char:EOLhyphen"/>ther figures, wherein you must e<g ref="char:EOLhyphen"/>recte
<figure/>
perpendicularly to the ground superficieces the two triangles BHK, and EMN, and then ele<g ref="char:EOLhyphen"/>uate the triangles DFM, &amp; ACH, in such sort that the angles M and H of these triangles, may concurre with the angles M and H of the o<g ref="char:EOLhyphen"/>ther erected triangles. And then imagining only a line to be drawen from the point G of the line AG to the point L in the ground superfi<g ref="char:EOLhyphen"/>cies, compare it with the former construction &amp; demonstration, and it will make it very easye to con<g ref="char:EOLhyphen"/>ceaue.</hi>
               </p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>By this it is manifest, that if there be two rectiline superficiall angles e<g ref="char:EOLhyphen"/>quall, and vpon those angles be eleuated on high equall right lines contay<g ref="char:EOLhyphen"/>ning together with the right lines put at the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ginning equall angles: per<g ref="char:EOLhyphen"/>pendicular lines drawen from those eleuated lines to the ground plaine su<g ref="char:EOLhyphen"/>perficieces wherein are the angles put at the beginning, are equall the one to the other. <hi>For it is manifest, that the perpendicular lines HK, &amp; MN, which are dra<g ref="char:EOLhyphen"/>wen from the endes of the equall eleuated lines AH, and DM, to the ground superficieces, are equall.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶The 31. Theoreme. The 36. Proposition. If there be three right lines proportionall: a Parallelipipedon described of those three right lines, is equall to the Parallelipipedon described of the middle line, so that it consiste of equall sides, and also be equiangle to the foresayd Parallelipipedon.</head>
               <p>SVppose that these three lines A, B, C, be proportionall, as A is to B, so let B be to C. Then I say, that the Parallelipipedon made of the lines A, B, C, is equall to the Pa<g ref="char:EOLhyphen"/>rallelipipedon made of the line B, so that the solide made of the line B consist of e<g ref="char:EOLhyphen"/>quall
<pb facs="tcp:7062:381"/>
sides, and be also equiangle to the solide made of the lines A, B, C. Describe (by the 23. of the eleuenth) a solide angle E contained vnder three superficiall angles,<note place="margin">Construction.</note> that is, DEG, GEF, and FED: and (by the 3. of the first) put vnto the line B euery one of these lines DE, GE, &amp; EF, equall:
<figure/>
and make perfecte the so<g ref="char:EOLhyphen"/>lide EK. And vnto the line A let the line LM be equall. And (by the 26. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) vnto the right line LM, and at the point in it L, describe vnto the solide angle E an equall so<g ref="char:EOLhyphen"/>lide angle, co<g ref="char:cmbAbbrStroke">̄</g>tained vnder these plaine superficiall an<g ref="char:EOLhyphen"/>gles NLX, XLM, and NLM, and vnto the line B put the line LX equall, &amp; the line LN to the line C. Now for that as the line A is to the line B,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> so is the line B to the line C: but the line A is equall to the line LM, and the line B to eue<g ref="char:EOLhyphen"/>ry one of these lines LX, EF, EG, and ED, and the line C to the line LN. Wherefore as LM is to EF, so is DE to LN: So then the sides about the e<g ref="char:EOLhyphen"/>qual angles MLN, &amp; D<g ref="char:EOLhyphen"/>EF, are reciprokall: Wher<g ref="char:EOLhyphen"/>fore (by the 14. of the sixt) the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me MN is equall to the parallelogramme DF. And forasmuch as two plaine superficiall angles, name<g ref="char:EOLhyphen"/>ly, DEF and NLM are equall the one to the other, and vpon them are erected vpward e<g ref="char:EOLhyphen"/>quall right lines, LX and EG; comprehending with the right lines put at the beginning e<g ref="char:EOLhyphen"/>quall angles the one to the other. Wherefore <note n="*" place="margin">It is euide<g ref="char:cmbAbbrStroke">̄</g>t that those per<g ref="char:EOLhyphen"/>pendiculars are all one with the sta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>ding lines of the solides, if their solide an<g ref="char:EOLunhyphen"/>gles be made of super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<g ref="char:EOLhyphen"/>c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>il right an<g ref="char:EOLhyphen"/>gles onely.</note> perpendicular lines drawen from the pointes X and G to the plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces wherin are the angles NLM, and DEF, are (by the Co<g ref="char:EOLhyphen"/>rollary of the former Proposition) equall the one to the other: and those perpendiculars are the altitudes of the Parallelipipedons LH and EK, by the 4. definition of the sixt. Wherfore the solides LH and EK, are vnder one and the selfe same altitude. But Parallelipipedons consisting vpon equall bases, and being vnder one and the selfe altitude, are (by the 31. of the eleuenth) equall the one to the other. Wherefore the solide LH is equall to the solide EK. But the solide LH is described of the lines A, B, C, and the solide EK is described of the line B. Wherefore the Parallelipipedon described of the lines A, B, C, is equall to the Pa<g ref="char:EOLhyphen"/>rallelipipedon made of the line B, which consisteth of equall sides, and is also equiangle to the foresaid Parallelipipedon. If therfore there be three right lines proportionall, a Parallelipipe<g ref="char:EOLhyphen"/>do<g ref="char:cmbAbbrStroke">̄</g> described of those three lines is equall to the Parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> described of the middle line, so that is consist of equall sides, and also be equiangle to the foresaid Parallelipipedon: which was required to be proued.</p>
               <p>
                  <pb n="352" facs="tcp:7062:381"/>
                  <hi>The construction and demonstration of this Proposition, and of the next Proposition following, may easily be conceaued and vnderstanded by the figures described in the plaine belonging to them. But ye may for the more full sight of them, describe such bodies of pasted paper, hauing their sides pro<g ref="char:EOLhyphen"/>portionall, as is required in the Propositions.</hi>
               </p>
               <div type="section">
                  <head>¶New inuentions (coincident) added by Master Iohn Dee.</head>
                  <div type="subsection">
                     <head>A Corollary. 1.</head>
                     <p>
                        <hi>Hereby it is euident, that if three right lines be proportionall: the Cube produced of the middle line, is equall to the rectangle Parallelipipedon made of those three lines.</hi>
                     </p>
                     <p>For a Cube is a Parallelipipedon of equall sides: and also rectangled: as we suppose the Parallelipi<g ref="char:EOLhyphen"/>pedon, made of the three lines to be likewise rectangled.</p>
                     <div type="problem">
                        <head>¶A Probleme. 1.</head>
                        <p>
                           <hi>A Cube being geuen, to finde three right lines proportionall, in any proportion geuen betwene two right lines: of which three lines, the rectangle Parallelipipedon produced, shall be equall to the Cube geuen.</hi>
                        </p>
                        <p>Suppose AC to be the
<figure/>
Cube geuen: whose roote, suppose to be AB. Let the proportion geuen, be that which in betwene the two right lines D and E, I say now, three right lines are to be found, proportionall, in the proportion of D to E, of which, the rectangle Pa<g ref="char:EOLhyphen"/>rallelipipedon produced, shall be equall to AC. By the 12. of the sixt let a line be found, which to AB haue that proportion that D hath to E. Let that line be F: and by the same 12. of the sixth, let an other line be found, to which, AB, hath that proportion that D hath to E: and let that line found be H. Let a rectangle Paral<g ref="char:EOLhyphen"/>lelipipedon mathematically be produced of the three right lines F, AB, and H, which suppose to be K: I say now, that F, AB, and H, are three right lines found pro<g ref="char:EOLhyphen"/>portionall in the proporti<g ref="char:EOLhyphen"/>on of D to E, of which, the rectangle Parallelipipedon K, produced, is equall to AC the Cube geuen. First it is euident that F, AB, and H, are proportionall in the proportion of D to E. For, by construction, as D is to E, so is F to AB: and by construction likewise, as D is to E, so is AB to H. Wherefore F is to AB, and AB is to H, as D is to E. So then it is manifest, F, AB, and H, to be proportionall in the proportion of D to E, and AB to be the middle line. By my former Corollary, therefore, the rectangle parallelipipedon made of F, AB, and H, is equall to the Cnbe made of AB. But AC, is (by supposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ion) the Cube made of AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and of the three lines F, AB, and H, the rectangle parallelipipedon produced, is K, by construction: Wher<g ref="char:EOLhyphen"/>fore, K, is equall to AC: A Cube being geuen, therefore, three right lines are found, proportionall in
<pb facs="tcp:7062:382"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ny prop<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>ion geue<g ref="char:cmbAbbrStroke">̄</g> be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>wene two right lines, of which three right lines the rectangle parallelipipedon p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>oduced, is equall to the Cube geuen. Which ought to be done.</p>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 2.</head>
                        <p>
                           <hi>A re<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>angle Parallelipipedon being geuen, to finde three right lines proportionall: of the which, the rectangle Parallelipipedon produced, is equall to the rectangle Parallelipipedon geuen.</hi>
                        </p>
                        <p>
                           <note place="margin">
                              <hi>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ling of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                                    <desc>〈◊〉</desc>
                                 </gap>. &amp;c.</hi>
                           </note>Listen to this new deuise, you couragious Mathematiciens: consider, how nere this crepeth to the famous Probleme of doubling the Cube. What hope may (in maner) any young beginner co<g ref="char:cmbAbbrStroke">̄</g>ceiue, by one meanes or other, at one time or other, to execute this Probleme? <note n="*" place="margin">
                              <hi>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>m<gap reason="illegible" resp="#PDCC" extent="4 letters">
                                    <desc>••••</desc>
                                 </gap>rati<g ref="char:EOLhyphen"/>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> 
                                 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>f p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ibilitie in the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>oblem.</hi>
                           </note> Seing to a Cube may in<g ref="char:EOLhyphen"/>finitely in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inite Parallelipipedons be found equall: all which Parallelipipedons shall be produced of three right lines proportionall, by the fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>obleme: but to any <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ectangle Parallelipipedon geuen, some one Cube is equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> as is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>asie to demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>at<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>: We can not doubt, but vnto our rectangle Paral<g ref="char:EOLhyphen"/>lelipipedon geuen, many other <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ectangle Parallelipipedons are also equall, hauing their three lines of p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>oduction, proportionall.<note place="margin">
                              <hi>An other argu<g ref="char:EOLhyphen"/>me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>t to com<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>t the studious.</hi>
                           </note> In th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> former Probleme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ini<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ly in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>inite Parallelipipedons may be found of three proportionall lines p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>oduced, equall to the Cube g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>uen: it is to wete, the three lines to be of all proportions, that a man can deuise betwene two right lines: and here any one will serue: where also i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> infinite varietie: though all of one quantitie: as be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ore in the Cube. I leaue as now, with thys marke here set vp to shoote at. Hit it who can.</p>
                     </div>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 32. Theoreme. The 37. Proposition. If there be fower right lines proportionall: the Parallelipipedons descri<g ref="char:EOLhyphen"/>bed of those lines, being like and in like sort described, shall be proportio<g ref="char:EOLhyphen"/>nall. And i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the Parallelipipedons described of them, being like and in like sort described, be proportionall: those right lines also shall be proportionall.</head>
               <p>SVppose that these fower right lines AB, CD, EF, and GH, be proportionall, as AB is to CD, so let EF be to GH, and vpon the lines AB, CD, EF, and GH, describe these Parallelipipedons KA, LC, ME, and NG, being like and in like sort desc<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>bed. Then I say, that as the solide KA is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the solide LC, so is the solide ME to the solide
<figure/>
NG.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the first part.</note> For forasmuch as the Pa<g ref="char:EOLhyphen"/>rallelipipedon KA is like to the Parallelipipedon LC: therfore (by the 33. of the eleuenth) the solide KA is to the solide LC in treble proportion of that which the side AB is to the side CD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>: and by the same reason the Parallelipipedon ME is to the Parallelipipedon NG in treble proportion of that which the side EF is to the side GH. Wherfore (by the 11. of the fift) as the Parallelipipedon KA is to the Parallelipipedon LC, so is the Parallelipipedon ME t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the Parallelipipedon NG.</p>
               <p>But now suppose, that as the Parallelipipedon KA is to
<pb n="353" facs="tcp:7062:382"/>
the Parallelipipedon LC, so is the Parallelipipedon ME to the Parallelipipedon NG. Then I say,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the second part, which is the conuerse of the first part.</note> that as the right line AB is to the right line CD, so is the right li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> EF to the right line GH. For againe forasmuch as the solide KA is to the solide LC in treble proportion of that which the side AB is to the side CD, and the solide ME also is to the solide NG in treble proportion of that which the line EF is to the line GH, and as the solide KA is to the solide LC, so is the solide ME to the solide NG. Wherefore also as the line AB is to the line CD, so is the line EF to the line GH. If therefore there be fower right lines proportionall: the Parallelipipedons described of those lines, being like &amp; in like sort described, shall be pro<g ref="char:EOLhyphen"/>portionall. And if the Parallelipipedons described of them, and being like and in like sort de<g ref="char:EOLhyphen"/>scribed, be proportionall: those right lines also shall be proportionall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 33. Theoreme. The 38. Proposition. If a plaine superficies be erected perpendicularly to a plaine superficies, and from a point taken in one of the plaine superficieces be drawen to the other plaine superficies, a perpendicular line: that perpendicular line shall fall vpon the common section of those plaine superficieces.</head>
               <p>SVppose that the plaine superficies CD be erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the plaine superfi<g ref="char:EOLhyphen"/>cies AB, and let their common section be the line DA: and in the superficies CD take a point at all aduentures, and let the same be E. Then I say, that a perpendicu<g ref="char:EOLhyphen"/>lar line drawen from the point E to the
<figure/>
plaine superficies AB, shall fall vpon the right line DA. For if not, then let it fall without the line DA, as the line EF doth, and let it fall vpon the plaine su<g ref="char:EOLhyphen"/>perficies AB in the point F.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> And (by the 12. of the first) from the point F draw vnto the line DA, being in the superficies AB a perpendicular line F<g ref="char:EOLhyphen"/>G, which line also is erected perpendicu<g ref="char:EOLhyphen"/>larly to the plaine superficies CD: by the third diffinitio<g ref="char:cmbAbbrStroke">̄</g>: by reason we presuppose CD and AB to be perpendicularly erec<g ref="char:EOLhyphen"/>ted ech to other. Draw a right line from the point E to the point G. And foras<g ref="char:EOLhyphen"/>much as the line FG is erected perpendi<g ref="char:EOLhyphen"/>cularly to the plaine superficies CD, and the line EG toucheth it being in the superficies CD. Wherefore the angle FGE is (by the 2. definition of the eleuenth) a right angle. But the line EF is also erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the superficies AB: wherefore the angle EFG is a right angle. Now therefore two angles of the triangle EFG, are equall to two right angles: which (by the 17. of the first) is impossible. Wherfore a perpendicular line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the point E to the s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>erficies AB, falleth not with<g ref="char:EOLhyphen"/>out the line DA. Wherefore it falleth vpon the line DA: which was required to be proued.</p>
               <div type="section">
                  <head>¶ Note.</head>
                  <p>
                     <hi>Campane</hi> maketh this as a Corollary, following vpon the 13: and very well, with small ayde of other Propositions, he proueth it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> whose demonstratio<g ref="char:cmbAbbrStroke">̄</g> there, <hi>Flussas</hi> hath in this place, and none other: though he sayth that <hi>Campane</hi> of such a Propositio<g ref="char:cmbAbbrStroke">̄</g>, as of <hi>Euclides,</hi> maketh no mention.</p>
                  <p>
                     <pb facs="tcp:7062:383"/>In this figure ye may more fully see the former Proposi<g ref="char:EOLhyphen"/>tion
<figure/>
and demonstration if ye erecte perpendicularly vnto the ground plaine superficies AB the superficies CD, and imagine a line to be extended from the point E to the point F, instede whereof ye may extend if ye will a thred.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 34. Theoreme. The 39. Proposition. If the opposite sides of a Parallelipipedon be deuided into two equall partes, and by their common sections be extended plaine superficieces: the commo<g ref="char:cmbAbbrStroke">̄</g> section of those plaine superficieces, and the diameter of the Paral<g ref="char:EOLhyphen"/>lelipipedon shall deuide the one the other into two equall partes.</head>
               <p>SVppose that AF be a Parallelipipedon, and let the opposite sides thereof CF and AH be deuided into two equall partes in the pointes K, L, M, N, and likewise let the opposite sides AD and GF be deuided into two equall partes in the point<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s X, P, O, R, and by those sections extend these two plaine superficieces KN &amp; XR, and let the common section of those plaine superficieces be the line VS, and let the diagonall line of the solide AB be the line
<figure/>
DG. Then I say, that the lines VS and DG do deuide the one the o<g ref="char:EOLhyphen"/>ther into two equall partes,<note place="margin">Construction.</note> that is, that the line VT is equall to the line TS, and the line DT to the line TG. Drawe these right lines DV, VE, BS, and SG.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now for<g ref="char:EOLhyphen"/>asmuch as the line DX is a parallel to the line OE, therfore (by the 29. of the first) the angles DXV and VOE being alternate angles, are equall the one to the other. And forasmuch as the line DX is equall to the line OE, and the line XV to the line VO, and they comprehend equall angles: Wherefore the base DV is equall to the base VE (by the 4. of the first) and the triangle DXV is equall to the triangle VOE, and the rest of the angles to the rest of the angles. Wherefore the angle XVD is equall to the angle OVE. Wherefore DVE is one right line, and by the same reason BSG is also one right line, and the line BS is equall to the line SG. And forasmuch as the line CA is equall to the line DB, and is vn<g ref="char:EOLhyphen"/>to it a parallel, but the line CA is equall to the line GE, and is vnto it also a parallel: wher<g ref="char:EOLhyphen"/>fore (by the firs<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> common sentence) the line DB is equall to the line GE, &amp; is also a parallel vnto it: but the right lines DE and BG do ioyne these parallel lines together: Wherefore (by the 33. of the first) the line DE is a parallel vnto the line BG. And in either of these lines
<pb n="354" facs="tcp:7062:383"/>
are taken pointes at all aduentures, namely, D, V, G, S, and a right line is drawen from the point D to the point G, and an other from the point V to the point S. Wherefore (by the 7. of the eleuenth) the lines DG and VS are in one and the selfe same plaine superficies. And for<g ref="char:EOLhyphen"/>asmuch as the line DE is a parallel to the line BG, therefore (by the 24. of the first) the an<g ref="char:EOLhyphen"/>gle EDT is equall to the angle BGT, for they are alternate angles, and likewise the angle DTV is equall to the angle GTS. Now then there are two triangles, that is, DTV and GTS, hauing two angles of the one equall to two angles of the other, and one side of the one equall to one side of the other, namely, the side which subtendeth the equall angles, that is, the side DV to the side GS, for they are the halfes of the lines DE and BG: Wherefore the sides remayning are equall to the sides remayning. Wherfore the line DT is equall to the line TG, and the line VT to the line TS. If therefore the opposite sides of a Parallelipipedon be de<g ref="char:EOLhyphen"/>uided into two equall partes, and by their sections be extended plaine superficieces, the com<g ref="char:EOLhyphen"/>mon section of those plaine superficieces, and the diameter of the Parallelipipedon, do deuide the one the other into two equall partes: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Euery playne superficies extended by the center of a parallelipipedon, diuideth that solide into two equall partes: and so doth not any other playne superficies not extended by the center.</hi>
                  </p>
                  <p>For euery playne extended by the center, cutteth the diameter of the parallelipipedon in the cen<g ref="char:EOLhyphen"/>ter into two equall partes. For it is proued, that playne superficieces which cutte the solide into two equall partes, do cut the dimetient into two equall partes in the center. Wherefore all the lines drawen by the center in that playne superficies shall make angles with the dimetient. And forasmuch as the di<g ref="char:EOLhyphen"/>ameter falleth vpon the parallel right lines of the solide, which describe the opposite sides of the sayde solide, or vpon the parallel playne superficieces of the solide, which make angels at the endes of the diameter: the triangles contayned vnder the diameter, and the right line extended in that playne by the center, and the right line, which being drawen in the opposite superficieces of the solide, ioyneth together the endes of the foresayde right lines, namely, the ende of the diameter, and the ende of the line drawen by the center in the superficies extended by the center, shall alwayes be equall, and equi<g ref="char:EOLhyphen"/>angle, by the 26. of the first. For the opposite right lines drawen by the opposite playne superficieces of the solide do make equall angles with the diameter, forasmuch as they are parallel lines, by the 16. of this booke. But the angles at the ce<g ref="char:cmbAbbrStroke">̄</g>ter are equall, by the 15. of the first, for they are head angles: &amp; one side is equall to one side, namely, halfe the dimetient. Wherefore the triangles contayned vnder e<g ref="char:EOLhyphen"/>uery right line drawen by the center of the parallelipipedon in the superficies, which is extended also by the sayd center, and the diameter thereof, whose endes are the angles of the solide, are equall, equi<g ref="char:EOLhyphen"/>later, &amp; equiangle (by the 26. of the first). Wherfore it followeth that the playne superficies which cut<g ref="char:EOLhyphen"/>teth the parallelipipedon, doth make the partes of the bases on the opposite side, equall, and equiangle, and therefore like, and equall both in multitude, and in magnitude: wherefore the two solide sections of that solide, shalbe equall and like, by the 8. diffinition of this booke. And now that no other playne superficies, besides that which is extended by the center, deuideth the parallelipipedon into two equall partes, it is manifest: if vnto the playne superficies which is not extended by the center, we extend by the center a parallel playne superficies (by the Corollary of the 15. of this booke). For forasmuch as that superficies which is extended by the center, doth deuide the parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> into two equall par<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>: it is manifest, that the other playne superficies (which is parallel to the superficies which deuideth the solide into two equall partes) is in one of the equall partes of the solide: wherefore seing that the whole is euer greater then his partes, it must nedes be that one of these sections is lesse then the halfe of the solide, and therefore the other is greater.</p>
                  <p>For the better vnderstanding of this former proposition, &amp; also of this Corollary added by <hi>Flussas,</hi> it shalbe very nedefull for you to describe of pasted paper or such like matter a parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> or a Cube, and to deuide all the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes therof into two equall parts, by drawing by the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:cmbAbbrStroke">̄</g>ters of the sayd parallelogrammes (which centers are the poynts made by the cutting of diagonall lines drawen fro<g ref="char:cmbAbbrStroke">̄</g> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> opposite angles of the sayd parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes) lines parallels to the sides of the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes: as in the former figure described in a plaine ye may see, are the sixe parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes DE, EH, HA, AD, DH, and CG, whom these parallel lines drawen by the ce<g ref="char:cmbAbbrStroke">̄</g>ters of the sayd parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes, namely, XO, OR, PR, and PX, do deuide into two equall parts: by which fower lines ye must imagine a playne superfi<g ref="char:EOLhyphen"/>cies to be extended, also these parallel lynes KL, LN, NM, and M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, by which fower lines likewise y<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> must imagine a playne superficies to be extended ye: may if ye will put within your body made thus of pasted paper, two superficieces made also of the sayd paper, hauing to their limites lines equall to the foresayde parallel lines: which superficieces must also be deuided into two equall partes by parallel
<pb facs="tcp:7062:384"/>
lines drawen by their centers, and must cut the one the other by these parallel lines. And for the dia<g ref="char:EOLhyphen"/>meter of this body, exte<g ref="char:cmbAbbrStroke">̄</g>d a thred from one angle in the base of the solide to his opposite angle, which shall passe by the center of the parallelipipedon, as doth the line DG in the figure before described in the playne. And draw in the base and the opposite superficies vnto it, Diagonall lines, from the angles from which is extended the diameter of the solide: as in the former description are the lines BG and DE. And when you haue thus described this body, compare it with the former demonstration, and it will make it very playne vnto you, so your letters agree with the letters of the figure described in the booke. And this description will playnely set forth vnto you the corollary following that proposition. For where as to the vnderstanding of the demonstration of the proposition the superficieces put within the body were extended by parallel lynes drawen by the ce<g ref="char:cmbAbbrStroke">̄</g>ters of the bases of the parallelipipe<g ref="char:EOLhyphen"/>don: to the vnderstanding of the sayd Corollary, ye may extende a superficies by any other lines dra<g ref="char:EOLhyphen"/>wen in the sayd bases, so that yet it passe through the middest of the thred, which is supposed to be the center of the parallelipipedon.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 35. Theoreme. The 40. Proposition. If there be two Prismes vnder equall altitudes, &amp; the one haue to his base a parallelogramme, and the other a triangle, and if the parallelogramme be double to the triangle: those Prismes are equall the one to the other.</head>
               <p>
                  <hi>SVppose that these two Prismes ABCDEF, GHKMON, be vnder equall altitudes, and let the one haue to his base the parallelogramme AC, and the o<g ref="char:EOLhyphen"/>ther the triangle GHK, and let the parallelogramme AC be double to the tri<g ref="char:EOLhyphen"/>angle GHK. Then I say, that the Prisme ABCDEF is equall to the Prisme GHKMON.</hi>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <hi>Make perfecte
<figure/>
the Parallelipipedons AX &amp; GO. And forasmuch as the parallelo<g ref="char:EOLhyphen"/>gramme AC is double to the tri<g ref="char:EOLhyphen"/>angle GHK,</hi>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> 
                  <hi>but the parallelo<g ref="char:EOLhyphen"/>gramme GH is also (by the 41. of the first) double to the triangle GHK, wherefore the parallelo<g ref="char:EOLhyphen"/>gramme AC is equall to the pa<g ref="char:EOLhyphen"/>rallelogramme GH. But Parallelipipedons consisting vpon equall bases and vnder one and the selfe same altitude, are equall the one to the other (by the 31. of the eleuenth). Wherefore the solide AX is equall to the solide GO. But the halfe of the solide AX is the Prisme AB<g ref="char:EOLhyphen"/>CDEF, and the halfe of the solide GO is the Prisme GHKMON. Wherfore the Prisme ABCDEF is equall to the Prisme GHKMON. If therefore there be two Prismes vn<g ref="char:EOLhyphen"/>der equall altitudes, and the one haue to his base a parallelogramme, &amp; the other a triangle, and if the parallelogramme be double to the triangle: those Prismes are equall the one to the other: which was required to be proued.</hi>
               </p>
               <p>This Proposition and the demonstration thereof are not hard to conceaue by the former figures: but ye may for your fuller vnderstanding of the<g ref="char:cmbAbbrStroke">̄</g> take two equall Parallelipipedons equilate<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and equi<g ref="char:EOLhyphen"/>angle the one to the other described of pasted paper or such like matter, and in the base of the one Pa<g ref="char:EOLhyphen"/>rallelipipedon draw a diagonall line, and draw an other diagonall line in the vpper superficies opposite vnto the said diagonall line drawen in the base. And in one of the parallelogrammes which are set vp<g ref="char:EOLhyphen"/>on the base of the other Parallelipipedon draw a diagonall line, and drawe an other diagonall line in the parallelogramme opposite to the same. For so if ye extend plaine superficieces by those diagonall lines there will be made two Prismes in ech body. Ye must take heede that ye put for the bases of eche of these Parallelipipedons equall parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes. And then note the<g ref="char:cmbAbbrStroke">̄</g> with letters according to the let<g ref="char:EOLhyphen"/>ters of the figures before described in the plaine. And co<g ref="char:cmbAbbrStroke">̄</g>pare the<g ref="char:cmbAbbrStroke">̄</g> with the demonstration, and they will make both it and the Proposition very clere vnto you. They will also geue great light to the Corollary following added by <hi>Flussas.</hi>
               </p>
               <div type="section">
                  <pb n="355" facs="tcp:7062:384"/>
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>By this and the former propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>itions it is manifest, that Prismes and solides</hi> 
                     <note n="*" place="margin">
                        <hi>Which of some are cal<g ref="char:EOLhyphen"/>led sided Co<g ref="char:EOLhyphen"/>lumnes.</hi>
                     </note> 
                     <hi>contayned vnder two poligo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on figures equall, like, and parallels, and the rest parallelogrammes: may be compared the one to the other after the selfe same maner that parallelipipedons are.</hi>
                  </p>
                  <p>For forasmuch as (by this proposition and by the second Corollary of the 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of this booke) it is manifest, that euery parallelipipedon may be resolued into two like, and equal Prismes, of one and the same altitude, whose base shalbe one and the selfe same with the base of the parallelipipedon or the halfe thereof, which Pris<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es also shalbe cont<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>yned vnder the selfe same side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> with the parallelipipedo<g ref="char:cmbAbbrStroke">̄</g>, the sayde side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> beyng also sides of like proportion: I say that Prisme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> may be compared together after the like maner that their Parallelipipedon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> For if we would deuide a Prisme like vnto his foli<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e by the 25. of this booke, ye shall finde in the Corollaryes of the 25. propo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>tio<g ref="char:cmbAbbrStroke">̄</g>, that that which is set forth touching a parallelipipedon, followeth not onely in a Prisme, but also in any sided columne whose op<g ref="char:EOLhyphen"/>posite bases are equall, and like, and his sides parallelogrammes.</p>
                  <p>If it be required by the 27. proposition vpon a right line geuen to describe a Prisme like and in like sorte situate to a Prisme geuen: describe <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>st the whole parallelipipedon whereof the prisme geuen is the halfe (which thing ye see by this 40. proposition may be done). And vnto that parallelipipedo<g ref="char:cmbAbbrStroke">̄</g> de<g ref="char:EOLhyphen"/>scribe vpon the right line geuen by the sayd 27. proposition an other parallelipipedon like: and the halfe thereof shalbe the prisme which ye seeke for, namely, shalbe a prisme described vpon the right line geuen, and like vnto the prisme geuen.</p>
                  <p>In deede Prismes can not be cut according to the 28. proposition. For that in their opposite sides can be drawen no diagonall lines: howbeit by that 28. proposition those Prismes are manifestly con<g ref="char:EOLhyphen"/>firmed to be equall and like, which are the halues of one and the selfe same parallelipipedon.</p>
                  <p>And as touching the 29. proposition, and the three following it, which proueth that parallelipi<g ref="char:EOLhyphen"/>pedons vnder one and the selfe same altitude, and vpon equall bases, or the selfe same bases, are equal: or if they be vnder one and the selfe same alti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> they are in proportion the one to the other, as their bases are<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to apply these comparisons vnto <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> it is to <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> required, that the bases of the Prismes compared together, be either all parallelogrammes, or all tria<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gles. For so one and the selfe altitude remayning, the comparison of thinges equall <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> one and th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> selfe same, and the halfes of the bases are euer the one to the other in the same proportion, that their wholes are. Wherfore Prismes which are the halues of the parallelipipedons, and which haue the same proportion the one to the other that the whole parallelipipedons haue, which are vnder one and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> self<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ame altitude: must needes cause that their bases being the halues of the base<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the parallelip<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>p<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>e in the same proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, that their whole parallelipip<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>don<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are. If there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> the w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ole parallelipipedons be in the proportion of the whole bases, their h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> also (which are Prismes) shalbe in the (proportion either of the wholes if their bases be parallel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>mm<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> or of the hal<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>f they be triangles, which is euer all one by the 15. of the fiueth.</p>
                  <p>And forasmuch as by the 33. proposition, like parallelipipedons which are the doubles of their Prismes are in treble proportion the one to the other that their sides of like proportion are, it is mani<g ref="char:EOLhyphen"/>fest, that Prismes being their halues (which haue the one to the other the same proportion that their wholes haue, by the 15 of the fiueth) and hauing the selfe same sides that thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> parallelipipedons haue, are the one to the other in treble proportion of that which the sides of like proportion are.</p>
                  <p>And for that Prismes are the one to the other in the same proportion that their parallelipipedons are, and the bases of the Prismes (being all either triangles or parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes) are the one to the other in the same proportion that the bases of the parallelipipedons are, whose altitudes also are alwayes e<g ref="char:EOLhyphen"/>quall, we may by the 34. proposition conclude, that the bases of the prismes and the bases of the paral<g ref="char:EOLhyphen"/>lelipipedons their doubles (being ech the one to the other in one and the selfe same proportion) are to the altitudes, in the same proportion that the bases of the double solides, namely, of the parallelipi<g ref="char:EOLhyphen"/>pedons are. For if the bases of the equall parallelipipedo<g ref="char:cmbAbbrStroke">̄</g>s be reciprokall with their altitudes, then their halues which are Prismes shall haue their bases reciprokall with their altitudes.</p>
                  <p>By the 36. proposition we may conclude, that if there be three right lines proportionall, the an<g ref="char:EOLhyphen"/>gle of a Prisme made of these three lines (being common with the angle of his parallelipipedon which is double) doth make a prisme, which is equall to the Prisme described of the middle line, and contay<g ref="char:EOLhyphen"/>ning the like angle, consisting also of equall sides. For a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in the parallelipipedon, so also in the Prisme, this one thing is required, namely, that the three dimensions of the proportionall lines do make an an<g ref="char:EOLhyphen"/>gle like vnto the angle contayned of the middle line taken three tymes. Now then if the solide angle of the Prisme be made of those three right lines, there shall of them be made an angle like to the angle of the parallelipipedon which is double vnto it. Wherefore it followeth of necessitie, that the Prismes which are alwayes the halues of the Parallelipipedons, are equiangle the one to the other, as also are their doubles, although they be not equilater: and therefore those halues of equall solides are equall the one to the other: namely, that which is described of the middle proportionall line is equall to that which is described of the three proportionall lines.</p>
                  <p>By the 37. proposition also we may conclude the same touching Prismes which was concluded
<pb facs="tcp:7062:385"/>
touching Parallelipipedons. For forasmuch as Prismes, described like &amp; in like sort of the lines geuen, are the halues of the Parallelipipedons which are like and in like sort described, it followeth that these Prismes haue the one to the other the same proportion that the solides which are their doubles haue. And therfore if the lines which describe them be porportionall, they shalbe proportionall, and so con<g ref="char:EOLhyphen"/>uersedly according to the rule of the sayd 37. proposition.</p>
                  <p>But forasmuch as the 39. proposition supposeth the opposite superficiall sides of the solide to be parallelogrammes, and the same solide to haue one diameter, which thinges a Prisme can not haue, therefore this proposition can by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o meanes by applyed to Prismes.</p>
                  <p>But as touching solides whose bases are two like, equall, and parallel poligonon figures, and their sides are parallelogrammes,<note place="margin">
                        <hi>Sided Co<g ref="char:EOLhyphen"/>lumnes.</hi>
                     </note> forasmuch as by the second Corollary of the 25. of this booke it hath bene declared, that such solides are composed of Prismes, it may easely be proued that their nature is such, as is the nature of the Prismes, whereof they are composed. Wherefore a paralle<g ref="char:EOLhyphen"/>lipipedon being by the 27. proposition of this booke described, there may also be de<g ref="char:EOLhyphen"/>scribed the halfe thereof, which is a Prisme: and by the description of Prismes, there may be composed a solide like vnto a solide geuen composed of Prismes. So that it is manifest, that that which the 29. 30. 31. 32. 33. 34. and 37. propositions set forth touching parallelipipedons, may well be applyed also to these kyndes of solides.</p>
               </div>
            </div>
            <trailer>The end of the eleuenth booke of Euclides Elementes.</trailer>
         </div>
         <div n="12" type="book">
            <pb n="353" facs="tcp:7062:385"/>
            <head>¶The twelueth booke of Eu<g ref="char:EOLhyphen"/>clides Elementes.</head>
            <p>IN THIS TVVELVETH BOOKE, EVCLIDE setteth forth the passions and proprieties of Pyramids, Prismes, Cones, Cylinders, and Spheres. And compareth Pyramids, first to Pyramids, then to Prismes: so likewise doth he Cones, and Cylinders. And lastly he compareth Spheres the one to the other. But before he goeth to the treatie of those bodies, he proueth that, like Poligonon figures inscribed in circles, and also the circles the<g ref="char:cmbAbbrStroke">̄</g> selues are in proportion the one to the other, as the squares of the diameters of those circles are. Because that was ne<g ref="char:EOLhyphen"/>cessary to be proued, for the confirmation of certayne passions and proprieties of those bodies.</p>
            <div type="part">
               <head>¶The 1. Theoreme. The 1. Proposition. Like Poligonon figures described in circles: are in that proportion the one to the other, that the squares of their diameters are.</head>
               <p>SVppose that there be two circles ABCDE, and FGHKL, and in them let there be described like Poligonon figures, name<g ref="char:EOLhyphen"/>ly, ABCDE, and FGHKL, and let the diameters of the circles be BM, and GN. Then I say, that as the square of the line BM is to the square of the line GN, so is the Poligonon fi<g ref="char:EOLhyphen"/>gure ABCDE to the Poligonon figure FGHKL.<note place="margin">Construction.</note> Drawe these right lines BE, AM, GL, and FN. And forasmuch as the Poligonon figure ABCDE is like to the Poligonon figure FGHKL,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore the angle BAE is equall to the angle
<figure/>
GFL, and as the line BA is to the line AE, so is the line GF to the line FL (by the defi<g ref="char:EOLhyphen"/>nition of like Poligonon figures). Now therefore there are two triangles BAE and GFL,
<pb facs="tcp:7062:386"/>
hauing one angle of the one equall to one angle of the other, namely, the angle BAE equall to the angle GFL, and the sides about the equall angles are proportionall. Wherefore (by the first definition of the sixt) the triangle ABE is equiangle to the triangle FGL. Wherefore the angle AEB is equall to the angle FLG. But (by the 21. of the third) the angle AEB is equall to the angle AMB, for they consiste vpon one and the selfe same circumference: and
<figure/>
by the same reason the angle FLG is equall to the angle FNG. Wherfore the angle AMB is equall to the angle FNG. And the right angle BAM, is (by the 4. petition) equall to the right angle GFN. Wherefore the angle remayning, is equall to the angle remayning. Wherefore the triangle AMB is equiangle to the triangle FNG. Wherefore proportionally as the line BM is to the line GN, so is the line BA to the line GF. But the square of the line BM is to the square of the line GN in double proportion of that which the line BM is to the line GN (by the Corollary of the 20. of the sixt). And the Poligonon figure ABCDE is to the Poligonon figure FGHKL in double proportion of that which the line BA is to the line GF (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. of the sixt). Wherefore (by the 11. of the fift) as the square of the line BM is to the square of the line GN, so is the Poligonon figure ABCDE, to the Poligonon fi<g ref="char:EOLhyphen"/>gure FGHKL. Wherefore like Poligonon figures described in circles, are in that pro<g ref="char:EOLhyphen"/>portion the one to the other, that the squares of the diameters are: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶ Iohn Dee his fruitfull instructions, with certaine Corollaries, and their great vse.</head>
                  <p>WHo can not easily perceaue, what occasion and ayde, <hi>Archimedes</hi> had, by these first &amp; second Pro<g ref="char:EOLhyphen"/>positions, to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inde the nere Area, or Content of a circle: betwene a Poligonon figure within the circle, and the like about the same circle, described? Whose precise quantities are most easily knowen: being comprehended of right lines. Where also (to auoyde all occasions of errour) it is good in num<g ref="char:EOLhyphen"/>bers, not hauing precise square rootes, to vse the Logisticall processe, according to the rules, with √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 12, √<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 19, and so, of such like. Who can not readily fall into <hi>Archimedes</hi> reckoning and ac<g ref="char:EOLhyphen"/>count, <milestone type="tcpmilestone" unit="unspecified" n="1"/> by his method? To finde the proportion of the circumference of any circle to his diameter, to be almost triple, and one seuenth of the diameter: but to be more then triple and ten one &amp; seuen<g ref="char:EOLhyphen"/>tithes: that is to be lesse then 3 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and more then 3 10/71. And where <hi>Archimedes</hi> vsed a Poligonon figure of 96. sides: he that, for exercise sake, or for earnest desire of a more nerenes, will vse Polygonon figures of 384. sides (or more) may well trauaile therein, till either wearines cause him stay, or els he finde his labour fruitles. In deede <hi>Archimedes</hi> concluded proportion, of the circumference to the dia<g ref="char:EOLhyphen"/>meter,
<pb n="357" facs="tcp:7062:386"/>
hath hitherto serued the vulgare and<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Mechanicall wor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> men: wherewith, who so is not con<g ref="char:EOLhyphen"/>cented, let his owne Methodicall trauaile satisfie his desire: or let him procure other therto. For, nar<g ref="char:EOLhyphen"/>rower termes (of greater and lesse) found, and appointed to the circumference, will also winne to the Area of the circle a nearer quantitie: seing, it is well dem<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>rated of <hi>Archimed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s,</hi> that a triangle rectan<g ref="char:EOLhyphen"/>gle,<milestone type="tcpmilestone" unit="unspecified" n="2"/> of whose two sides (contayning the right angle) one is equall to the semidiameter of the circle, and the other to the circumference of the same, is equall to the Area of that circle. Vpon which two Theoremes, it followeth, that the square made of the diameter, is in that proportion to the circle (ve<g ref="char:EOLhyphen"/>ry <milestone type="tcpmilestone" unit="unspecified" n="3"/> neare) in which, 14, is to 11<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wherefore euery circle is—eleuen fowertenthes (well neare) of the square about him described. The one side, then, of that square, deuide into 14. equall partes:<note place="margin">
                        <hi>The squaring of the circle.</hi>
                     </note> and from that point which endeth the eleuenth part, drawe to the opposite side, a line, parallel to the other sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s, and so make pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fecte the parallelogramme. Then, by the last Proposition of the second booke, vnto that parallelogramme (whose one side hath those 11. equall partes), make a square equall. Then is it euident, that square to be equall to the circle, about which the first square is described. As ye may here beholde in these figures.</p>
                  <figure/>
                  <p>Gentle frend, the great desire, which I haue, that both with pleasure and also profite, thou mayest spend thy time in these excellent studies, doth cause me here to furnishe thee somewhat (extraordina<g ref="char:EOLhyphen"/>rily) about the circle: not onely by pointing vnto thee, the welspring of <hi>Archimedes,</hi> his so much won<g ref="char:EOLhyphen"/>dre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> at; and iustly commended trauaile (in the former 3. Theoremes, here repeated), but also to make thee more apt, to vnderstand and practise this and other bookes following, where, vse of the circle may be had in any consideration: as in Cones, Cylinders, and Spheres, &amp;c.</p>
                  <div type="subsection">
                     <head>¶A Corollary. 1.</head>
                     <p>By <hi>Archimedes</hi> second Theoreme (as I haue here alleaged them) it is manifest, that a pa<g ref="char:EOLhyphen"/>rallelogramme contained, either vnder the semidiameter and halfe the circumference, or vnder the halfe semidiameter and whole circumference of any circle, is equall to the circle: by the <hi>41.</hi> of the first: and first of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ A Corollary. 2.</head>
                     <p>Likewise it is euident, that the parallelogramme contayned vnder the semidiameter, and halfe of any portion of the circumference of a circle geuen, is equall to that sector of the same circle, to which the whole portion of the circumference geuen, doth belong. Or you may vse the halfe semidiameter, and the whole portion of the circumference as sides of the said parallelogramme.</p>
                     <p>
                        <hi>The farther winning, and inferring, I commi<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap> to your skill, care, and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>udy<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>. But in an other sort will I geue you newe ayde, and instruction here.</hi>
                     </p>
                  </div>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:387"/>
                  <head>¶ A Theoreme.</head>
                  <p>
                     <hi>Of all circles, the circumferences to their owne diameters, haue one and the same proportion: in what one circle soeuer, they are assigned.</hi>
                  </p>
                  <p>That is (as <hi>Archimedes</hi> hath demonstrated) almost, as 22. to 7: or nearer, if nearer be fou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d: vn<g ref="char:EOLhyphen"/>till the very precise proportion be demonstrated. Which, what soeuer it be, in all circumferences to their proper diameters, will be demonstrated one and the same.</p>
                  <div type="subsection">
                     <head>A Corollary. 1.</head>
                     <p>
                        <hi>Wherefore if two circles be propounded, which suppose to be A and B, as the circumference of A is to the circumference of B, so is the
<figure/>
diameter of A to the diameter of B.</hi>
                     </p>
                     <p>For by the former Theoreme, as the circumference of A, is to his own diameter, so is the circumference of B, to his own diameter: Wherfore, alternately, as the circumference of A, is to the circumference of B, so is the diameter of A, to the diameter of B: Which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>A Corollary. 2.</head>
                     <p>
                        <hi>It is now then euident, that we can geue two circles whose circumferences one to the other, shall haue any proportion geuen in two right lines.</hi>
                     </p>
                     <p>The great Mechanicall vse (besides Mathematicall considerations) which, these two Corollaryes may haue in Wheeles of Milles, Clockes, Cranes, and other engines for water workes, and for warres, and many other purposes, the earnest and wittie Mechanicien will soone boult out, &amp; gladly practise.</p>
                     <p>
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> Iohn Dee.</hi>
                     </p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 2. Theoreme. The 2. Proposition. Circles are in that proportion the one to the other, that the squares of their diameters are.</head>
               <p>SVppose that there be two circles ABCD, and EFGH, and let their diame<g ref="char:EOLhyphen"/>ters be BD and FH. Then I say, that as the square of the line DB is to the square of the line FH,<note place="margin">
                     <p>Demonstrati<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>eading to an impossibi<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>itie.</p>
                     <p>Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</p>
                     <p>The first case.</p>
                  </note> so is the circle ABCD to the circle EFGH. For if the circle ABCD be not vnto the circle EFGH, as the square of the line BD is to the square of the line FH: then the square of the line BD shall be to the square of the line FH, as the circle ABCD is to a superficies, either lesse then the circle EFGH, or grea<g ref="char:EOLhyphen"/>ter. First let the square of the line BD be to the square of the line FH, as the circle ABCD is to a superficies lesse then the circle EFGH, namely, to the superficies S. Describe (by the 6. of the fourth) in the circle EFGH a square EFGH. Now this square thus described is grea<g ref="char:EOLhyphen"/>ter then the halfe of the circle EFGH. For if by the pointes E, F, G, H, we drawe right lines
<pb n="358" facs="tcp:7062:387"/>
touching the circle,<note place="margin">
                     <p>That a square within any circle descri<g ref="char:EOLhyphen"/>bed is bigger than halfe the circle.</p>
                     <p>That the Isos<g ref="char:EOLhyphen"/>celes trian<g ref="char:EOLhyphen"/>gles, without the square, are greater then halfe the seg<g ref="char:EOLhyphen"/>ments wherin they are.</p>
                  </note> the square EFGH, is the halfe of the square described about the circle, but the square described about the circle, is greater then the circle. Wherefore the square EFGH, which is inscribed in the circle, is greater then the halfe of the circle EFGH. Deuide the circumferences EF, FG, GH, and HE, into two equall partes in the pointes K, L, M, N. And drawe these right lines EK, KF, FL, LG, GM, MH, HN, and NE. Wherefore euery one of these triangles EKF, FLG, GMH, and HNE, is greater then the halfe
<figure/>
of the segme<g ref="char:cmbAbbrStroke">̄</g>t of the cir<g ref="char:EOLhyphen"/>cle which is described a<g ref="char:EOLhyphen"/>bout it. For if by the pointes K, L, M, N, be drawen lines touching the circle, and then be made perfecte the paral<g ref="char:EOLhyphen"/>lelogra<g ref="char:cmbAbbrStroke">̄</g>mes made of the right lines EF, FG, G<g ref="char:EOLhyphen"/>H, &amp; HE, euery one of the triangles EKF, F<g ref="char:EOLhyphen"/>LG, GMH, &amp; HNE, is the halfe of the paral<g ref="char:EOLhyphen"/>lelogragra<g ref="char:cmbAbbrStroke">̄</g>me which is de<g ref="char:EOLhyphen"/>scribed about it (by the 41. of the first): but the segme<g ref="char:cmbAbbrStroke">̄</g>t described about it is lesse then the paral<g ref="char:EOLhyphen"/>lelogramme. Wherefore euery one of these triangles EKF, FLG, GMH, and HNE, is greater then the halfe of the segment of the circle which is described about it. Now then deuiding the circumferences remaining into two equall partes, and drawing right lines from the pointes where those diuisions are made, &amp; so continually doing this, we shall at the length (by the 1. of the tenth) leaue certaine segmentes of the circle, which shall be lesse then the ex<g ref="char:EOLhyphen"/>cesse, wherby the circle EFGH excedeth the superficies S. For it hath bene proued in the first Proposition of the tenth booke, that two vnequall magnitudes being geuen, if from the greater be taken away more then the halfe, and likewise againe from the residue more then the hal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e, and so continually, there shall at the length be left a certaine magnitude which shall be lesse then the lesse magnitude geuen. Let there be such segmentes left, &amp; let the segmentes of the circle EFGH, namely, which are made by the lines EK, KF, FL, LG, GM, MH, HN, and NE, be lesse then the excesse, whereby the circle EFGH excedeth the superficies S. Wherefore the residue, namely, the Poligonon figure EKFLGMHN, is greater then the superficies S. Inscribe in the circle ABCD a Poligonon figure like to the Poligonon figure EKFLGMHN, and let the same be AXBOCPDR. Wherefore (by the Proposition next going before) as the square of the line BD is to the square of the line FH, so is the Po<g ref="char:EOLhyphen"/>ligonon figure AXBOCPDR to the Poligonon figure EKFLGMHN. But as the square of the line BD is to the square of the line FG, so is the circle ABCD supposed to be to the superficies S. Wherefore (by the 11. of the fift) as the circle ABCD is to the superfi<g ref="char:EOLhyphen"/>cies S, so is the Poligonon figure AXBOCPDR to the Poligonon figure EKFLGM<g ref="char:EOLhyphen"/>HN. Wherefore alternately (by the 16. of the fift) as the circle ABCD is to the Poligonon figure described in it, so is the superficies S to the Poligonon figure EKFLGMHN. But the circle ABCD is greater then the Poligonon figure described in it. Wherefore also the superficies S is greater then the Poligonon figure EKFLGHMN: but it is also lesse: which is impossible. Wherefore as the square of the line BD is to the square of the line FH,
<pb facs="tcp:7062:388"/>
so is not the circle ABCD to any superficies lesse then the circles EFGH.</p>
               <p>
                  <note place="margin">Second case.</note>In like sort also may wproue, that as the square of the line FH is to the square of the line BD, so is not the circle EFGH to any superficies lesse then the circle ABCD. I say, namely, that as the square of the
<figure/>
line BD is to the square of the line FH, so is not the circle ABCD to a<g ref="char:EOLhyphen"/>ny superficies greater the<g ref="char:cmbAbbrStroke">̄</g> then the circle EFGH. For if it be possible, let it be to a greater, namely, to the superficies S. Wher<g ref="char:EOLhyphen"/>fore by conuersion, as the square of the line FH is to the square of the line BD, so is the su<g ref="char:EOLhyphen"/>perficies S to the circle ABCD. <note n="*" place="margin">This As<gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> afte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ward at the end of the dem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>stra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion proued.</note> But as the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>perficies S is to the cir<g ref="char:EOLhyphen"/>cle ABCD, so is the cir<g ref="char:EOLhyphen"/>cle EFGH to some su<g ref="char:EOLhyphen"/>pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ficies l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>sse the<g ref="char:cmbAbbrStroke">̄</g> the cir<g ref="char:EOLhyphen"/>cle ABCD. Wherefore (by the 11. of the fift) as the square of the line FH is to the square of the line BD, so is the circle EFGH, to some superficies lesse then the circle ABCD: which is in the first case proued to be impossible. Wherefore as the square of the line BD is to the square of the line FH, so is not the circle ABCD to any superficies greater then the circle EFGH. And it is also proued that it is not, to any lesse. Wherefore as the square of the l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne BD is to the square of the line FH, so is the circle ABCD to the circle EFGH. Wherefore circles are in that proportion the one to the other, that the squares of their diame<g ref="char:EOLhyphen"/>ters are: which was required to be proued.</p>
               <div type="section">
                  <head>¶ An Assumpt.</head>
                  <p>I say now, that the superficies S be<g ref="char:EOLhyphen"/>ing
<figure/>
greater then the circle EFGH, as the superficies S is to the circle ABC<g ref="char:EOLhyphen"/>D, so is the circle EFGH to some su<g ref="char:EOLhyphen"/>perficies lesse then the circle ABCD. For, as the superficies S is to the circle ABCD, so let the circle EFGH be to the superficies T. Now I say, that the superficies T is lesse then the circle ABCD. For for that as the superficies S is to the circle ABCD, so is the cir<g ref="char:EOLhyphen"/>cle EFGH to the superficies T, there<g ref="char:EOLhyphen"/>fore alternately (by the 16. of the fift) as the superficies S is to the circle EF<g ref="char:EOLhyphen"/>GH, so is the circle ABCD to the su<g ref="char:EOLhyphen"/>perficies T. But the superficies S is grea<g ref="char:EOLhyphen"/>ter then the circle EFGH (by supposition). Wherefore also the circle ABCD is greater then the superficies T (by the 14. of the fift). Wherefore as the superficies S is to the circle
<pb n="359" facs="tcp:7062:388"/>
ABCD, so is the circle EFGH to some superficies lesse then the circle ABCD: which was required to be demonstrated.</p>
               </div>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Circles haue the one to the other, that proportion, that like Poligonon figures and in like sort de<g ref="char:EOLhyphen"/>scribed in them haue.</hi> For, it was by the first Proposition proued, that the Poligonon figures haue that proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, that the squares of the diameters haue, which proportion likewise, by this Proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the circles haue.</p>
               </div>
               <div type="section">
                  <head>¶ Very needefull Problemes and Corollaryes by Master Ihon Dee inuented: whose wonderfull vse also, be partely declareth.</head>
                  <div type="subsection">
                     <head>A Probleme. 1.</head>
                     <p>
                        <hi>Two circles being geue<g ref="char:cmbAbbrStroke">̄</g>: to finde two right lines, which haue the same proportion, one to the other, that the geuen circles haue, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </p>
                     <p>Suppose A and B, to be the diameters of two circles geuen: I say that two right lines are to be fou<g ref="char:cmbAbbrStroke">̄</g>de, hauing that proportio<g ref="char:cmbAbbrStroke">̄</g>, that the circle of A hath to the circle of B. Let to A &amp; B (by the 11 of the sixth) a third proportionall line be found, which suppose to be C.<note place="margin">
                           <hi>Construction.</hi>
                        </note> I say now that A hath to C, that pro<g ref="char:EOLhyphen"/>portion which the circle of A hath to the cir<g ref="char:EOLhyphen"/>cle
<figure/>
of B. For forasmuch as A, B, and C, are (by construction) three proportionall lines,<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> the square of A is to the square of B, as A is to C, (by the Corollary of the 20. of the sixth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> but as the square of the line A is to the square of the line B, so is the circle whose diameter is the line A, to the circle whose diameter is the line B, by this second of the eleue<g ref="char:cmbAbbrStroke">̄</g>th. Wherfore the circles of the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> A and B, are in the proportion of the right lines A and C. Therefore two circles be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ng geuen, we haue found two right lines hauing the same proportion betwene the<g ref="char:cmbAbbrStroke">̄</g>, that the circles geuen, haue one to the other: which ought to be done.</p>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 2.</head>
                     <p>
                        <hi>Two circles being geuen, and a
<figure/>
right line: to finde an other right line, to which the line geue<g ref="char:cmbAbbrStroke">̄</g> shall haue that proportion, which the one circle hath to the other.</hi>
                     </p>
                     <p>Suppose two circles geue<g ref="char:cmbAbbrStroke">̄</g>: which let be A &amp; B, &amp; a right line geue<g ref="char:cmbAbbrStroke">̄</g>, which let be C: I say that an other right line is to be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ounde, to which the line C shall haue that proportion that the circle A,<note place="margin">
                           <hi>Construction</hi>
                        </note> hath to the circle B. As the diameter of the circle A, is to the dia<g ref="char:EOLhyphen"/>meter of the circle B, so let the line C be to a fourth line, (by the 12. of the <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> line be D. And, by the 11. of the sixth, let a thirde line proportionall be found, to the lines C &amp; D, which let be E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> I say now, that the line C hath to the line E, that proportion which the circle A, hath to the circle B. For (by construction) the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> C, D, and E,<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> are proportionall: therefore the square of C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to the square of D, as C is to E, by the Corollary of the 10. of the sixth. But by construction, as the diameter of the circle A, is to the diameter of the circle B, so is C, to D: wherefore as the square of the diameter of the circle A, is to the square of the diameter of the circle B, so is the square of the line C to the square of the line D, by the 22. of the sixth. But as the square of the diameter of A, the circle, is to the square of the diameter of the circle D, so is the cir<g ref="char:EOLhyphen"/>cle A, to the circle B, by the second of the twelfth: wherefore by 11. of the fiueth, as the circle A, is to
<pb facs="tcp:7062:389"/>
the circle B, so is the square of the line C, to the square of the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e D. But it is proued, that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s the square of the line C, is to the square of the line D, so is the line C to the line E. Wherefore by the 11. of the fiueth, as the circle A, is to the circle B, so is the line C to the line E. Two circles being geuen there<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ore, and a right line, we haue found a right line, to which the right line geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n, hath that proportion, which the one circle hath to the other. Which ought to be done.</p>
                     <div type="note">
                        <head>Note.</head>
                        <p>The difference betwene this Probleme, and that next before, is this: there, although we had two circles geuen,<note place="margin">
                              <hi>Di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>reence betwene the first probleme and the se<g ref="char:EOLhyphen"/>cond.</hi>
                           </note> and two lines were found in that proportion the one to the other, in which the geuen circles were,: and here likewise are two circles geue<g ref="char:cmbAbbrStroke">̄</g>, and two line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> also are had in the same proportio<g ref="char:cmbAbbrStroke">̄</g>, that the geue<g ref="char:cmbAbbrStroke">̄</g> circles are: yet there we tooke at pleasure the first of the two lines, wherunto we framed the second proportionally, to the circles geuen. But here the first of the two lines, is assigned, poynted, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>d determined to vs: and not our choyse to be had therein, as was in the former Probleme.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 3.</head>
                     <p>
                        <hi>A circle being geuen, to finde an other circle, to which the geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rcle is in any proportion geue<g ref="char:cmbAbbrStroke">̄</g> in two right lines.</hi>
                     </p>
                     <p>Suppose the circle A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>C geuen, and therefore his semidiameter is geuen: whereby his diameter also is geuen: which diameter let be AC. Let the proportion geuen, be that which is betwene <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, F, two right lines. I say, a circle is to be found, vnto which A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>C hath that proportion that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> hath to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. As <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,<note place="margin">
                           <hi>Construction.</hi>
                        </note> so let AC the diameter, be to an o<g ref="char:EOLhyphen"/>ther
<figure/>
right line, by the 12. of the sixth. Which line suppose to be 11. Betwene AC and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> finde a middle proportionall line, by the 13. of the sixth: which let be LN. By the 10. of the first, deuide <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>N, into two equall pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tes: and let that be done in the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. Now vpon <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>, (o being made the center) describe a circle: which let be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> I say that ABC, is to LMN, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to F. For seing that AC, LN, and H, are three right lines in con<g ref="char:EOLhyphen"/>tinuall proportion (by construction) therefore (by the Corollary of the 20. of the sixth) as AC is to H, so is the square of A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to the square of LN. But AC is to H, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, by construction. Wherefore the square of AC is to the square of LN, as E is F: but as the square of the diameter AC, is to the square of the diameter LN, so is the circle ABC to the circle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>MN, by this 2. of the twelueth, wherefore by the 11. of the fiueth, the circle ABC is to the circle LMN, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is to F. A circle being geuen (therefore) an other circle is founde, to which the geuen circle is in any proportion geuen betwene two right lines: which ought to be done.</p>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 4.</head>
                     <p>
                        <q>Two circles being geuen, to finde one circle equall to them both.</q>
                     </p>
                     <p>Suppose the two circles geue<g ref="char:cmbAbbrStroke">̄</g>, haue their diameters A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp; CD. I say that a circle must be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ound equall to the two circles whose diameters are A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and CD: vnto the
<figure/>
line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>,<note place="margin">
                           <hi>Cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ruction.</hi>
                        </note> at the point A, erect a perpendicular line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>: from which (sufficiently produced) cut a line equall to CD, which let be AF. By the first peticion draw from F to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> a right line: so is FA<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> made a tri<g ref="char:EOLhyphen"/>angle rectangle. I say now that a circle whose diameter is F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, is equal to the two circles whose diameters are A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>D.<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> For by the 47. of the first, the square of F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is equall to the squares of A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp; AF. Which AF is (by construction) equall to CD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> wherefore the square of F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is equall to the squares of AB and CD. But circles are one to an other, as the squares of their diameters are one to the other, by this second of the twelueth. Therefore the circle whose diameter is <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> is equall to the circles whose diameters are A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and CD. Therefore two circles being geuen we haue found a circle equall to them both. Which was required to be done.</p>
                     <div type="corollary">
                        <pb n="360" facs="tcp:7062:389"/>
                        <head>A Corollary. 1.</head>
                        <p>
                           <hi>Hereby it is made euident, that in all triangles rectangle, the circles, semicircles, quadrants, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> any other portions of circles described vpon the subtendent line, is equall to the two circles, semicircles, quadrants, or any two other like portions of circles, described on the two lines comprehending the right angle, like to like being compared.</hi>
                        </p>
                        <p>For like partes haue that proportion betwene them selues, that their whole magnitudes haue, of which they are like partes, by the 15. of the fifth. But of the whole circles, in the former probleme it is euident: and therefore in the fornamed like portions of circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s, it is a true consequent.</p>
                     </div>
                     <div type="corollary">
                        <head>A Corollary. 2.</head>
                        <p>
                           <hi>By the former probleme, it is also manifest, vnto circles three, fower, fiue, or to how many so euer one will geue, one circle may be geuen equall.</hi>
                        </p>
                        <p>For if first, to any two, by the former probleme, you finde one equall, and then vnto your found circle and the third of the geuen circles, as two geuen circles, finde one other circle equall, and then to that second found circle, and to the fourth of the first geuen circles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> as two circles, one new circle be found equall, and so proceede till you haue once cuppled orderly, euery one of your propou<g ref="char:cmbAbbrStroke">̄</g>ded circles (except the first and second already doone) with the new circle thus found for so the last found circle is equall to all the first geuen circles. If ye doubt, or sufficiently vnderstand me not: helpe your selfe by the discourse and demonstration of the last proposition in the second booke, and also of the 31. in the sixth booke.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>¶ A Probleme 5.</head>
                     <p>
                        <hi>Two vnequall circles being geuen, to finde a circle equall to the excesse of the greater to the lesse.</hi>
                     </p>
                     <p>Suppose the two vnequal circles geue<g ref="char:cmbAbbrStroke">̄</g>, to be ABC &amp; DEF, &amp; let ABC be the greater:<note place="margin">
                           <hi>Cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ruction.</hi>
                        </note> whose dia<g ref="char:EOLhyphen"/>meter suppose to be AC: &amp; the diameter of DEF suppose to be DF. I say a circle must be found equal to that excesse in magnitude, by which ABC is greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> DEF. By the first of the fourth, in the circle ABC. Apply a right line equall to DF:
<figure/>
whose one end let be at C, and the other let be at B. Fro<g ref="char:cmbAbbrStroke">̄</g> B to A draw a right line. By the 30. of the third, it may appeare,<note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> that ABC is a right angle: and thereby ABC, the triangle is rectangled: wher<g ref="char:EOLhyphen"/>fore by the first of the two corollaries, here before, the circle ABC is equall to the circle DEF, (For BC by constructi<g ref="char:EOLhyphen"/>on is equall to DF) and more ouer to the circle whose diameter is AB. That circle therefore whose diameter is AB, is the circle conteyning the magnitude, by which ABC is greater then DEF. Wherefore two vnequal circles being geuen, we haue found a circle equall to the excesse of the greater to the lesse: which ought to be doone.</p>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 6.</head>
                     <p>
                        <hi>A Circle being geuen to finde two Circles equall to the same: which found Circles, shall haue the one to the other, any proportion geuen in two right lines.</hi>
                     </p>
                     <p>Suppose ABC, a circle geuen: and the proportion
<figure/>
geuen, let it be that, which is betwene the two right lines D and E. I say, that two circles are to be found equall to ABC: and with al, one to the other,<note place="margin">
                           <hi>Cons<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>uction.</hi>
                        </note> in the proportio<g ref="char:cmbAbbrStroke">̄</g> of D to E. Let the diameter of ABC be AC. As D is to E, so let AC be deuided, by the 10. of the sixth, in the poynt F. At F, to the line AC let a perpe<g ref="char:cmbAbbrStroke">̄</g>dicular be drawne FB, and let it mete the circu<g ref="char:cmbAbbrStroke">̄</g>fere<g ref="char:cmbAbbrStroke">̄</g>ce at the poynt B. From the poynt B to the points A and C, let right lines be drawne: BA and BC. I say that the circles whose diamete<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> are the lines BA and BC are equall to the circle ABC: and that those circles hauing to their diameters the lines BA and BC are one to the other in the proportion of the line D to the line E. For, first that they are equal, it is e<g ref="char:EOLhyphen"/>uident:
<pb facs="tcp:7062:390"/>
by reason that ABC is a triangle rectangle: wher<g ref="char:EOLunhyphen"/>fore
<figure/>
by the 47. of the first the squares of BA and BC are equall to the square of AC: And so by this second it is mani<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>est, the two circles to be equall to the circle ABC. Secondly as D is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, so is AF to FC: by construction.
<q>
                           <hi>And as the line AF is to the line FC, so is the square of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>A to the square of the line BC.</hi> [Which thing, we will briefely p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>oue thus. The parallelogramme contayned vnder AC and AF,<note place="margin">
                              <hi>Note this well: for it i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> of great vse.</hi>
                           </note> is equall to the square of BA: by the Lemma after the 32. of the tenth booke: and by the same Lemma or Assumpt, the parallelogramme contayned vn<g ref="char:EOLhyphen"/>der AC and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>C, is equall to the square of the line BC. Wherfore as the first parallelogramme hath it selfe to the second<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> so hath the square of BA (equall to the first pa<g ref="char:EOLhyphen"/>rallelogramme) it selfe, to the square of BC, equall to the second parallelogramme. But both the parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes haue one heigth, namely, the line AC: and bases, the lines AF and FC: wherefore as AF is to FC, so is the parallelog<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>amme contayned vnder AC, AF, to the parallelogramme contayned vnder AC, FC, by the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>st of the sixth.</q>
And therefore as AF is to FC, so is the square of BA to the square of BC.] And as the square of BA is to the square of BC: so is the circle whose diameter is BA, to the circle whose diameter is BC, by this second of the twelfth. Wherefore the circle whose diameter is BA, is to the circle whose diameter is BC, as D is to E. And before we proued them equall to the circle ABC. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>fore a circle being geuen, we haue found two circles equall to the same: which haue the one to the other any proportion geuen in two right lines. Which ought to be done.</p>
                     <div type="note">
                        <head>Note.</head>
                        <p>
                           <note place="margin">
                              <hi>An other way of demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>at<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>o<g ref="char:cmbAbbrStroke">̄</g> of the f<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>st <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>ro<g ref="char:EOLhyphen"/>bleme of th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap> addition.</hi>
                           </note>He<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e may you per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>eiue an other way how to execute my first probleme, for if you make a right an<g ref="char:EOLhyphen"/>gle conteyned of the diameters geue<g ref="char:cmbAbbrStroke">̄</g>, as in this figure suppose them BA and BC: and then subtend the right angle with the line AC: and from the right angle, let fall a line perpendicular to the base AC: that perp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ndicular at the point of his fall, deuideth AC into AF and FC, of the proportion required.</p>
                     </div>
                     <div type="corollary">
                        <head>A Corollary.</head>
                        <p>It followeth of thinges manifestly proued in the demonstration of this probleme, that in a trian<g ref="char:EOLhyphen"/>gle rectangle, if from the right angle to the base, a perpendicular be let fall: the same perpendicular cutteth the base into two partes,<note place="margin">Note this pro<g ref="char:EOLhyphen"/>per<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ie of a tri<g ref="char:EOLhyphen"/>angle rectangle.</note> in that proportion, one to the other, that the squares of the righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> lines, conteyning the right angle, are in, one to the other: those on the one side the perpendicular, being compared to those on the other, both square and segment.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Probleme 7.</head>
                     <p>
                        <hi>Betwene two circles geuen, to finde a circle middell proportionall.</hi>
                     </p>
                     <p>Let the two circles geuen, be
<figure/>
ACD and BEF: I say, that a cir<g ref="char:EOLhyphen"/>cle is to be fou<g ref="char:cmbAbbrStroke">̄</g>d, which betwene ACD and BEF is middell pro<g ref="char:EOLhyphen"/>portionall.<note place="margin">
                           <hi>Construction.</hi>
                        </note> Let the diameter of ACD, be AD, and of BEF, let B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> be the diameter: betwene AD and BF, finde a line middell proportionall, by the 13. of the sixth: which let be HK: I say that a circle, whose diameter is HK is middell proportionall betwene ACD and BEF. To AD, HK, and BF, (three right lines in con<g ref="char:EOLunhyphen"/>tinuall proportion, by construc<g ref="char:EOLhyphen"/>tion) let a fourth line be found: to which BF shal haue that pro<g ref="char:EOLhyphen"/>portion, that AD hath to HK:
<q>by the 12. of the sixth, &amp; let that line be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>.<note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> 
                           <hi>It is manifest that the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ower lines AD, HK, BF, and L, are in continuall proportion.</hi> [For by co<g ref="char:cmbAbbrStroke">̄</g>struction, as AD is to HK, so is B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> to L. And by constructi<g ref="char:EOLhyphen"/>on,
<pb n="361" facs="tcp:7062:390"/>
on, as AD is to HK, so is HK to BF: wherefore HK is to BF, as BF is to L: by the 11. of the fifth, wher<g ref="char:EOLhyphen"/>fore the 4. lines are in continuall proportion.] Wherefore as the first is to the third, that is AD to BF, so is the square of the first to the square of the second:</q>
that is, the square of AD, to the square of HK: by the corollary of the 20. of the sixth. And by the same corollary, as HK is to L, so is the square of HK to the square of BF. But by alternate proportion, the line AD is to BF, as HK is to L: wherefore the square of AD is to the square of HK, as the square of HK is to the square of BF. Wherefore the square of HK, is middell proportionall, betwene the square of AD and the square of BF. But as the squares are one to the other, so are the circles (whose diameters produce the same squares) one to the other, by this seco<g ref="char:cmbAbbrStroke">̄</g>d of the twelfth: wherfore the circle whose diameter is the line HK, is middel proportional, betwene the circles whose diameters are the line a AD and BF. Wherefore betwene two circles geuen, we haue found a circle middell proportionall: which was requisite to be doone.</p>
                     <div type="corollary">
                        <head>¶A Corollary.</head>
                        <p>
                           <hi>Hereby it is manifest, three lines or more, being in continuall proportion, that the circles hauing those lines to their diameters, are also in continuall proportion.</hi>
                        </p>
                        <p>As of three, our demonstration hath already proued: so of fower, will the proufe go forward<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> if you adde a fifth line in continuall proportion to the fower geuen: as we did to the three, adde the fourth: namely, the line L. And so, if you haue 6, by putting to one more, the demonstration will be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>asie and plaine. And so of as many as you will.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 8.</head>
                     <p>
                        <hi>To a circle being geuen, to finde three circles equall: Which three circles shall be in continuall proportion, in any proportion geuen betwene two right lines.</hi>
                     </p>
                     <p>Suppose the circle geuen to be <hi>ABC:</hi> and the proportion geuen to be that which is betwene the lines <hi>X</hi> and <hi>Y.</hi> I say, that three circles are to be geuen, which three, together, shall be equall to the circle <hi>ABC:</hi> and withall in continuall proportion, in the same proportion which is betwene the right lines <hi>X</hi> and <hi>Y.</hi> Let the diameter of <hi>ABC,</hi> be <hi>AC.</hi> Of <hi>AC,</hi> make a square: by the 46. of the first: which which let be <hi>ACDE.</hi> From the point <hi>D</hi>
                        <figure/>
drawe a line, sufficiently long (any way, <note n="*" place="margin">
                           <p>
                              <hi>Though I say, without the square, yet you must thinke, that it may be also within the square: &amp; that diuersly. Wher<g ref="char:EOLhyphen"/>fore this Pro<g ref="char:EOLhyphen"/>bleme may haue diuerse cases, so, but briefly, to a<g ref="char:EOLhyphen"/>
                                 <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>yde all, may thus be said: cut any side of that square into 3. parts<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> in the pro<g ref="char:EOLunhyphen"/>portion of X to Y.</hi>
                           </p>
                           <p>
                              <hi>Note the maner of the drift in this demonstra<g ref="char:EOLhyphen"/>tion and con<g ref="char:EOLhyphen"/>struction, mixt<g ref="char:EOLhyphen"/>ly and with no determination to the construc<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g>
                                 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> as commo<g ref="char:cmbAbbrStroke">̄</g>ly i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> in probleme<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap> which is here of me so vsed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> for an example to young stude<g ref="char:cmbAbbrStroke">̄</g>tes of variety in art.</hi>
                           </p>
                        </note> without the square): which let be <hi>DO.</hi> At the point <hi>D,</hi> and from the line <hi>DO,</hi> cut a line equall to <hi>X:</hi> which let be <hi>DM.</hi> At the point <hi>M,</hi> and from the line <hi>MO,</hi> cut a line equall to <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> which let be <hi>MN.</hi> At the point <hi>N,</hi> to the two lines <hi>DM</hi> and <hi>MN,</hi> set a third line proportionall, by the 12. of the sixt: which let be <hi>NO.</hi> From <hi>E</hi> (one of the angles of the square <hi>ACDE,</hi> next to <hi>D</hi>) draw a right line to <hi>O:</hi> making per<g ref="char:EOLhyphen"/>fecte the triangle <hi>DEO.</hi> Now from the pointes <hi>M</hi> and <hi>N,</hi> drawe lines, to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ide <hi>DE,</hi> parallel to the side <hi>EO:</hi> by the 31. of the first: which let be <hi>MF</hi> and <hi>NG.</hi> Wher<g ref="char:EOLhyphen"/>fore, by the 2. of the sixt, the side <hi>DE,</hi> is proportionally cut in the pointes <hi>F</hi> and <hi>G,</hi> as <hi>DO</hi> is cut in the pointes <hi>M</hi> and <hi>N:</hi> ther<g ref="char:EOLhyphen"/>fore, as <hi>DM</hi> is to <hi>MN,</hi> so is <hi>DF,</hi> to <hi>FG:</hi> and as <hi>MN</hi> is to <hi>NO,</hi> so is <hi>FG</hi> to <hi>GE.</hi> Wherefore, seing <hi>DM, MN,</hi> and <hi>NO,</hi> are, by construction, continually proportioned, in the proportion of <hi>X</hi> to <hi>Y:</hi> So likewise, are <hi>DF, FG,</hi> and <hi>GE,</hi> in continuall propor<g ref="char:EOLhyphen"/>tion, in the proportion of <hi>X</hi> to <hi>Y,</hi> by the 11. of the fift. From the pointes <hi>F</hi> and <hi>G,</hi> to the opposite <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ide <hi>AC,</hi> let right lines be drawen parallel to the other sides: which lines, suppose to be <hi>FI,</hi> and <hi>GK:</hi> making thereby, three parallelogrammes <hi>DI, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>K,</hi> and <hi>GC,</hi> equall to the whole square <hi>ACDE.</hi> Which three parallelogrammes, by the first of the sixt, are one to an other, as their bases, <hi>DF, FG,</hi> and <hi>GE,</hi> are. But <hi>DF, FG,</hi> and <hi>GE,</hi> were proued to be in continuall propor<g ref="char:EOLhyphen"/>tion, in the proportion of <hi>X</hi> to <hi>Y:</hi> Wherefore, the three parallelogrammes <hi>DI, FK,</hi> and <hi>GC,</hi> by the 11. of the fifth, are also in continuall proportion, and in the same, which <hi>X</hi> is in, to <hi>Y.</hi> Let three squares be made, equall to the three parallelogrammes <hi>DI, FK,</hi> and <hi>GC:</hi> by the last of the second: Let the sides of those squares be, orderly, <hi>S, T,</hi> and <hi>V.</hi> Forasmuch as, it was last concluded that the three paral<g ref="char:EOLhyphen"/>lelogrammes,
<pb facs="tcp:7062:391"/>
                        <hi>DI, FK,</hi> and <hi>GC</hi> (which are equall to the square <hi>ACDE</hi>) are also in continuall pro<g ref="char:EOLhyphen"/>portion, in the proportion of <hi>X</hi> to <hi>Y,</hi> therefore their equalls, namely, the three squares of <hi>S, T,</hi> &amp; <hi>V,</hi> are also equall to the whole square <hi>ACDE,</hi> and in continuall proportion, in the proportion of <hi>X</hi> to <hi>Y.</hi> Wherefore the three circles, whose diameters are <hi>S, T,</hi> and <hi>V,</hi> are equall to the circle, whose diameter is <hi>AC,</hi> the side of the square <hi>ACDE,</hi> and also in continuall proportion, in the proportion of <hi>X</hi> to <hi>Y:</hi> by this second of the twelfth. But, by construction, <hi>AC</hi> is the diameter of the circle <hi>ABC.</hi> Wherefore we haue found three circles, equall to <hi>ABC:</hi> namely, the circle, whose diameter is <hi>S:</hi> and the circle, whose diameter is <hi>T:</hi> and the circle, whose diameter is <hi>V:</hi> which three circles, also, are in continuall propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tion, in the proportion of <hi>X</hi> to <hi>Y.</hi> Wherefore to a circle being geuen, we haue found three cir<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>les equall in any proportion, geuen, betwene two right lines: which was requisite to be done.</p>
                     <div type="corollary">
                        <head>¶A Corollary.</head>
                        <p>
                           <hi>Hereby, it is euident, that a circle geuen, we may finde circles 4, 5, 6, 10, 20, 100, 1000, or how many soeuer shall be appointed, being in continuall proportion, in any proportion, geuen betwene two right lines: which circles, all together, shall be equall to the circle geuen.</hi>
                        </p>
                        <p>For, euermore deuiding the one side of the chiefe square (which is made of the diameter of the circle geuen) into so many partes, as circles are to be made: so that betwene those partes be continued the proportion geuen betwene two right lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and from the pointes of those diuisions, drawe paral<g ref="char:EOLhyphen"/>lels, perpendiculars to the other side of the said chiefe square: making so many parallelogrammes of the chiefe square, as are circles to be made: and to those parallelogrammes (orderly) making equall squares: it is manifest that the sides of those squares, are the diameters of the circles required to be made.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Probleme. 9.</head>
                     <p>
                        <hi>Three circles being geuen, to finde three equall to them: which three found circles shall be in con<g ref="char:EOLhyphen"/>tinuall proportion, in any proportion geuen betwene two right lines.</hi>
                     </p>
                     <p>Suppose the three circles
<figure/>
geuen to be <hi>A, B,</hi> and <hi>C,</hi> and let the proportion geuen, be that which is betwene the right lines <hi>X</hi> &amp; <hi>Y.</hi> I say, three other circles are to be found, equall to <hi>A, B,</hi> and <hi>C,</hi> &amp; with all, in continuall proportion, in the proportion of <hi>X</hi> to <hi>Y.</hi> By the 2. Corollary of my 4. Probleme,<note place="margin">
                           <hi>Construction.</hi>
                        </note> make one circle e<g ref="char:EOLhyphen"/>quall to the three circles <hi>A, B,</hi> and <hi>C.</hi> Which one circle sup<g ref="char:EOLhyphen"/>pose to be <hi>D:</hi> And by the pro<g ref="char:EOLhyphen"/>bleme next before, let three circles be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ound, equall to <hi>D,</hi> and with all, in co<g ref="char:cmbAbbrStroke">̄</g>tinuall pro<g ref="char:EOLhyphen"/>portion, in the same propor<g ref="char:EOLhyphen"/>tion which is betwene <hi>X</hi> and <hi>Y.</hi> Which three circles, sup<g ref="char:EOLhyphen"/>pose to be <hi>E, F,</hi> and <hi>G.</hi> I say, that <hi>E, F, G,</hi> are equall to <hi>A, B, C:</hi> and with all, in continuall proportion, in the propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tio<g ref="char:cmbAbbrStroke">̄</g> of <hi>X</hi> to <hi>Y.</hi>
                        <note place="margin">
                           <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                        </note> For, by co<g ref="char:cmbAbbrStroke">̄</g>struction, the circle <hi>D</hi> is equall to the circles <hi>A, B,</hi> &amp; <hi>C:</hi> and by con<g ref="char:EOLhyphen"/>struction likewise, the circles <hi>E, F,</hi> and <hi>G,</hi> are equall to the same circle <hi>D:</hi> Wherfore the three circles <hi>E, F,</hi> &amp; <hi>G,</hi> are e<g ref="char:EOLhyphen"/>quall to the three circles <hi>A, B,</hi> &amp; <hi>C:</hi> and by construction <hi>E, F,</hi>
                        <pb n="362" facs="tcp:7062:391"/>
and <hi>G,</hi> are in conti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uall proportion, in the proportion of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to the line <hi>Y.</hi> Wherefore <hi>E, F,</hi> and <hi>G,</hi> are equall to <hi>A, B,</hi> and <hi>C:</hi> and in continuall proportion, in the proportion of <hi>X</hi> to <hi>Y.</hi> Three circles, therefore, being geuen, we haue found three circles, equall to them, and also in continuall proportion, in any proportion geuen, betwene two right lines. Which was requisite to be done.</p>
                     <div type="corollary">
                        <head>¶A Corollary. 1.</head>
                        <p>
                           <hi>It is hereby very manifest, that vnto</hi> 4.5.6.10.20. 100. <hi>or how many circles soeuer, shall be geuen, we may finde</hi> 3.4.5.8.10. <hi>or how many soeuer, shall be appointed: which, all together, shall be equall to the circles geuen, how many soeuer they are: and with all, our found circles, to be in continu<g ref="char:EOLhyphen"/>all proportion, in any proportion assigned betwene two right lines geuen.</hi>
                        </p>
                        <p>For, euermore, by the Corollary of the 4. Probleme, reduce all your circles to one: and by the Co<g ref="char:EOLhyphen"/>rollary of my 8. Probleme, make as many circles as you are appointed, equall to the circles geuen, and continuall in proportion, in the same, wherein, the two right lines geuen, are<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> And so haue you perfor<g ref="char:EOLhyphen"/>med, the thing required.</p>
                     </div>
                     <div type="note">
                        <head>Note.</head>
                        <p>What incredible fruite in the Science of proportions may hereby grow, no mans tounge can suffi<g ref="char:EOLhyphen"/>ciently expresse. And sory I am, that vtterly l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ysure is taken from me, somewhat to specifie in particu<g ref="char:EOLhyphen"/>lar hereof.</p>
                     </div>
                     <div type="note">
                        <head>¶The key of one of the chiefe treasure houses, belonging to the State Mathematicall.</head>
                        <p>THat<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> which in thes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 9. Problemes, is said of circles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> is much more sayd of squares, by whose meanes, circles, are thus handled. And therefore seing to all Polygonon right lined figures, equall squares <milestone type="tcpmilestone" unit="unspecified" n="1"/> may be made, by the falt of the second: and contrariwise, to any square, a right lined figure may be made <milestone type="tcpmilestone" unit="unspecified" n="2"/> equall, and withall, like to any right lined figure geuen, by the 25 of the sixt. And fourthly, seing vpon <milestone type="tcpmilestone" unit="unspecified" n="3"/> the said plaine figures<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> as vpo<g ref="char:cmbAbbrStroke">̄</g> base<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> may Pri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>mes, Parallelipipedons, Pyramids, sided Columnes, Cones,<milestone type="tcpmilestone" unit="unspecified" n="4"/> and Cylinders, be reared: which being <note n="*" place="margin">
                              <hi>Note and remember one <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>th in these so<g ref="char:EOLhyphen"/>lids.</hi>
                           </note> all of one height, shall haue that proportion, one to the o<g ref="char:EOLhyphen"/>ther, that their bases haue, one to the other. And fiftly, seing Spheres, Cones, and Cylinders are one to other in <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> knowen proportions: and so may be made, one to the other in any proportion as<g ref="char:EOLhyphen"/>signed.<milestone type="tcpmilestone" unit="unspecified" n="5"/> And <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap>, seing vnder euery one of the kindes of figures, both plaine, and solide, infinite cases may chaunce, by the ayde of these Problemes, to be soluted and executed: How infinite (then) vpon <milestone type="tcpmilestone" unit="unspecified" n="6"/> infinite, is the number of practises, either Mathematicall, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> Mechanicall, to be performed, of compa<g ref="char:EOLhyphen"/>risons betwene diuers kindes, of plaines to plaines, and solides to solides<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </p>
                        <p>Fa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>h<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>ore, to speake of playne superficiall figures, in respect of the con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>r, or Area of the circle,<milestone type="tcpmilestone" unit="unspecified" n="7"/> sundry mixt line figures, Anular and Lunular figures: and also of circles to be geuen equall to the sayd <gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>sed figures<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and in all proportions els: and euermore thinking of solides, (like high) set vpon a<g ref="char:EOLhyphen"/>ny of those vnused figures, (O Lord) in co<g ref="char:cmbAbbrStroke">̄</g>sideration of al the premisses, how infinite, how straunge and <milestone type="tcpmilestone" unit="unspecified" n="8"/> 
                           <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>credible <gap reason="illegible" resp="#PDCC" extent="6 letters">
                              <desc>••••••</desc>
                           </gap>ation<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and practise<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>, may (by the ayde and direction of these few problemes) <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> redi<g ref="char:EOLhyphen"/>ly into the imagination<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and handes of them, that will bring their minde and intent wholy and fixedly to such mathematicall discourses? In these Elementes, I entend but to geue to young beginners some <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap>, and courage to exercise <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>heir owne witts, and talent, in this most pleasant and profitable sci<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>nc<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>. All <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hinge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> not, neither y<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>an, in euery place be sayd. <hi>Opportunitie, and S<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ff<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>ci<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>ty, best are to be allowed.</hi>
                        </p>
                     </div>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 3. Theoreme. The 3. Proposition. Euery Pyramis hauing a triangle to his base: may be deuided into two Py<g ref="char:EOLhyphen"/>ramids equall and like the one to the other, and also like to the whole, ha<g ref="char:EOLhyphen"/>uing also triangles to their bases, and into two equall prismes: and those two prismes are greater then the halfe of the whole Pyramis.</head>
               <p>SVppose that there be a Pyramis, whose base let be the triangle ABC, and his toppe the point D. Then I say, that the Pyramis ABCD may be deuided into two Pyramids equall and like the one to the other, and also like to the whole, ha<g ref="char:EOLhyphen"/>uing also triangles to their bases, and into two equall prismes, and those two
<pb facs="tcp:7062:392"/>
prismes are greater then the halfe of the whole Pyramis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Deuide (by the 10. of the first) the lines AB, BC, CA, AD, BD, &amp; CD, into two equall partes in the pointes E, F, G, H, K, and L.<note place="margin">Cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ruction.</note> And drawe these right lines EH, EG, GH, HK, KL, LH, EK, KF, and FG. Now forasmuch as the line AE is equall to the line
<figure/>
EB, and the line AH to the line HD:<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the first part, namely, that the whole Pyra<g ref="char:EOLhyphen"/>mids is de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<g ref="char:EOLhyphen"/>ded into two Pyramis e<g ref="char:EOLhyphen"/>quall and like the one to the other, and also to the whole, &amp; hauing tri<g ref="char:EOLunhyphen"/>angles to their bases.</note> therfore (by the 2. of the sixt) the line EH is a parallel to the line DB. And by the same reason, the line HK is a parallel to the line AB: Wherefore HEKB is a parallelogramme. Wherefore the line HK is equall to the line EB. But the line EB is equall to the line AE: Wherefore the line AE is equall to the line HK. And the line AH is equall to the line HD. Now therfore there are two lines AE and AH, equall to two lines KH and HD, the one to the other, and the angle EAH is (by the 29. of the first) equall to the an<g ref="char:EOLhyphen"/>gle KHD: Wherefore (by the 4. of the first) the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ase EH is equall to the base KD. Wherefore the triangle AEH, is equall and like to the triangle HKD. And by the same reason also the triangle AHG, is equall and like to the triangle HLD. And forasmuch as two right lines EH &amp; HG touching the one the other, are parallels to two right lines KD and DL touching also the one the other, and not being in one and the selfe same plaine superficies with the two first: those lines (by the 10. of the ele<g ref="char:EOLhyphen"/>uenth) containe equall angles. Wherefore the angle EHG is equall to the angle KDL. And forasmuch as two right lines EH and HG, are equall to two right lines KD &amp; DL, the one to the other, and the angle EHG is (by the 10. of the eleuenth) equall to the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DL, therefore (by the 4. of the first) the base EG is equall to the base LK. Wherefore the trian<g ref="char:EOLhyphen"/>gle EHG is equall and like to the triangle KDL. And by the same reason also the triangle AEG is equall and like to the triangle HKL. Wherefore the Pyramis, whose ba<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e is the tri<g ref="char:EOLhyphen"/>angle AEG, and toppe the point H, is equall and like to the Pyramis, whose base is the trian<g ref="char:EOLhyphen"/>gle HKL, and toppe the point D.</p>
               <p>And forasmuch as to one of the sides of the triangle ADB, namely, to the side AB, is drawen <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> parallel line HK, therefore the whole triangle A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DB is equiangle to the triangle DHK, and their sides are proportionall (by the Corollary of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the sixt). Wherefore the triangle ADB is like to the triangle DHK. And by the same reason also the triangle DBC is like to the triangle DKL, and the triangle ADC to the triangle DHL. And forasmuch as two right lines BA and AC touching the one the other, are parallels to two right lines KH and HL, touching also the one the other, but not being in one and the selfe same super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies with the two first lines, therefore (by the 10. of the eleuenth) they containe e<g ref="char:EOLhyphen"/>quall angles. Wherefore the angle BAC is equall to the angle KHL. And as the line BA is to the line AC, so is the line KH to the line HE. Wherefore the triangle ABC is like to the triangle KHL. Wherfore the whole Pyramis whose base is the triangle ABC &amp; top the point D, is like to the pyramis whose base is the triangle HKL, and toppe the point D. But the pryamis whose base is the triangle HKL and toppe the point D, is proued to be like to the pyramis whose base is the triangle AEG and toppe the poynt H. Wherefore also the pyramis whose base is the triangle ABC, and toppe the poynt D, is like to the pyramis whose base is the triangle AEG and toppe the poynt H (by the 21. of the sixth).<note place="margin">The conclusi<g ref="char:EOLhyphen"/>on of the first part.</note> Wherefore either of these py<g ref="char:EOLhyphen"/>ramids AEGH and HKLD is like to the whole pyramis ABCD.</p>
               <p>
                  <note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the second part, namely, that it is deui<g ref="char:EOLhyphen"/>ded moreouer into two equal Prismes.</note>
                  <hi>[And forasmuch as the lines BE, KH, are parallel lines and equall, as it hath bene proued, there<g ref="char:EOLhyphen"/>fore the right lines BK and EH, which ioyne them together, are equall and parallels, by the 33. of the first. Againe fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asmuch as the lines BE and FG are parallel lines and equall, therefore lines EG
<pb n="363" facs="tcp:7062:392"/>
and BF, which ioyne them together, are also equall and parallels: and by the same reason forasmuch as the lines FG and KH are equall parallels, the lines FK and GH, which ioyne them together, are also equall parallels. Wherefore BEHK, DEGF, and KHGF, are parallelogrammes. And forasmuch as their opposite sides are equall, by the 34. of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore the triangles FHG, &amp; BKF, are equi<g ref="char:EOLhyphen"/>angle, by the 8. of the first: and therefore, by the 4. of the same, they are equall: and moreouer, by the 15. of the eleuenth, their superficiec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s are parallels. Wherefore the solide BKFEHG is a Prisme, by the 11. definition of the eleuenth. Likewise forasmuch as the sides of the triangle HKL are equall and parallels to the sides of the triangle GFC, as it hath before bene proued: It is manifest, that CFKL, FKHG, and CLHG, are parallelogrammes, by the 33. of the first. Wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the whole solide KL<g ref="char:EOLhyphen"/>HFCG, is a Prisme, by the 11. de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>inition of the eleuenth, and is contayned vnder the sayd parallelo<g ref="char:EOLhyphen"/>grammes CFKL, FKHG, and CLHG, and the two triangles HKL and GFC, which are opposite and parallels.]</hi>
               </p>
               <p>And forasmuch as the line BF is equall to the line FC, therefore (by the 41. of the first) the parallelogramme EBFG is double to the triangle GFC. And forasmuch as if there be two Prismes of equall altitudes, and the one haue to his base a parallelogramme, and the other a triangle, and if the parallelogramme be double to the triangle, those Prismes are (by the 40. of the eleuenth) equall the one to the other: therefore the Prisme contained vn<g ref="char:EOLhyphen"/>der the two triangles BKF and EHG, and vnder the three parallelogrammes EBFG, EBKH, and KHFG,<note place="margin">Conclusion of the second part.</note> is equall to the Prisme contained vnder the two triangles GFC, and HKL, and vnder the three parallelogrammes KFCL, LCGH, and HKFG.</p>
               <p>And it is manifest, that both these Prismes, of which the base of one is the parallelo<g ref="char:EOLhyphen"/>gramme EBFG,<note place="margin">Demonstratio<g ref="char:cmbAbbrStroke">̄</g> of the last part that the two Prismes are greater then the halfe of the whole Py<g ref="char:EOLhyphen"/>ramis.</note> and the opposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e vnto it the line KH, and the base of the other is the tri<g ref="char:EOLhyphen"/>angle GFC, and the opposite side vnto it the triangle KLH, are greater then both these Pyramids, whose bases are the triangles AGE, and HKL, and toppes the pointes H &amp; D. For if we drawe these right lines EF and EK, the Prisme whose base is the parallelogramme EBFG, and the opposite vnto it the right line HK, is greater then the Pyramis whose base is the triangle EBF, &amp; toppe the point K. But the Pyramis whose base is the triangle EBF, and toppe the point K, is equall to the Pyramis whose base is the triangle AEG and toppe the point H, for they are contained vnder equall and like plaine superficieces. Wherefore also the Prisme whose base is the parallelogramme EBFG and the opposite vnto it the right line HK, is greater then the Pyramis, whose base is the triangle AEG, and toppe the point H. But the prisme whose base is the parallelogramme EBFG, and the opposite vnto it the right line HK, is equall to the prisme, whose base is the triangle GFC, and the opposite side vnto it the triangle HKL: And the Pyramis whose base is the triangle AEG, and toppe the point H, is equall to the Pyramis, whose base is the triangle HKL, and toppe the point D.<note place="margin">Conclusion of the last part.</note> Wherefore the foresaid two prismes are greater then the foresaid two Pyramids, whose bases are the tri<g ref="char:EOLhyphen"/>angles AEG, HKL, and toppes the pointes H and D. Wherefore the whole Pyramis whose base is the triangle ABC,<note place="margin">Conclusion of the whole proposition.</note> and toppe the point D, is deuided into two Pyramids equall and like the one to the other, and like also vnto the whole Pyramis, hauing also triangles to their bases, and into two equall prismes, and the two prismes are greater then halfe of the whole Pyramis: which was required to be demonstrated.</p>
               <p>
                  <hi>If ye will with diligence reade these fower bookes following of Euclide, which concerne bodyes, and clearely see the demonstrations in them conteyned, it shall be requisite for you when you come to any proposition, which concerneth a body or bodies, whether they be regular or not, first to describe of p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed paper (according as I taught you in the end of the definitions of the eleuenth booke) such a body or bodyes, as are there required, and hauing your body, or bodyes thus described, when you haue no<g ref="char:EOLhyphen"/>ted it with letters according to the figure set forth vpo<g ref="char:cmbAbbrStroke">̄</g> a plaine in the propositio<g ref="char:cmbAbbrStroke">̄</g>, follow the constructi<g ref="char:EOLhyphen"/>on required in the proposition. As for example, in this third propositio<g ref="char:cmbAbbrStroke">̄</g> it is sayd that,</hi> Euery pyramis ha<g ref="char:EOLhyphen"/>uing a triangle to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>is base, may be deuided into two pyramids. &amp;c. <hi>Here first describe a pyramis of pasted paper ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ing his base triangled (it skilleth not whether it be equilater, or equiangled, or not, only in this pro<g ref="char:EOLhyphen"/>position is required that the base be a triangle. Then for that the proposition supposeth the base of the pyramis to be the triangle ABC, note the base of your pyramis which you haue described with the let<g ref="char:EOLhyphen"/>ters ABC, and the toppe of your pyramis with the letter D: For so is required in the proposition. And thus haue you your body ordered ready to the construction. Now in the construction it is required that ye deuide the lines, AB, BC, CA. &amp;c, namely, the sixe lines which are the sids of the fower triangles con<g ref="char:EOLhyphen"/>tayning
<pb facs="tcp:7062:393"/>
the piramis, into two equall partes in the poyntet <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, F, G, &amp;c. That is, ye must deuide the line AB of your pyramis into two equall partes, and note the poynt of the deuision with the letter E, and so the line BC being deuided into two equall partes, note the poynt of the deuision with the letter F. And so the rest, and this order follow ye as touching the rest of the construction there put, and when ye haue finished the construction, compare your body thus described with the demonstration: and it will make it very playne and easy to be vndersta<g ref="char:cmbAbbrStroke">̄</g>ded. Whereas without such a body described of matter, it is hard
<figure/>
for young beginners (vnlesse they haue a very deepe imagination) fully to conceaue the demonstration by the sig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e as it is described in a plaine. Here for the better declaration of that which I haue sayd, haue I set a figure, whose forme if ye describe vpon pasted paper noted with the like letters, and cut the lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>A, DA, DC, and folde it accordingly, it will make a Pyramis described according to the construction required in the proposition. And this order follow ye as touching all other propositions which con<g ref="char:EOLhyphen"/>cerne bodyes.</hi>
               </p>
               <div type="section">
                  <head>¶An other demonstration after <hi>Campane</hi> of the 3. proposition.</head>
                  <p>Suppose that there be a Pyramis <hi>ABCD</hi> hauing to his base the triangle <hi>BCD,</hi> and let his toppe be the solide angle <hi>A</hi>: from which let there be drawne three subtended lines <hi>AB, AC,</hi> and <hi>AD</hi> to the three angles of the base, and deuide all the sides of the base into two equall partes in the three poyntes <hi>E, F, G</hi>: deuide also the three subte<g ref="char:cmbAbbrStroke">̄</g>ded lines <hi>AB, AC,</hi> and <hi>AD</hi> in two equall partes in the three points <hi>H, K, L.</hi> And draw in the base these two lines <hi>EF</hi> and <hi>EG</hi>:
<figure/>
So shall the base of the pyramis be deuided into three su<g ref="char:EOLhyphen"/>perficieces: whereof two are the two triangles <hi>BEF,</hi> and <hi>EGD,</hi> which are like both the one to the other, and also to the whole base, by the 2 part of the seco<g ref="char:cmbAbbrStroke">̄</g>d of the sixth, &amp; by the definitio<g ref="char:cmbAbbrStroke">̄</g> of like super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iciec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s, &amp; they are equal the one to the other, by the 8. of the first: the third superficies is a quadrangled parallelogramme, namely, <hi>EFGC</hi>: which is double to the triangle <hi>EGD,</hi> by the 40. and 41. of the first. Now then agayne from the poynt <hi>H</hi> draw vnto the points <hi>E</hi> and <hi>F</hi> these two subtendent lines <hi>HE</hi> and <hi>HF</hi>: draw also a subtended line from the poynt <hi>K</hi> to the poynt <hi>G.</hi> And draw these lines <hi>HK, KL,</hi> and <hi>LH.</hi> Wherefore the whole pyramis <hi>ABCD</hi> is deuided into two pyramids, which are <hi>HBEF,</hi> and <hi>AHKL,</hi> and into two prismes of which the one is <hi>EHFGKC,</hi> and is set vpon the quadrangled base <hi>CFGE</hi>: the other is <hi>EGDHKL,</hi> and hath to his base the triangle <hi>EGD.</hi> Now as touching the two pyramids <hi>HBEF</hi> and <hi>AHKL,</hi> that they are equall the one to the other, and also that they are like both the one to the o<g ref="char:EOLhyphen"/>ther and also to the whole, it is manifest by the definition of equall and like bodyes, and by the 10. of the eleuenth, and by 2. part of the second of the sixth. And that the two Prismes are equall it is manifest by the last of the eleuenth. And now that both the prismes taken together are greater then the halfe of the whole pyramis, hereby it is manifest, for that either of them may be deuided into two pyramids, of which the one is a triangular pyramis equall to one of the two pyramids into which together with the two prismes is deuided the whole pyramis, and the other is a quadrangled pyramis double to the other pyramis. Wherefore it is playne that the two prismes taken together are three quarters of the whole
<pb n="364" facs="tcp:7062:393"/>
pyramis deuided. But if ye are desirous to know the proportio<g ref="char:cmbAbbrStroke">̄</g> betwen them, reade the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, of this booke. But now here to this purpose it shall be sufficient to know, that the two prismes taken together do ex<g ref="char:EOLhyphen"/>ceede in quantity the two partial pyramids taken together, into which together with the two prismes the whole pyramis was deuided.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 4. Theoreme. The 4. Proposition. If there be two Pyramids vnder equall altitudes, hauing triangles to their bases, and either of those Pyramids be deuided into two Pyramids equall the one to the other, and like vnto the whole, and into two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall Prismes, and againe if in either of the Pyramids made of the two first Pyramids be still obserued the same order and maner: then as the base of the one Pyra<g ref="char:EOLhyphen"/>mis is to the base of the other Pyramis, so are all the Prismes which are in the one Pyramis to all the Prismes which are in the other, being equall in multitude with them.</head>
               <p>SVppose that there be two Pyramids vnder equall altitudes, hauing triangles to their bases, namely, ABC, and DEF, and hauing to their toppes the pointes G and H. And let either of these pyramids be diuided into two pyramids equall the one to the other, and like vnto the whole, and into two equall prismes (accor<g ref="char:EOLhyphen"/>ding to the methode of the former Proposition). And againe, let either of those pyramids so<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> made of the two first pyramids, be imagined to be after the same order deuided, and so do con<g ref="char:EOLhyphen"/>tinually. Then I say, that as the base ABC is to the base DEF, so are all the prismes which are in the pyramis ABCG, to all
<figure/>
the prismes which are in the py<g ref="char:EOLhyphen"/>ramis DEFH being equall in multitude with them. For for<g ref="char:EOLhyphen"/>asmuch as the line BX is equall to the line XC, and the line AL to the line LC: (For as we saw in the construction pertayning to the former Proposition, al the sixe sides of the whole pyramids, are ech deuided into two equall parts, the like of which construc<g ref="char:EOLhyphen"/>tion is in this proposition also supposed): therefore the line X<g ref="char:EOLhyphen"/>L is a parallel to the line AB, &amp; the triangle ABC, is like to the triangle LXC, (by the Corolla<g ref="char:EOLhyphen"/>ry of the second of the sixth): and by the same reason the tri<g ref="char:EOLhyphen"/>angle DEF is like to the triangle RWF. And forasmuch as the line BC is double to the line CX, and the line FE to the line FW: therefore as the line BC is to the line CX, so is the line EF to the line FW. And vpon the lines BC and CX are described rectiline figures like and in like sort set, namely, the triangles ABC and LXC, and vpon the lines EF and FW are also described rectiline figures, like and in like sort set, namely, the triangles DEF &amp; RWF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But if there be fower right lines proportionall, the rectiline figures described of them being like und in like sort set, shall also be proportionall (by the 22. of the sixt). Wherefore as the
<pb facs="tcp:7062:394"/>
triangle ABC is to the triangle LXC, so is the triangle DEF to the triangle RWF. Wherefore alternately (by the 16. of the fift) as the triangle ABC is to the triangle DEF, so is the triangle LXC to the triangle RWF.<note n="*" place="margin">An Assu<g ref="char:cmbAbbrStroke">̄</g>pt.</note> But as the triangle LXC is to the triangle RWF, so is the prisme whose base is the triangle LXC, and the opposite side vnto it the tri<g ref="char:EOLhyphen"/>angle OMN, to the prisme whose base is the triangle RWF, and the opposite side vnto it the triangle STV (by the Corollary of the 40. of the eleuenth). For these prismes are vnder one &amp; the selfe same altitude, namely, vnder the halfe of the altitude of the whole Pyramids, which Pyramids are supposed to be vnder one and the selfe same altitude: this is also proued in the Assumpt following). Wherefore (by the 11. of the fift) as the triangle ABC is to the triangle DEF, so is the
<figure/>
prisme whose base is the triangle LXC, and the opposite side vn<g ref="char:EOLhyphen"/>to it the triangle OMN, to the prisme whose base is the trian<g ref="char:EOLhyphen"/>gle RWF, and the opposite side vnto it the triangle STV. And forasmuch as there are two prismes in the pyramis ABCG equal the one to the other, &amp; two prismes also in the pyramis DE<g ref="char:EOLhyphen"/>FH equall the one to the other: therefore as the prisme, whose base is the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me BK<g ref="char:EOLhyphen"/>LX, and the opposite side vnto it the line MO, is to the prisme, whose base is the triangle LXC, and the opposite side vnto it the triangle OMN, so is the prisme, whose base is the parallelogramme PERW, and the opposite vnto it the line ST, to the prisme, whose base is the triangle RWF, and the opposite side vnto it the triangle STV. Wherfore by co<g ref="char:cmbAbbrStroke">̄</g>position (by the 18. of the fift) as the prismes KBXLMO, &amp; LXCMNO, are to the prisme LXCMNO, so are the prismes PEWRST, and RWFSTV, to the prisme RWFSTV. Wherefore alternately (by the 16. of the fift) as the two prismes KB<g ref="char:EOLhyphen"/>XLMO, and LXCMNO, are to the two prismes PEWRST, and RWFSTV, so is the prisme LXCMNO to the prisme RWFSTV. But as the prisme LXCMNO is to the prisme RWFSTV, so haue we proued that the base LXC is to the base RWF, and the base ABC to the base DEF. Wherefore (by the 16. of the fift) as the triangle ABC is to the triangle DEF, so are both the prismes which are in the pyramis ABCG, to both the prismes which are in the pyramis DEFH. And in like sort if we diuide the other pyramids after the selfe same maner, namely, the pyramis OMNG, and the pyramis STVH: as the base OMN is to the base STV, so shall both the prismes that are in the pyramis OMNG, be to both the prismes which are in the pyramis STVH. But as the base OMN is to the base STV, so is the base ABC to the base DEF. Wherefore (by the 11. of the fift) as the base ABC is to the base DEF, so are the two prismes that are in the pyramis ABCG, to the two prismes that are in the pyramis DEFH, and the two prismes that are in the pyramis OMNG, to the two prismes that are in the pyramis ST<g ref="char:EOLhyphen"/>VH, and the fower prismes to the fower prismes. And so also shall it followe in the prismes made by diuiding the two pyramids AKLO, and DPRS, and of all the other pyramids in generall, being equall in multitude.</p>
               <p>
                  <pb n="365" facs="tcp:7062:394"/>[And that as the triangle LXC is to the triangle RWF, so is the prisme whose base is the triangle LXC and the opposite side OMN,<note place="margin">An Assumpt.</note> to the prisme whose base is the triangle RWF and the opposite side the triangle STV, may thus be proued. For in the selfe same construc<g ref="char:EOLhyphen"/>tion, imagine perpendicular lines to be drawne from the poyntes G and H to the two playne superficieces wherein are the triangles ABC and DEF. Now those perpendicular lines shall be equall the one to the other, for that the two pyramids are supposed to be of equall altitude. And forasmuch as two right lines, namely, GC, and the perpendicular line drawne from the poynt G, are deuided by two parallel playne superficieces, namely, ABC and OMN, ther<g ref="char:EOLhyphen"/>fore (by the 17. of the eleuenth) the partes of the lines deuided are proportionall. But the line GC is by the playne superficies OMN deuided into two equall partes in the poynt N. Where<g ref="char:EOLhyphen"/>fore also the perpendicular line drawne from the poynt G to the playne superficies wherein is the triangle ABC, is deuided into two equall partes by the superficies OMN: &amp; by the same reason also the perpendicular line which is drawne from the poynt H to the playne superficies DEF is deuided into two equall partes, by the playne superficies STV. And the perpendicu<g ref="char:EOLhyphen"/>lars drawne from the poyntes G and H to the playne superficieces ABC and DEF are equal. Wherefore also the perpendicular lines which are drawne from the triangles OMN and STV to the playne superficieces ABC and DEF are equal the one to the other. Wherfore also the prismes whose bases are the triangles LXC and RWF and the opposite sides the tri<g ref="char:EOLhyphen"/>angles OMN and STV, are of equall altitude. Wherefore also the parallelipipedons which are described of the forsayd prismes and being equall in altitude with them, are the one to the other, as the base of the one is to the base of the other. Wherefore also as the halfe of the bases of those parallelipipedons, namely, as the base LXC is to the base RWF, so are the forsayd prismes the one to the other.] If therfore there be two pyramis vnder equall altitudes, hauing triangles to their bases and either of those pyramids be deuided into two pyramids equall the one to the other and like vnto the whole, and into two equall prismes,<note place="margin">Conclusion of the whole.</note> and agayne if in either of the pyramids made of the two first pyramids be still obserued the same order and maner: then as the base of the one pyramis is to the base of the other piramis, so all the prismes which are in the one pyramis to all the prismes which are in the other being equall in multitude: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 5. Theoreme. The 5. Proposition. Pyramids consisting vnder one and the selfe same altitude, hauing tri<g ref="char:EOLhyphen"/>angles to their bases: are in that proportion the one to the other that their bases are.</head>
               <p>SVppose that these two Pyramids, whose bases are the triangles ABC &amp; DEF, and toppes the pointes G and H, be vnder equall altitudes. Then I say, that as the base ABC is to the base DEF, so is the pyramis ABCG to the pyramis DEFH. For if the pyramis ABCG be not to the pyramis DEFH, as the base ABC is to the base DEF, then as the base ABC is to the base DEF,<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> so is the pyra<g ref="char:EOLhyphen"/>mis ABCG to a solide, either lesse then the pyramis DEFH, or greater. First let it be to some lesse, and let the same be X. And (by the 3. of the twelfth) let the pyramis DEFH be deuided into two pyramids equall the one to the other, and like vnto the whole, and into two equall prismes. Now the two prismes are greater then the halfe of the whole pyramis.
<pb facs="tcp:7062:395"/>
And againe (by the same) let the
<figure/>
pyramids which are made of the diuision, be in like sort diuided, and do this continually, vntill there remaine some pyramids made of the pyramis DEFH, which are lesse then the excesse, whereby the pyramis DEFH excedeth the solide X. Let such pyramids be taken, and for ex<g ref="char:EOLhyphen"/>ample sake, let those pyramids be DPRS, &amp; STVH. Wherfore the prismes remayning which are in the pyramis DEFH, are greater then the solide X. De<g ref="char:EOLhyphen"/>uide (by the Proposition next go<g ref="char:EOLhyphen"/>ing before) the pyramis ABCG in like sort, &amp; as many times as the pyramis DEFH <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s deuided. Wherefore (by the same) as the base ABC is to the base DEF, so are all the prismes which are in the pyramis AB<g ref="char:EOLhyphen"/>CG, to all the prismes which are in the pyramis DEFH. But as the base ABC is to the base DEF, so is the pyramis ABCG to the solide X. Wherefore (by the 11. of the fift) as the pyramis ABCG, is to the solide X, so are the prismes which are in the pyramis ABCG, to the prismes which are in the pyramis DEFH. Wherefore alternately (by the 16. of the fift) as the pyramis ABCG is to the prismes which are in it, so is the solide X, to the prismes which are i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the pyramis DEFH. But the pyramis ABCG is greater then the prismes which are in it. Wherefore also the solide X is greater then the prismes which are in the pyramis DEFH (by the 14. of the fift). But it is supposed to be lesse which is impossible. Wherefore as the base ABC is to the base DEF, so is not the pyramis ABCG to any solide lesse then the pyramis DEFH.</p>
               <p>I say moreouer, that as the base ABC is to the base DEF, so is not the pyramis ABCG, <milestone type="tcpmilestone" unit="unspecified" n="1"/> to any solide greater then the pyramis DEFH. For if it be possible, let it be vnto some greater, namely, to the solide X. Wherefore (by conuersion, by the Corollary of the 4. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>) as the base DEF is to the base ABC, so is the solide X to the pyramis ABCG. But as the solide X is to the pyramis ABCG, so is the pyramis DEFH to some solide lesse then the pyramis ABCG, <note n="*" place="margin">In the As<g ref="char:EOLhyphen"/>su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>llow<g ref="char:EOLhyphen"/>in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the second <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rop<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ition of this b<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ke.</note> as we haue before proued. Wherefore also (by the 11. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift) as th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> base DEF is to the base ABC, so is the pyramis DEFH, to some solide lesse then the pyramis ABCG which thing we haue proued to be impossible. Wherfore as the base ABC is to the base DEF, so is not the pyramis ABCG to any solide greater then the pyramis DEFH: and it is als<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proued that it is not in that proportion to any lesse then the pyramis DEFH. Wherefore as the base ABC is to the base DEF, so is the pyramis ABCG to the pyramis DEFH. Wherefore pyramids consisting vnder one and the selfe same altitude, and hauing triangles to their bases, are in that proportion the one to the other, that their bases are: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <pb n="366" facs="tcp:7062:395"/>
               <head>¶ The 6. Theoreme. The 6. Proposition. Pyramids consisting vnder one and the selfe same altitude, and hauing P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ligo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>on figures to their bases: are in that proportion the one to the other, that their bases are.</head>
               <p>SVppose that there be two Pyramids, hauing to their bases these Polig<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>non figures ABCED, and FGHKL, and let their <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oppes be the pointes M and N, which let be of one and the selfe same altitude. Then I say, that as the base ABCED is to the base FGHKL, so is the pyramis ABCEDM, to the pyramis FGH<g ref="char:EOLhyphen"/>KLN.<note place="margin">Construction.</note> Diuide the base
<figure/>
ABCED into these tri<g ref="char:EOLhyphen"/>angles ABC, ACD, &amp; CDE, and likewise the base FGHKL into these triangles FGH, FHL, and HKL. And imagine that vpon euery one of those triangles be set a pyramis of equall al<g ref="char:EOLhyphen"/>titude with the two pyra<g ref="char:EOLhyphen"/>mids put at the begin<g ref="char:EOLhyphen"/>ning.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And for that as the triangle ABC is to the triangle ADC, so is the pyramis ABCM to the pyramis ADCM (by the 5. of this boke). Wher<g ref="char:EOLhyphen"/>fore, by composition (by the 18. of the fift) as the fower sided figure ABCD is to the triangle ACD, so is the pyramis ABCDM to the py<g ref="char:EOLunhyphen"/>ramis ACDM. But as the triangle ACD is to the triangle CDE, so is the pyramis ACDM to the pyramis CDEM. Wherefore of equalitie (by the 22. of the fift) as the base ABCD is to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> base CDE, so is the pyramis ABCDM to the pyramis CDEM. Wher<g ref="char:EOLhyphen"/>fore againe by composition (by the 18. of the fift) as the base ABCDE is to the base CDE, so is the pyramis ABCEDM to the pyramis CDEM. And by the same reason also as the base FGHKL is to the base HKL, so is the pyramis FGHKLN to the pyramis HKLN. And forasmuch as there are two pyramids CDEM and HKLN, hauing tri<g ref="char:EOLhyphen"/>angles to their bases, and being vnder one and the selfe same altitude, therefore (by the 5. of the twelfth) as the base CDE is to the base HKL, so is the pyramis CDEM to the pyra<g ref="char:EOLhyphen"/>mis HKLN. Now for that as the base ABCED is to the base CDE, so is the pyramis ABCEDM to the pyramis CDEM. But as the base CDE is to the base HKL, so is the pyramis CDEM to the pyramis HKLN. Wherefore of equalitie (by the 22. of the fift) as the base ABCED is to the base HKL, so is the pyramis ABCEDM to the pyramis HKLN. But also as the base HKL is the base FGHKL, so is the Pyramis HKLN to to the pyramis FGHKLN. Wherefore againe of equalitie (by the 22. of the fift) as the base ABCED is to the base FGHKL, so is the pyramis ABCEDM to the pyramis FGHKLN. Wherefore pyramids consisting vnder one and the selfe same altitude, and ha<g ref="char:EOLhyphen"/>uing Polygonon figures to their bases, are in that proportion the one to the other, that their bases are: which was required to be proued.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:396"/>
               <head>The 7. Theoreme. The 7. Proposition. Euery prisme hauing a triangle to his base, may be deuided into three py<g ref="char:EOLhyphen"/>ramids equall the one to the other, hauing also triangles to their bases.</head>
               <p>SVppose that ABCDEF be a prisme, hauing to his base the triangle ABC, and the opposite side vnto it, the triangle DEF. Then I say that the prisme ABCDEF, may be deuided into three piramids equall the one to the other, and hauing triangles to their bases.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Draw these right lines BD, EC, and CD. And foras<g ref="char:EOLhyphen"/>much as ABED is a parallelogramme, and his di<g ref="char:EOLhyphen"/>ameter
<figure/>
is the line BD, therefore the triangle ABD is equall to the triangle EDB. Wherefore also the pyramis whose base is the triangle ABD, and toppe the poynt C, is equall to the pyramis whose base is the triangle EDB, &amp; toppe the point C, by the 5. of this booke. But the pyramis whose base is the triangle EDB, and toppe the poynt C, is one and the same which the pyramis whose base is the triangle EBC, and toppe the poynt D, for they are comprehended of the selfe same playne superficieces, namely, of y<hi rend="sup">e</hi> triangles BDEDEC, DBC, and EBC. Wherefore also the pyra<g ref="char:EOLhyphen"/>mis whose base is the triangle ABD and toppe the poynt C, is equall to the pyramis whose base is the triangle EBC and toppe the point D. A<g ref="char:EOLhyphen"/>gaine forasmuch as BCFE is a parallelogramme, and the diameter thereof is EC, therefore the triangle ECF is equall to the triangle CBE. Wherefore also the pyramis whose base is the triangle EBC and toppe the poynt D, is equall to the pyramis, whose base is the triangle ECF and toppe the poynt D, by the 5. of this booke. But the pyramis whose base is the trian<g ref="char:EOLhyphen"/>gle BEC and toppe the poynt D, is proued to be equall to the pyramis whose base i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the trian<g ref="char:EOLhyphen"/>gle ABD, and toppe the poynt C. Wherfore also the pyramis whose base is the triangle CEF and toppe the poynt D, is equall to the pyramis whose base is the triangle ABD &amp; toppe the poynt C. Wherefore the prisme ABDEF is deuided into three equall pyramids hauing tri<g ref="char:EOLhyphen"/>angles to their bases. And forasmuch as the pyramis whose base is the triangle ABD and toppe the poynt C, is one &amp; the selfe same with the pyramis whose base is the triangle CAB &amp; toppe the poynt D (for they are contayned vnder the selfe same playne superficieces) but it hath bene proued that the pyramis whose base is the triangle ABD and toppe the poynt C, is the third pyramis of the prisme whose base is the triangle ABC a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d the opposite side vnto it the triangle DEF. Wherefore the pyramis whose base is the triangle ABC and toppe the poynt D is the third pyramis of the prisme, whose base is the triangle ABC, and opposite side the triangle DEF. Wherefore euery prisme hauing a triangle to his base, may be deuided in<g ref="char:EOLhyphen"/>to three pyramids equall the one to the other, hauing also triangles to their bases: which was required to be proued.</p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>
                     <q>Hereby it is manifest that euery pyramis is the third part of a prisme ha<g ref="char:EOLhyphen"/>uing one and the same base with it and also being vnder the selfe same al<g ref="char:EOLhyphen"/>titude with it. For if the base of the prisme be any other rectiline figure the<g ref="char:cmbAbbrStroke">̄</g> a triangle, that also may be deuided into prismes which shal haue triangles to their bases.</q>
                  </p>
                  <p>
                     <pb n="367" facs="tcp:7062:396"/>Here <hi>Campane</hi> and <hi>Flussas</hi> adde certayne Corollaryes.</p>
                  <div type="subsection">
                     <head>First Corollary.</head>
                     <p>
                        <hi>Euery Prisme is treble to the Piramis, which hath the selfe same triangle to his base that the Prisme hath, and the selfe same altitude.</hi> As it is manifest by this propo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tion, where the Prisme is deuided into three equall Pyramids, of which, two are vpo<g ref="char:cmbAbbrStroke">̄</g> one and the selfe same base, and vnder one and the selfe same altitude. But if the Prisme haue to his <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ase a parallelogramme, and if the Pyramis haue to his base the halfe of the same parallelogramme, and their altitudes be equall, then agayne the Pyramis shalbe the third part of the Prisme. For it was manifest, by the 40. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>leu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>th, that Prismes, being vnder equall altitudes, and the one hauing to his base a triangle, and the other a parallelogramme double to the same triangle, are equall the one to the other. Wherof followeth the former conclusion.</p>
                  </div>
                  <div type="subsection">
                     <head>Second Corollary.</head>
                     <p>
                        <hi>If there be many Prismes vnder one and the same altitude, and hauing triangles to their bases,</hi>
                        <note place="margin">
                           <p>
                              <hi>Note.</hi>
                           </p>
                           <p>
                              <hi>Sided Co<g ref="char:EOLhyphen"/>lumnes (some<g ref="char:EOLhyphen"/>time called prismes) are triple to pyra<g ref="char:EOLhyphen"/>mids, hauing one base and equall he<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>th with them.</hi>
                           </p>
                        </note> 
                        <hi>and if the triangular bases be so ioyned together vpon one and the same playne, that they compose a Po<g ref="char:EOLhyphen"/>ligonon figure: A pyramis set vpon that base being a Poligonon figure, and vnder the same altitude, is the third part of that solide, which is compos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>d of all the Prismes added together.</hi>
                     </p>
                     <p>For forasmuch as eu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ry one of the Prismes which hath to his base a triangle, to euery one of the Pyramids set vpon the same base (the altitude being alwayes one and the same) is treble, it is manifest by the 12. of the fiueth, that all the Prismes are to all the Pyramids treble. Wherefore Parallelipipe<g ref="char:EOLhyphen"/>dons are treble to Pyramids set vpon the selfe same base with them, and vnder the same altitude,<note place="margin">
                           <hi>Note: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>aral<g ref="char:EOLhyphen"/>lelipipedons treble to pyra<g ref="char:EOLhyphen"/>mids of one base and heith with them.</hi>
                        </note> for that they contayne two Prismes.</p>
                  </div>
                  <div type="subsection">
                     <head>Third Corollary.</head>
                     <p>
                        <hi>If two Prismes being vnder one and the selfe same altitude, haue to their bases either, both tri<g ref="char:EOLhyphen"/>angles, or both parallelogrammes, the Prismes are the one to the other, as their bases are.</hi>
                     </p>
                     <p>For forasmuch as those Prismes are equemultiqlices vnto the Pyram<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ds vpon the selfe same bases, and vnder the same altitude, which Pyramids are in proportion as their bases, it is manifest (by the 15. of the fift), that the Prismes are in the proportion of the bases. For by the former Corollary, the Pris<g ref="char:EOLhyphen"/>mes are treble to the Pyramids s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t vpon the triangular bases.</p>
                  </div>
                  <div type="subsection">
                     <head>Fourth Corollary.</head>
                     <p>
                        <hi>Prismes are in sesquealtera proportion to Pyramids which haue the selfe same quadrangled base that the Prismes haue, and are vnder the selfe same altitude.</hi>
                     </p>
                     <p>For, that Pyramis contayneth two Pyramids set vpon a triangular base of the same Prisme, for it is proued, that that Prisme is treble to the Pyramis which is set vpon the halfe of his quadrangled base, vnto which the other which is set vpon the whole base is double, by the sixth of this booke.</p>
                  </div>
                  <div type="subsection">
                     <head>Fiueth Corollary.</head>
                     <p>
                        <hi>Wherefore we may in like sort conclude, that solides mencioned in the second Corollary (which solids Campane calleth sided Columnes) being vnder one and the selfe same altitude, are in proportion the one to the other, as their bases, which are poligonon figures.</hi>
                     </p>
                     <p>For they are in the proportion of the Pyramids or Prismes, set vpon the selfe same bases, and vn<g ref="char:EOLhyphen"/>der the selfe same altitude, that is, they are in the proportio<g ref="char:cmbAbbrStroke">̄</g> of the bases of the sayde Pyramids or Pris<g ref="char:EOLhyphen"/>mes. For those solids may be deuided into Prismes hauing the selfe same altitude, when as their oppo<g ref="char:EOLhyphen"/>site bases may be deuided into triangles, by the 20 of the sixth. Vpon which triangles Prismes beyng set, are in proportion as their bases.</p>
                     <p>By this 7. Proposition it playnely appeareth that <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>lide,</hi> as it was before noted in the diffinition<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> vnder the diffinition of a Prisme, comprehended also those kinds of solids, which <hi>Campane</hi> calleth sided Columnes. For in that he sayth, <hi>Euery Prisme hauing a triangle to his base, may be deuided<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> &amp;c.</hi> he neded not (taking a Prisme in that sense which <hi>Campane</hi> and most men take it) to haue added that particle, <hi>hauing to his base a triangle.</hi> For by their sense, there is no Prisme, but it may haue to his base a triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and so it may seeme that <hi>Euclide</hi> ought without exceptio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> haue sayd, that, euery prisme whatsoeuer, may
<pb facs="tcp:7062:397"/>
be deuided into three pyramids equall the one to the other, hauing also triangles to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>heir bases. For so do <hi>Campane</hi> and <hi>Flussas</hi> put the proposition, leauing out the former particle <hi>hauing to his base a triangle,</hi> which yet is red in the Greeke copye, &amp; not le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t out by any other interpreters knowne abroade except by <hi>Campane</hi> and <hi>Flussas.</hi> Yea and the Corollary following of this proposition added by <hi>Theon</hi> or <hi>Euclide,</hi> and ame<g ref="char:cmbAbbrStroke">̄</g>ded by <hi>M. Dee</hi> semeth to confirme this sence.</p>
                     <p>
                        <hi>Of this, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> made manifest, that euery pyramis is the third part of the prisme, hauing the same base with it, and equall altitude. For, and if the base of the prisme haue any other right lined figure (then a triangle) and also the superficies opposite to the base, the same figure: that prisme may be de<g ref="char:EOLhyphen"/>uided into prismes, hauing triangled bases: and the superficieces to those bases opposite, also triangled a <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ike and equally.</hi>
                     </p>
                     <p>For there, as we see are put these wordes, <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>or and if the base of the prisme be any other right lined figure<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> &amp;c.</hi> whereof a man may well inferre that the base may be any other rectiline figure what<g ref="char:EOLhyphen"/>soeuer, &amp; not only a triangle or a parallelogramme, and it is true also in that sence, as it is plaine to see by the second corollary added out of <hi>Flussas,</hi> which corollary, as also the first of his corollaries, is in a maner all one with the Corollary added by <hi>Theon</hi> or <hi>Euclide.</hi> Farther <hi>Theon</hi> in the demonstration of the 10. proposition of this booke (as we shall a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>terward see) most playnely calleth not onely sided co<g ref="char:EOLhyphen"/>lumnes prismes, but also parallelipipedons. And although the 40. proposition of the eleuenth booke may seme hereunto to be a l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t. For that it can be vnderstanded of those prismes onely which haue triangles to their like, equall, opposite, and parallel sides, or but of some sided columnes, and not of all: yet may that let be thus remoued away, to say that <hi>Euclide</hi> in that propositio<g ref="char:cmbAbbrStroke">̄</g> vsed <hi>genus pro specie,</hi> that is, the ge<g ref="char:EOLhyphen"/>nerall word for some special kinde therof: which thing also is not rare, not only with him, but also with other learned philosophers. Thus much I thought good by the way to note in farther defence of <hi>Euclide</hi> definition of a Prisme.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>The 8. Theoreme. The 8. Proposition. Pyramids being like &amp; hauing triangles to their bases, are in treble propor<g ref="char:EOLunhyphen"/>tion the one to the other, of that in which their sides of like proportion are.</head>
               <p>SVppose that these pyramids whose bases are the triangles GBC and HEF and toppes, the poyntes A and D be like, and in like sort described, and let AB and DE be sides of like proportion. Then I say that the pyramis ABCG is to the pyramis DEFH in treble proportio<g ref="char:cmbAbbrStroke">̄</g>, of that in which the side AB is to the side DE. Make perfect the parallelipipedons, namely, the solides BCKL &amp; EFXO. And foras<g ref="char:EOLhyphen"/>much as the pyramis ABCG is like to the pyramis DEFH,<note place="margin">Construction.</note> therfore the angle ABC is equall to the angle DEF,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> &amp; the
<figure/>
angle GBC to the angle HEF, and moreouer the angle ABG to the angle DEH, and as the line AB is to the line DE, so is the line BC to the line EF, and the line BG to the line EH. And for that as the line AB is to the line DE, so is the line BC to the line EF, and the sides about the equall an<g ref="char:EOLhyphen"/>gles, are proportionall, therefore the parallelo<g ref="char:EOLhyphen"/>gramme BM is like to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me EP: and by the same reason the pa<g ref="char:EOLhyphen"/>rallelogramme BN is like
<pb n="368" facs="tcp:7062:397"/>
to the parallelogramme ER, and the parellelogramme BK is like vnto the parallelogramme EX. Wherefore the three parallelogrammes BM, KB and BN are like to the three paral<g ref="char:EOLhyphen"/>lelogrammes EP, EX, and ER. But the three parallelogrammes BM, KB, and BN are equall and like to the three opposite parallelogrammes, and the three parallelogrammes EP, EX, and ER are equall and like to the three opposite parallelogrammes. Wherefore the pa<g ref="char:EOLhyphen"/>rallelipipedons BCKL and EFXO are comprehended vnder playne superficieces like and equall in multitude. Wherefore the solide BCKL is like to the solide EFXO. But like paral<g ref="char:EOLunhyphen"/>lelipipedons are (by the 33. of the eleuenth) in treble proportion the one to the other of that in which side of like proportion is to side of like proportion. Wherefore the solide BCKL is to the solide EFXO in treble proportion of that in which the side of like proportion AB is to the side of like proportion DE. But as the solide BCKL is to the solide EFXO, so is the py<g ref="char:EOLhyphen"/>ramis ABCG to the pyramis DEFH (by the 15. of the fifth) for that the pyramis is the sixth part of this solide: for the prisme, being the halfe of the parallelipipedon is treble to the pyramis. Wherefore the pyramis ABCG is to the pyramis DEFH in treble proportion of that in which the side AB is to the side DE. Which was required to be proued.</p>
               <div type="section">
                  <head>Corollary.</head>
                  <p>
                     <q>Hereby it is manifest that like pyramids hauing to their bases poligonon figures, are in treble proportion the one to the other, of that in which side of like proportion, is to side of like proportion.</q>
                  </p>
                  <p>For if they be deuided into pyramids hauing triangles to their bases (for like poligono<g ref="char:cmbAbbrStroke">̄</g> figures are deuided into like triangles, and equal in multitude, and the sides are of like propor<g ref="char:EOLhyphen"/>tion) as one of the pyramids of the one, hauing a triangle to his base, is to one of the pyramids of the other, hauing also a triangle to his base, so also are all the pyramids of the one pyramis hauing triangles to their bases to all the pyramids of the other pyramis hauing also triangles to their bases. That is, the pyramis hauing to his base a poligono<g ref="char:cmbAbbrStroke">̄</g> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igure, to the pyramis hauing also to his base a poligono<g ref="char:cmbAbbrStroke">̄</g> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>igure. But a pyramis hauing a triangle to his base, is to a pyramis ha<g ref="char:EOLunhyphen"/>uing also a triangle to his base, &amp; being like vnto it, in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that in which side of like proportio<g ref="char:cmbAbbrStroke">̄</g> is to side of like proportio<g ref="char:cmbAbbrStroke">̄</g>. Wherfore a pyramis hauing to his base a poligono<g ref="char:cmbAbbrStroke">̄</g> figure, is to a pyramis hauing also a poligonon figure to his base, the sayd pyramids being like the one to the other, in treble proportion of that in which side of like proportion is to side of like proportion. <hi>Likewise Prismes and sided columnes, being set vpon the bases of those pyramids,</hi>
                     <note place="margin">An addition by Campane and Flussas.</note> 
                     <hi>and vnder the same altitude (forasmuch as they are equemultiplices vnto the pyramids, namely, triples, by the corollary of the 7. of this booke) shal haue the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ormer porportion that the pyramids haue, by the 15, of the fifth, and therefore they shall be in treble proportion of that in which the sides of like propor<g ref="char:EOLhyphen"/>tion are.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 9. Proposition. In equall pyramids hauing triangles to their bases, the bases are recipro<g ref="char:EOLhyphen"/>kall to their altitudes. And pyramids hauing triangles to their bases, whose bases are reciprokall to their altitudes, are equall the one to the other.</head>
               <p>SVppose that BCGA and EFHD be equall pyramids, hauing to their bases the triangles BCG and EFH, and the tops the pointes A and D. Then I say that the bases of the two pyramids BCGA and EFHD are reciprokall to their al<g ref="char:EOLhyphen"/>titudes: that is, as the base BCG is to the base EFH, so is the altitude of the py<g ref="char:EOLhyphen"/>ramis EFHD to the altitude of the pyramis BCGA. Make perfect the parallelipipedons,
<pb facs="tcp:7062:398"/>
namely, BGML and EHPO. And forasmuch as the pyramis BCGA is equall to the pyra<g ref="char:EOLhyphen"/>mis EFHD,<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on of the first part.</note> &amp; the solide BGML is sextuple to the pyramis BCGA. (For the paralleli<g ref="char:EOLhyphen"/>pipedon is duple to the Prisme set vpon the base of the Pyramis, &amp; the Prisme is triple to the pyramis): and likewise the solide EHPO is sextuple to the pyramis EFHD. Wherefore the solide BGML is equal to the solide EHPO. But in equall parallelipipedons, the bases are (by the 34. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) reci<g ref="char:EOLhyphen"/>prokall
<figure/>
to their altitudes. Wherfore as the base BN is to the base EQ, so is the alti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>de of the solide EHP<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, to the altitude of the solide BGML. But as the base BN is to the base E<g ref="char:EOLhyphen"/>Q, so is the triangle GBC to the triangle HEF (by the 15. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ifth, for the triangles GBC &amp; HEF are the halues of the paral<g ref="char:EOLhyphen"/>lelogrammes BN and E<g ref="char:EOLhyphen"/>Q)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherfore (by the 11. of the fifth) as the triangle GBC is to the triangle H<g ref="char:EOLhyphen"/>EF, so is the altitude of the solide EHPO to the altitude of the solide BG<g ref="char:EOLhyphen"/>ML. But the altitude of the solide EHPO is one and the same with the altitude of the pyra<g ref="char:EOLhyphen"/>mis EFHD, and the altitude of the solide BGML is one and the same with the altitude of the pyramis BCGA. Wherefore as the base GBC is to the base HEF, so is the altitude of the pyramis EFHD to the altitude of the pyramis BCGA. Wherefore the bases of the two pyramids BCGA and EFHD are reciprokall to their altitudes.</p>
               <p>But now suppose that the bases of the pyramids BCGA and EFHD, be reciprokall to their altitudes,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>f the second part, which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the conue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se of the first.</note> that is, as the base GBC is to the base HEF, so let the altitude of the pyramis EFHD be to the altitude of the pyramis BCGA. Then I say that the pyramis BCGA is e<g ref="char:EOLhyphen"/>quall to the pyramis EFHD. For (the selfe same order of construction remaining), for that as the base GBC is to the base <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>EF, so is the altitude of the pyramis EFHD to the alti<g ref="char:EOLhyphen"/>tude of the pyramis BCGA. But as the base GBC is to the base HEF, so is the parallelo<g ref="char:EOLhyphen"/>gramme GC to the parallelogramme HF. Wherefore (by the 11. of the fifth) as the parallelo<g ref="char:EOLhyphen"/>gramme GC is to the parallegoramme HF, so is the altitude of the pyramis EFHD to the altitude of the pyramis BCGA. But the altitude of the pyramis EFND and of the solide EHPO, is one and the selfe same, and the altitude of the pyramis BCGA and of the solide BGML, is also one and the same. Wherefore as the base GC is to the base HF, so is the alti<g ref="char:EOLhyphen"/>tude of the solide EHPO to the altitude of the solide BGML. But parallelipipedons, whose bases are reciprokall to their altitudes are (by the 34. of the eleuenth) equall the one to the o<g ref="char:EOLhyphen"/>ther. Wherefore the parallelipipedon BGML is equall to the parallelipipedon EHPO. But the pyramis BCGA is the sixth part of the solide BGML, and likewise the pyramis EFHD is the sixth part of the solide EHPO. Wherefore the pyramis BCGA is equall to the pyra<g ref="char:EOLhyphen"/>mis EFHD. Wherefore in equall pyramids hauing triangles to their bases, the bases are reci<g ref="char:EOLhyphen"/>prokall to their altitudes. And pyramids hauing triangles to their bases, whose bases are reci<g ref="char:EOLhyphen"/>procall to their altitudes, are equall the one to the other: which was required to be demonstra<g ref="char:EOLhyphen"/>ted.</p>
               <div type="section">
                  <pb n="369" facs="tcp:7062:398"/>
                  <head>A Corrollary added by <hi>Campane</hi> and <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Hereby it is manifest that equall pyramids hauing to their bases Poligonon figures, haue their bases reciprokall with their altitudes. And Pyramids whose bases being poligonon figures are recipro<g ref="char:EOLhyphen"/>kall with their altitudes, are equall the one to the other.</hi>
                  </p>
                  <p>Suppose that vpon the poligonon figures A
<figure/>
and B, be set equall pyramids. Then I say that their bases A and B are reciprokall with their al<g ref="char:EOLhyphen"/>titudes. Describe by the 25. of the sixth, triangles equall to the bases A and B. Which let be C and D. Vpon which let there be set pyramids equall in altitude with the pyramids A and B. Wherfore the pyramids C and D, being set vpo<g ref="char:cmbAbbrStroke">̄</g> bases equall with the bases of the pyramids A and B, and ha<g ref="char:EOLhyphen"/>uing also their altitudes equall with the altitudes of the sayd pyramids A and B, shall be equall by the 6. of this booke. Wherefore by the first part of this proposition, the bases of the pyramids, C to D are reciprokall with the altitudes of D to C. But in what proportion are the bases C to D, in the same are the bases A to B, forasmuch as they are equall. And in what proportion are the altitudes of D to C, in the same are the altitudes of B to A, which altitudes are likewise equall. Wherefore by the 11. of the fifth, in what proportion the bases A to B are, in the same reciprokally are the altitudes of the pyramids B to A. In like sort by the second part of this proposition may be proued the conuerse of this corollary. The same thing followeth also in a Prisme, and in a sided columne, as hath before at large bene declared in the corollary of the 40. propo<g ref="char:EOLhyphen"/>sition of the 11. booke. For those solides are in proportio<g ref="char:cmbAbbrStroke">̄</g> the one to the other, as the pyramids or paral<g ref="char:EOLhyphen"/>lelipipedons, for they are either partes of equemultiplices or equemultiplices to partes.</p>
               </div>
            </div>
            <div type="part">
               <head>The 10. Theoreme. The 10. Proposition. Euery cone is the third part of a cilinder, hauing one and the selfe same base and one and the selfe same altitude with it.</head>
               <p>SVppose that there be a cone hauing to his base the circle ABCD; and let there be a cilinder hauing the selfe same base, and also the same altitude that the cone hath. Then I say that the cone is the third part of the cilinder, that is, that the cilinder is in treble proportion to the cone. For if the cilinder be not in treble proportion to the cone, then the cilinder is either in greater proportions then triple to the cone, or els in lesse. First let it be in greater then triple.<note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="7 letters">
                        <desc>〈7 letters〉</desc>
                     </gap>
                  </note> And describe (by the 6. of the fourth) in the circle ABCD a <milestone type="tcpmilestone" unit="unspecified" n="1"/> square ABCD. Now the square ABCD,
<figure/>
is greater then the halfe of the circle ABC<g ref="char:EOLhyphen"/>D. For if about the circle ABCD, we de<g ref="char:EOLhyphen"/>scribe a square, the square described in the circle ABCD is the halfe of the square described about the circle. And let there be Parallelipipedon prismes described vpon those squares,<note place="margin">Parallelipipe<g ref="char:EOLhyphen"/>dons called Prismes.</note> equall in altitude with the cilinder. But prismes are in that proporti<g ref="char:EOLhyphen"/>on the one to the other, that their bases are (by the 32. of the eleuenth, and 5. Co<g ref="char:EOLhyphen"/>rollary of the 7. of this booke). Wherefore the prisme described vpon the square A<g ref="char:EOLhyphen"/>BCD is the halfe of the prisme described vpon the square that is described about the circle. Now the clinder is lesse then the
<pb facs="tcp:7062:399"/>
prisme which is made of the square described abou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the circle ABCD, being equal in altitude with it, for it contayneth it. Wherfore the prisme described vpon the square ABC<g ref="char:EOLhyphen"/>D and being equall in altitude with the cylinder, is greater then half the cylinder. Deuide (by the 30. of the third) the circumferences AB, BC, CD and DA into two equall parts in the points E, F, G, H, And draw these right lines AE, EB, BF, FC, CG, GD, DH &amp; HA. Wherfore euery one of these triangles AEB, BFC, CGD and DHA is greater then halfe of that segment of the circle ABCD which is described about it, as we haue before in the 2. proposition declared. Describe vpon euery one of these triangles AEB, BFC, CGD, and DHA a prisme of equall altitude with the cylinder. Wherefore euery one of these prismes so described is greater then the halfe part of the segment of the cylinder that is set vpon the sayd segments of the circle. For if by the pointes E, F, G, H, be drawen parallell lines to the lines AB, BC, CD and DA, and then be made perfect the parallelogrammes made by those parallell lines, and moreouer vpon those parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes be erected parallelipipedons equall in altitude with the cylinder, the prismes which are described vpon eche of the triangles AEB, BFC, CGD, and DHA are the halfes of euery one of those parallelipipedons. And the segments of the cylinder are lesse then those parallelipipedons so described. Wherefore also euery one of the prismes which are described vpon the triangles AEB, BFC, CGD and DHA is greater then the halfe of the segment of the cylinder set vpon the sayd segment. Now therefore deuiding euery one of the circumferences remaining into two equall partes, and drawing right lines, and raysing vp vpon euery one of these triangles prismes equall in altitude with the cylinder, and doing this continually, we shall at the length (by the first of the tenth) leaue certaine segments of the cylinder which shalbe lesse then the excesse whereby the cylinder excedeth the cone more then thrise. Let those segments be AE, EB, BF, FC, CG, GD, DH and HA. Wherfore the prisme remayning, whose base is the poligonon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure AEBFCGDH, and altitude the selfe
<figure/>
same that the cylinder hath, is greater then the cone taken three tymes.<note n="*" place="margin">By this it is manifest that Euclide com<g ref="char:EOLhyphen"/>prehended si<g ref="char:EOLhyphen"/>ded Columnes also vnder the name of a Prisme.</note> But the prisme whose base is the poligonon figure AEBFCGDH and altitude the selfe same that the cylinder hath, is treble to the pyramis whose base is the poligonon figure AEBFCGDA and altitude the selfe same that the cone hath, by the corollary of the 3. of this booke. Wherfore also the pyramis whose base is the poligonon figure AEBFCGDH and toppe the self same that the cone hath, is greater then the cone which hath to his base the circle ABCD. But it is also lesse, for it is contayned of it which is impossible. Wherefore the cylinder is not in greater proportion then triple to the cone.</p>
               <p>I say moreouer that the cylinder is not in lesse proportion then triple to the cone<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> For if it be possible let the cylinder be in lesse proportion then triple to the cone. Wherefore by con<g ref="char:EOLhyphen"/>uersion, the cone is greater then the third part of the cylinder. Describe now (by the sixth of the fourth) in the circle ABCD a square ABCD. Wherefore the square ABCD is grea<g ref="char:EOLhyphen"/>ter then the halfe of the circle ABCD vpon the square ABCD describe a pyramis hauing one &amp; the selfe same altitude with the cone. Wherfore the pyramis so described is greater the<g ref="char:cmbAbbrStroke">̄</g> halfe of the cone. (For if as we haue before declared we describe a square about the circle, the square ABCD is the halfe of the square described about the circle, and if vppon the squares be described parallelipipedons equall in altitude with the cone, which solides are also
<pb n="371" facs="tcp:7062:399"/>
called prismes, the prisme or parallelipipedon described vpo<g ref="char:cmbAbbrStroke">̄</g> the square ABCD is the halfe of the prisme which is described vpo<g ref="char:cmbAbbrStroke">̄</g> the square described about the circle, for they are the one to the other in that proportio<g ref="char:cmbAbbrStroke">̄</g> that their bases are (by the 32. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th, &amp; 5. corollary of the 7. of this booke.) Wherfore also their third parts are in the self same proportion (by the 15. of the fift). Wherfore the pyramis whose base is the square ABCD is the halfe of the pyramis set vpon the square described about the circle. But the pyramis set vpon the square described a<g ref="char:EOLhyphen"/>bout the circle is greater then the cone whome it comprehendeth. Wherfore the pyramis whose base is the square ABCD, and altitude the self same that the cone hath; is greater then the halfe of the cone.) Deuide (by the 30. of the third) euery one of the circumferences AB, BC, CD, and DA into two equall partes in the pointes E, F, G, and H: and drawe these right lines AE, EB, BF, FC, CG, GD, DH, and HA. Wherefore euery one of these triangles AEB, BFC, CGD, and DHA is greater then the halfe part of the segment of the circle described about it. Vppon euery one of these triangles AEB, BFC, CGD, and DHA de<g ref="char:EOLhyphen"/>scribe a pyramis of equall altitude with the cone and after the same maner euery one of those pyramids so described is greater then the halfe part of the segment of the cone set vpon the segment of the circle. Now therefore diuiding (by the 30, of the third) the circumferences re<g ref="char:EOLhyphen"/>maining into two equall parts, &amp; drawing right lines &amp; raysing vp vpon euery one of those triangles a pyramis of equall altitude with the cone, and doing this continually, we shal at the length (by the first of the tenth) leaue certayne segmentes of the cone, which shalbe lesse then the excesse whereby the cone excedeth the third part of the cylinder. Let those segmentes be AE, EB, BF, FC, CG, GD, DH, and HA. Wherefore the pyramis remayning, whose base is the poligono<g ref="char:cmbAbbrStroke">̄</g> figure AEBFCGDH and altitude the self same with the cone, is grea<g ref="char:EOLhyphen"/>ter then the third part of the cylinder. But the pyramis whose base is the poligonon figure AEBFCGDH and altitude the self same with the cone, is the third part of the prisme whose base is the poligono<g ref="char:cmbAbbrStroke">̄</g> figure AEBFCGDH and altitude the self same with the cylin<g ref="char:EOLhyphen"/>der. Whefore <note n="*" place="margin">A prisme hauing for his base a po<g ref="char:EOLhyphen"/>ligonon figure as we haue often before noted vnto you.</note> the prisme whose base is the poligonon figure AEBFCGDH, and altitude the self same with the cylinder, is greater then the cylinder whose base is the circle ABCD. But it is also lesse, for it is contayned of it, which is impossible. Wherfore the cylinder is not in lesse proportion to the cone then in treble proportion. And it is proued that it is not in grea<g ref="char:EOLhyphen"/>ter proportion to the cone then in treble proportion, wherefore the cone is the third part of the cylinder. Wherfore euery cone is the third part of a cylinder, hauing one &amp; the self same base, and one and the selfe same altitude with it: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶ Added by M. Iohn Dee.</head>
                  <div type="subsection">
                     <head>¶ A Theoreme. 1.</head>
                     <p>The superficies of euery vpright Cylinder, except his bases, is equall to that circle whose semidia<g ref="char:EOLhyphen"/>meter is middell proportionall betwene the side of the Cylinder, and the diameter of his base.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ A Theoreme. 2.</head>
                     <p>
                        <hi>The superficies of euery vpright, or Isosceles Cone, except the base, is equall to that circle, whose semidiameter is middell proportionall betwene the side of that Cone, and the semidiameter of the cir<g ref="char:EOLhyphen"/>cle: which is the base of the Cone.</hi>
                     </p>
                     <p>My entent in additions is not to amend <hi>Euclide<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> Method, (which nedeth little adding or none at all). But my desire is somwhat to furnish you, toward a more general art Mathematical the<g ref="char:cmbAbbrStroke">̄</g> 
                        <hi>Euclides</hi> Eleme<g ref="char:cmbAbbrStroke">̄</g>ts,<note place="margin">
                           <hi>Note: M. Dee his chiefe purpose in his additions.</hi>
                        </note> (remayning in the termes in which they are written) can sufficiently helpe you vnto. And though <hi>Eu<g ref="char:EOLhyphen"/>clides</hi> Elementes with my Additions, run not in one Methodicall race toward my marke: yet in the meane space my Additions either geue light, where they are annexed to <hi>Euclides</hi> matter, or geue some ready ayde, and shew the way to dilate your discourses Mathematicall; or to inuent and practise thinges
<pb facs="tcp:7062:400"/>
Mechanically. And (in deede) if more leysor had happened, many more straunge matters Mathemati<g ref="char:EOLhyphen"/>call had, (according to my purpose generall) bene presently published to your knowledge: but want of due leasour cau<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>eth you to want, that, which my good will toward you, most hartely doth wish you.</p>
                     <p>As concerning the two Theoremes here annexed, their veritie, is by <hi>Archimedes,</hi> in his booke of the Sphere and Cylinder manifestly demonstrated, and at large: you may therefore boldly trust to them, and vse them, as suppositions, in any your purposes: till you haue also their demo<g ref="char:cmbAbbrStroke">̄</g>strations. But if you well remember my instructions vpon the first proposition of this booke, and my other addition, vpon the second, with the suppositions how a Cylinder and a Cone are Mathematically produced, you will not neede <hi>Archimedes</hi> demonstration: nor yet be vtterly ignoraunt of the solide quantities of this Cylin<g ref="char:EOLunhyphen"/>der and Cone here compared: (the diameter of their base, and heith being knowne in any measure) nei<g ref="char:EOLhyphen"/>ther can their croked superficies remayne vnmeasured. Whereof vndoubtedly great pleasure and com<g ref="char:EOLhyphen"/>moditie may grow to the sincere student, and precise practiser.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 11. Theoreme. The 11. Proposition. Cones and Cylinders being vnder one and the selfe same altitude, are in that proportion, the one other that their bases are.</head>
               <p>LEt there be take<g ref="char:cmbAbbrStroke">̄</g> cones &amp; cylindres vnder one and the selfe same altitude whose bases let be t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e circles ABCD and EFGH, and axes the lines KL &amp; MN, and let the diameters of their bases be AC and EG. The<g ref="char:cmbAbbrStroke">̄</g> I say that as the circle ABCD is to the circle EFGH, so is the cone AL to the cone EN, and also the cylinder AL to the cylinder EN.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion as tou<g ref="char:EOLhyphen"/>ching Cones.</note> For if the cone AL be not to the cone EN as the circle ABCD is to the circle EFGH, then as the circle ABCD is to the circle E<g ref="char:EOLhyphen"/>F
<figure/>
GH, so is the cone AL to some solide either lesse then the cone EN, or greater.<note place="margin">First case. 1.</note> First let it be vnto a lesse, namely, to the solide X. And vnto that which the solide X is lesse then the cone EN, let the solide Y be equall. Wherefore the cone EN is equal to the solides X &amp; Y. Describe (by the 6. of the fourth) in the circle EFGH a square EFGH. Wherefore the square is greater then halfe the circle. Rayse vp vpon the square EFGH a pyramis of equall altitude with the cone. Wherefore the pyramis, so raysed vp is greater then halfe of the cone. For if we describe a square about the cir<g ref="char:EOLhyphen"/>cle, and vpon that square rayse vp a pyramis of equall al<g ref="char:EOLhyphen"/>titude
<pb n="374" facs="tcp:7062:400"/>
with the cone, the pyramis which is set vpon the square described with in the circle is the halfe of the pyramis set vpon the square described about the circle, for they are in propor<g ref="char:EOLhyphen"/>tion the one to the other as their bases. But the cone is lesse then the pyramis which is set vpon the square described about the circle. Wherefore the pyramis whose base is the square EF<g ref="char:EOLhyphen"/>GH and top one and the same with the cone is greater then halfe of the cone: deuide (by the 30. of the third) the circumferences EF, FG, GH and HE into two equal partes in the points O, P, R, S, and draw these right lines HO, OE, EP, PF, FR, RG, GS, and SH. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore euery one of these triangles HOE, EPF, FRG, and GSH is greater then halfe the segme<g ref="char:cmbAbbrStroke">̄</g>t of the circle described about the<g ref="char:cmbAbbrStroke">̄</g>. Rayse vp vpo<g ref="char:cmbAbbrStroke">̄</g> euery one of the triangles HOE, EPF, FRG and GSH a pyramis equall in altitude with the cone. Wherefore euery one of the pyramids so raysed vp is greater then the halfe part of the segment of the cone described about it. Now then deuiding (by the 30. of the third) the circumferences remayning into two equall partes, and drawing right lines, and raysing vp vppon euery one of those triangles a pyramis of equall altitude with the cone, and thus doing continually, we shall at length by the first of the tenth leue certayne segmentes of the cone which shall be lesse then the solide Y. Let those segmentes be HOE, EPP, FRG, and GSH. Wherefore the pyramis remay<g ref="char:EOLhyphen"/>ning whose base is the poligonon figure HOEPFRGS &amp; top the selfe same with the cone, is greater then the solide X. Inscribe in the circle ABCD a poligonon figure like and in like sort situate to the poligonon figure HOEPFRGS, and let the same be DTAVBZCW, and vpon it rayse a pyramis of equall altitude with the cone AL. Now for that as the square of the line AC is to the square of the line EG, so is the poligonon figure DTAVBZCW to the poligonon figure HOEPFRGS (by the first of this booke): But as the square of the line AC is to the square of the line EG, so is the circle ABCD to the circle EFGH (by the second of this booke). Wherefore (by the 11. of the fift) as the circle ABCD is to the circle E<g ref="char:EOLhyphen"/>FGH, so is the poligonon figure DTAVBZCW to the poligonon figure HOEPFRG<g ref="char:EOLhyphen"/>S (this foloweth also of the corollary of the 2. of this booke): moreouer as the circle ABCD is to the circle EFGH, so is the cone AL to the solide X. And as the poligonon figure DT<g ref="char:EOLhyphen"/>AVBZCW, is to the poligonon figure HOEPRGS, so is the pyramis whose base is the poligonon figure DTAVBZCW, and toppe the poynt L, to the pyramis whose base is the poligonon figure HOEPRS, and toppe the poynt N. Wherefore (by the 11. of the fift) as the cone AL is to the solide X, so is the pyramis whose base is the poligono<g ref="char:cmbAbbrStroke">̄</g> figure DTAV<g ref="char:EOLhyphen"/>BZCW, and toppe the poynt L, to the pyramis whose base is the poligonon figure HEOP<g ref="char:EOLhyphen"/>FRGS, and toppe the poynt N. Wherefore alternately (by the 16. of the fift) as the cone AL is to the pyramis which is in it, so is the solide X, to the pyramis which is in the cone EN. But the cone AL, is greater then the pyramis which is in it. Wherefore also the solide X is greater then the pyramis which is in the cone EN. But it is also lesse by construction. Which is impossible. Wherfore as the circle ABGD, is to the circle EFGH, so is not the cone AL to any solide lesse then the cone EN.<note place="margin">Second case.</note>
               </p>
               <p>In like sorte also may we proue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that as the circle EFGH, is to the circle ABCD, so is not the cone EN to any solide lesse then the cone AL. Now I say that as the circle ABCD, is to the circle EFGH, so is not the cone AL to any solide greater then the cone EN. For if it be possible let it be vnto a greater, namely to the solide X. Wherefore by conuersion, as the circle EFGH, is to the circle ABCD, so is the solide X to the cone AL: but as the solide X is to the cone AL, so is the cone EN, to some solide lesse then the cone AL (as we may see by the assumpt put after th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> second of this booke): Wherefore (by the 11. of the fift) as the circle EFGH is to the circle ABCG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the cone EN to some solide lesse then the cone A<g ref="char:EOLhyphen"/>L, which we haue proued to be impossible. Wherefore as the circle ABCD, is to the circle E<g ref="char:EOLhyphen"/>FGH, so is not the cone AL to any solide greater then the cone EN. And it is also proued that it is not to any lesse. Wherefore as the circle ABCD, is to the circle EFGH, so is the cone AL to the cone EN.</p>
               <p>But as the cone is to the cone, so is the cylinder to the cylinder, (by the 15. of the fift) for
<pb facs="tcp:7062:401"/>
the one is in treble proportion to the other.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion touching cylinders.</note> W<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>erefore) by the 11. of the fift) as the circle A<g ref="char:EOLhyphen"/>BCD is to the circle EFGH, so are the cylinders which are set vpon them the one to the other, the said cylinders being vnder equall altitudes with the cones. Cones therefore and cy<g ref="char:EOLhyphen"/>linders, being vnder one &amp; the self same altitude, are in that proportion the one to the other, that their bases are: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 12. Proposition. Like Cones and Cylinders, are in treble proportion of that in which the di<g ref="char:EOLhyphen"/>ameters of their bases are.</head>
               <p>SVppose that these Cones and Cylinders, whose bases are the circles ABCD, and EFGH, and diameters of their bases BD and FH, be like, and let the axes of the Cones or Cylinders be KL and MN, and let the centres o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> their bases be the pointes K and M. Then I say, that the cone, whose base is the circle ABCD, &amp; toppe the point L,<note place="margin">First part which concer<g ref="char:EOLhyphen"/>neth cones.</note> is to the Cone, whose base is the circle EFGH, and toppe the point N, in treble proportion of that in which the diameter BD is to the diameter FH. For if the cone ABCDL be not to the cone EFGHN in treble proportion of that in which the diameter
<figure/>
BD is to the diameter FH, the cone ABCDL shall be in treble proportion of that in which the diameter BD is to the diameter FH, either to some solide lesse then the cone EFGHN, or to some solide greater.<note place="margin">First case<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Construction.</note> First let it be vnto a lesse, namely, to the solide X. Describe (by the 6. of the fourth) in the circle EFGH, a square EFGH. Wherefore the square EFGH is greater then the halfe of the circle EF<g ref="char:EOLhyphen"/>GH. Ra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se vp from the square EFGH a pyramis of equall altitude with the cone. Wherfore the pyramis so raised vp, is greater then the halfe part of the cone. Diuide now (by the 30. of the third) the circumferences EF, FG, GH, &amp; HE, into two equall partes in the pointes O, P, R, S, and drawe these right lines EO, OF, FP, PG, GR, RH, HS, and SE. Wherefore euery one of these
<pb n="372" facs="tcp:7062:401"/>
triangles EOF, FPG, GRH, and HSE, is greater then the halfe of the segment of the circle EFGH, described about ech of them. Erecte vpon euery one of these triangles EOF, FPG, GRH, and HSE, a pyramis, hauing one and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> selfe same altitude with the cone. Wherefore euery one of the pyramids so raised vp, is greater then the halfe of the segment of the cone described about them. Now therefore diuiding the circumferences remayning into two equall partes, and drawing right lines, and raising vp vpon euery one of the triangles a pyramis, hauing one and the selfe same altitude with the cone, and, thus doing continually, we shall at the length (by the 1. of the tenth) leaue certaine segmentes of the cone, which shall be lesse then the excesse whereby the cone EFGHN excedeth the solide X. Let those segmentes left be EO, OF, FP, PG, GR, RH, HS, and SE. Wherefore the pyramis remayuing, whose base is the Poligonon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure EOFPGRHS, and toppe the point N, is greater then the solide X.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>eadi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>g to an impossibi<g ref="char:EOLhyphen"/>litie.</note> Describe (by the 18. of the sixt) in the circle ABCD, vn<g ref="char:EOLhyphen"/>to the Poligonon figure EOFPGRHS, a Poligonon figure like and in like sort situate, and let the same be ATBVCZDW, and from it raise vp a pyramis, hauing one and the selfe same altitude with the cone ABCL. And let one of the triangles comprehending the pyra<g ref="char:EOLhyphen"/>mis, whose base is the Poligonon figure ATBVCZDW, and toppe the point L, be LBT, and let also one of the triangles comprehending the pyramis, whose base is the Poligonon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<g ref="char:EOLhyphen"/>gure EOFPGRHS, and toppe the point N, be NFO, and drawe these right lines KT and MO. And forasmuch as the cone ABCDL is like to the cone EFGHN, therefore (by the 20. definition of the eleuenth) as the diameter BD is to the diameter FH, so is the axe KL to the axe MN: But as the diameter BD is to the diameter FH, so (by the 15. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift) is the semidiameter BK to the semidiameter FM. Wherefore (by the 11. of the fift) as BK is to FM, so is KL to MN. Wherefore alternately also (by the 16. of the fift) as BK is to KL, so is FM to MN. Wherefore the sides about the equall angles BKL and FMN (which angles are equall, for that they are right angles (by the 18. definition of the eleuenth) are proportionall. Wherefore (by the 1. definition of the sixt) the triangle BKL is like to the triangle FMN. Againe for that as BK is to KT, so is FM to MO, and they comprehend equall angles, namely, BKT and FMO, for what part the angle BKT is of those fower right angles which are made at the centre K, the selfe same part is the angle FMO of the fower right angles which are made at the centre M: forasmuch therefore as the sides about the equall angles are proportionall, the triangle BKT is like to the triangle FMO. Againe, forasmuch as it was proued, that as BK is to KL, so is FM to NM, but BK is equall to KT, and FM to MO, therefore as TK is to KL, so is OM to MN. Wher<g ref="char:EOLhyphen"/>fore the sides about the equall angles TKL and OMN (which angles are equall, for that they are right angles) are proportionall. Wherefore the triangle LKT is like to the trian<g ref="char:EOLhyphen"/>gle MNO. And for that (by the 6. of the sixt) and by reason of the likenes of the triangles LKB and NMF, as LB is to BK, so is NF to FM, and againe by reason of the likenes of the triangles BKT and FMO, as KB is to BT, so is MF to FO, therefore of equalitie (by the 22. of the fift) as LB is to BT, so is NF to FO. Againe for that by reason of the likenes of the triangles LKT and NOM, as LT is to TK, so is NO to OM, and by rea<g ref="char:EOLhyphen"/>son of the likenes of the triangles TKB &amp; OMF, as KT is to TB, so is MO to OF: ther<g ref="char:EOLhyphen"/>fore of equalitie (by the 22. of the fift) as LT is to TB, so is NO to OF. And it was proued that as TB is to BL, so is OF to FN. Wherefore againe of equalitie, as TL is to LB, so is ON to NF.Wherefore the sides of the triangles LTB &amp; NOF, are proportionall. Wher<g ref="char:EOLhyphen"/>fore (by the 5. of the sixt) the triangles LTB and NOF, are equiangle. Wherefore also they are like. Wherefore the pyramis, whose base is the triangle BKT, and toppe the point L, is like vnto the pyramis, whose base is the triangle FMO, and toppe the point N. For they are comprehended vnder like plaine superficieces, and equall in multitude. But pyramids be<g ref="char:EOLhyphen"/>ing like, and hauing triangles to their bases, are (by the 8. of the twelfth) in treble proporti<g ref="char:EOLhyphen"/>on the one to the other of that in which sides of like proportion are. Wherefore the pyramis
<pb facs="tcp:7062:402"/>
BKTL is to the pyramis FMON, in treble proportion of that in which the line BK is to the line FM. And in like sort if we draw right lines from the pointes A, W, D, Z, C, and V, to the point K, and likewise from the pointes E, S, H, R, G, and P, to the point M, and raise vp vpon the triangles pyramids, hauing the selfe same altitudes with the cones, we may proue that euery one of those pyramids of one &amp; the selfe same order, is to euery one of the pyramids of the selfe same order, in treble proportion of that in which the side of like proportion BK is to side of like proportion FM, that is, of that which the line BD hath to the line FH. But as one of the antecedentes is to one of the consequentes, so are all the antecedentes to all the con<g ref="char:EOLhyphen"/>seque<g ref="char:cmbAbbrStroke">̄</g>tes (by the 12. of the fift). Wherfore as the pyramids BKTL is to the pyramis FMON, so is the whole pyramis, whose base is the Poligonon figure ATBVCZDW, and toppe the point L, to the whole pyramis, whose base is the Poligonon figure EOFPGRHS, and toppe the point N. Wherefore the pyramis, whose base is the Poligonon figure ATBVCZDW, and toppe the point L, is to the pyramis, whose base is the Poligonon figure EOFPGRHS, and toppe the point N, in treble proportion of that in which the line BD is to the line FH.
<figure/>
And it is supposed also that the cone, whose base is the circle ABCD, and toppe the point L, is to the solide X, in treble proportion of that in which the line BD is to the line FH. Wherefore as the cone, whose base is the circle ABCD, and toppe the point L, is to the solide X, so is the pyramis, whose base is the Poligonon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure ATBVCZDW, &amp; toppe the point L, to the pyramis, whose base is the Poligonon figure EOFPGRHS, and toppe the point N. Wherefore alter<g ref="char:EOLhyphen"/>nately (by the 16. of the fift) as the cone, whose base is the circle ABCD, &amp; toppe the point L, is to the pyramis which is in it, whose base is the Poligono<g ref="char:cmbAbbrStroke">̄</g> figure ATBVCZDW, &amp; toppe the point L, so is the solide X, to the pyramis, whose base is the Poligonon figure EOFPGRHS, and toppe the point N. But the foresaid cone is greater then the pyramis which is in it, for it containeth it. Wherefore the solide X is greater then the pyramis, whose base is the Poligonon figure EOFPGRHS, and toppe the point N. But it was supposed to be lesse: which is impossible. Wherfore the cone ABCDL is not to any solide lesse then the cone EFGHN, in treble proportion of that in which the
<pb n="373" facs="tcp:7062:402"/>
diameter BD is to the diameter FH. In like sort also may we proue, that the cone EFGHN is not to any solide lesse then the cone ABCDL, in treble proportion of that in which FH is to BD.</p>
               <p>Now also I say, that the cone ABCDL is not to any solide greater then the cone EFGHN,<note place="margin">Second case.</note> in treble proportion of that in which the diameter BD is to the diameter FH. For if it be possible, let it be to a greater, namely, to the solide X. Wherefore by conuersion (by the Corollary of the 4. of the fift) the solide X is to the cone ABCDL, in treble proportion of that in which the diameter FH is to the diameter BD. But as the solide X is to the cone ABCDL, so is the cone EFGHN to some solide lesse then the cone ABCDL (as it is easie to see by the Assumpt put after the 2. Proposition). Wherefore also the cone EFGHN is vnto some solide lesse then the cone ABCDL, in treble proportion of that in which the diameter FH is to the diameter BD: which is proued to be impossible. Wherefore the cone ABCDL is not to any solide greater then the cone EFGHN in treble proportion of that in which the diameter BD is to the diameter FH, and it is also proued that it is not to any lesse. Wherefore the cone ABCD is to the cone EFGHN in treble proportion of that in which the diameter BD is to the diameter FH.</p>
               <p>But as cone is to cone, so is cylinder to cylinder (by the 15. of the fift) for the cylinder is triple to the cone which is described on the one and selfe same base,<note place="margin">Second par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which concer<g ref="char:EOLhyphen"/>neth Cillin<g ref="char:EOLhyphen"/>ders.</note> and hauing one and the selfe same altitude with the cone. For it is proued (by the 10. of the twelfth) that euery cone is the third part of a cylinder, hauing one and the selfe same base with it, and one &amp; the selfe same altitude. Wherefore the cylinder is vnto the cylinder in treble proportion of that in which the diameter BD is to the diameter FH. Wherfore like cones &amp; cylinders are in treble proportion of that in which the diameters of their bases are: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The 13. Theoreme. The 13. Proposition. If a Cylinder be diuided by a playne superficies being a parallell to the two opposite playne super<g ref="char:EOLhyphen"/>ficieces: then as the one Cylinder is to the other Cylinder, so is the axe of the one to the axe of the other.</head>
               <p>SVppose that there be a cylinder AD, whose axe let
<figure/>
be EF, and let the opposite bases be the circles A<g ref="char:EOLhyphen"/>EB, and CFD, and let AD be diuided by the su<g ref="char:EOLhyphen"/>perfices GH being a parallell to the two opposite playne superficieces, AB, and CD. Then I say that as the cylin<g ref="char:EOLhyphen"/>der BG is to the cylinder GD, so is the axe EK to the axe KF. Extend the axe EF on either side to the pointes L &amp; M. And vnto the axe EK put as many axes equall as you will, namely,<note place="margin">Construction.</note> EN, &amp; NL, &amp; likewise vnto the axe FK put as many axes e<g ref="char:EOLhyphen"/>qual as you wil, namely, FX &amp; XM. And by the points L, N, &amp; X, M, exte<g ref="char:cmbAbbrStroke">̄</g>de playne superficieces parallels to the two superficieces AB, &amp; CD (by the corollary of the 15. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th): &amp; in the plaine superficieces thus extended by the pointes L, N, X, M, imagine to be be drawne these circles, namely, OP, RS, TV, and ZW, hauing to their centers the pointes L, N, M, X, and le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> them be equal to either of the circles AB, and CD and vp<g ref="char:EOLhyphen"/>pon those circles imagine those cylinders PR, RB, DT, TW to
<pb facs="tcp:7062:403"/>
be set.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> Now forasmuch as the axes LN, NE, and EK are
<figure/>
equall the one to the other: Therefore the cylinders PR, RB, and BG are (by the 11. of the twelueth) in proportion the one to the other as their bases are. But the bases are equall. Wherefore also the cylinders PR, RB, and BG are equall the one to the o<g ref="char:EOLhyphen"/>ther. And forasmuch as the axes LN, NE, and EK are equal the one to the other, and the cylinders PR, RB, and BG are al<g ref="char:EOLhyphen"/>so equall the one to the other, and the multitude of the axes L<g ref="char:EOLhyphen"/>N, NE, and EK is equall to the multitude of the cylinders P<g ref="char:EOLhyphen"/>R, RB, and BC: therefore how multiplex the whole axe KL is to the axe EK, so multiplex is the whole cylinder PG to the cylinder BG. And (by the same reason) also how multiplex the whole axe MK is the axe KF, so multiplex is the whole cylin<g ref="char:EOLhyphen"/>der WG to the cylinder GD. Wherfore if the axe KL be equal the axe KM, the cylinder PG is equall to cilinder GW. And if the axe KL be greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n the axe KM, the cylinder PG is greater then the cylinder GW. And if it be lesse it is lesse. Now therefore there are foure magnitudes, namely, the two axes EK, and KF, and the two cylinders BG, and GD, and vnto the axe EK, and to the cylinder BG, namely, to the first and the third are taken equemultiplices, namely, the axe KL, and the cylinder PG. And likewise vnto the axe GF, and vnto the cy<g ref="char:EOLhyphen"/>linder GD, namely, the second and the fourth, are taken other equemultiplices, namely, the axe KM, and the cylinder GW. And it is proued that if the axe KL, excede the axe KM, the cylinder PG excedeth the cylinder GW and that if it be equall it is equall, and if it be lesse it is lesse. Wherfore (by the 6. defini<g ref="char:EOLhyphen"/>tion of the fift) as the axe EK is to the axe KF, so is the cylinder BG to the cylinder GD. If therfore a cylinder be diuided by a plaine superficies being a parallel to the two opposite plaine superficieces: then as the one cylinder is to the other cilinder, so is the axe of the one to the axe of the other: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 14. Theoreme. The 14. Proposition. Cones and Cylinders consisting vpon equall bases, are in proportion the one to the other as their altitudes.</head>
               <p>SVppose that the cylinders FD, and EB, and the cones AGB, and CKD, do consiste vpon equall bases, namely, vpon the circles AB, and CD.<note place="margin">Construction.</note> Then I say that as the cylinder EB is to the cylinder FD, so is the axe GH to the axe KL. Extende the axe KL directly to the poynte N, and vnto the axe GH, put the axe LN equall: and about the axe LN imagine a cylinder CM.<note place="margin">Demonstrati<g ref="char:EOLhyphen"/>on touching Cylinders.</note> Now forasmuch as the cilinders EB, and CM, are vnder equall altitudes, therefore (by the 11. of the twelueth) they are in the proportion the one to the other as their bases are. But the bases are equall the one to the other. Wherefore also the cylinders EB, and CM are equall the one to the other. And forasmuch as the whole cylinder FM, is diuided by a playne superficies CD being
<pb n="374" facs="tcp:7062:403"/>
a parallell to either of the opposite plaine su<g ref="char:EOLhyphen"/>perficieces:
<figure/>
therefore (by the 13. of the twel<g ref="char:EOLhyphen"/>ueth) as the cylinder CM is to the cylinder FD, so is the axe LN to the axe LK. But the cylinder CM is equal to the cylinder EB, and the axe LN to the axe GH. Wherefore as the cylinder EB is to the cylinder FD, so is the axe GH to the axe KL.</p>
               <p>But as the cylinder EB is to the cylinder FD, so (by the 15. of the fift) is the cone ABG to the cone CDK,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion touching Cones.</note> for the cylinders are in treble proportion to the cones (by the 10. of the twelueth). Wherefore (by the 11. of the fift) as the axe GH is to the axe KL, so is the cone ABG to the cone CDK, &amp; the cylinder EB to the cylinder FD. Wherfore cones &amp; cy<g ref="char:EOLunhyphen"/>linders consisting vpon equal bases are in pro<g ref="char:EOLhyphen"/>portion the one to the other as their altitudes: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶The 15. Theoreme. The 15. Proposition. In equall Cones and Cylinders, the bases are reciprokall to their alti<g ref="char:EOLhyphen"/>tudes. And cones and Cylinders whose bases are reciprokall to their alti<g ref="char:EOLhyphen"/>tudes, are equall the one to the other.</head>
               <p>SVppose that these cones ACL, EGN or these cylinders AX, EO, whose bases are the circles ABCD, EFGH,
<figure/>
and axes KL, and MN (which axes are also the altitudes of the cones &amp; cylinders) be equall the one to the other.<note place="margin">First part of the propositio<g ref="char:cmbAbbrStroke">̄</g> demonstrated touching Cones.</note> The<g ref="char:cmbAbbrStroke">̄</g> I say that the bases of the cylinders XA &amp; EO are reci<g ref="char:EOLunhyphen"/>prokal to their altitudes, that is, that as the base A<g ref="char:EOLhyphen"/>BCD is to the base EFGH,<note place="margin">Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</note> so the altitude MN to the altitude KL. For the altitude KL is either equall to the altitude MN or not. First let it be e<g ref="char:EOLhyphen"/>quall.<note place="margin">The first case.</note> But the cylinder AX is equal to the cylinder EQ. But cones and cylinders consisting vnder one and the selfe same altitude, are in proportion the one to the other as their bases are (by the 11. of the twelueth). Wherfore the base ABCD is equall to the base EFGH. Wherefore also they are recipro<g ref="char:EOLhyphen"/>kal: as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL.</p>
               <p>But now suppose that the altitude LK be not equall to the altitude M. N,<note place="margin">
                     <p>Second case.</p>
                     <p>Construction.</p>
                  </note> but let the altitude
<pb facs="tcp:7062:404"/>
MN be greater. And (by the 3. of the first) from
<figure/>
the altitude MN take away PM equall to the al<g ref="char:EOLhyphen"/>titude KL, so that let the line PM, be put equal to the line KL. And by the point P let there be extended a playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies TVS, which let cut the cylinder EO, and be a parallell to the two opposite playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icieces, that is, to the circles EFGH, and RO.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion touching cylinders.</note> And making the base the circle EFGH, &amp; the altitude MP imagine a cylinder ES. And for that the cylinder AX is equall to the cylinder EO, and there is an other cylinder ES, therfore (by the 7. of the fift) as the cylinder AX is to the cylinder ES, so is the cylinder EO to the cy<g ref="char:EOLhyphen"/>linder ES. But as the cylinder AX is to the cylin<g ref="char:EOLhyphen"/>der ES, so is the base ABCD to the base EF<g ref="char:EOLhyphen"/>GH. For the cylinders AX, and ES are vnder one and the selfe same altitude. And as the cylinder EO, is to the cylinder ES, so is the alti<g ref="char:EOLhyphen"/>tude MN to the altitude MP. For cylinders co<g ref="char:cmbAbbrStroke">̄</g>si<g ref="char:EOLhyphen"/>sting vpo<g ref="char:cmbAbbrStroke">̄</g> equall bases are in proportion the one to the other as their altitudes. Wherfore as the base ABCD is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the base EFGH, so is the al<g ref="char:EOLhyphen"/>titude MN to the altitude MP. But the altitude PM is equall to the altitude KL. Where<g ref="char:EOLhyphen"/>fore as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. Wherefore in the equall cylinders AX, and EO the bases are reciprokall to their altitudes. But now suppose that the bases of the cylinders AX, and EO be reciprokal to their altitudes, that is, as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL.<note place="margin">Second part demonstra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed.</note> Then I say that the cylinder AX is equall to the cylinder EO. For the selfe same order of constructio<g ref="char:cmbAbbrStroke">̄</g> remayning, for that as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL, but the altitude KL is equall to the altitude PM. Wherefore as the base ABCD is to the base EFGH, so is the altitude MN to the altitude PM. But as the base ABCD is to the base EFGH, so is the cylinder AX to the cylinder ES, for they are vnder equall altitudes: and as the altitude MN is to the altitude PM, so is the cylinder EO to the cylinder ES (by the 14. of the twelueth). Wherefore also as the cylinder AX is to the cylinder ES, so is the cylinder EO to the cylinder ES. Wherefore the cylinder AX is equall to the cylinder EO (by the 9. of the fift). And so also is it in the cones which ha<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the selfe same bases and altitudes with the cylinders. Wherefore in equall cones and cylinders, the bases are reciprokall to their altitudes &amp;c. which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corrollary added by <hi>Campane</hi> and <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Hitherto hath bene shewed the passions and proprieties of cones and cylinders whose altitudes fall perpendicularly vpon the bases. Now will we declare that cones and cilinders whose altitudes fall ob<g ref="char:EOLhyphen"/>liquely vpon their bases haue also the selfe same passions and proprieties which the foresayd cones and cilinders haue.</hi>
                  </p>
                  <p>Forasmuch as in the tenth of this booke it was sayd, that euery Cone is the third part of a cilinder hauing one, and the selfe same base, &amp; one &amp; the selfe same altitude with it, which thing was demo<g ref="char:cmbAbbrStroke">̄</g>stra<g ref="char:EOLhyphen"/>ted by a cilinder geuen, whose base is cut by a square inscribed in it, and vpon the sides of the square are described Isosceles triangles, making a poligonon figure, and againe vpon the sides of this poligonon figure are infinitely after the same maner described other Isosceles triangles taking away more the<g ref="char:cmbAbbrStroke">̄</g> the halfe, as hath ofte<g ref="char:cmbAbbrStroke">̄</g>times bene declared: therfore it is manifest, that the solides set vpon these bases, being
<pb n="375" facs="tcp:7062:404"/>
vnder the same altitude that the cilinder inclined is, and being also included in the same cilinder, do take away more then the halfe of the cilinder, and also more the<g ref="char:cmbAbbrStroke">̄</g> the halfe of the residue, as it hath bene proued in erected cylinders. For these inclined solides being vnder equall altitudes and vpon equall ba<g ref="char:EOLhyphen"/>ses with the erected solides are equall to the erected solides by the corollary of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>0. of the eleuenth. Wherfore they also in like sort as the erected, take away more then the halfe. If therfore we co<g ref="char:cmbAbbrStroke">̄</g>pare the inclined cilinder, to a cone set vpon the selfe same base, and hauing his altitude erected, and reason by an argument leading to an impossibilitie by the demonstration of the tenth of this booke, we may proue that the sided solide included in the inclined cylinder is greater then the triple of his pyramis, and it is also equall to the same which is impossible. And this is the first case, wherein it was proued that the cilinder not being equall to the triple of the cone is not greater then the triple of the same. And as touching the second case, we may after the same maner conclude that that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ided solide contayned in the cylin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>er is greater then the cylinder: which is very absurd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Wherefore if the cylinder be neither greater then the triple of the cone, nor lesse, it must nedes be equall to the same. The demonstration of these inclined cylinders most playnely followeth the demonstration of the erected cylinders: for it hath already bene proued, that pyramids, and sided solides (which are also called generally Prismes) being set vpon equall bases and vnder one and the selfe same altitude, whether the altitude be erected or in<g ref="char:EOLhyphen"/>clined, are equall the one to the other, namely, are in proportion as their bases are by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of this booke. Wherefore a cylinder inclined shall be triple to euery cone (although also the cone be erected) set vpon one and the same base with it, and being vnder the same altitude. But the cilinder erected was the triple of the same cone by the tenth of this booke. Wherefore the cilinder inclined is equall to the cilinder erected being both set vpon one and the selfe same base, and hauing one and the selfe same al<g ref="char:EOLhyphen"/>titude. The same also cometh to passe in cones, which are the third partes of equall cilinders, &amp; there<g ref="char:EOLhyphen"/>fore are equall the one to the other.</p>
                  <p>Wherefore according to the eleuenth of this booke it followeth, that cylinders and cones incli<g ref="char:EOLhyphen"/>ned or erected, being vnder one and the selfe same altitude, are in proportion the one to the other as their bases are. For forasmuch as the erected are in proportion as their bases are, and to the erected ci<g ref="char:EOLhyphen"/>linders the inclined are equall: therefore they also shall be in proportion as their bases are.</p>
                  <p>And therefore by the 12. of this booke like cones and cylinders being inclined are in triple propor<g ref="char:EOLhyphen"/>tion of that in which the diameters of the bases are. For forasmuch as they are equall to the erected which haue the proportion by the 12. of this booke, and their bases also are equall with the bases of the erected, therefore they also shall haue the same proportion.</p>
                  <p>Wherefore it followeth by the 13. of this booke, tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> cylinder inclined, being cut by a playne su<g ref="char:EOLhyphen"/>perficies parallel to the opposite playne superficieces therof shall be cut according to the proportions of the axes. For suppose that vpon one and the selfe same base <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e set an erected cylinder and an inclined cylinder, being both vnder one and the selfe same altitude, which <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> a playne superficies pa<g ref="char:EOLhyphen"/>rallel to the opposite bases. Now it is manifest that the sections of the one cylinder are equall to the sec<g ref="char:EOLhyphen"/>tion of the other cylinder, for they are set vpon equall bases, and vnder one and the selfe same altitude, namely, betwene the parallel playne superficieces. And their axes also are by those parallel playne su<g ref="char:EOLhyphen"/>perfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> proportionally by the 1<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>leuen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h. Wherefore the inclined cylinders (being equall to the erected cylinders) shall haue the proportion of thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> axes, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> also haue the erected. For in ech<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the proportion of the axes is one and the same.</p>
                  <p>Wherefore inclined Cones and Cylinders being set vpon equall bases, shall by the 14. of this booke be in <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> as their altitude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>clined are equall to the erected which haue the selfe same bases and altitude, and the erected are i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportion as their altitudes: therfore the inclined shall be in proportion the one to the other as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he selfe same alti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>udes which are common to ech, namely, to the inclined and to the erected.</p>
                  <p>And therefore in equall cones and cylinders whether they be inclined or erected, the bases shall be reciprokally proportionall with the altitudes, and contrariwise by the 15. of this booke. For foras<g ref="char:EOLhyphen"/>much as we haue oftentimes shewed that the inclined cones and cylinders are equall to the erected, ha<g ref="char:EOLhyphen"/>uing the self<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> same bases and altitudes with them, and the erected vnto whome the inclined are equall, ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e their bases r<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ipro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>all proportionally with their altitudes, therefore it followeth, that the incli<g ref="char:EOLhyphen"/>ned (being equall to the erected) haue also their bases and altitudes (which are common to eche) reci<g ref="char:EOLhyphen"/>prokally proportionall. Likewise if thei<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> altitudes &amp; bases be reciprokally proportionall, they the<g ref="char:cmbAbbrStroke">̄</g>selues also shall be equall, for that they are equall to the erected cylinders and cones set vpon the same bases and being vnder the same altitude<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which erected cylinders are equall the one to the other by the same 15. of this booke. Wherefore we may conclude, that those passions &amp; proprieties which in this twelfth booke haue bene proued to be in cones and cylinders whose altitudes are erected perpendicularly to the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> set obliquely vpo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> their bases. How<g ref="char:EOLhyphen"/>beit this is to be noted that such inclined cones or cylinders are not perfect rou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d as are the erected:
<pb facs="tcp:7062:405"/>
so that if they be cut by a playne superficies passing at right angles with their altitude, this section is a Conicall section, which is called <hi>Ellipsis,</hi> and shall not describe in their superficies a circle as it doth in e<g ref="char:EOLhyphen"/>rected cylinders &amp; cones, but a certaine figure, whose lesse diameter is in cylinders equall to the dime<g ref="char:EOLhyphen"/>tient of the base: that is, is one and the same with it. And the same thing happeneth also in cones incli<g ref="char:EOLhyphen"/>ned, being cut after the same maner.</p>
               </div>
            </div>
            <div type="part">
               <head>The 1. Probleme. The 16. Proposition. Two circles hauing both one and the selfe same centre being geuen, to in<g ref="char:EOLhyphen"/>scribe in the greater circle a poligonon figure, which shall consist of equall and euen sides, and shall not touch the superficies of the lesse circle.</head>
               <p>SVppose that there be two circles ABCD, and EFGH hauing one &amp; the selfe same centre, namely, K. It is required in the greater circle which let be ABCD to inscribe a poligonon figure which shalbe of equal and euen sides and not touch the circle EFGH. Drawe by the centre K a right line BD.<note place="margin">Construction.</note> And (by the 11. of the first) from the point G rayse vp vnto the right line BD a perpendicular line AG, and ex<g ref="char:EOLhyphen"/>tend it to the point C. Wherefore the line AC toucheth the circle EFGH (by the 15. of the third). Now therfore if (by the 30. of the third) we diuide the circumference BAD into two equall partes, and againe the halfe of that into
<figure/>
two equal partes, and thus do co<g ref="char:cmbAbbrStroke">̄</g>tinually, we shall (by the corollary of the 1. of the tenth) at the length leaue a certayne circumference lesse then the circumference AD. Let the circumference left be LD. And from the point L. Drawe (by the 12. of the first) vnto the line BD a perpendi<g ref="char:EOLhyphen"/>culare line<note n="*" place="margin">Note this L<g ref="char:EOLhyphen"/>M, because of KZ in the next proposition, and here the point M: for the point Z in the next demonstration.</note> LM,<note place="margin">I. Dee.</note> and extende it to the point N. And draw these right lines LD and DN. And forasmuch as the angles DML, and DMN are right angles, therfore (by the 3. of the third) the right line BD diuideth the right line LN into two equall parts in the pointe M. Wherfore (by the 4. of the first) the rest of the sides of the triangles DML, and DMN, namely, the lines DL, and DN shalbe equall. And forasmuch as the line AC is a parallell to the LN (by the 28. of the first): But AC toucheth the circle EFGH, wherfore the line LM toucheth not the circle EFGH, and much lesse do the lines LD, and DN touch the circle EFGH. If therefore there be applied right lines equall to the line LD continually into the circle A<g ref="char:EOLhyphen"/>BCD (by the 1. of the fourth) there shalbe described in the circle ABCD a poligonon figure which shalbe of equall, and<note n="*" place="margin">For that the sections were made by the number two: that is, by ta<g ref="char:EOLhyphen"/>king halues, and of the resi<g ref="char:EOLhyphen"/>due the hal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and so, to LD, being an halfe, and a residue which shall be a co<g ref="char:cmbAbbrStroke">̄</g>mon measure backe againe, to make sides of the Poligonon figure.</note> euen sides, and shall not touch the lesse circle, namely, EFGH: (by the 14. of the third or by the 29.) which was required to be done.</p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>Hereby it is manifest that a perpendicular line drawen from the poynt L to the line BD toucheth not one of the circles.</p>
               </div>
               <div type="section">
                  <head>¶An Assumpt added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>If a Sphere be cut of a playne superficies, the common section of the superficieces, shall be the cir<g ref="char:EOLhyphen"/>cumference of a circle.</hi>
                  </p>
                  <p>
                     <pb n="376" facs="tcp:7062:405"/>Suppose that the sphere ABC be cut by the playne superficies AEB, and let the centre of the sphere be the poynt D.<note place="margin">
                        <hi>Construction.</hi>
                     </note> And from the poynt D, let there be drawne vnto the playne superficies AEB a perpendicular line (by the 11. of the eleuenth) which let be the line DE. And from the poynt E draw in the playne superficies AEB vnto the common section of the sayd superficies and the sphere, lines how many so euer, namely, EA and EB. And draw these lines DA and DB. Now forasmuch as the right angles DEA and DEB are equall (for the line DE is e<g ref="char:EOLhyphen"/>rected
<figure/>
perpendicularly to the playne superficies).<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> And the right lines DA &amp; DB which subtend those angles are by the 12. defini<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion of the eleuenth equall: which right lines moreouer (by the 47. of the first) do contayne in power the squares of the lines DE, EA, and DE, EB: if therfore from the squares of the lines DE, EA, and DE, EB ye take away the square of the line DE which is co<g ref="char:cmbAbbrStroke">̄</g>mon vnto them, the residue namely the squares of the lines EA &amp; EB shall be equal. Wherfore also the lines EA and EB are equall. And by the same reason may we proue, that all the right lines drawne from the poynt E to the line which is the co<g ref="char:cmbAbbrStroke">̄</g>mon section of the superficies of sphere and of the playne su<g ref="char:EOLhyphen"/>perficies, are equall. Wherefore that line shall be the cir<g ref="char:EOLhyphen"/>cumference of a circle, by the 15. definition of the first. <note n="*" place="margin">
                        <hi>The circles so made: or so con<g ref="char:EOLhyphen"/>sidered in the sphere, are cal<g ref="char:EOLhyphen"/>led the greatest circles: All o<g ref="char:EOLhyphen"/>ther, not ha<g ref="char:EOLhyphen"/>uing the center of the sphere, to be their center also<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> are called lesse circles.</hi>
                     </note> But if it happen the plaine superficies which cutteth the sphere, to passe by the centre of the sphere, the right lines drawne from the centre of the sphere to their common section, shal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> be equall by the 12. definition of the eleuenth. For that common section is in the superficies of the sphere. Wherefore of neces<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>itie the playne superficies comprehended vnder that line of the common section shall be a circle, and his centre shall be one and the same with the centre of the sphere.</p>
                  <div type="subsection">
                     <head>Iohn Dee.</head>
                     <p>
                        <hi>Euclide</hi> hath among the definition of solides omitted certayne, which were easy to conceaue by a kinde of Analogie. As a segment of a sphere, a sector of a sphere, the vertex, or toppe of the segment of a sphere: with such like. But that (if nede be) some farthe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> light may be geue<g ref="char:cmbAbbrStroke">̄</g>, in this figure next before,<note place="margin">
                           <hi>Note these de<g ref="char:EOLhyphen"/>scriptions.</hi>
                        </note> vn<g ref="char:EOLhyphen"/>ders<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>and a segment of the sphere ABC to be that part of the sphere contayned betwen the circle AB, (whose center is E) and the sphericall superficies AFB. To which (being a lesse segment) adde the cone ADB (whose base is the former circle: and toppe the center of the sphere) and you haue DAFB a sector of a sphere, or solide sector (as I call it). DE extended to F, sheweth the top or vertex of the seg<g ref="char:EOLhyphen"/>ment, to be the poynt F and EF is the altitude of the segment sphericall. Of segmentes, some are grea<g ref="char:EOLhyphen"/>ter the<g ref="char:cmbAbbrStroke">̄</g> the halfe sphere, some are lesse. As before ABF is lesse, the remanent, ABC is a segment greater then the halfe sphere.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶A Corollary added by the same <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>By the foresayd assumpt it is manifest, that if from the centre of a sphere the lines drawne per<g ref="char:EOLhyphen"/>pendicularly vnto the circles which cutte the sphere, be equall: those circles are equall. And the perpendicular lines so drawne fall vpon the centres of the same circles.</hi>
                  </p>
                  <p>For the line which is drawne fro<g ref="char:cmbAbbrStroke">̄</g> the centre of the sphere to the circumference, containeth in pow<g ref="char:EOLunhyphen"/>er, the power of the perpendicular line, and the power of the line which ioyneth together the endes of those lines. Wherfore fro<g ref="char:cmbAbbrStroke">̄</g> that square or power of the line from the center of the sphere to the circum<g ref="char:EOLhyphen"/>ference or co<g ref="char:cmbAbbrStroke">̄</g>mon sectio<g ref="char:cmbAbbrStroke">̄</g> drawne, which is the semidiameter of the sphere, taking away the power of the perpendicular, which is co<g ref="char:cmbAbbrStroke">̄</g>mon to them <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ound about, it followeth, that the residues how many so euer they be, be equall powers, and therefore the lines are equall the one to the other. Wherefore they will describe equall circles, by the first definition of the third. And vpon their centers fall the perpendicular lines by the 9. of the third.</p>
                  <p>
                     <note n="*" place="margin">
                        <hi>An other Co<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note>And those circles vpon which falleth the greater perpendicular lines are the lesse circles. For the pow<g ref="char:EOLhyphen"/>ers of the lines drawne from the centre of the sphere to the circumference being alwayes one and e<g ref="char:EOLhyphen"/>quall, to the powers of the perpendicular lines and also to the powers of the lines drawne from the cen<g ref="char:EOLhyphen"/>tres of the circles to their circumference, the greater that the powers of the perpendicular lines taken away from the power contayning them both are, the lesse are the powers and therefore the lines re<g ref="char:EOLhyphen"/>mayning, which are the semidiameters of the circles and therefore the lesse are the circles which they describe. Wherefore if the circles be equall, the perpendicular lines falling from the centre of the sphere vpon the<g ref="char:cmbAbbrStroke">̄</g>, shall also be equall. For if they should be greater or lesse, the circles should be vnequall as it is before manifest. But we suppose the perpendiculars to be equall.<note n="*" place="margin">
                        <hi>An other Co<g ref="char:EOLhyphen"/>rollary.</hi>
                     </note> Also the perpendicular lines falling vpon those bases are the least of all, that are drawne from the centre of the sphere: for the other drawne
<pb facs="tcp:7062:406"/>
from the centre of the sphere to the circumference of the circles, are in power equall both to the pow<g ref="char:EOLhyphen"/>ers of the perpendiculars and to the powers of the lines ioyning these perpendiculars and these subten<g ref="char:EOLhyphen"/>dent lines together: making triangles rectangleround about: as most easily you may conceaue of the figure here annexed.</p>
                  <p>A the Center of the Sphere.</p>
                  <figure/>
                  <p>AB the lines from the Center of the Sphere to the Circumfe<g ref="char:EOLhyphen"/>rence of the Circles made by the Section.</p>
                  <p>BCB the Diameters of Circles made by the Sections.</p>
                  <p>AC the perpendiculars from the Center of the Sphere to the Circles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> whose diameters BC<g ref="char:EOLhyphen"/>B are one both sides or in any situation els.</p>
                  <p>CB the Semidiameters of the Cir<g ref="char:EOLhyphen"/>cles made by the Sections.</p>
                  <p>AO a perpendicular longe then AC: and therefore the Semi<g ref="char:EOLhyphen"/>diameter OB is lesse.</p>
                  <p>ACB, &amp; AOB triangles rectangle.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 2. Probleme. The 17. Proposition. Two spheres consisting both about one &amp; the selfe same ce<g ref="char:cmbAbbrStroke">̄</g>tre, being geue<g ref="char:cmbAbbrStroke">̄</g>, to inscribe in the greater sphere a solide of many sides (which is called a Polyhedron) which shall not touch the superficies of the lesse sphere.</head>
               <p>SVppose that there be two spheres about one &amp; the selfe same ce<g ref="char:cmbAbbrStroke">̄</g>tre, namely, about A. It is required in the greater sphere to inscribe a Polyhedron, or a solide of ma<g ref="char:EOLhyphen"/>ny sides, which shal not with his superficies touch the superficies of the lesse sphere. Let the spheres be cut by some one plaine superficies passing by the center A.<note place="margin">Construction.</note> Then shall their sectio<g ref="char:cmbAbbrStroke">̄</g>s be circles.<note n="*" place="margin">This is also proued in the As<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>umpt be<g ref="char:EOLhyphen"/>fore added out o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> Flus<g ref="char:EOLhyphen"/>sas.</note> [For (by the 12. definition of the eleuenth) the Diameter remai<g ref="char:EOLhyphen"/>ning fixed, and the semicircle being turned round about maketh a sphere. Wherefore in what positio<g ref="char:cmbAbbrStroke">̄</g> so euer you imagine the semicircle to be, the playne superficies which passeth by it shal make in the superficies of y<hi rend="sup">e</hi> sphere a circle. And it is manifest that is also a greater circle, for the diameter of the sphere which is also the diameter of the semicircle, and therefore also of the circle is (by the 15. of the third) greater then all the right lines drawne in the circle or sphere] which circles shall haue both one center being both also in that one playne superficies,<note place="margin">Note what a greater or greatest circle in a Spere is.</note> by which the spheres were cut. Suppose that that section or circle in the greater sphere be BC<g ref="char:EOLhyphen"/>DE, and in the lesse, be the circle FGH. Drawe the diameters of those two circles in such sorte that they make right angles, and let those diameters be BD, and CE.<note place="margin">First part of the Construc<g ref="char:EOLhyphen"/>tion.</note> And let the line AG, being part of the line AB, be the semidiameter of the lesse sphere and circle, as AB is the semidiameter of the greater sphere and greater circle: both the spheres and circles hauing one and the same center. Now two circles that is, BCDE, and FGH consisting both about one, and the selfe same centre being geuen, let there be described (by the proposition nexte going before) in the greater circle BCDE a poligonon figure consisting of equall and euen sides, not touching the lesse circle FGH. And let the sides of that figure in the fourth part of the circle, namely, in BE, be BK, KL, LM, and ME. And draw a right line from the
<pb n="377" facs="tcp:7062:406"/>
point K to the point A, and extende it to the point N. And (by the 12. of the eleuenth) from the A rayse vp to the superficies of the circle BCDE a perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line AX, and let it light vppon the superficies of the greater sphere in the point X.<note place="margin">Note<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </note> And by the line AX, and by either of these lines BD, and KN extend playne superficieces. Now by that which was before spoke<g ref="char:cmbAbbrStroke">̄</g>, those plaine superficieces shal in the <note n="*" place="margin">
                     <p>You know full well, that in the su<g ref="char:EOLhyphen"/>perficies of the sphere, <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>ly the circumferences of the cir<g ref="char:EOLhyphen"/>cles are: but by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>se cir<g ref="char:EOLhyphen"/>cumferences the limitatio<g ref="char:cmbAbbrStroke">̄</g> and assigning of circles is vsed: and so, the circumfe<g ref="char:EOLhyphen"/>rence of a circle, vsually called a circle, which in this place can not offend.</p>
                     <p>This figure is restored by M. Dee his diligence. For in the greeke and Latine Euclides, the line GL, the line AG, and the line K<g ref="char:EOLhyphen"/>Z, (in which three lynes the chiefe pinch of both the demonstrations doth stand), are vntruely drawen: as by comparing, the studious may per<g ref="char:EOLhyphen"/>ceaue.</p>
                  </note> superficies of y<hi rend="sup">e</hi> sphere make two greater circles. Let their semicircles consisting vpon the diameters BD, and KN be BXD. and KXN. And foras<g ref="char:EOLhyphen"/>much
<figure/>
                  <note place="margin">
                     <p>Note.</p>
                     <p>You must imagine <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> right line AX, to be perpe<g ref="char:cmbAbbrStroke">̄</g>dicular vpon the di<g ref="char:EOLhyphen"/>ameters BD and CE: though here AC the se<g ref="char:EOLhyphen"/>midiater, seme to be part of AX. And so in other pointes, in this figure, and many other, strengthen your imagination, accor<g ref="char:EOLhyphen"/>ding to the tenor of con<g ref="char:EOLhyphen"/>structions: though in the delineatio<g ref="char:cmbAbbrStroke">̄</g> in plaine, sense be not satisfied.</p>
                  </note>
as the line XA is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rected perpendicularly to the playne su<g ref="char:EOLhyphen"/>perficies
<figure/>
of the circle BCDE. Therfore al the plaine superficieces which are drawne by the line XA are erected perpendicularly to the superficies of the circle BCDE (by the 18. of the eleuenth). Wherefore the semicircles BXD, and KXN are erected perpen<g ref="char:EOLhyphen"/>dicularly to the playne superficies of the circle BCDE. And for<g ref="char:EOLhyphen"/>asmuch as the semicircles BED, BXD, and KXN are equall, for they consist vpon equall diameters <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D, and KN: therefore also the fourth parts or quarters of those circles, namely, BE, BX and KX are equal the one to the other. Wherefore how many sides of a poligonon figure there are in the fourth parte or quarter BE, so many also are there in the other fourth partes or quarters BX, and KX, equall to the right lines BK, KL, LM, and ME. Let those sides be described, and let them be BO, OP, PR, RX, KS, ST, TV, and VX:<note place="margin">
                     <p>Note.</p>
                     <p>BO equall to BK, in respect of M. Dee his demonstration following.</p>
                  </note> and drawe these right lines SO, T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and V<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And from the pointes O and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Drawe to the playne superficies of the circle BCDE perpendicular lines, which perpendicular lines will fall vpon the com<g ref="char:EOLhyphen"/>mon section of the plaine superficieces, namely, vpon the lines BD, &amp; KN, (by the 38. of the
<pb facs="tcp:7062:407"/>
eleuenth) for that the playne superficieces of the semicircles BXD, KXN are erected per<g ref="char:EOLhyphen"/>pendicularly to the playne superficies of the circle BCDE. Let those perpendicular lines be GZ, <note n="†" place="margin">Note <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>his point Z that you may the better vn<g ref="char:EOLhyphen"/>derstand M. Dee his demo<g ref="char:cmbAbbrStroke">̄</g>stration.</note> and SW. And drawe a right line from the point Z to the point W. And forasmuch as in the equall semicircles BXD, and KXN the right lines BO, and KS are equall, from the ends wherof are drawne perpe<g ref="char:cmbAbbrStroke">̄</g>dicular lines OZ, and SW, therfore (by the corollary of the 35. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) the line OZ is equall to the line SW, &amp; the line BZ is equal to the line KW. <hi>[Flussas proueth this an other way thus: Forasmuch as in the triangles SWK, and OZB, the two an<g ref="char:EOLhyphen"/>gles SWK, and OZB are equal, for that by co<g ref="char:cmbAbbrStroke">̄</g>struction they are right angles, and by the 27. of third the angles WKS, and ZBO are equall, for they subtend equal circumferences SXN and OXD, and the side SK is equall to the side OB as it hath before ben proued. Wherefore (by the 26. of the first) the o<g ref="char:EOLhyphen"/>ther sides &amp; angles are equall, namely, the line OZ to the lines SW, and the line BZ to the line KW,]</hi>
But the whole line BA is equall to the whole line KA: by the defi<g ref="char:EOLhyphen"/>nition
<figure/>
of a circle Wherfore the residue ZA is equall to the residue WA. Wherfore the line ZW is a parallel to the line BK (by the 2. of the sixt). And forasmuch as either of these lines OZ, &amp; SW is erected perpendicularly to the playne superficies of the circle BCDE, therefore the line OZ is a parallell to the line SW, (by the 6. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) and it is proued that it is also equal vnto it. Wherfore the lines WZ and SO are also equall and parallels (by the 7, of the eleuenth, and the 33. of the first, and by the 3. of the first) And for<g ref="char:EOLhyphen"/>asmuch as WZ is a parallell to SO: But ZW is a parallell is to KB: Wherefore SO is also a parallell to KB, (by the 9. of the eleuenth). And the lines BO, and KS do knit them toge<g ref="char:EOLhyphen"/>ther. Wherefore the fower sided figure BOKS is in one and the selfe same playne superficies.
<pb n="378" facs="tcp:7062:407"/>
For (by the 7. of the eleuenth) if there be any two parallell right lines, and if in either of them be taken a point at allauentures, a right line drawn by these points is in one, &amp; the selfe same playne superficies with the parallels: And by the same reason also euery one of the fower sided figures SOPT, and TPKV is in one and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> selfe same playne superficies. And the triangle VRX is also in one and the selfe same plaine superficies (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the eleuenth). Now if we imagine right lines drawne fro<g ref="char:cmbAbbrStroke">̄</g> the pointes O, S, P, T, R, V, to the point <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, there shalbe descri<g ref="char:EOLhyphen"/>bed
<figure/>
a Polyhedro<g ref="char:cmbAbbrStroke">̄</g> or a solide figure of many sides, betwene the circu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLunhyphen"/>ferences BX and KX composed of pyramids, whose bases are the fower sided figures BKOS, SOPT, TPRV, and the triangle VRX, and toppe the point A. And if in euery one of the sides K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>L, LM and ME we vse the selfe same construction that we did in BK, and moreouer in the other three quadrants or quarters, and also in the other halfe of the sphere, there shall then be made a Polyhedron or solide figure consisting of many sides described in the sphere, which Polyhedron is made of the pyramids whose bases are the foresayd fower sided figur<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, and the triangle VRK, and others which are in the selfe same order with them: and common toppe to them all in the point A. Now I say that the forsayd polihedron solide of many sides toucheth not the superficies of the lesse sphere in which is the circle FGH:<note place="margin">Second part of the con<g ref="char:EOLhyphen"/>struction.</note> Draw (by the 11. of the eleuenth) fro<g ref="char:cmbAbbrStroke">̄</g> the poynt A to the playne superficies of the fower sided figure KBOS a perpendicular line AY, and let it fall vpon the playne superficies in the point Y. And drawe these right lines BY &amp; YK. And forasmuch as the line AY is erected perpendicularly to the playne superficies of BKOS,
<pb facs="tcp:7062:408"/>
therefore the same AY is erected perpendicularly to all the right lines that touch it, and are in the plaine superficies of the fower sided figure (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. definition of the eleuenth).<note place="margin">Second part of the demon<g ref="char:EOLhyphen"/>stration.</note> Where<g ref="char:EOLhyphen"/>fore the line AY is erected perpendicularly to either of these lines BY, and YK. And foras<g ref="char:EOLhyphen"/>much as (by the 15. definition of the first) the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is equall to the line AK, therfore the square of the line AB is equall to the square of the line AK. And to the square of the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are equall the squares of the lines AY, and YB (by the 47. of the first) for the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.
<figure/>
YA is a right angle. And to the square of the line AK are equal the squares of the lines AY and YK. Wherefore the squares of the lines AY, and YB are equal to the squares of the lines AY, and TK: take away the square of the line AY which is common to them both. Wherefore the residue, namely, the square of the B<g ref="char:EOLhyphen"/>Y is equall to the residue, namely, to the square of the line YK. Wherefore the line BY is equall to the line YK. In like sort also may we proue that right lines drawne from the point Y to the pointes O, and S, are equall to either of the lines BY, and YK. Wherefore making the center the poynt Y, and the space either the line BY or the line YK describe a circle, and it shal passe by the poyntes O, and S, and the fower sided figure KBOS shalbe inscribed in the circle. And forasmuch as the line KB is greater then the line WZ (by the 2. of the sixt, because AK is greater then AW), but the line WZ is equall to the line SO. Wherfore the line BK is greater the<g ref="char:cmbAbbrStroke">̄</g> the line SO. But the line BK is equal to either
<pb n="372" facs="tcp:7062:408"/>
of the lines KS, and BO by construction. Wherefore either of the lines KS, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>O is grea<g ref="char:EOLhyphen"/>ter th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n the line SO. And forasmuch <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s in the circle is a fower sided figure KBOS, and the sides BK, BO and KS are equal, and the side OS is lesse then any one of them, and the line BY is drawne from the centre of the circle: therefore the square of the line KB is greater then the double of the square of the line BY (by the 12. of the second) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>for that it sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>endeth an angle greater then a right angle contayned of the two equall <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and YK, which angle BYK is an obtuse angle. For the 4. angles at the ce<g ref="char:cmbAbbrStroke">̄</g>ter Y are equal to 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> right angles: of which three, namely, the angles BYK, KYS, and BYO are equall by the 4. of the first, and the fourth namely, y<hi rend="sup">e</hi> angle SYO is lesse then any of those three angles, by the 25. of the first.) Drawe (by the 12. of the first) from the point K to the line BZ, a perpendicular line <note n="✚" place="margin">Which of necessity shall fall vpon Z, as M. Dee proueth it: and his profe is set af<g ref="char:EOLhyphen"/>ter at this marke <g ref="char:cross">✚</g> following.</note> KZ. And forasmuch as the line BD is lesse then the double to the line DZ (for the line BD is double to the line DA, which is lesse then the line DZ): but as the line BD is to the line DZ, so is the parallelogramme contained vnder the lines DB and BZ, to the parallelo<g ref="char:EOLhyphen"/>gramme contained vnder the lines DZ and ZB
<figure/>
(by the 1. of the sixt): therefore if ye describe vp<g ref="char:EOLhyphen"/>on the line BZ a square, and making perfecte the parallelogramme contained vnder the lines ZD, and ZB, that which is contained vnder the lines DB &amp; BZ, shall be lesse then the double to that which is contained vnder the lines DZ and ZB. And if ye drawe a right line from the point K to the point D, that which is contained vnder the lines DB and BZ, equall to the square of the line BK (by the Corollary of the 8. of the sixt) for the angle BKD is a right angle, by the 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the third, for it is in the se<g ref="char:EOLhyphen"/>micircle BED): and that which is contained vnder the lines DZ and ZB, is equall to the square of the line KZ (by the same Corollary). Wherefore the square of the line KB, is lesse then the double to the square of the line KZ. But the square of the line KB is greater then the double to the square of the line BY, as before hath bene proued. Wherfore the square of the line KZ, is greater then the square of the line BY (by the 10. of the fift). And foras<g ref="char:EOLhyphen"/>much as (by the 15. definition of the first) the line BA is equall to the line KA: therefore al<g ref="char:EOLhyphen"/>so the square of the line BA is equall to the square of the line KA. But (by the 47. of the first) vnto the square of the line AB, are equall the squares of the lines BY &amp; YA: (for the an<g ref="char:EOLhyphen"/>gle BYA is by construction, a right angle). And (by the same reason) to the square of the line KA, are equall the squares of the lines KZ and ZA: (for the angle KZA is also by construction, a right angle). Wherefore the squares of the lines BY and YA, are equall to the squares of the lines KZ and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Of which the square of the line KZ is greater then the square of the line BY, as hath before bene proued. Wherefore the residue, namely,<note place="margin">I. Dee.</note> the square of the line ZA, is lesse then the square of the line YA. Wherfore the line YA is grea<g ref="char:EOLhyphen"/>ter then the line AZ. <note n="*" place="margin">But AZ is greater the<g ref="char:cmbAbbrStroke">̄</g> A<g ref="char:EOLhyphen"/>G, as in the for<g ref="char:EOLhyphen"/>mer propositio<g ref="char:cmbAbbrStroke">̄</g>, KM was eui<g ref="char:EOLhyphen"/>dent to be grea<g ref="char:EOLhyphen"/>ter then KG: so may it also be made mani<g ref="char:EOLhyphen"/>fest that KZ doth neyther touch nor cut the circle FG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>H.</note> Wherefore the line AY is much greater the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line AG. But the line AY falleth vpon one of the bases of the Polihedron, and the line AG falleth vpon the superficies of the lesse sphere. Wherefore the Polihedron toucheth not the superficies of the lesse sphere.</p>
               <p>An other and more ready demonstration to proue that the line AY is greater then the line AG.<note place="margin">An other proue that the line AY is greater the<g ref="char:cmbAbbrStroke">̄</g> the line AG.</note> Raise vp (by the 11. of the first) from the poynt G to the line AG a perpendicular line GL. And draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point A to the poynt L. Now the<g ref="char:cmbAbbrStroke">̄</g> deuiding (by the 30. of the third) the circumference E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> into halues, &amp; agayne that halfe into halues, &amp; thus do<g ref="char:EOLhyphen"/>ing continually, we shall at the length by the corollary of the first of the tenth, leaue a certayne circumference, which shal be lesse then the circumference of the circle BCD which is subten<g ref="char:EOLhyphen"/>ded of a line equall to the line GL. Let the circumference left be KB. Wherfore also the right line KB is lesse then the right line GL. And forasmuch as the fower sided figure BKOS is
<pb facs="tcp:7062:409"/>
in a circle, and the lines OB, BK, and KS are equall, and the line OS is lesse, therefore the an<g ref="char:EOLunhyphen"/>gle BYK is an obtuse angle. Wherefore the line BK is greater then the line BY. But the line GL is greater then the line KB. Wherefore the line GL is much greater then the line BY. Wherefore also the square of the line GL is greater then square of the line BY. And foras<g ref="char:EOLhyphen"/>much as (by the <hi>15.</hi> definition of the first) the line AL is equall to the line AB, therefore the square of the line AL is equall to the square of the line AB. But vnto the square of the line AL are equall the squares of the lines AG and GL, and to the square of the line AB are e<g ref="char:EOLhyphen"/>quall the squares of the lines BY and YA. Wherefore the squares of the lines AG and GL are equall to the squares of the lines BY and YA, of which the square of the line BY is lesse then the square of the line GL. Wherefore the residue, namely, the square of the line YA is greater the<g ref="char:cmbAbbrStroke">̄</g> the square of the line AG. Wherfore also the line AY is greater the<g ref="char:cmbAbbrStroke">̄</g> the line AG. Wherfore two spheres consisting both about one and the selfe same center, being geuen, there is inscribed in the greater sphere a polihedron or solide of many sides which toucheth not the superficies of the lesse sphere<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was requi<g ref="char:cmbAbbrStroke">̄</g>red to be done.</p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>And if in the other sphere, namely, in the lesse sphere be inscribed a Polihedron or so<g ref="char:EOLhyphen"/>lide of many sides like to the polihedron inscribed in the sphere BCDE, then the polihedron inscribed in the sphere BCDE is to the polihedron inscribed in the other sphere in treble proportion of that in which the diameter of the sphere BCDE is to the diameter of the other sphere. For those solides being deuided into pyramids equall in number and equall in order, the pyramids shall be like. But like pyramids are (by the 8. of the twelfth) the one to the other in treble proportion of that in which side of like proportion is to side of like proportion. Wher<g ref="char:EOLhyphen"/>fore the pyramis whose base is the fower sided figure KBOS and toppe the poynt A is to that py<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>amis which is of like order in the other sphere, in treble proportion of that in which side of like proportio<g ref="char:cmbAbbrStroke">̄</g> is to side of like proportio<g ref="char:cmbAbbrStroke">̄</g>, that is of that in which the side AB which is drawne fro<g ref="char:cmbAbbrStroke">̄</g> the ce<g ref="char:cmbAbbrStroke">̄</g>ter of the sphere which is about the ce<g ref="char:cmbAbbrStroke">̄</g>ter A, is to the side which is drawn fro<g ref="char:cmbAbbrStroke">̄</g> the ce<g ref="char:cmbAbbrStroke">̄</g>ter of the other sphere. And in like sort also euery one of y<hi rend="sup">e</hi> pyramids which is in y<hi rend="sup">e</hi> sphere which is about the ce<g ref="char:cmbAbbrStroke">̄</g>tre A is to euery one of the pyramids of the selfe same order in the other sphere in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that in which the side AB is to the side which is drawne from the center of the other sphere. But as one of the antecedentes is to one of the consequen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>es, so are all the antecedents to all the conse<g ref="char:EOLhyphen"/>quentes by the 12. of the fifth. Wherefore the whole polihe<g ref="char:EOLhyphen"/>dron solide of many sides which is in the sphere which is about the center A, is to the whole polihedron or solide of many sides which is in the other sphere, in treble proportion of that in which the side AB is to the side which is drawne from the center of the other sphere, that is, of that which the diameter BD is to the diameter of the other sphere, by the 15. of the fifth: which was required to be demonstrated.</p>
                  <p>
                     <pb n="380" facs="tcp:7062:409"/>
                     <figure/>
                     <hi>M. Dee</hi> his deuise, to helpe the imagina<g ref="char:EOLhyphen"/>tion to young stude<g ref="char:cmbAbbrStroke">̄</g>ts in Geometry<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and to make his demonstration more euident as concerning the errors by hym corrected in Euclides figure, by the ignorant, mislined.</p>
               </div>
               <div type="section">
                  <head>I. D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e.</head>
                  <p>This figure is answerable to the firste plaine: which, cutting the two Spheres by their common center A, made two concen<g ref="char:EOLhyphen"/>tricall circles (hauing the same center with the two Spheres) namely BCDE, and FG<g ref="char:EOLhyphen"/>H. Vppon which, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> you aptly reare perpen<g ref="char:EOLhyphen"/>dicularly, the second figure contayning two concentricall circles, (to the first e<g ref="char:EOLhyphen"/>quall) and make the pointes noted with like letters to agree, and afterward vppon the se<g ref="char:EOLhyphen"/>co<g ref="char:cmbAbbrStroke">̄</g>d figure, set on the third figure being here for the better handling made a semicircle: which vppon the first figure must also be e<g ref="char:EOLhyphen"/>rected perpendicularly: And lastly if you take the little quadrangled figure BOKS, and make euery point to touch, his like: &amp; then reade the construction &amp; wey the de<g ref="char:EOLhyphen"/>monstratio<g ref="char:cmbAbbrStroke">̄</g> (twise o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> thrise being red ouer) shall you in this del<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>eatio<g ref="char:cmbAbbrStroke">̄</g> in apt pastborde, or like ma<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ter framed, finde al things in this probleme very euident.</p>
                  <p>I neede not warne you, that the line AY may easely be imagined, or with a fine thred supplyed: or of the right lines imaginable betwene P and T, and betwene R and V, I neede say nothing, trusting that the great exercise past, by that tyme you are orderly come to this place, will haue made you suf<g ref="char:EOLhyphen"/>ficient perfect to supply any farther thinge herein to be considered.</p>
                  <p>The little fowercornerd peeces remay<g ref="char:EOLunhyphen"/>ning to the semicircle, are to be let through the first ground playne: therby to stay this se<g ref="char:EOLunhyphen"/>micircle the better in his apt place and situ<g ref="char:EOLhyphen"/>ation: which it will the more aptly doo, if ye do <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>, the contrary arasses of the slitt of it, and of the slitt of the second figure, into which it is to be let: abating the<g ref="char:cmbAbbrStroke">̄</g> alike much: a litle will serue. Experience, by aduise, will teach sufficiently.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:410"/>
                  <head>¶ Master Dee his aduise and demonstration, reforming a great errour in the designation of the former figure of Euclides second Probleme: with two Corollaries (by him inferred) vpon his said demonstration.</head>
                  <div type="subsection">
                     <head>A Theoreme.</head>
                     <p>
                        <g ref="char:cross">✚</g>
                        <hi>That a right line dr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>wen from the point K, perpendicularly vpon the line BZ, doth fall vpon the point Z: we will, thus, make euident.</hi>
                     </p>
                     <p>BY the premisses, it is manifest, that the point Z is that point where a right line from the point O be<g ref="char:EOLhyphen"/>ing perpendicularly let <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>all to the circle BCDE, doth touch the same circle. Which point Z also is proued to be in the right line BAD, the common section of two circles cutting eche other: being one to the other perpendicularly erected. These thinges, with other, before demonstrated, I here make my suppositions. Consider now the two triangles rectangles OZB and KZB:
<q>Of which, the angle OZB i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> equall to the angle KZB. For, by construction, they are both right angles: <note n="*" place="margin">
                              <hi>This, as an assumpt, is presently proued.</hi>
                           </note> and the angle Z<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>O is equall to the angle ZBK. For, if from D to K you imagine a right line: and the like from D to O: you haue two triangles in equall semicircles, rectangles, namely, DKB and DOB: which haue the diameter BD common: and BK, the Chord, equall to BO the Chord, by construction. Where<g ref="char:EOLhyphen"/>fore (by the 47. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>irst) the third side, namely, DK, is equall to the third, namely, DO: Wherefore (by the 5. of the sixt) the angle ZBO, which is DBO, is equall to ZBK, which is DBK. (For the line ZB, by con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ruction, is part of DB).</q>
And seing two angles of OZB, are proued equall, to two angles of KZB, of necessitie the third, namely, ZOB, is equall to the third, namely, ZKB, by the 32. of the first. Wherefore the two triangles rectangles OZB and KZB, are proued equiangled. By the fourth, there<g ref="char:EOLhyphen"/>fore, of the sixt, their sides are proportionall: therefore by the premisses, proued, as BO is to BK, so is OZ to KZ, and the third line, which subtendeth the angle ZOB, to the third line which subten<g ref="char:EOLhyphen"/>deth the angle ZKB. But, by construction, BO is equall to BK: therefore OZ is equall to KZ: And the third al<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o is equall to the third: Wherefore the point Z, in respecte of the two triangles rectangles, OZB and KZB determineth one and the same magnitude, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the line BZ. Which can not be: if any other point, in the line BZ, were assigned, nearer, or farther of, from the point B. One onely poynt therefore, is that, at which the two perpendiculars KZ and OZ fall: But, by construction, OZ fal<g ref="char:EOLhyphen"/>leth at Z the point and therefore at the same Z, doth the perpendicular, drawen from K, fall likewyse: Which was required to be demonstrated.</p>
                     <p>Although a briefe monition, mought herein haue serued for the pregnant or the humble learner, yet for them that are well pleased to haue thinges made plaine, with many wordes, and for the stiffenecked busie body, it was necessary, with my controlment of other, to annexe the cause &amp; reason therof, both inuincible and also euident.</p>
                  </div>
                  <div type="subsection">
                     <head>A Corollary. 1.</head>
                     <p>Hereby it is manifest, that two equall circles cutting one the other by the whole diameter, if from one and the same end of their common diameter, equall portions of their circumferences be taken: and from the pointes ending those equall portions, two perpendiculars be let downe to their common diame<g ref="char:EOLhyphen"/>ter, those perpendiculars shall fall vpon one and the same point of their common diameter.</p>
                  </div>
                  <div type="subsection">
                     <head>2.</head>
                     <p>Secondly it followeth that those perpendiculars are equall.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ Note.</head>
                     <p>From circles in our first supposition eche to other perpendicularly erected, we procede and inferre now these Corollaries, whether they be perpendicularly erected or no: by reasou the demonstration hath a like force, vpon our suppositions here vsed.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 16. Theoreme. The 18. Proposition. Spheres are in treble proportion the one to the other of that in which their diameters are.</head>
               <p>
                  <pb n="381" facs="tcp:7062:410"/>SVppose that there be two spheres ABC and DEF, and let their diameters be BC and EF. Then I say that the sphere ABC is to the sphere DEF in treble proportion of that in which the diameter BC is to the diameter EF.<note place="margin">Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</note> For if not, then the sphere ABC is in treble proportion of that in which BC is to EF, ei<g ref="char:EOLhyphen"/>ther to some sphere lesse then the sphere DEF, or to some sphere greater. First let it be vnto a lesse, namely, to GHK.<note place="margin">The first case.</note> And imagine that the spheres DEF and GHK be both about one and the selfe same centre.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion leading to an impossibi<g ref="char:EOLhyphen"/>litie.</note> And (by the proposition next going before) describe in the grea<g ref="char:EOLhyphen"/>ter sphere DEF a polihedron or a solide of many sides not touching the superficies of the lesse sphere GHK. And suppose also that in the sphere ABC be inscribed a polihedron like to the polihedron which is in the sphere DEF. Wherefore (by the corollary of the same) the po<g ref="char:EOLhyphen"/>lihedron which is in the sphere ABC, is to the polihedron which is in the sphere DEF in treble proportion of that in which the diameter BC is to the diameter EF. But by supposition the sphere ABC is to the sphere GHK in treble proportion of that in which the diameter B<g ref="char:EOLhyphen"/>C is to the diameter EF. Wherefore as the sphere ABC is to the sphere GHK, so is the poli<g ref="char:EOLhyphen"/>hedro<g ref="char:cmbAbbrStroke">̄</g> which is described in the sphere ABC to the polihedro<g ref="char:cmbAbbrStroke">̄</g> which is described in the sphere DEF by the 11. of the fift. Wherfore alternately (by the 16. of the fift) as the sphere ABC is to the polihedron which is described in it, so is the sphere GHK to the polihedron which is in
<figure/>
the sphere DEF. But the sphere ABC is greater then the polihedro<g ref="char:cmbAbbrStroke">̄</g> which is described in it. Wherfore also the sphere GHK is greater then the polihedro<g ref="char:cmbAbbrStroke">̄</g> which is in the sphere DEF by the 14. of the fift. But it is also lesse, for it is contayned in it, which impossible. Wherefore the sphere ABC is not in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that in which the diameter BC, is to y<hi rend="sup">e</hi> diameter EF, to any sphere lesse then the sphere DEF. In like sort also may we proue that the sphere DEF is not in treble proportion of that in which the diameter EF is to the diameter BC, to any sphere lesse then the sphere ABC.<note place="margin">Second case.</note> Now I say that the sphere ABC is not in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that in which the diameter BC is to the diameter EF to any sphere greater the<g ref="char:cmbAbbrStroke">̄</g> the sphere DEF. For if it be possible, let it be to a greater, namely, to LMN. Wherfore by conuersion the sphere LMN is to the sphere ABC in treble proportion of that in which the diameter EF is to the diameter BC. But as the sphere LMN is to the sphere ABC, so is the sphere DEF to some sphere lesse the<g ref="char:cmbAbbrStroke">̄</g> the sphere ABC,<note n="*" place="margin">As it is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to gather by the <gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>umpt put after the seco<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of this boo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>.</note> as it hath before bene proued, for the sphere LMN is greater then the sphere DEF. Wherfore the sphere DEF is in treble proportio<g ref="char:cmbAbbrStroke">̄</g> of that in
<pb facs="tcp:7062:411" rendition="simple:additions"/>
which the diameter EF is to the diameter BC to some sphere lesse the<g ref="char:cmbAbbrStroke">̄</g> the sphere ABC, which is proued to be impossible. Wherefore the sphere ABC is not in treble proportion of that in which BE is to EF to any sphere greater the<g ref="char:cmbAbbrStroke">̄</g> the sphere DEF. And it is also proued that it is not to any lesse. Wherefore the sphere ABC is to the sphere DEF in treble proportion of that in which the diameter BC is to the diameter EF: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corrollary added by <hi>Flussas.</hi>
                  </head>
                  <p>Hereby it is manifest, that spheres are the one to the other, as like Polihedrons and in like sort des<g ref="char:EOLhyphen"/>cribed in them are: namely, eche are in triple proportion of that in which the diameters.</p>
                  <div type="subsection">
                     <head>A Corollary added by M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> Dee.</head>
                     <p>
                        <hi>It is then euident, how to geue two right lines, hauing that proportion betwene them, which, any two spheres geuen, haue the one to the other.</hi>
                     </p>
                     <p>For, if to their diameters, as to the first and second lines (of fower in continuall proportion) you adioyne a third and a fourth line in continu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ll proportion (as I haue taught before): The first and fourth lines, shall aunswere the P<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>obleme.<note place="margin">
                           <hi>Note: a gene<g ref="char:EOLhyphen"/>rall rule.</hi>
                        </note> How generall this rule is, in any two like solides, with their correspondent (or Omologall) lines, I neede not, with more wordes, declare.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶ Certaine Theoremes and Problemes (whose vse is manifolde, in Spheres, Cones, Cylinders, and other solides) added by Ioh. Dee.</head>
                  <div type="subsection">
                     <head>A Theoreme. 1.</head>
                     <p>
                        <hi>The whole superficies of any Sphere, is quadrupla, to the greatest circle, in the same sphere con<g ref="char:EOLhyphen"/>tayned.</hi>
                     </p>
                     <p>It is needeles to bring <hi>Archimedes</hi> demonstration hereof, into this place: seing his boke of the Sphere and Cylinder, with other his wo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>kes, are euery where to be had, and the demo<g ref="char:cmbAbbrStroke">̄</g>stration therof, easie.</p>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 2.</head>
                     <p>Euery sphere, is quadrupl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, to that Cone, whose base is the greatest circle, &amp; height, the semidia<g ref="char:EOLhyphen"/>meter of the same sphere.</p>
                     <p>
                        <hi>This is the 32. Proposition of</hi> Archimedes <hi>fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>st booke of the Sphere and Cylinder.</hi>
                     </p>
                     <div type="problem">
                        <head>A Probleme. 1.</head>
                        <p>
                           <hi>A Sphere being geuen, to make an vpright Cone, equall to the same: or in any other proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>, geuen, betwene two right lines.</hi>
                        </p>
                        <p>Suppose the Sphere geuen, to be <hi>A,</hi> his diameter being <hi>BC,</hi> and center <hi>D:</hi> with a line equall to the semidiameter <hi>BD</hi> (which let be <hi>NO</hi>) describe a circle <hi>NRP:</hi>
                           <note place="margin">
                              <hi>Construction.</hi>
                           </note> whose diameter let be <hi>NP,</hi> and center <hi>O,</hi> it is euident, that <hi>NRP</hi> is equall to the greatest circle in <hi>A,</hi> contayned. At the center <hi>O,</hi> let a perpen<g ref="char:EOLhyphen"/>dicular be reared equall to <hi>BD</hi> (the semidiameter of <hi>A</hi>) which suppose to be <hi>OQ:</hi> It is now plaine that to the Cone, whose base is the circle <hi>NRR,</hi> and height <hi>OQ,</hi> the Sphere <hi>A,</hi> is quadrupla: by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 2. Theoreme here, and by construction. Take a line equall to <hi>NP,</hi> which let be <hi>FE:</hi> and with the se<g ref="char:EOLhyphen"/>midiameter <hi>FE</hi> (making th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> point <hi>F</hi> center) describe a circle: which suppose to be <hi>EKG,</hi> and dia<g ref="char:EOLhyphen"/>meter <hi>EG.</hi> At the center <hi>F,</hi> reare a line perpendicular to <hi>EKG,</hi> by the 12. of the eleuenth: and make it equall to <hi>OQ.</hi> Let that line be <hi>
                              <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>L.</hi> I say that the Cone, whose base is the circle <hi>EKG,</hi> and height the line <hi>FL</hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> is equall to <hi>A.</hi> For seing <hi>FE,</hi> the semidiameter of <hi>EKG,</hi> is equall to <hi>N<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>
                           </hi> (the diameter of <hi>NRP</hi>) by construction:<note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> 
                           <hi>EG,</hi> the diameter of <hi>EKG,</hi> shall be double to <hi>NP.</hi> Wherfore the square of <hi>EG,</hi> is quadrupla to the square of <hi>NP:</hi> by the 4. of the second. But as the square of <hi>EG,</hi> is to the
<pb n="382" facs="tcp:7062:411"/>
square of <hi>NP,</hi> so is the circle <hi>EKG</hi> to the circle <hi>NRP,</hi> by the 2. of this twelueth. Wherefore the circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <hi>EKG,</hi> is quadruple to the circle <hi>NRP.</hi> And <hi>FL,</hi> the height is (by construction) equall to <hi>OQ</hi> the height. Wherefore the cone, whose base is the circle <hi>EKG,</hi> and height <hi>FL,</hi> is quadruple to the cone, whose base is <hi>NRP,</hi> and the height <hi>OQ:</hi> by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>1. of this twelfth. But vnto the same cone whose base is <hi>NRP,</hi> and height <hi>OQ,</hi>
                           <figure/>
the Sphere <hi>A</hi> is likewise proued quadrupla<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>. Wherefore the cone whose base is <hi>EKG</hi> and height <hi>FL,</hi> is equall to the Sphere <hi>A:</hi> by the 7. of the fift. To a Sphere being geuen therefore, we haue made an vpright cone equall.</p>
                        <p>And as concerning the other part of this Probleme,<note place="margin">
                              <hi>The second part of the Probleme two wayes executed.</hi>
                           </note> it is now easie to execute, and that two wayes: I meane to <hi>A</hi> the sphere geuen, to make an vpright cone in any proportion geuen betwen two right lines. For, let the pro<g ref="char:EOLhyphen"/>portion geuen, be that which is betwene <hi>X</hi> and <hi>Y.</hi> By the order <milestone type="tcpmilestone" unit="unspecified" n="1"/> of my additions, vpon the 2. of this twelfth booke: to the circle <hi>EKG</hi> make an other circle in that proportion that <hi>X</hi> is to <hi>Y:</hi> which let be <hi>Z.</hi> Vpon the center of <hi>Z,</hi> reare a line perpendicular and equall to <hi>FL.</hi> I say that the cone, whose base is <hi>Z,</hi> and the height equall to <hi>FL,</hi> is to <hi>A,</hi> in the proportion of <hi>X</hi> to <hi>Y.</hi> For the cone vppon <hi>Z,</hi> by construction, hath height equall to the height of the cone <hi>LEKG:</hi> and <hi>Z,</hi> by construction, is to <hi>EKG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>
                           </hi> as <hi>X</hi> is to <hi>Y:</hi> Wherefore, by the 11. of this twelfth, the cone vpon <hi>Z,</hi> is to the cone <hi>LEKG,</hi> as <hi>X</hi> is to <hi>Y.</hi> But the cone <hi>LEKG</hi> is proued equall to the sphere <hi>A.</hi> Wherfore the cone vpon <hi>Z,</hi> is to <hi>A,</hi> as <hi>X</hi> is to <hi>Y,</hi> by the 7. of the fift. To a Sphere geuen therefore, we haue made a cone in any proportion geuen, betwene two right lines. Secondly, as <hi>X</hi> is to <hi>Y,</hi> so to <hi>FL,</hi> let there be a fourth <milestone type="tcpmilestone" unit="unspecified" n="2"/> line, by the 12. of the sixt: and suppose it to be <hi>W.</hi> I say that a cone, whose base is equall to <hi>EKG,</hi> and height the line <hi>W,</hi> is to <hi>A,</hi> as <hi>X</hi> is to <hi>Y.</hi> For, by the 14. of this twelfth, cones being set on equall bases, are one to the other, as their heightes are: But, by construction, the height <hi>W,</hi> is to the height <hi>FL,</hi> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <hi>X</hi> is to <hi>Y.</hi> Wherefore the cone which hath his base equall to <hi>EKG,</hi> and height the line <hi>W,</hi> is to the cone <hi>LEKG,</hi> as <hi>X</hi> is to <hi>Y.</hi> And it is proued, that to the cone <hi>LEKG,</hi> the Sphere <hi>A</hi> is equall: Wher<g ref="char:EOLhyphen"/>fore, by the 7. of the fift, the cone, whose base is equall to <hi>EKG,</hi> and height the line <hi>W,</hi> is to <hi>A,</hi> as <hi>X</hi> is to <hi>Y.</hi> Therefore a Sphere being geuen, we haue made an vpright cone,<note place="margin">
                              <hi>An vpright Cone.</hi>
                           </note> in any proportion geuen be<g ref="char:EOLhyphen"/>twene two right lines. And before, we made an vpright cone, equall to the Sphere geuen. Wherfore a Sphere being geuen, we haue made an vpright cone, equall to the same, or in any other proportion, geuen betwene two right lines. I call that an vpright cone, whose axe is perpendicular to his base.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>¶ A Corollary.</head>
                     <p>Of the first part of the demonstration, it is euident: A Sphere being propounded, that a Cone, whose base hath his semidiameter, equall to the diameter therof, and height equall to the semidiame<g ref="char:EOLhyphen"/>ter of the same Sphere, is equall to that sphere propounded.</p>
                     <div type="problem">
                        <head>¶ A Probleme. 2.</head>
                        <p>
                           <hi>A Sphere being geuen, and a circle, to re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>re an vpright Cone, vpon that circle (as a base) equall to the Sphere geuen: or in any proportion betwene two right lines assigned.</hi>
                        </p>
                        <p>Suppose the Sphere geuen, to be Q: and the circle geuen to be C. By the first Probleme make an vpright cone equall to Q the Sphere geuen: which cone suppose to be A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and (by the 2. Probleme of my additions vpon the second of this twelfth booke) as C the circle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>euen, is to the base of A, so let
<pb facs="tcp:7062:412"/>
the height of A, be to a line found: which let be D.
<figure/>
Then it is euident, that the cone, which hath for his base C, the circle geuen, and height the line D, last found, shall be equall to Q the Sphere geuen: which cone let be F. For, by construction, F hath his base and height in reciprokall proportion with the cone A, made equall to Q the Sphere geuen: Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>fore by the 15. of this twelfth, and 7. of the fifth, this vp<g ref="char:EOLhyphen"/>right cone F, reared vpon C, the circle geuen, is e<g ref="char:EOLhyphen"/>quall to Q, the Sphere geuen: which thing the Probleme first required.</p>
                        <p>And the second part of this Probleme is thus per<g ref="char:EOLhyphen"/>formed. Suppose the proportion geuen to be that which is betwene X &amp; Y.<note place="margin">
                              <hi>The second part of the Probleme.</hi>
                           </note> Then, as X is to Y, so let an other right line found, be to the h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ight of F: which line let be G. For this G, the found height (by con<g ref="char:EOLhyphen"/>struction) being to the height of F, as X is to Y, doth cause this cone (which let be M) vpon C, the circle geue<g ref="char:cmbAbbrStroke">̄</g> (or an other to it equall) duely reared, to be vn<g ref="char:EOLhyphen"/>to the cone F, as X is to Y, by the 14. of this twelfth. But F is proued equall to the Sphere geuen: Wher<g ref="char:EOLhyphen"/>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e M, is to the Sphere geuen, as X is to Y. And M, is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>eared vpon the circle geuen: or his equall. Wher<g ref="char:EOLhyphen"/>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e, a Sphere being geuen, &amp; a circle, we haue rea<g ref="char:EOLhyphen"/>red an vp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ight cone, vpon that geuen circle (as a base) equall to the Sphere geuen: or in any propor<g ref="char:EOLhyphen"/>tion, betwene two right lines assigned: which was required to be done.</p>
                     </div>
                     <div type="problem">
                        <head>¶ A Probleme. 3.</head>
                        <p>
                           <hi>A Sphere being geuen, and a right line, to make an vpright cone, equall to the Sphere geuen, or in any other proportion geuen betwene two right lines: which made cone, shall haue his height equall to the right line geuen.</hi>
                        </p>
                        <p>Suppose the Sphere geuen, to be R: and the
<figure/>
right line geuen, to be S. To R the Sphere geuen, make an vpright cone, equall: by the first Probleme: which cone suppose to be A. Then as S, the line ge<g ref="char:EOLhyphen"/>uen, is to the height of A, so let the base of A, be to an other circle, which let be K, by my additions, vp<g ref="char:EOLhyphen"/>on the second Proposition of this twelfth booke. I say that an vpright cone, hauing his height, equall to S, the right line geuen, and his base K, is equall to the Spheres geuen. Let this cone be noted by L: for by construction, thys cone L, and A, haue theyr heightes and bases reciprokall in proportio<g ref="char:cmbAbbrStroke">̄</g>: Wher<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ore this cone L, and the cone A, are equall, by the 15. of the twelfth). But A is equall to the Sphere ge<g ref="char:EOLhyphen"/>uen by construction. Wherefore L is equall to the Spheres geuen. And the height of L, is equall to the right line geuen, by construction: which ought to be done.</p>
                        <p>
                           <note place="margin">
                              <hi>The second <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>a<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>robleme.</hi>
                           </note>For the second part: sinde a circle, which shall haue to the base of L, any proportion appointed in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ight lines: as the proportion of X to Y: which, by my additions, vpo<g ref="char:cmbAbbrStroke">̄</g> the second of this booke, ye haue l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>arned to do. Then, with the height, equall to the heigth of L reared vpon this last found circle, which l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>t be T, as a base, you shall satisfie the Probleme. L<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>t that Cone be V. For this last cone V, is to L, as his base is to the base of L, by the 11. of this twelfth. But <hi>L</hi> is proued equall to the Sphere geuen: Wher<g ref="char:EOLhyphen"/>fore by the 7. of the fift, this l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>t cone V, hath to R,
<pb n="383" facs="tcp:7062:412"/>
the Sphere geuen, that proportion which is betwene X and Y assigned: and forasmuch as the height of this cone V, is equall to the height of L: and the height of L, equall to S, the right line geuen (by construction): it is euident, that a Sphere being geuen, &amp; a right line, we haue made an vpright cone, equall to the Sphere geuen, or in any other proportion geuen betwene two right lines: which made cones, haue their height equall to the right line geuen: which ought to be done.</p>
                        <p>Vuwilling I am to vse thus many wordes, in matters so plain<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and ease.<note place="margin">☜</note> But this (I thinke) can not hinder them, that by nature are not so quicke of inuention, as to lead euery thing, gen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rally spo<g ref="char:EOLhyphen"/>ken, to a particular execution.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>¶ A Theoreme. 3.</head>
                     <p>
                        <hi>Euery Cylinder which hath his base, the greatest Circle in a Sp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>er<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>, &amp; heith equall to the diame<g ref="char:EOLhyphen"/>ter of that Sphere, is Sesquialtera to that Sphere. Also the superficies of that Cylinder, with his two bases is Sesquilatera to the superficies of the Sphere: and without his two bases, is equall to the superfi<g ref="char:EOLhyphen"/>cies of that Spher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </p>
                     <p>Suppose, a sphere to be signified by A, and an vpright cylinder hauing to his base a circle equall to the greatest circle in A contayned, and his heith equall to the diameter of A, let be signified by FG. I say that FG, is sesquialter to A: Secondly I say that the croked cylindricall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>icies of FG, together <milestone type="tcpmilestone" unit="unspecified" n="1"/> with the superfici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ces of his two opposit bases, is sesquialtera to the whole superficies sphericall of A.<milestone type="tcpmilestone" unit="unspecified" n="2"/> Thirdly I say that the cylindricall superficies of FG, omitting his two opposite bases, is equall to the superficies of the spere A. Let the base of FG, be the circle FLB: whose center, sup<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ose M, and <milestone type="tcpmilestone" unit="unspecified" n="3"/> diameter FB. And the axe of the same FG, let be, MH. Which is his heith (for we suppose the cy<g ref="char:EOLhyphen"/>linder to be vpright): and suppose H, to be his toppe or vertex. Forasmuch as, by supposition MH is equall to the diameter of A. Let MH be deuided into two equall partes in the point N, by a playne su<g ref="char:EOLhyphen"/>perficies passing by the point N, and being parallell to the opposit bases of FG. By the thirtenth of this twelfth booke, it then foloweth, that the cylinder FG, is also deuided into two equall parts: being cy<g ref="char:EOLhyphen"/>linders: which two equall cylinders let be IG, and FK: the axe of IG suppose to be HN: and of FK the axe to be NM. And for that, FG, is
<figure/>
an vpright cylinder, and at the poynt N, cut by a playne Superficies parallell to his opposite bases, the common se<g ref="char:EOLhyphen"/>ction of that playne superficies and the cylind<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r FG, must be a <note n="*" place="margin">
                           <hi>This may ea<g ref="char:EOLhyphen"/>sely be demon<g ref="char:EOLhyphen"/>strated, as in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> 17. proposition the section of a sphere was pro<g ref="char:EOLhyphen"/>ued to be a cir<g ref="char:EOLhyphen"/>cle.</hi>
                        </note> circle, equall, to his base FLB, and haue his center, the point N. Which circle, let be IO<g ref="char:EOLhyphen"/>K: And seing that FLB is, by supposi<g ref="char:EOLhyphen"/>tion, equall to the greatest circle in A, IOK, also, shall be equall to the grea<g ref="char:EOLhyphen"/>test circle, in A, contained: Also, by rea<g ref="char:EOLhyphen"/>son MH, is by supposition, equal to the diameter of A: and NH, by constru<g ref="char:EOLhyphen"/>ctio<g ref="char:cmbAbbrStroke">̄</g>, half of MH, it is manifest that N<g ref="char:EOLhyphen"/>H is equall to the semidiameter of A. If, therefore, you suppose a cone to haue the circle IOK to hi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> base: and NH to his heith, the sphere A, shall be to that Cone, quadrupla, by the 2. Theoreme. Let that cone be HIOK. Wherefore A, is quadrupla to HIOK. And the Cylinder IG hauing the same base, with HIOK (the circle IOK) and the same heith, (the right line NH) is triple to the cone HIOK by the 10. of this twelfth booke. But to IG, the whole cylinder FG, is double, as is proued: Wherefore FG, is triple and triple, to the cone HIOK, that is, sextuple. And A is proued quadrupla to the same HIOK. Wherefore FG is to HIOK, as 6. to 1: and A, is to HIOK, as 4. to 1: <note n="*" place="margin">
                           <hi>For taking a<g ref="char:EOLhyphen"/>way all doubt, this, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> a Lem<g ref="char:EOLhyphen"/>ma, afterward is dem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>strated.</hi>
                        </note> Therfore FG is to A, as 6, to 4: which in the least termes, is, as 3 to 2. but 3 to 2, is the termes of sesquialtera proportion. Wherefore the cylinder FG, is to A sesquialtera in proportion. Secondly, forasmuch as the superficies of a cylinder (his two oppo<g ref="char:EOLhyphen"/>site bases excepted) is equall to that circle whose semidiameter is middell proportionall betwene the side of the cylinder, and the diameter of his base: (as vnto the 10. of this booke, I haue added.) But of FG, the side BG, being parallell and equall to the axe MH, must also be equall to the diameter of A. And the base FLB, being (by supposition) equall to the greatest circle in A contained, must haue his di<g ref="char:EOLhyphen"/>ameter (FB) equal to the sayd diameter of A. The middle proportional therfore betwene BG and FB, being equall eche to other, shal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> be a line, equall to either of them.
<q>[As i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> set BG and FB together, as one line, and vpon that line composed, as a diameter make a semicircl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and from the center, to the circumference draw a lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> perpendicular to the sayd diameter:<note place="margin">
                              <hi>A Lemma (as it were) presently de<g ref="char:EOLhyphen"/>monstrated.</hi>
                           </note> by the <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> of the sixth, that perpendicular, is middel proportional betwene FB and BG, the semidiame<g ref="char:EOLhyphen"/>ters<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and he him selfe also a semidiameter: and therfore by the definition of a circle, equall to FB, and likewise, to BG.] And a circle, hauing his semidiameter, equall to the diameter FB, is quadruple to the circle FLB. [For the square of euery whole line, is quadruple to the squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>re of his halfe line, as may
<pb facs="tcp:7062:413"/>
be proued by the 4. of the second: and by the second of this twelfth, circles are one to the other, as the squares of their diameters, are.]</q>
Wherfore the superficies cylindricall of FG, alone, is quadrupla to his base FLB. But if a certayne quantity be dupla to one thing, and an other, quadrupla to the same one thing, those two quantities together are sextupla to the same one thing. Therefore seing the base, oppo<g ref="char:EOLhyphen"/>site to FLB, (being equall to to FLB)
<figure/>
added to FLB, maketh that co<g ref="char:cmbAbbrStroke">̄</g>pound, double to FLB: that double added to the cylindricall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>icies of FG, doth make a superficies sextupla to FLB. And the superficies of A, is quadrupla to the same FLB, by the first Theo<g ref="char:EOLhyphen"/>reme. Therefore the cylindricall super<g ref="char:EOLhyphen"/>ficies of FG, with the superficieces of his two bases is to the superficies FLB, as 6 to 1, and the superficies of A to F<g ref="char:EOLhyphen"/>LB, is as 4 to 1. Wherfore the cylindri<g ref="char:EOLhyphen"/>call superficies of FG, &amp; his two bases, together, are to the superficies of A, as 6 to 4: that is, in the smallest termes, as 3 to 2. Which is proper to ses<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uialtera proportion. Thirdly, it is already made euident that the superficies cylindrical, of FG (onely by it self) <milestone type="tcpmilestone" unit="unspecified" n="3"/> is quadrupla to FLB. And also it is proued, that the superficies of the sphere A, is quadrupla to the same FLB, Wherefore by the 7. of the fifth, the cylindricall superficies of FG, is equall to the superfi<g ref="char:EOLhyphen"/>cies of A. Therfore, euery cylinder, which hath his base the greatest circle in a sphere, and heith equal to the diameter of that sphere, is sesquialtera to that spere: Also the superficies of that cylinder with his two bases, is sesquialtera, to the superficies of the sphere: and without his two bases is equall to the su<g ref="char:EOLhyphen"/>perficies of the sphere: which was to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>The Lemma.</head>
                     <p>
                        <hi>If A be to C, as</hi> 6, <hi>to</hi> 1: <hi>and B, to C, as</hi> 4 <hi>to</hi> 1: <hi>A, is to B, as</hi> 6, <hi>to</hi> 4.</p>
                     <p>For, seing, B is to C, as 4 to 1, by supposition: therefore backward, by the 4. of the fifth, C is to B, as 1, to 4. Imagine now two orders of qnantities, the first,
<figure/>
A, C, and B the second, 6, 1, and 4. Forasmuch as, A, is to C, as 6, to 1, by supposition: and C is to B, as 1, to 4, as we haue proued: wherfore, A is to B, as 6 to 4, by the 22 of the fift. Therfore, if A be to C as 6 to 1, and B to C, as 4 to 1: A is to B, as 6, to 4. which was to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>Note.</head>
                     <p>Sleight things (some times) lacking euide<g ref="char:cmbAbbrStroke">̄</g>t proufe, brede doubt or ignorance. And, I nede not warn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> you, how gen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rall, this demonstration is: for if you put in the place of 6 and 4, any other numbers, the like manner of conclusion will follow. So likewise, in place of 1. any other one number may be, as, if A be to C as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to 5: and B vnto C, be as 7 to 5: A, shall be to B, as 6 to 7. &amp;c.</p>
                     <div type="problem">
                        <head>A Probleme. 4.</head>
                        <p>
                           <hi>To a Sphere geuen, to make a cylinder equall, or in any proportion geuen betwene two right lines.</hi>
                        </p>
                        <p>Suppose the geuen Sphere to be A: and the proportion geuen to be that betwene X and Y. I say that a cylinder is to be made, equall to A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> or els in the same proportion to A, that is betwene X to Y. Let a cylinder be made (such one as the Theoreme next before supposed) that shall haue his base equall to the greatest circle in A,<note place="margin">
                              <hi>Construction.</hi>
                           </note> and height equall to the diameter of A: Let that cylinder b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the vpright cy<g ref="char:EOLhyphen"/>linder BC. Le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the one side of BC, be the right line QC. Deuide QC into three equal part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> of which, let QE containe two, and let the third part be CE. By the point E suppose a plaine (parallel to the bases of BC) to passe through the cylinder BC, cutting the same by the circle DE. I say that the cy<g ref="char:EOLhyphen"/>linder BE is equall to the Sphere A. For seing BC, being an vpright cylinder, is cu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> by a plaine, pa<g ref="char:EOLhyphen"/>rallel to his bases, by construction: therefore as the cylinder DC, is to the cylinder BE, so is the axe of DC, to the axe of BE, by the 13. of this twelfth.<note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> Wherefore as the axe is to the axe, so is cylinder to cylinder. But axe is to axe, as side to side, namely, CE to QE, because the axe is parallel to any side
<pb n="384" facs="tcp:7062:413"/>
of an vpright cylinder: by the definition of a cylin<g ref="char:EOLhyphen"/>der.
<figure/>
And the circle of the section, is parallel to the bases, by construction. Wherefore in the parallelo<g ref="char:EOLhyphen"/>gramme (made of the axe, and of two semidiame<g ref="char:EOLhyphen"/>ters, on one side parallels, one to the other, being coupled together by a line drawen betwene their endes in their circumferences, which line is the side QC) it is euident, that the axe of BC is cut in like proportion, that the side QC is cut. Wherfore the cylinder DC, is to the cylinder BE, as EC is to QE. Wherefore, by composition, the cylinders DC and BE, that is, whole BC, are to the cylinder BE, as CE and QE (the whole right line QC) are to QE. But by co<g ref="char:cmbAbbrStroke">̄</g>struction, QC is of 3. such partes, as QE containeth 2. Wherefore the cylinder BC, is of 3. such partes, as BE contayneth 2. Wherefore BC the cylinder, is to BE, as 3. to 2: which is sesquialtera proportion. But by the former Theoreme, BC is sesquialtera to the Sphere A: Wherefore, by the 7. of the fift, BE is equall to A. Therefore to a Sphere geuen, we haue made a cylinder equall.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>Or thus more briefely omitting all cutting of the Cylinder.</head>
                     <p>Forasmuch as BC is an vpright Cylinder: his sides are equall to his axe or heith: therefore the two cylinders, whereof one hath the heith QC and the other the heith QE, hauing both their bases, the greatest circle in the Sphere A, are one to the other as QC is to QE, by the 14. of this twelfth, but QC is to QE, as 3. to 2, by construction: and 3. to 2. is in Sesquialtera proportion: therefore the cylinder BC hauing his heith QC, &amp; his base the greatest circle in A co<g ref="char:cmbAbbrStroke">̄</g>teyned, is Sesquialtera to the cylinder which hath his base the greatest circle in A conteyned, and heith the line QE. But by the former Theoreme, BC, is also Sesquialtera to A: wherfore the cylinder hauing the base BQ (which by supposition,<note place="margin">
                           <hi>The second part of the Probleme.</hi>
                        </note> is equal to the greatest circle in A conteyned) and heith, QE, is equall to the sphere A, by the 7. of the fift. And now it can not be hard, to geue a cylinder, to A, in that proportion, which is betwene X and Y. For let the side QE, be to QP, as Y is to X, by the 12. of the sixt. Therefore backeward, QP is to QE as X to Y. Wherefore the cylinder hauing the base the greatest circle in A and heith the line QP,<note place="margin">*</note> is to the cy<g ref="char:EOLhyphen"/>linder hauing the same base, and heith the line QE, as X is to Y, by the 14. of this twelfth: but the cylin<g ref="char:EOLhyphen"/>der hauing the heith QE, &amp; his base the greatest circle in A, conteyned, is proued equall to the Sphere A. Wherefore by the 7. of the fift, the cylinder whose heith is QP and base the greatest circle in A, con<g ref="char:EOLhyphen"/>teyned, is to the sphere A, as X to Y. Therefore to a sphere geuen: we haue made a cylinder, in any pro<g ref="char:EOLhyphen"/>portion geuen b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>twene two right lines, and also, before we haue to a sphere geuen, made a cylinder e<g ref="char:EOLhyphen"/>quall. Therefore to a sphere geuen, we haue made a cylinder equall, or in any proportion geuen betwene two right lines.</p>
                     <div type="problem">
                        <head>A Probleme. 5.</head>
                        <p>A Sphere being geuen, and a circle, vpon that circle as a base, to rere a cylinder, equall to the sphere geuen: or in any proportion, geuen betwene two right lines.</p>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 6.</head>
                        <p>
                           <hi>A Sphere being geuen, and a right line, to make a cylinder, equall to the sphere geuen, or in any o<g ref="char:EOLhyphen"/>ther proportion, betwene two right lines, geuen.</hi>
                        </p>
                        <p>In this 5. and 6. probleme, first make a cylinder equall to the sphere geuen, by the 4. probleme: and then by the order of the 2. and 3. problemes, in cones, execu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e these accordingly in cylinders.</p>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 7.</head>
                        <p>
                           <hi>Two vnlike Cones or Cylinders, being geuen, to finde two right lines, which haue the same propor<g ref="char:EOLhyphen"/>tion one to the other, that the two geuen cones or cylinders haue one to the other.</hi>
                        </p>
                        <p>Vpon one of their bases rere a cone (if cones be compared) or a cylinder (if cylinders be compared) equall to the other: by the order of the second and third probleme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>: and the heith of the cone or cy<g ref="char:EOLhyphen"/>linder,
<pb facs="tcp:7062:414"/>
on whose base you rered an equall cone or cylinder, with the new heith found, haue that pro<g ref="char:EOLhyphen"/>portion, which the cones or cylinders haue, one to the other, by the 14. of this twelfth booke.</p>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 8.</head>
                        <p>
                           <hi>An vpright Cone, and Cylinder, being geuen: to finde two right lines, hauing that proportion, the one to the other, which the Cone and Cylinder haue, one to the other.</hi>
                        </p>
                        <p>Suppose, <hi>QEK</hi> an vpright
<figure/>
cone and <hi>AB</hi> an vpright cylinder geuen. I say two right lines are to be geue<g ref="char:cmbAbbrStroke">̄</g> which shall haue that pro<g ref="char:EOLhyphen"/>portion one to the other,<note place="margin">
                              <hi>Construction.</hi>
                           </note> which <hi>QEK</hi> and <hi>AB</hi> haue one to the o<g ref="char:EOLhyphen"/>ther. Vpon the base <hi>BH,</hi> erecte a Cone, equall to <hi>QEK</hi>: by the or<g ref="char:EOLhyphen"/>der of the seco<g ref="char:cmbAbbrStroke">̄</g>d probleme: which let be <hi>OBH,</hi> and his heith let be <hi>OC,</hi>
                           <note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> and let the heith of <hi>AB,</hi> (the cylinder), be <hi>CS,</hi> produce <hi>CS</hi> to <hi>P</hi>: so that <hi>CP,</hi> be tripla to <hi>CS,</hi> &amp; make pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>fect the cone <hi>PBH.</hi> I say that <hi>FC</hi> and <hi>OC</hi> haue that pro<g ref="char:EOLunhyphen"/>portion, which <hi>AB</hi> hath to <hi>QEK.</hi> For, by constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ction, <hi>OBH</hi> is e<g ref="char:EOLhyphen"/>quall to <hi>QEK</hi>: and <hi>PBH</hi> is equal to <hi>AB,</hi> as we will proue. (Assumpt wise). And <hi>PBH,</hi> and <hi>OBH</hi> are vpon one base, namely <hi>BH</hi>: wher<g ref="char:EOLhyphen"/>fore by the 14. of this twelfth, <hi>
                              <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>B<g ref="char:EOLhyphen"/>H</hi> and <hi>OBH</hi> are one to the other as their heithes <hi>PC,</hi> and <hi>OC,</hi> are one to the other) wherefore the cylinder and cone equall to <hi>PBH</hi> and <hi>OBH</hi> are as <hi>PC</hi> is to <hi>OC:</hi> by the 7. of the fifth. But <hi>AB</hi> the cylin<g ref="char:EOLhyphen"/>der <hi>QEK</hi> the cone, are equall to <hi>PBH</hi> and <hi>OBH:</hi> by construction: wherefore <hi>AB</hi> the cylinder, is to <hi>Q<g ref="char:EOLhyphen"/>EK</hi> the cone, as <hi>PC,</hi> is to <hi>OC.</hi> Wherefore we haue found two right lines hauing that proportion that a cone and a cylinder geuen, haue one to the other. Which thing we may execute vpon the base of the cone geuen, as we did vpon the base of the cylinder geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n, on this maner. Vpon the base of the cone <hi>QEK,</hi>
                           <note place="margin">
                              <hi>An other way of executing this probleme.</hi>
                           </note> which base, let be <hi>EK,</hi> erect a cylinder, equall to <hi>AB,</hi> by the order of my second probleme. Which cylinder let be <hi>ED,</hi> and <hi>GT</hi> his heith, and let the heith of the cone <hi>QEK,</hi> be <hi>QG.</hi> Take the line <hi>GR,</hi> the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hird part o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <hi>QG,</hi> (by the 9. of the sixth): and with a playne passing by <hi>R,</hi> parallel to <hi>EK,</hi> cut of the cylinder <hi>E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>:</hi> which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hall be equall to the cone <hi>QEK,</hi> by the assumpt following<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> I say now, that <hi>A<g ref="char:EOLhyphen"/>B,</hi> the cylinder, is to <hi>QEK</hi> the cone, as <hi>GT,</hi> is to <hi>GR.</hi> For the cylinder <hi>ED</hi> is to the cylinder <hi>EF,</hi> as <hi>GT</hi> is to <hi>GR,</hi> by the 14. of the twelfth: and to <hi>ED</hi> is the cylinder <hi>AB</hi> equall: by construction: and to <hi>EF,</hi> we haue proued the cone <hi>QEK,</hi> equall, wherefore by the 7. of the fifth, <hi>AB</hi> is to <hi>QEK,</hi> as <hi>GT</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> to <hi>GR.</hi> Wherfore an vpright cone, &amp; a cylinder being geuen, we haue found two right lines hauing the same proportion betwene them, which the cone and the cylinder, haue one to the other: which was requisite to be done.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>An assumpt.</head>
                     <p>
                        <hi>If a co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e, and a cylinder, being both on one base, are equall one to the other: the heith of the cone <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s tripla to the heith of the cylinder. And if a cone and a cylinder being both on one base, the heith of the cone be tripla to the heith of the cylinder, the cone and the cylinder are equall.</hi>
                     </p>
                     <p>We will vse the cylinder AB &amp; the cone PBH in the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>mer probleme: with their base &amp; heithes so noted as before. I say if PBH be equall to AB, that CP the heith of PBH, is tripla to CS the heith of the cylinder AB. Suppose vpon the base BH, a cone to be rered of the heith of CS, which let be S<g ref="char:EOLhyphen"/>BH: it is manifest that AB is tripla to that cone SBH, by the 10. of this twelfth. Wherfore a cone equal to AB the cylinder is tripla to SBH the cone, by the 7. of the fifth, but PBH is supposed equall to AB. Therefore PBH is tripla to SBH, therefore the heith of PBH shall be tripla to the heith of SBH by the 14. of the twelfth. But the heith of PBH, is CP: and of SBH, the heith is CS: wherefore CP is tripla to CS. And CS is the heith of the cylinder AB by supposition. Therefore a cone and a cylinder,
<pb n="385" facs="tcp:7062:414"/>
being both on one base, and equall, the heith of the cone is tripla to the heith of the cylinder. And the second part, as easely may be confirmed. For if AB a cylinder, and PBH a cone, haue one base, as the circle about BH: and the heith of PBH be tripla to the heith of AB, I say that PBH, and AB are equal.<note place="margin">
                           <hi>The conuerse of the assu<g ref="char:cmbAbbrStroke">̄</g>pt.</hi>
                        </note> The heith of AB let be (as afore) CS: and of PBH, the heith, let be CP: of the heith CS, imagine a Cone vpon the same base BH: by the 10. of this twelfth, AB shall be triple to that cone. And the cone PBH hauing heith CP, (by supposition) tripla to CS, shall also be tripla to that cone SBH: by the 14. of this twel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>th. Wherefore by the 7. of the fifth AB and PBH are equall. Therefore, if a cone and a cy<g ref="char:EOLhyphen"/>linder being both on one base, the heith of the cone be tripla to the heith of the cylinder, the cone and the cylinder are equall. So haue we demonstrated both partes: as was required.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ A Theoreme. 4.</head>
                     <p>
                        <hi>The superficies of the segment or protion of any sphere, is equall to the circle, whose semidiameter, is equall to that right line which is drawne from the toppe of that segment to the circumference of the circle, which is the base of that portion or segment.</hi>
                     </p>
                     <p>As in the Sphere
<figure/> A, a Segment being cut of by the circle, whose diameter is CE: &amp; the same circle being the base of the Segment, whose top also is <hi>D:</hi> the croked superficies sphe<g ref="char:EOLhyphen"/>ricall of the same Seg<g ref="char:EOLhyphen"/>ment, is equall to a cir<g ref="char:EOLhyphen"/>cle whose Semidiame<g ref="char:EOLhyphen"/>ter is equall to the right line DC. As is the cir<g ref="char:EOLhyphen"/>cle B.</p>
                     <p>This hath <hi>Archimedes</hi> demonstrated in this first booke of the Sphere and Cylinder, in his 40. and 41 propositions: and I remitte them thether, that will herein demonstratiuely be certified: I would wish all Mathematiciens, as well of verities easy, as of verities rare and obscure, to seeke the causes demon<g ref="char:EOLhyphen"/>stratiue, the finall fruite thereof, is perfection in this art.</p>
                     <div type="note">
                        <head>Note.</head>
                        <p>Besides all other vses and commodities, that are of the Croked superficieces of the Cone, Cylin<g ref="char:EOLhyphen"/>der, and Sphere, so easely and certaynely, of vs to be dealt with all: this is not the least, that a notable Er<g ref="char:EOLhyphen"/>ror, which among Sophisticall brablers, and vngeometricall Masters and Doctors, hath a long time bene vpholden: may most euidently, hereby be confuted, and vtterly rooted out of all mens fantasies for e<g ref="char:EOLhyphen"/>uer. The Error is this, <hi>Curui, ad rectum, nulla est proportio,</hi> that is:<note place="margin">
                              <hi>A great error commonly maintained.</hi>
                           </note> Betwene croked and straight, is no pro<g ref="char:EOLhyphen"/>portion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> This error, in line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>, sup<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>ficieces, and solides, may with more true demonstrations be ouer<g ref="char:EOLhyphen"/>throwne, then the fauourers of that fond fa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>tasie are able, with argument, either probable or Sophi<g ref="char:EOLhyphen"/>sticall to make shew or pretence to the contrary. In lines, I omitte (as now) <hi>Archimedes</hi> two wayes, for the finding of the proportion of the circles circumference to a straight line. I meane, by the inscription and ci<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>cumscription of like poligonon figures, and that other, by spirall lines. And I omitte likewise (as now) in solides, of a parallelipipedon, equall to a Sphere, Cone, or Cylinder<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> or any segment or sector of the sayd solides. And onely, here require you to consider in this twelfth booke, the wayes brought to your knowledge, how to the croked superficies of a cone and cylinder, and of a sphere,<note place="margin">
                              <hi>Betwene straight and croked all maner of propor<g ref="char:EOLhyphen"/>tio<g ref="char:cmbAbbrStroke">̄</g> may be geuen.</hi>
                           </note> (the whole, any segment or sector thereof) a playne and straight superficies may be geuen equall:
<q>Namely, a Circle to be geuen equall, to any of the sayd croked superficieces assigned, and geuen. And then farther by my Additions vpon the second proposition, you haue meanes to proceede in all proportions, that any man can in right lines geue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> or assigne. The<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>fore, <hi>Curui ad rectum, proportio omnimoda potest dari.</hi> One thing it is, to demonstrate, that betwene a croked line and a straight or a croked superficies and a playne or straight superficies, &amp;c. there is proportion. And an other thing it is, to demonstrate a particular and speciall kinde of proportion, being betwene a croked superficies and a straight or playne superficies. For this al<g ref="char:EOLhyphen"/>so confi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>meth the first.</q>
This short warning will cause you to auoyde the sayd error, and make you also hable to cure them, which are infected therewith.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 5.</head>
                     <p>
                        <hi>Any two Spheres being geuen, as the Sphericall Superficies of the one, is to the Sphiricall Super<g ref="char:EOLhyphen"/>ficies
<pb facs="tcp:7062:415"/>
of the other: So is the greatest Circle conteyned in that one, to the greatest Circle conteyned in that other. And as greatest Circle, is to greatest Circle, so is Sphericall superficies, to Sphericall superficies.</hi>
                     </p>
                     <p>For the superficies of euery sphere is quadrupla to his greatest circle, by my first Theoreme: wher<g ref="char:EOLhyphen"/>fore, of two geuen spheres, as the sphericall super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>icies of the one, is to his greatest circle, so is the sphericall superficies of the other, to his greatest circle: therfore by alternate proportion, as spherical su<g ref="char:EOLhyphen"/>perficies is to spherical superficies, so is greatest circle to greatest circle. And therfore also as greatest cir<g ref="char:EOLhyphen"/>cle, is to greatest circle, so is spherical superficies to sphericall superficies: which was to be demo<g ref="char:cmbAbbrStroke">̄</g>strated.</p>
                     <div type="problem">
                        <head>A Probleme. 9.</head>
                        <p>
                           <hi>A Sphere being geuen, to geue an other Sphere, to whose Sphericall superficies, the superficies Sphe<g ref="char:EOLhyphen"/>ricall of the Sphere geuen shall haue any proportion, betwene two right lines geuen.</hi>
                        </p>
                        <p>Suppose A, to be a sphere geuen, and the proportion geuen, to be that, which is betwene the right lines X and Y. I say that a sphere is to be g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>uen to whose sphericall superficies, the superficies sphericall of A, shall haue that proportion which X hath to Y. Let the greatest circle, conteyned in A the sphere be the circle BCD.<note place="margin">
                              <hi>Construction.</hi>
                           </note> And by the probleme of my additions, vpon the second proposition of this booke, as X is to Y, so let the circle BCD be to an other circle found, let that other circle be EFG: and his dia<g ref="char:EOLhyphen"/>meter EG. I say that
<figure/>
the sphericall superfici<g ref="char:EOLhyphen"/>es of the sphere A, hath to the sphericall super<g ref="char:EOLhyphen"/>ficies of the sphere, whose greatest circle is EFG, (or his equall) that proportion, which X hath to Y.<note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> For (by con<g ref="char:EOLunhyphen"/>struction) BCD is to E<g ref="char:EOLhyphen"/>FG, as X is to Y: and by the theoreme next be<g ref="char:EOLhyphen"/>fore<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> as BCD is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>F<g ref="char:EOLhyphen"/>G, so is the spherical su<g ref="char:EOLunhyphen"/>perficies of A (whose greatest circle is BCD, by supposition) to the sphericall superficies, of the sphere, whose grea<g ref="char:EOLhyphen"/>test circle is EFG: wherefore, by the 11. of the fifth as X is to Y: So is the sphericall su<g ref="char:EOLhyphen"/>perficies of A, to the sphericall superficies of the sphere, whose greatest circle is EFG: wherefore, the sphere whose diame<g ref="char:EOLhyphen"/>ter is EG, (the diameter also of EFG) is the sphere, to whose sphericall superficies, the sphericall su<g ref="char:EOLhyphen"/>perficies of the sphere A, hath that proportion which X hath to Y. A sphere being geuen therefore, we haue geuen an other sphere, to whose sphericall superficies, the superficies sphericall of the sphere geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> hath any proportion geuen, betwene two right lines: which ought to be done.</p>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 10.</head>
                        <p>
                           <hi>A sphere being geuen, and a Circle lesse then the greatest Circle, in the same Sphere conteyned. to coapt in the Sphere geuen, a Circle equall to the Circle geuen.</hi>
                        </p>
                        <p>
                           <q>Suppose A, to be the sphere geuen: and the circle geuen lesse then the greatest circle in A contey<g ref="char:EOLhyphen"/>ned, to be FKG.<note place="margin">
                                 <hi>The diffini<g ref="char:EOLhyphen"/>
                                    <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                       <desc>•</desc>
                                    </gap>io<g ref="char:cmbAbbrStroke">̄</g> of a cir<g ref="char:EOLhyphen"/>c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                       <desc>•</desc>
                                    </gap>e <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                       <desc>••</desc>
                                    </gap>ap<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                       <desc>••</desc>
                                    </gap>d in a sp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                       <desc>•</desc>
                                    </gap>er<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                       <desc>•</desc>
                                    </gap>.</hi>
                              </note> I say, that in the Sphere A, a circle, equall to the circle FKG, is to be coapted. First vn<g ref="char:EOLhyphen"/>derstand, what we meane here, by coapting of a circle in a Sphere. We say, that circle to be coapted in a Sphere, whose whole circumference is in the superficies of the same Sphere.</q>
Let the greatest circle in the Sphere A conteyned, be the circle BCD. Whose diameter suppose to be BD, and of the circle FKG, let FG be the diameter.<note place="margin">
                              <hi>Construction.</hi>
                           </note> By the 1. of the fourth, let a line equall to FG, be coapted in the circle BCD. Which line coapted, let be BE. And by the line BE, suppose a playne to passe, cutting the Sphere A, and
<pb n="386" facs="tcp:7062:415"/>
to be perpendicularly erected to the superficies
<figure/>
of BCD.<note place="margin">
                              <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                           </note> Seing that the portion of the playne remayning in the sphere, is called their common section: the sayd section shall be a circle, as be<g ref="char:EOLhyphen"/>fore is proued. <hi>And the common section of the sayd playne, and the greatest circle BCD, (which is BE by supposition) shall be the diameter of the same cir<g ref="char:EOLhyphen"/>cle,</hi> as we will proue. For, let that circle be BL<g ref="char:EOLhyphen"/>EM. Let the center of the sphere A, be the point H: which H, is also the ce<g ref="char:cmbAbbrStroke">̄</g>ter of the circle BCD, because BCD is the greatest circle in A contey<g ref="char:EOLhyphen"/>ned. From H, the center of the sphere A, let a line perpendicularly be let fall to the circle BL<g ref="char:EOLhyphen"/>EM. Let that line be HO: and it is euident that HO shall fall vpon the common section BE, by the 38. of the eleuenth. And it deuideth BE, into two equall parts, by the second part of the third proposition of the third booke: by which poynt O, all other lines drawne in the circle BLEM, are, at the same pointe O, deuided into two e<g ref="char:EOLhyphen"/>quall parts. As if from the poynt M, by the point O, a right line be drawne one the other side com<g ref="char:EOLunhyphen"/>ming to the circumference, at the poynt N: it is manifest that NOM is deuided into two equall partes at the poynt O:<note place="margin">
                              <hi>This is mani<g ref="char:EOLhyphen"/>fest: if you con<g ref="char:EOLunhyphen"/>sider the two triangles rect<g ref="char:EOLhyphen"/>angles, HOM and HON: and then with all, vse the</hi> 47. <hi>of the first of Euclide.</hi>
                           </note> by reason, if from the cen<g ref="char:EOLhyphen"/>ter H, to the poyntes N and M, right lines be drawne, HN and HM, the squares of HM, and HN are equall: for that all the semidiameters of the sphere are equal: and therefore their squares are equall one to the other: and the square of the perpendicular HO, is common: wherefore the square of the third line MO is equall to the square of the third line NO: and therefore the line MO to the line NO. So therefore is NM equally deuided at the poynt O. And so may be proued of all other right lines, drawne in the circle BLEM, passing by the poynt O, to the circumference one both sides. Wherefore O is the center of the circle BLEM: and therefore BE passing by the poynt O is the diameter of the circle BLEM. Which circle (I say) is equal to FKG: for by construction BE is e<g ref="char:EOLhyphen"/>quall to FG: and BE is proued the diameter of BLEM, and FG is by supposition the diameter of the circle FKG: wherefore BLEM is equall to FKG the circle geuen: and BLEM is in A the sphere geue<g ref="char:cmbAbbrStroke">̄</g>. Wherfore we haue in a sphere geuen coapted a circle equall to a circle geuen: which was to be done.</p>
                        <div type="corollary">
                           <head>A Corollary.</head>
                           <p>Besides our principall purpose, in this Probleme, euidently demonstrated, this is also made mani<g ref="char:EOLhyphen"/>fest: that if the greatest circle in a Sphere, be cut by an other circle, erected vpon him at right angles,<milestone type="tcpmilestone" unit="unspecified" n="1"/> that the other circle is cut by the center, and that their common section is the diameter of that other <milestone type="tcpmilestone" unit="unspecified" n="2"/> circle: and therefore that other circle deuided is into two equall partes.<milestone type="tcpmilestone" unit="unspecified" n="3"/>
                           </p>
                        </div>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 11.</head>
                        <p>
                           <hi>A Sphere being geuen, and a circle, lesse then double the greatest circle in the same Sphere con<g ref="char:EOLhyphen"/>tained, to cut of, a segment of the same Sphere, whose Sphericall superficies, shall be equall to the cir<g ref="char:EOLhyphen"/>cle geuen.</hi>
                        </p>
                        <p>Suppose K to be a Sphere geuen, whose greatest circle let be ABC: and the circle geuen, suppose to be DEF.<note place="margin">
                              <hi>Construction.</hi>
                           </note> I say, that a segment of the Sphere K, is to be cut of, so great, that his Sphericall superfi<g ref="char:EOLhyphen"/>cies, shall be equall to the circle DEF. Let the diameter of the circle ABC, be the line AB. At the point A, in the circle ABC, coapt a right line equall to the semidiameter of the circle DEF (by the first of the fourth). Which line suppose to be AH. From the point H, to the diameter AB, let a per<g ref="char:EOLhyphen"/>pendicular line be drawen: which suppose to be HI. Produce HI to the other side of the circumfe<g ref="char:EOLhyphen"/>rence, and let it come to the circumference at the point L. By the right line HIL (perpendicular to AB) suppose a plaine superficies to passe, perpendicularly erected vpon the circle ABC: and by this
<pb facs="tcp:7062:416"/>
plaine superficies, the Sphere to be cut into two seg<g ref="char:EOLhyphen"/>mentes:
<figure/>
one lesse then the halfe Sphere, namely, HA<g ref="char:EOLhyphen"/>LI: and the other greater then the halfe Sphere, name<g ref="char:EOLhyphen"/>ly, HBLI. I say, that the Sphericall superficies of the segment of the Sphere K, in which the segment (of the greatest circle) HALI, is contayned, (whose base is the circle passing by HIL, and toppe the point A) is equall to the circle DEF.<note place="margin">
                              <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                              </hi>
                           </note> For the circle, whose semidiame<g ref="char:EOLhyphen"/>ter is equall to the line AH, is equall to the Sphericall su<g ref="char:EOLhyphen"/>perficies of the segment HAL, by the 4. Theoreme here added. And (by construction) AH is equall to the semi<g ref="char:EOLhyphen"/>diameter of the circle DEF: therefore the Sphericall superficies of the segment of the Sphere K (cut of by the circle passing by HIL) whose toppe is the point A, is equall to the circle DEF. Wherefore, a Sphere being geuen, and a circle lesse then double the greatest circle in the same Sphere, we haue cut of, a segment of the same Sphere, whose Sphericall superficies, is equall to the circle geuen: which was requisite to be done.</p>
                        <div type="note">
                           <head>¶ An aduise.</head>
                           <p>In noting or signifying of Spheres, sometimes we vse by one and the same circle, in plaine designed, to re<g ref="char:EOLhyphen"/>present a Sphere and also the greatest circle in the same contained: and likewise, by a segment of that circle, sig<g ref="char:EOLhyphen"/>nifie a segment of the same Sphere, as by a straight line, we often signifie the circle, which is the base of a segment of a Sphere, Cone, or Cylinder: and so in such like. Wherin, consider our suppositions: and take heede when we shift from one signification to an other, in one and the same designation: and withall remember the principall intent of our drift: and such light thinges, can not either trouble or offend thee. Compendiousnes and artificiall custome, procureth such meanes: sufficient, to stirre vp i<g ref="char:EOLhyphen"/>magination Mathematicall: or to informe the practiser Mechanicall.</p>
                        </div>
                     </div>
                     <div type="problem">
                        <head>A Probleme. 12.</head>
                        <p>
                           <hi>To cut a Sphere geuen, into two such segmentes, that the Sphericall superficieces of the segment<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap> shall haue one to the other, any proportion geuen between two right lines.</hi>
                        </p>
                        <p>Suppose F to be a Sphere geuen: and the proportion geuen, to be that, which is betwene GH and HI. I say, that the Sphere F, is to be cut into two such segmentes, that the Sphericall superficies of those segmentes, shall haue that proportion, one to the other, which, the right line GH, hath to the right line HI.<note place="margin">
                              <hi>Construction.</hi>
                           </note> Suppose ABCE to be a greatest circle, in the Sphere F, contained: and his diameter, to be AB. Deuide ABinto two such partes, as GI is di<g ref="char:EOLhyphen"/>uided
<figure/>
into, in the point H (by the 10. of the sixt) Let those partes be AD, and DB. So that, as GH is to HI, so is AD to DB. By the point D, let a plaine superficies passe, cutting the Sphere F, and the diameter AB: So, that vnto that cutting plaine, the diameter AB, be perpendi<g ref="char:EOLhyphen"/>cular: and the Sphere also thereby by deuided into two seg<g ref="char:EOLhyphen"/>mentes, whose co<g ref="char:cmbAbbrStroke">̄</g>mon base suppose to be the circle CE, hauing the center, the point D: and the toppe of the one to be the point A, and the toppe of the other to be the point B: and the segmentes them selues, to be noted by EAC, and EBC: Drawe from the two toppes, A and B, to C (a point in the circumference of their common base) two right lines AC and BC. I say now, that the Sphericall superficies of the segment EAC, hath to the Sphericall superficies of the segment EBC, the same proportion,<note place="margin">
                              <hi>Demonstratio<g ref="char:cmbAbbrStroke">̄</g>
                              </hi>
                           </note> which GH hath to HI. For, forasmuch as circles haue that proportion, one to the other, that the squares of their diameters haue one to the other (by the 2. of this twelfth).<note place="margin">
                              <hi>This in maner of a Lemm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap> is presently pro<g ref="char:EOLhyphen"/>ued.</hi>
                           </note> 
                           <hi>And the squares of theyr semidi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>me<g ref="char:EOLhyphen"/>ters, have the same proportion one to the other, which the squares of theyr diameters haue.</hi> [For like partes haue
<pb n="387" facs="tcp:7062:416"/>
that proportion one to the other, that the whole magnitudes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> whose like partes they are, haue the one to the other: by the 15. of the fift.
<q>But the square of euery diameter is quadruple to the the square of his semidiameter: as hath often before<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> bene proued: therefore, circle haue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> one to an other, that proportion, that the squares of their semidiameters haue one to the other].</q>
Wherefore, seing AC and BC are semidiameters of two circles, whereof eche is equall to the Sphericall superficies of the seg<g ref="char:EOLhyphen"/>mentes, betwene whose toppes and circumference of their base, they are drawen: by the 4. Theoreme of these additions: it followeth that both those circles whose semidiameters they are: and also those Sphericall superficieces, which are equall to those circles, haue the one to the other, the same propor<g ref="char:EOLhyphen"/>tion, which the square of AC hath to the square of BC. But AC is drawen betwene the circumference of the base, and toppe of the segment Sphericall, EAC, by construction: and likewise BC is drawen betwene the <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> and the <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap> of the base Sphericall segments <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>BC, by construction: Wherefore the Sphericall superficies of the segment EAC, is to the Sphericall superficies of the seg<g ref="char:EOLhyphen"/>ment EBC, as the square of AC is to the square of BC. But the square of AC is to the square of BC, as AD is to DB: by the Corollary of the Probleme of my additions vpon the second of this twelfth: And AD is to DB, as GH is to HI: by construction. Wherefore the Sphericall superficies of the seg<g ref="char:EOLhyphen"/>ment EAC, is to the Sphericall superficies of the segment EBC, as GH is to HI. We haue therfore, cut the Sphere geuen, into two such segmentes, that the Sphericall superficieces of the segmentes, haue one to the other any proportion geuen betwene two right lines: which was to be done.</p>
                        <div type="corollary">
                           <head>¶ A Corollary. 1.</head>
                           <p>
                              <hi>Here it app<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>areth demonstrated, that, circles are one to the other, as the squares of their semidi<g ref="char:EOLhyphen"/>ameter are, one to the other.</hi>
                           </p>
                           <p>Wherby (as occasion shall serue) you may, by force of the former argument, vse other like partes of the diameter, as well as halues.</p>
                        </div>
                        <div type="corollary">
                           <head>¶ A Corollary. 2.</head>
                           <p>It is also euident, that the Sphericall superficieces of the two segment<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>s of any Sphere, to whose common base, the diameter (passing to their two topp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>s) is perpendicular, haue that proportion the o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>e to the other, that the portions of the sayd diameter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> haue the one to the other: that superficies and that portion of the diameter on the one side of the common base, being compared to that superficies, and that portion of the diameter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> on the other side of the common base.</p>
                        </div>
                        <div type="corollary">
                           <head>¶ A Corollary. 3.</head>
                           <p>
                              <hi>It likewise euidently followeth, that the two Sphericall superficieces of two segmentes of a Sphere: which two segmentes are equall to the Sphere, are in that proportion the one to the other, that their axes (perpendicularly erected to their bases) are in, one to the other: where soeuer in the Sphere those seg<g ref="char:EOLhyphen"/>mentes be taken.</hi>
                           </p>
                           <p>I say that the Sphericall superficies of the segment
<figure/>
CAE, and the Sphericall superficies of the segment FGH, hauing their axes AD and GI (perpendicular to their bases): are in proportion one to the other, as AD is to GI: if the segment of the Sphere contai<g ref="char:EOLhyphen"/>ning CAE with (the segment of the same Sphere) FGH be equall to the whole Sphere. For seing the diameter (of axe) AD, extended to the other pole or toppe, opposite to A (which opposite toppe, let be Q) doth make with the segment CAE, the comple<g ref="char:EOLhyphen"/>ment of the whole Sphere:<note place="margin">
                                 <hi>Note here of Axe, base, &amp; soliditie, more then I nede to bring any far<g ref="char:EOLunhyphen"/>ther proofe for.</hi>
                              </note> and by supposition, the segment FGH, whith the segment CAE, are equall to the whole Sphere: Wherefore from equall, taking CAE (the segment common) remayneth the seg<g ref="char:EOLhyphen"/>ment CQE, equall to the segment FGH. And ther<g ref="char:EOLhyphen"/>by, <hi>Axe, Base, Solitie,</hi> and superficies Sphericall of the segment FGH, must (of ncessitie) be equall to
<pb facs="tcp:7062:417"/>
the Axe, Base, Soliditie, and superficies Sphericall of the segment CQE: Wherefore, by the second Corollary here, and the 7. of the fift, our conclusion is inferred, the superficies Sphericall, of the seg<g ref="char:EOLhyphen"/>ment CAE, to be, to the superficies Sphericall of the segment FGH, as AD is to GI.</p>
                        </div>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 6.</head>
                     <p>
                        <hi>To any solide sector of a Sphere, that vpright C<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>e is equall, whose base is equall to the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>nnex Sphericall superficies of that sector, and heith equall to the semidiameter of the same Sphere.</hi>
                     </p>
                     <p>Hereof the demonstration in respect of the premi<g ref="char:EOLhyphen"/>ses:
<figure/>
and the common argument of inscriptio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and cir<g ref="char:EOLhyphen"/>cumscription of figures is easy: and neuerthelesse, if your owne write will not helpe you sufficiently: you may take helpe at <hi>Archimedes</hi> hand, in his first booke &amp; last proposition of the sphere and cylinder. Whether if ye haue recourse, you shal<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> perceaue how your Theo<g ref="char:EOLhyphen"/>reme here amendeth the common translation there: and also our delin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ation geueth more s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y shew of the chiefe circumstances necessary to the construction, then there you shall finde. Of the sphere here imagined to be A, we note a solide sector by the letter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> PQRO. So that PQR doth signifie the sphericall superficies, to that solide sector belonging:<note place="margin">
                           <hi>Note.</hi>
                        </note> (<hi>which is also common to the segment of the same sphere PRQ</hi>) and therefore a line drawne from the toppe of that segment<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (which toppe suppose to be Q) is the semidiameter of the circle, which is equall to the sphericall superficies of the sayd solide sector, or segment<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> as before is taught. Let that line be QP. By Q draw a line contingent: which let be SQT. At the poynt Q from the line QS, cut a line equall to PQ which let be SQ. And vnto SQ, make QT equall, then draw the right lines OSO<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and OQ. About which OQ (as an axe fa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ened) if you imagine the triangle OST, to make an <note n="*" place="margin">
                           <hi>I say halfe a circ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>lar reuo<g ref="char:EOLhyphen"/>lution for that su<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>iseth in the whole diameter ST, to describe a circle by: i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> it be moued <gap reason="illegible" resp="#PDCC" extent="2 letters">
                                 <desc>••</desc>
                              </gap>out his center Q &amp;c.</hi>
                        </note>halfe circular reuolution, you shall haue the vpright cone OST: (whose heith is OQ, the semidia<g ref="char:EOLhyphen"/>meter of the sphere, and base the circle, whose diameter is ST,) equall to the solide sector PQRO.</p>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 7.</head>
                     <p>
                        <hi>To any segment, or portion of a Sphere, that cone
<figure/>
i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> equall, which hath that circle to his base, which is the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>se of the segme<g ref="char:cmbAbbrStroke">̄</g>t, and heith, a right line, which vnto the heith of the segme<g ref="char:cmbAbbrStroke">̄</g>t hath that proportio<g ref="char:cmbAbbrStroke">̄</g>, which the semidiameter of the Sphere together with the heith of the other segment remay<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>g hath to the heith of the same other segment remayn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ng.</hi>
                     </p>
                     <p>This is well demonstrated by <hi>Archi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>des</hi> &amp; there<g ref="char:EOLhyphen"/>fore nedeth no inuention of myne,<note place="margin">
                           <hi>Lib 2 prop 2. de Sphe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>a &amp; Cylindr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>.</hi>
                        </note> to confirme the same: and for that the sayd demonstration is ouer long here to be added, I will refere you thether for the de<g ref="char:EOLhyphen"/>monstration: and here supply that which to <hi>Archimedes</hi> demonstration shall geue light, and to your farther spe<g ref="char:EOLhyphen"/>culation and practise, shal be a great ayde and direction. Suppose K to be a sphere: &amp; the greatest circle K in co<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>teyned, let be ABCE, and his diameter BE, &amp; ce<g ref="char:cmbAbbrStroke">̄</g>ter D. Let the sphere K, be cutte by a playne superficies, per<g ref="char:EOLhyphen"/>p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ndicularly erected vpon the sayd greatest circle AB<g ref="char:EOLhyphen"/>CE: let the section be the circle about AC: And let the
<pb n="388" facs="tcp:7062:417"/>
segmentes of the sphere be the one that wherein is A<g ref="char:EOLhyphen"/>BC,
<figure/>
whose <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>oppe is B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and the other let be that where<g ref="char:EOLhyphen"/>in is AEC and his toppe, let be E: I say that a cone which hath his base the circle about AC, &amp; held a line which to BF (the heith of the segment, whose toppe is B,) hath that proportion that a line compo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ed of DE, the semidiameter of the sphere, and EF (the heith of the other remayning segment, whose toppe is E) hath to EF, (the heith of that other segment remayning), is equall to the segment of the sphere K, whose toppe is B. To make this cone, take my easy order thus. Frame your worke for the find<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ng of the fourth proportionall line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> by making EF the first and a line composed of D<g ref="char:EOLhyphen"/>E and EF, the second<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and the third, let be BF: then by the 12. of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>h, let the fourth proportionall line be found: which let be FG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> vpon F the center of the base of the segment, whose toppe is B, erect a line perpendi<g ref="char:EOLhyphen"/>cular equall to FG found and drawe the lines GA and GC: and so make perfect the cone GAC. I say, that the cone GAC, is equall to the segment (of the sphere K) whose toppe is B. In like maner, for the other segme<g ref="char:cmbAbbrStroke">̄</g>t whose toppe is E, to finde the heith due for a cone equal to it: by the order of the Theoreme you must thus frame your lines: let the first be BF: the second DB and BF, composed in one right line, and the third must be EF: where by the 12. of the sixth, finding the fourth, it shall be the heith, to rere vpon the base, (the circle a<g ref="char:EOLhyphen"/>bout AC,) to make an vpright cone, equall to the seg<g ref="char:EOLhyphen"/>ment, whose toppe is E.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ Logistically. ¶</head>
                     <p>The Logisticall finding hereof is most easy: the diameter of the sphere being geuen, and the por<g ref="char:EOLhyphen"/>tions of the diameter in the segmentes conteyned (or axes of the segmentes) being knowne. Then or<g ref="char:EOLhyphen"/>der your numbers in the rule of proportion, as I here haue made most playne, in ordring of the lines: for the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ought heith will be the producte.</p>
                     <div type="corollary">
                        <head>A Corollary. 1.</head>
                        <p>Hereby, and other the premises it is euident that to any segment of a Sphere,<note place="margin">Note.</note> whose whole diame<g ref="char:EOLhyphen"/>ter is knowne and the Axe of the segment geuen, An vpright cone may be made equall: or in any pro<g ref="char:EOLhyphen"/>portion, betwene two right lines assigned<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and therefore also a cylinder may to the sayd segment of the Sphere, be made equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>r in any proportion geuen, betwene two right lines.</p>
                     </div>
                     <div type="corollary">
                        <head>A Corollary. 2.</head>
                        <p>Manifestly also, of the former theoreme, it may be inferred that a Sphere, and his diameter be<g ref="char:EOLhyphen"/>ing deuided, by one and the same playne superficies, to which the sayd diameter is perpendicular<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the two segmentes of the Sphere, are one to the other in that proportion, in which a rectangle parallelipipe<g ref="char:EOLhyphen"/>don hauing for his base the square of the greater part of the diameter, and his heith a line composed of the lesse portion of the diameter, and the semidiameter: to the rectangle parallelip<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>pedon, hauing for his base the square of the lesse portion of the diameter, &amp; his heith a line composed of the semidiameter &amp; the greater part of the diameter.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 8.</head>
                     <p>
                        <hi>Euery Sphere, to the cube, made of his diameter, is (in maner) as</hi> 11. <hi>to</hi> 21.</p>
                     <p>
                        <pb facs="tcp:7062:418"/>As vpon the first and second propositio<g ref="char:cmbAbbrStroke">̄</g>s of this booke, I began my additions with the circle (be<g ref="char:EOLhyphen"/>ing the chiefe among playne figures) and therein brought manifold considerations, about circles: as of the proportion betwene their circumferences and their diameters of the content or Area of circles: of the proportion of circles to the squares described of their diameters: &amp; of circles to be geuen in al pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>portions, to other circles: with diuerse other most necessary problemes (whose vse is partly there speci<g ref="char:EOLhyphen"/>fied): So haue I in the end of this booke, added some such Problemes &amp; Theoremes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> about the sphere (being among solides the chiefe) as of the same, either in it selfe considered, or to cone and cylinder, compared (by reason of superficies, or soliditie, in the hole, or in part<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>) such certaine knowledge demon<g ref="char:EOLunhyphen"/>stratiue may arise, and such mechanical exercise thereby be deuised, that (sure I am) to the sincere &amp; true student great light, ayde, and comfortable courage (farther to wade) will enter into his hart; and to the Mechanicall, witty, and industrous deuiser, new maner of inuentions, &amp; executions in his workes will (with small trauayle for fete application) come to his perceiueraunce and vnderstanding. Therefore, e<g ref="char:EOLhyphen"/>uen a, manifolde speculations &amp; practises may be had with the circle, his quantitie being not knowne in any kinde of smallest certayne measure: So likewise of the sphere many Problemes may be executed and his precise quantitie, in certaine measure, not determined, or knowne: yet, because, both one of the first (humane) occasio<g ref="char:cmbAbbrStroke">̄</g>s of inuenting and stablishing this Arte, was measuring of the earth (and therfore called Geometria, that is, Earthmeasuring), and also the chiefe and generall end (in deede) is mea<g ref="char:EOLhyphen"/>sure: and measure requireth a determination of quantitie in a certayne measure by nu<g ref="char:cmbAbbrStroke">̄</g>ber expressed: It was nedefull for Mechanicall earthmeasures, not to be ignorant of the measure and contents of the circle, neither of the sphere his measure and quantitie, as neere as sense can imagine or wish. And (in very deede) the quantitie and measure of the circle, being knowne, maketh not onely, the cone and cy<g ref="char:EOLhyphen"/>linder, but also the sphere his quantitie to be as precisely knowne, and certayne. Therefore seing in re<g ref="char:EOLhyphen"/>spect of the circles quantitie (by <hi>Archimedes</hi> specified) this Theoreme is noted vnto you: I wil, by order, vpon that (as a supposition) inferre the conclusion of this our Theoremes.</p>
                     <p>Suppose a sphere to be
<figure/>
signified by A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> whose diame<g ref="char:EOLhyphen"/>ter let be RS. To RS, let a line equall be taken, which let be TV: of TV, by the 46. of the first, describe a square. Let that square be TY. With in TY let a circle be inscri<g ref="char:EOLhyphen"/>be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>: by the 2. of the fourth, which cir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> suppose to be OZW. That OZW is e<g ref="char:EOLhyphen"/>quall to the greatest circle in the sphere A conteyned, it is euident by the diameter, e<g ref="char:EOLhyphen"/>qual to TV. If vpo<g ref="char:cmbAbbrStroke">̄</g> the square TY, as a base, be erected, a pa<g ref="char:EOLunhyphen"/>rallelipipedo<g ref="char:cmbAbbrStroke">̄</g> recta<g ref="char:cmbAbbrStroke">̄</g>gle, whose h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ith is equall to TV, it is e<g ref="char:EOLhyphen"/>uident that that parallelipi<g ref="char:EOLhyphen"/>pedon is a cube. Which let be done: and that cube pro<g ref="char:EOLhyphen"/>duced, let be noted by TX. Likewise, if vpon the circle OZW, as a base, and of the heith equall to the line TV, a cylinder be erected, it is ma<g ref="char:EOLunhyphen"/>nifest that the cylinder hath his base equall to the greatest circle, in the sphere A, con<g ref="char:EOLhyphen"/>teined: &amp; heith, a line equall to the diameter of the same sphere A. Which cylinder let be produced and noted by Z<g ref="char:EOLhyphen"/>M. I say now that the sphere A, is to the cube TX, (in ma<g ref="char:EOLhyphen"/>ner) as the number 11. is to the number 21. For seing the cube TX, was produced of his base, (the square TY), be<g ref="char:EOLhyphen"/>ing brought into the heith of a line equall to TV: &amp; like<g ref="char:EOLhyphen"/>wise seing the cylinder ZM,
<pb n="389" facs="tcp:7062:418"/>
is produced, of his base (the circle OZW) being brought into a line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> equal to the said TV: it followeth, seing their heithes is all one, that the cube TX shall be to the cylinder ZM, as the base of TX, (which is the square TY) is to the base of ZM, that is the circle OZW. But the square TY, is to the circle OZW, as the number 14. is to the nu<g ref="char:cmbAbbrStroke">̄</g>ber 11. (in maner), by <hi>Archimedes</hi> demo<g ref="char:cmbAbbrStroke">̄</g>stratio<g ref="char:cmbAbbrStroke">̄</g>: wherfore, the cube TX is to the cylinder ZM, as the number 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, is to the number 11. (well nere). And by my third Theoreme (here added) the cylinder ZM, is to the Sphere A, in sesquialtera proportion: that is, as 3. to 2. Where<g ref="char:EOLhyphen"/>fore the cylinder ZM, hauing the same 11. equall partes (which he conteyneth in respect of the cube T<g ref="char:EOLhyphen"/>X, being 14. of the same partes) deuided into 3. equall portions, euery one of those portions is 3⅔. And allowing to the Sphere A, two of those portions: it is euident, that the Sphere A shall be 71/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>; such partes as are 14. in the cube TX: and 11. in the cylinder ZM. Wherefore the Sphere A, is to the cube TX, as 71/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> to 14. The fraction being reduced, maketh <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>: and the number 14. being brought to the same name, and denomination of thirds, maketh 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. Put away now theyr common denominator<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and then remay<g ref="char:EOLhyphen"/>neth, for the Sphere A, 22. such partes, as the cube TX hath 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. And then depressing them, to the smal<g ref="char:EOLhyphen"/>lest termes: for the Sphere A, you shall haue 11. such partes as the cube TX conteyneth 21. Wherefore euery Sphere, to the cube made of his diameter is, as 11. to 21. which was requisite to be demonstrated.</p>
                     <div type="note">
                        <head>Note. 1.</head>
                        <p>Wherfore if you deuide the
<figure/>
one side (as TQ) of the cube TX into 21. equall partes, and where 11. partes do end, recke<g ref="char:EOLhyphen"/>ning from T, suppose the point P: and by that point P, imagine a plaine (passing parallel to the opposite bases) to cut the cube TX: and therby, the cube TX, to be deuided into two rectan<g ref="char:EOLhyphen"/>gle parallelipipedons, namely, TN, and PX: It is manifest, <note n="*" place="margin">
                              <hi>A rectangle parallelipipe<g ref="char:EOLhyphen"/>don geu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>n, e<g ref="char:EOLhyphen"/>quall to a Sphere geuen.</hi>
                           </note> TN, to be equall to the Sphere A, by construction: and the 7. of the fift.</p>
                     </div>
                     <div type="note">
                        <head>Note. 2.</head>
                        <p>Secondly, the whole quan<g ref="char:EOLhyphen"/>titie, of the Sphere A,<note place="margin">
                              <p>
                                 <hi>To a Sphere, or to any part of a Sphere assig<g ref="char:EOLhyphen"/>ned: as a third, fourth, fifth &amp;c to geue a paral<g ref="char:EOLhyphen"/>lelipipedon e<g ref="char:EOLhyphen"/>quall.</hi>
                              </p>
                              <p>
                                 <hi>Sided Columes Pyramids, and prismes to be geuen equall to a Sphere, or to any certayne part thereof.</hi>
                              </p>
                              <p>
                                 <hi>To a Sphere or any segment, or sector of the same, to geue a cone or cylinder equall or in any proportion as<g ref="char:EOLhyphen"/>signed.</hi>
                              </p>
                           </note> being co<g ref="char:cmbAbbrStroke">̄</g>
                           <g ref="char:EOLhyphen"/>tayned in the rectangle paralle<g ref="char:EOLhyphen"/>lipipedon TN, you may easilie transforme the same quantitie, into other parallelipipedons rectangles, of what height, and of what parallelogramme base you list: by my first and second Problemes vpon the 34. of this booke. And the like may you do, to any assigned part of the Sphere A: by the like meanes deuiding the parallelipipedon TN: as the part assigned doth require. As if a third, fourth, fifth, or sixth, part of the Sphere A, were to be had in a parallelipipedon, of any parallelogra<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>e base assigned, or of any heith assigned: then deuiding TP, in<g ref="char:EOLhyphen"/>to so many partes (as into 4. if a fourth part be, to be transformed: or into fiue, if a fifth part, be to be transformed &amp;c.) and then proceede, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s you did with cutting of TN, from TX. And that I say of paral<g ref="char:EOLhyphen"/>lelipipedons, may in like sort (by my <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>yd two problemes, added to the 34. of this booke) be done in any sided columnes, pyramids, and prisme<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>: so th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> in pyramids and some prismes you vse the cautions ne<g ref="char:EOLhyphen"/>cessary, in respect of their quan<gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap>odyes hauing parallel, equall, and opposite bases: whose partes <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap>re in their propositions, is by <hi>Eu<g ref="char:EOLhyphen"/>clide</hi> demonstrated. And finally, <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap> additions, you haue the wayes and orders how to geue to a Sphere, or any segme<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the same, Cones, or Cylinders equall, or in any proportion be<g ref="char:EOLhyphen"/>twene two right lines, geuen: with many other most necessary speculations and practises about the Sphere. I trust that I haue sufficiently <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>raughted your imagination, for your honest and profitable stu<g ref="char:EOLhyphen"/>die herein, and also geuen you rea<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>, <gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>tter, whe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> with to s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>p the mouthes of the malycious, igno<g ref="char:EOLhyphen"/>rant,
<pb facs="tcp:7062:419"/>
and arrogant, despisers of the most excellent discourses, trauayles, and inuentions mathematicall. Sting aswel the heauenly spheres, &amp; sterres their sphericall soliditie,<note place="margin">
                              <hi>Farther vse of Sphericall Geometrie.</hi>
                           </note> with their conue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e spherical super<g ref="char:EOLhyphen"/>ficies, to the earth at all times respecting, and their distances from the earth, as also the whole earthly Sphere and globe it selfe, and infinite other cases, concerning Spheres or globes, may hereby with as much ease and certainety be determined of, as of the quantitie of any bowle, ball, or bullet, which we may gripe in our handes (reason, and experience, being our witnesses): and without these aydes, such thinges of importance neuer hable of vs, certainely to be knowne, or attayned vnto.</p>
                        <p>Here ende <hi>M. Iohn Dee</hi> his additions vpon the last proposition of the twelfth booke.</p>
                     </div>
                  </div>
               </div>
               <div type="section">
                  <head>A proposition added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>If a Sphere touche a playne superficies<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> a right line drawne from the center to the touche, shall be erected perpendicularly to the playne superficies.</hi>
                  </p>
                  <p>Suppose that there be a Sphere BCDL: whose centre let be the poynt A. And let the playne su<g ref="char:EOLhyphen"/>perficies GCI touch the Spere in the poynt C, and extend a right line from the centre A to the poynt C. Then I say that the line AC is erected per<g ref="char:EOLhyphen"/>pendicularly
<figure/>
to t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e playne GIC. Let the sphere be cutte by playne superficieces passing by the right line LAC: which playnes let be ABCD<g ref="char:EOLhyphen"/>L and ACEL, which let cut the playne GCI by the right lines GCH and KCI. Now it is manifest (by the assumpt put before the 17. of this booke) that the two sections of the sphere shall be circles, hauing to their diameter the line LAC, which is also the diameter of the sphere. Wherefore the right lines GCH and KCI which are drawne in the playne GCI, do at the poynt C fall without the circles BCDL and ECL. Wherefore they touch the circles in the poynt C, by the second definition of the third. Wherefore the right line LAC maketh right angles with the lines GCH and KCI by the 16. of the third. Wherefore by the 4. of the eleuenth the right line AC is erected perpendicularly to to the playne superficies GCI wherein are drawne the lines GCH and KCI. If therefore a Sphere touch a playne superficies, a right line drawne from the centre to the touche, shall be erected perpendi<g ref="char:EOLhyphen"/>cularly to the playne superficies: which was required to be proued.</p>
               </div>
            </div>
            <trailer>The ende of the twelfth booke of Euclides Elementes.</trailer>
         </div>
         <div n="13" type="book">
            <pb n="390" facs="tcp:7062:419"/>
            <head>¶ The thirtenth booke of Euclides Elementes.</head>
            <p>IN THIS THIRTENTH BOOKE are set forth certayne most wonderfull and excellent passions of a lyne deuided by an extreme and meane proporti<g ref="char:EOLhyphen"/>on:<note place="margin">
                  <hi>The argument of the thir<g ref="char:EOLhyphen"/>tenth booke.</hi>
               </note> a matter vndoubtedly of great and infinite vse in Geometry, as ye shall both in thys booke, and in the other bookes following most euidently perceaue. It teacheth moreouer the composition of the fiue re<g ref="char:EOLhyphen"/>gular solides, and how to inscribe them in a Sphere geuen, and also setteth forth certayne comparisons of the sayd bodyes both the one to the other, and also to the Sphere, wherein they are described.</p>
            <div type="part">
               <head>The 1. Theoreme. The 1. Proposition. If a right line be deuided by an extreme and meane proportion, and to the greater segment, be added the halfe of the whole line: the square made of those two lines added together shalbe quintuple to the square made of the halfe of the whole lyne.</head>
               <p>SVppose that the right
<figure/>
line AB be deuided by an extreme and meane pro<g ref="char:EOLhyphen"/>portio<g ref="char:cmbAbbrStroke">̄</g> in the point C. And let the greater segment therof, be AC. And vnto AC, adde directly a ryght line AD, and let AD be equall to the halfe of the line AB.<note place="margin">Construction.</note> Then I say that the square of the line CD is quintuple to the square of the line DA. Describe (by the 46. of the first) vpon the lines AB and DC squares, namely, AE &amp; DF. And in the square DF, describe and make complete the figure. And ex<g ref="char:EOLhyphen"/>tend the line FC, to the point G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AB is deuided by an extreme and meane proportion in the point C, therefore that which is contayned vnder the lines AB and BC is equall to the square of the line AC. But that which is contayned vn<g ref="char:EOLhyphen"/>der the lines AB and BC, is the pa<g ref="char:EOLhyphen"/>rallelogramme CE, and the square of
<pb facs="tcp:7062:420"/>
the line AC is the square HF. Wherefore the parallelogramme CE is equall to the square HF. And forasmuch as the line BA, is double to the line AD, by construct<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>on: <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the lyne BA is equall to the line KA, and the line AD, to the lyne AH: therefore also, the lyne KA, is double to the line AH. But as the lyne KA is to the line AH, so is the parallelogramme C<g ref="char:EOLhyphen"/>K to the parallelogramme CH: Wherefore the parallelogramme CK is double to the paralle<g ref="char:EOLhyphen"/>logramme CH. And the parallelogrammes LH and CH are double to the parallelo<g ref="char:EOLhyphen"/>gramme CH (for supplementes of parallelogrammes are b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first equall the one to the other). Wherefore the parallelogramme CK is equall to the parallelogrammes LH &amp; CH. And it is proued that the parallelogramme CE is equall to the square FH. Wherefore the whole square AE is equall to the gn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mon MXN. And forasmuch as the line BA, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> dou<g ref="char:EOLhyphen"/>ble to the line AD, therefore the square of the line BA is, by the 20. of the sixth, quadruple to the square of the line DA, that is, the square AE to the square DH. But the square AE is equall to the gnomo<g ref="char:cmbAbbrStroke">̄</g> MXN, wherefore the gnomo<g ref="char:cmbAbbrStroke">̄</g> MXN, is also quadruple to the square DH. Wherefore the whole square DF is quintuple to the square DH. But the square DF, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the square of the line CD, and the square DH is the square of the line DA. Wherefore the square of the line CD, is quintuple to the square of the line DA. If therefore a right line be deuided by an extreame and meane proportion, and to the greater segment, be added the halfe of the whole line: the square made of those two lines added together shalbe quintuple to the square made of the halfe of the whole line: Which was required to be demonstrated.</p>
               <p>
                  <hi>Thys proposition is an other way demonstrated after the fiueth proposition of this booke.</hi>
               </p>
            </div>
            <div type="part">
               <head>The 2. Theoreme. The <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Proposition. If a right line, be in power quintuple to a segment of the same line: the dou<g ref="char:EOLhyphen"/>ble of the sayd segment is deuided by an extreame and meane proportion, and the greater segment thereof is the other part of the line geuen at the be<g ref="char:EOLunhyphen"/>ginning.</head>
               <p>
                  <note place="margin">This proposi<g ref="char:EOLhyphen"/>tion is the con<g ref="char:EOLhyphen"/>uerse of the former.</note>SVppose that the right line
<figure/>
DC be in power quin<g ref="char:EOLhyphen"/>tuple to a segment of the same line, namely, to A<g ref="char:EOLhyphen"/>D, and let the double of the line AD be the line AB. Then I say that the line AB is deuided by an extreme and meane proportion, and the greater segment thereof is the lyne AC.<note place="margin">Construction.</note> Describe on either of the lines AB and CD squares, namely AE and D<g ref="char:EOLhyphen"/>F. And in the square DF make per<g ref="char:EOLhyphen"/>fect the figure, and extend the line FC to the point G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the square DF is quintuple to the square DH, by suppossition, therfore the gno<g ref="char:EOLhyphen"/>mon MNX is quadruple to the square DH. And forasmuch as the line AB is double to the line AD, therefore the square of the line AB is quadruple to the square of the line AD (by the 20.
<pb n="391" facs="tcp:7062:420"/>
of the sixt), that is, the square AE to the square DH. And it is proued that the gnomon M<g ref="char:EOLhyphen"/>NX is quadruple to the square DH. Wherefore the gnomon MNX is equall to the square AE. And forasmuch as the line AB is double to the line AD, but the line AB is equall to the line AK, and the line AD to the line AH: therefore the line AK is double to the line A<g ref="char:EOLhyphen"/>H: wherefore also (by the first of the sixth) the parallelogramme AG is double to the paral<g ref="char:EOLhyphen"/>lelogramme CH. But the parallelogrammes LH and CH are double to the parallelogramme CH (by the 43. of the first): wherefore the parallelogramme AG is equall to the parallelo<g ref="char:EOLhyphen"/>grammes LH and CH. And it is proued that the whole gnomon MNX is equall to the whole square AE. Wherefore the residue HF is equall to the parallelogramme CE. And CE is that which is contained vnder the lines AB and CB, for the line AB is equall to the line BE, and HF is the square made of the line AC. Wherefore that which is contayned vnder the lines AB and BC, is equall to the square of the line AC. Wherfore as the line AB is to the line AC, so is the line AC to the line CB. <note n="*" place="margin">An Assu<g ref="char:cmbAbbrStroke">̄</g>pt.</note> But the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> greater then the line AC, wherefore the line AC is greater then the line CB. Wherefore the line AB is deuided by an extreme and meane proportion, and the greater segment thereof is the line AC. If therfore a right line be in power quintuple to a segment of the same line, the double of the sayd segment is deuided by an extreame &amp; meane proportion, and the greater segment thereof is the other part of the line geuen at the beginning: Which was required to be proued.</p>
               <p>
                  <note n="*" place="margin">The Assu<g ref="char:cmbAbbrStroke">̄</g>pt proued.</note> Now, that the double of the line AD (that is AB) is greater then the line AC may thus be proued. For if not, then if if it be possible let the line AC be double to the line AD, wherefore the square of the line AC is quadruple to the square of the line AD. Wherefore the squares of the lines AC and AD are quintuple to the squares of the line AD. And it is supposed that the square of the line DC is quintuple to the square of the line AD, wherefore the square of the line DC is equall to the square of the lines AC and AD: which is impos<g ref="char:EOLhyphen"/>sible (by the 4. of the second). Wherefore the line AC is not double to the line AD. In like sorte also may we proue that the double of the line AD is not lesse then the line AC, for this is much more absurd: wherefore the double of the line AD is greater the<g ref="char:cmbAbbrStroke">̄</g> the line AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to be proued.</p>
               <p>
                  <hi>This proposition also is an other way demonstrated after the fiueth proposition of this booke.</hi>
               </p>
               <div type="section">
                  <head>Two Theoremes, (in <hi>Euclides</hi> Method necessary) added by <hi>M. Dee.</hi>
                  </head>
                  <div type="subsection">
                     <head>A Theoreme. 1.</head>
                     <p>
                        <hi>A right line can be deuided by an extreame and meane proportion, but in one onely poynt.</hi>
                     </p>
                     <p>Suppose a line diuided by extreame and meane proportion, to be AB. And let the greater segment be AC. I say, that AB can not be deuided by the sayd proportion, in any other point then in the point C. If an aduersary woulde contend that it may, in like sort, be deuided in an other point: let his other point, be supposed to be D: making AD; the greater segment of his imagined diuision. Which AD, also, let be lesse then our AC: for the first discourse. Now, forasmuch as by our aduersaries opinion, AD, is the greater segment, of his diuided line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the parallelogramme conteyned vnder AB, and DB, is equall to the square of AD, by the third definition and 17. proposition of the sixth Booke. And by the same definition and proposition, the parallelogramme vnder AB, and CB, conteyned, is equall to the square of our greater segment AC. Wherefore, as the parallelogramme, vnder AB, and D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, is to the square of AD: so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> parallelogramme, vnder AB, and CB, to the square of AC. For proportion of equality, is con<g ref="char:EOLhyphen"/>cluded
<figure/>
in them both. But, foras<g ref="char:EOLhyphen"/>much as D<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (by<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <note n="*" place="margin">
                           <hi>Because AC is supposed grea<g ref="char:EOLhyphen"/>ter then AD: therefore his residue is lesse, then the residue of AD, by the common sen<g ref="char:EOLhyphen"/>tence. Where<g ref="char:EOLhyphen"/>fore, by the sup<g ref="char:EOLhyphen"/>position: DB is greater then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>C.</hi>
                        </note> supposition) greater the<g ref="char:cmbAbbrStroke">̄</g> CB, the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me vnder AB, and DB, is greater then the parallelogramme vnder AC,
<pb facs="tcp:7062:421"/>
and CB: by the first of the sixth (for
<figure/>
AB is their equall heith.) Where<g ref="char:EOLhyphen"/>fore, the square of AD, shalbe grea<g ref="char:EOLhyphen"/>ter then the square of AC: by the 14. of the fifth. But the line AD, is lesse then the line AC, by supposi<g ref="char:EOLhyphen"/>tion: wherefore the square of AD is lesse then the square of AC. And it is concluded also to be greater then the square of AC: Where<g ref="char:EOLhyphen"/>fore the square of AD, is both greater, then the square of AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and also lesse. Which is a thing impos<g ref="char:EOLhyphen"/>sible. The square therefore of AD, is not equall to the parallelogramme vnder AB, and DB. And there<g ref="char:EOLhyphen"/>fore by the third definition of the sixth, AB is not deuided by an extreame and meane proportion, in the point D: as our aduersary imagined. And (Secondly) in like sort will the inconueniency fall out: if we assigne AD, our aduersaries greater segment, to be greater then our AC. Therefore seing neither on the one side of our point C: neither on the other side of the same point C, any point can be had, at which the line AB can be deuided by an extreame and meane proportion, it followeth of nec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ssitie, that AB can be deuided by an extreame and meane proportion in the point C, onely. Therefore, a right line can be deuided by an extreame and meane proportion, but in one, onely point: which was requi<g ref="char:EOLhyphen"/>site to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 2.</head>
                     <p>
                        <hi>What right line so euer, being deuided into two partes, hath those his two partes, proportionall, to the two segmentes of a line deuided by extreame and meane proportion: is also it selfe deuided by an ex<g ref="char:EOLhyphen"/>treame and meane proportion: and those his two partes, are his two segments, of the sayd proportion.</hi>
                     </p>
                     <p>Suppose, AB, to be a line deuided by an extreame and meane proportion in the point C, and AC to be the greater segment. Suppose also the right line DE, to be deuided into two partes, in the point F: and that the part DF, is to FE, as the segment AC, is to CB: or DF, to be, to AC, as FE is to CB. For so these pa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tes are proportionall, to the sayd segmentes. I say now, that DE is also deuided by an extreame and meane proportion in the point F. And that DF, FE, are his segmentes of the sayd pro<g ref="char:EOLhyphen"/>portion. For, seing, as AC, is to CB: so is DF, to FE: (by supposition). Therfore, as AC, and CB (which is AB) are to CB: so is DF, and FE, (which is DE) to FE: by the 18. of the fifth. Wherefore (alternate<g ref="char:EOLhyphen"/>ly) as AB is to DE: so is CB, to FE. And
<figure/>
therefore, the residue AC, is to the residue DF, as AB is to DE, by the fifth of the fift. And then alternately, AC is to AB, as DE, is to DF. Now therefore backward, AB is to AC, as DE is to DF. But as AB is to A<g ref="char:EOLhyphen"/>C, so is AC to CB: by the third definition of the sixth booke. Wherefore DE is to DF, as AC is to CB: by the 11. of the fifth. And by suppositi<g ref="char:EOLhyphen"/>on, as AC is to CB, so is DF to FE: wherefore by the 11. of the fifth, as DE is to DF: so is DF to FE. Wherefore by the 3. definition of the sixth, DE is deuided by an extreame and meane proportion, in the point F. Wherefore DF, and FE are the segmentes of the sayd proportion. Therefore, what right line so euer, being deuided into two partes, hath those his two partes, proportionall to the two seg<g ref="char:EOLhyphen"/>mentes of a line deuided by extreame and meane proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> is also it selfe deuided by an extreme and meane proportion, and those his two partes are his two segmentes, of the sayd proportion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> which was requisite to be demonstrated.</p>
                     <div type="note">
                        <head>Note.</head>
                        <p>Many wayes, these two Theoremes, may be demonstrated: which I leaue to the exercise of young studentes. But vtterly to want these two Theoremes, and their demonstrations: in so principall a line, or rather the chiefe piller of <hi>Euclides</hi> Geometricall pallace,<note place="margin">
                              <hi>The chie<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>e line in all Euclides Geometrie.</hi>
                           </note> was hetherto, (and so would remayne) a great disgrace. Also I thinke it good to note vnto you, what we meant, <hi>by one onely poynt.</hi> We m<gap reason="illegible" resp="#PDCC" extent="4 letters">
                              <desc>••••</desc>
                           </gap>, that the quantities of the two segmentes, can not be altered, the whole line being once geuen. And though, from either end of the whole line, the greater segment may begin:<note place="margin">
                              <hi>What is ment here by, A secti<g ref="char:EOLhyphen"/>on in one onely poi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>t.</hi>
                           </note> And so as it were the point of secti<g ref="char:EOLhyphen"/>on may seeme to be altered: yet with vs, that is no alteration: forasmuch as the quantities of the seg<g ref="char:EOLhyphen"/>mentes, remayne all one. I meane, the quantitie of the greater segment, is all one: at which end so euer it be taken: And therefore, likewise the quantitie of the lesse segment is all one, &amp;c. The like confidera<g ref="char:EOLhyphen"/>tion may be had in <hi>Euclides</hi> tenth booke, in the Binomiall lines. &amp;c.</p>
                        <p>
                           <hi>Io<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>n Dee. 1569. Decemb. 18.</hi>
                        </p>
                     </div>
                  </div>
               </div>
            </div>
            <div type="part">
               <pb n="392" facs="tcp:7062:421"/>
               <head>The 3. Theoreme. The 3. Proposition. If a right line be deuided by an extreme and meane proportion, and to the lesse segment be added the halfe of the gerater segment: the square made of those two lines added together, is quintuple to the square made of the half line of the greater segment.</head>
               <p>SVppose that the right line AB be deuided by an extreme and meane proportion in the point C. And let the greater segment thereof be AC. And deuide AC into two equall partes in the point D. Then I say that the square of the line BD, it quintuple to the square of the line DC.<note place="margin">Construction.</note> Describe (by the 46. of the first) vpon the line AB a square AE. And describe and make perfect the figure (that is diuide the lyne AT like vnto the diuision of the line AB, by the 10. of
<figure/>
the sixth, in the pointes R, H, by which pointes drawe (by the 31. of the first) vnto the line AB parallel lines RM and HN. So likewise draw by the pointes D, C, vnto the line BE these parallel lines DL and CS, &amp; draw the diameter BT).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the line AC is double to the line DC, therefore the square of AC, is quadruple to the square of DC, by the 20. of the sixth, that is, the square RS to the square F<g ref="char:EOLhyphen"/>G. And forasmuch as that which is contayned vnder the lines AB and BC is equall to the square of the line AC, and that which is contayned vnder the lines AB and BC is equall to the parallelogramme CE, &amp; the square of the line AC is the square RS: wherefore the parallelogramme CE is equall to the square RS. But the square RS is quadruple to the square FG: wherefore the parallelo<g ref="char:EOLhyphen"/>gramme CE also is quadruple to the square FG. Agayne forasmuch as the line AD is equall to the line DC, therfore the line HK is equall the line KF, wherefore also the square GF is equall to the square HL: wherefore the line GK is equall to the line KL, that is, the line MN to the line NE: wherefore the parallelogramme MF is equall to the parallelo<g ref="char:EOLhyphen"/>gramme FE. But the parallelogramme MF is equall to the parallelogramme CG, wherfore the parallelogramme CG is also equal to the parallelogramme FE. Put the parallelogramme CN, common to the<g ref="char:cmbAbbrStroke">̄</g> both: <note n="*" place="margin">Note, how CE and the gnonom XOP, are pro<g ref="char:EOLhyphen"/>ued equall, for it serueth in the conuerse de<g ref="char:EOLhyphen"/>monstrated by M. Dee, here next after.</note> Wherefore the gnomon XOP is equall to the parallelogramme CK. But the parallelogramme CE is proued to be quadruple to G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the square, wherefore the gnomon X<g ref="char:EOLhyphen"/>OP is quadruple to the square GF. Wherefore the square DN is quintuple to the square F<g ref="char:EOLhyphen"/>G. And DN is the square of the line DB, and GF the square of the line DC. Wherefore the square of the line DB is quintuple to the square of the line DC. If therefore a right line be deuided by an extreme and meane proportion, and to the lesse segment be added the halfe of the greater segment: the square made of those two lines added together, is quintuple to the square made of the halfe line of the greater segme<g ref="char:cmbAbbrStroke">̄</g>t. Which was required to be demonstrated.</p>
               <p>
                  <hi>Ye shall finde this proposition an other way demonstrated after the fiueth proposition of this booke.</hi>
               </p>
               <div type="section">
                  <head>Here foloweth M. Dee, his additions.</head>
                  <div type="subsection">
                     <head>¶A Theoreme. 1.</head>
                     <p>
                        <hi>If a right line, geuen, be quintuple in power, to the powre of a segment of him self: the double of that
<pb facs="tcp:7062:422"/>
segment,</hi>
                        <note place="margin">
                           <hi>This proposi<g ref="char:EOLhyphen"/>tion, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>the conuerse of the former.</hi>
                        </note> 
                        <hi>and the other part remayning, of the first geuen line, make a line, diuided by extreme and meane proportion: and that double of the segment, is the greater part thereof.</hi>
                     </p>
                     <p>Forasmuch as, this, is the conuerse of Euclides third proposition: we will vse the same suppositions and constructions there specified: so farre, as they shall serue our purpose. Beginning therefore at the conclusion, we must infer the part of the proposition, before graunted. It was concluded, that the square of the line DB, is quintuple, to the square of the line DC, his owne segment. Therefore DN (the square of DB) is quintuple, to GF, (the square of DC). But the squa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e of AC (the dou<g ref="char:EOLhyphen"/>ble of DC) which is RS, is quadruple to GF, (by the second Corollary of the 20. of the sixth): and therefore RS, with GF, are quintuple to GF: and so it is euident, that the square DN, is equall to the square RS, together with the square GF. Wherefore, from those two equalles, taking, the square GF, (common to them both): remayneth the square RS, equall to the Gnomon XOP. But to the Gno<g ref="char:EOLhyphen"/>mon XOP, the parallelogramme <note n="*" place="margin">
                           <hi>As we ha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>e noted the place of the peculier pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>e there <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>in the demo<g ref="char:cmbAbbrStroke">̄</g>stration of the</hi> 3.</note> CE is equall: Wherefore the square of the line AC, which is RS, is equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ll to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. Which parallelogamme is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder BE, (equall to AB:) and CB, the part remayning of the first line g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uen: which was DB. And the line AB, is made of the double of the segment DC, and of CB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the other part of the line DB, first gouen. Wherefore the double of the segment DC, with CB, the part remayning (which altogether, is the whole line AB) is to AC, (the double of the segment DC) as that same, AC, is to CB: by the second part of the 16. of the sixth. Ther<g ref="char:EOLhyphen"/>fore by the 3. definitio<g ref="char:cmbAbbrStroke">̄</g> of the sixth booke, the whole line AB, is deuided by an extreme and meane pro<g ref="char:EOLhyphen"/>portion, &amp; AC, (the double of the segme<g ref="char:cmbAbbrStroke">̄</g>t DC) being middell proportionall, is the greater part therof. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>efore, if a right line, be quintuple in power, &amp;c. (as in the proposition) which was to be demon<g ref="char:EOLhyphen"/>strated.</p>
                     <div type="proof">
                        <head>Or, thus it may be demonstrated.</head>
                        <p>Forasmuch as the square, DN is quin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>uple to the square GF, (I meane the square of DB the line geue<g ref="char:cmbAbbrStroke">̄</g>, to the square o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> DC the segme<g ref="char:cmbAbbrStroke">̄</g>t). And the same square DN, is equall to the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me vnder AB, CB, with the square made of the line DC: by the sixth of the second: (for vnto the line AC, equally deuided: the line, CB, is, as it were adioyned). Wherefore the parallelogramme vnder AB, CB, toge<g ref="char:EOLhyphen"/>ther with the square of DC, which is GF, is quintuple to
<figure/>
the square GF, made o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> line DC. Taking then, that square GF, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>rom the parallelogramme vnder AB, CB: that parallelogramme (vnder AB, CB) remayning alone, is but quadruple to the sayd square of the line DC. But, (by the 4. of the second, or the second Corollary of the 20. of the sixth) RS, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>he square of the line AC, is quadrupla to the same square GF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> Wherfore by the 7. of the fifth, the square of the line AC, is equall to the parallelogramme vnder A<g ref="char:EOLhyphen"/>B, CB, and so, by the second part of the 16. of the sixth: A<g ref="char:EOLhyphen"/>B, AC, and CB, are three lines in continuall proportion. And seing AB is greater the<g ref="char:cmbAbbrStroke">̄</g> AC, the same AC, the double of the line DC, shall be greater then the part BC, remay<g ref="char:EOLhyphen"/>ning: Wherfore by the 3. definition of the sixth, AB, (com<g ref="char:EOLhyphen"/>posed or made of the double of DC, and the other part of DB remaining) is deuided by an extreme and middel pro<g ref="char:EOLhyphen"/>portion: and also his greater segment is AC the double of the segment DC. Wherfore, If a right line be quintuple in power &amp;c. as in the proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> which was to be demonstrated.</p>
                     </div>
                  </div>
                  <div type="subsection">
                     <head>A Theoreme. 2.</head>
                     <p>
                        <hi>If a right line, deuided by an extreme and meane proportion, be geuen, and to the great segment <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>herof, he directly adioyned a line equal to the whole line geuen, that adioyned line, and the said greater segment, do make a line diuided by extreme and meane proportion, whose greater segment is the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>dioyned.</hi>
                     </p>
                     <p>Suppose the line geuen, deuided by extreame and meane proportion, to be AB deuided in the point C, and his greater segment, let be AC: vnto AC directly adioyne a line equall to AB: let that be AD: I say, that AD, together with AC, (that is DC) is a deuided by extreme and middel proportion, whose greater segment is AD, the line adioyned. Deuide AD, equally in the point E. Now, foras<g ref="char:EOLhyphen"/>much as AE, is the halfe of AD, (by construction,) it is also, the halfe of AB (equall, to AD, by con<g ref="char:EOLhyphen"/>struction): Wherfore by the 1. of the thirtenth, the square of the line composed of AC and AE (which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ne is EC) is quintuple to the square of the line AE. Wherefore the double of AE, and the line AC
<pb n="393" facs="tcp:7062:422"/>
composed, (as in one right line) is a line de<g ref="char:EOLhyphen"/>uided
<figure/>
by extreme and meane proportion, by the conuerse of this third (by me demonstra<g ref="char:EOLhyphen"/>ted): and the double of AE, is the greater segment. But DC is the line composed of the double of AE, &amp; the line AC: and with all, AD is the double of AE. Wherfore, DC, is a line deuided by extreme and meane pro<g ref="char:EOLhyphen"/>portion, and AD, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> hi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> greater segment. If a right line, therefore, deuided by extreme and meane pro<g ref="char:EOLhyphen"/>portion, be geuen, and to the greater segment thereof, be directly adioyned a line equall to the whole line geuen, that adioyned line, and the sayd greater segment, do make a line diuided by extreame and meane proportion, whose greater segment, is the line adioyned: Which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
                     <div type="proof">
                        <head>Two other briefe demonstrations of the same.</head>
                        <p>Forasmuch as, AD is to AC: as AB, is to AC (because AD is equall to AB, by construction): but as AB is to AC, so is AC to CB: by supposition. Therefore by the 11. of the fifth, as AC, is to CB, so is AD to AC. <note n="*" place="margin">
                              <hi>Therefore, by my second The<g ref="char:EOLhyphen"/>oreme added v<g ref="char:EOLhyphen"/>pon the second proposition, DC is deuided by ex<g ref="char:EOLunhyphen"/>treame and meane proporti<g ref="char:EOLhyphen"/>on in the point A. And because AC is bigger then CB: ther<g ref="char:EOLhyphen"/>fore DA is greater then A<g ref="char:EOLhyphen"/>C: wherefore if a right line &amp;c. as in the propo<g ref="char:EOLhyphen"/>sition. Which was to be de<g ref="char:EOLhyphen"/>monstrated.</hi>
                           </note> Wherefore, as AC, and CB, (which is AB) is to CB: so is AD, and AC (which is DC) to AC. Therefore, euersedly, as AB, is to AC: so is DC to AD. And it is proued, AD, to be to AC: as AC is to CB. Wherefore as AB is to AC, and AC, to CB: so is DC, to AD, and AD, to AC. But AB, AC, and CB are in continuall proportion, by supposition: Wherfore DC, AD, and AC, are in continuall proportion. Wherefore, by the 3. definition of the sixth booke, DC, is deuided by extreme and middell proportion, and his greatest segment, is AD. Which was to be demonstrated. Note from the marke <note n="*" place="margin">
                              <hi>Therefore, by my second The<g ref="char:EOLhyphen"/>oreme added v<g ref="char:EOLhyphen"/>pon the second proposition, DC is deuided by ex<g ref="char:EOLunhyphen"/>treame and meane proporti<g ref="char:EOLhyphen"/>on in the point A. And because AC is bigger then CB: ther<g ref="char:EOLhyphen"/>fore DA is greater then A<g ref="char:EOLhyphen"/>C: wherefore if a right line &amp;c. as in the propo<g ref="char:EOLhyphen"/>sition. Which was to be de<g ref="char:EOLhyphen"/>monstrated.</hi>
                           </note>, how this hath two demonstrations. One I haue set in the margent by.</p>
                     </div>
                     <div type="corollary">
                        <head>¶A Corollary. 1.</head>
                        <p>
                           <hi>Vpon Euclides third proposition demonstrated, it is made euident: that, of a line deuided by ex<g ref="char:EOLhyphen"/>treame and meane proportion, if you produce the lesse segment, equally to the length of the greater: the line therby adioyned, together with the sayd lesse segment, make a new line deuided by extreame and middle proportion: Whose lesse segment, is the line adioyned.</hi>
                        </p>
                        <p>For, if AB, be deuided by extreme and middell proportion in the point C, AC, being the greater segment, and CB be produced, from the poynt B, making a line, with CB, equall to AC, which let be CQ: and the line thereby adioyned, let be BQ: I say that CQ, is a line also deuided by an extreame and meane proportion, in the point B: and that BQ (the line adioyned) is the lesse segment. For by the thirde, it is proued, that halfe AC, (which, let be, CD) with CB, as one line, composed, hath his powre or square, quintuple to the powre of the
<figure/>
segment CD: Wherfore, by the second of this booke, the double of C<hi>D,</hi> is de<g ref="char:EOLhyphen"/>uided by extreme and middell propor<g ref="char:EOLhyphen"/>tion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> and the greater segment thereof, shalbe CB. But, by construction, CQ, is the double of CD, for it is equall to AC. Wherefore CQ is deuided by extreme and middle propor<g ref="char:EOLhyphen"/>tion, in the point B: and the greater segment thereof shalbe, CB. Wherefore BQ, is the lesse segment, which is the line adioyned. Therefore, a line being deuided, by extreme and middell proportion, if the lesse segment, be produced equally to the length of the greater segment, the line thereby adioyned to<g ref="char:EOLhyphen"/>gether with the sayd lesse segment, make a new line deuided, by extreme &amp; meane proportion, who<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e lesse segment, is the line adioyned. Which was to be demonstrated.</p>
                     </div>
                     <div type="corollary">
                        <head>¶A Corollary. 2.</head>
                        <p>
                           <hi>If<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> from the greater segment, of a line diuided, by extreme and middle proportion, a line, equall to the lesse segment be cut of: the greater segment, thereby, is also deuided by extreme and meane propor<g ref="char:EOLhyphen"/>tion, whose greater segment<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> shall be <gap reason="illegible" resp="#PDCC" extent="1 word">
                                 <desc>〈◊〉</desc>
                              </gap> that part of it, which is cut of.</hi>
                        </p>
                        <p>For, taking from AC, a line equall to CB: let AR remayne. I say, that AC, is deuided by an ex<g ref="char:EOLhyphen"/>treme and meane proportion in the point R: and that CR, the line cut of, is the greater segment. For it is proued in the former Corollary that CQ is deuided by extreme and meane proportion in the point B. But AC, is equall to CQ, by construction: and CR is equall to CB by construction: Wherefore the
<pb facs="tcp:7062:423"/>
re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>idue, AR is equall to BQ the residue. Seing therfore the whole AC is equall to the whole CQ: and the greater part of AC, which is C<g ref="char:EOLhyphen"/>R
<figure/>
is equal to CB the greater part of CQ and the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t also equall to the lesse: and withall seing CQ is proued to be diuided by extreme &amp; meane proportion in the point B, it foloweth of necessity, that, AC, is diuided by extreme and meane proportion in the point R. And seing CB, is the greater segment of CQ: CR shall be the greater segment of AC. Which was to be de<g ref="char:EOLhyphen"/>monstrated.</p>
                     </div>
                     <div type="corollary">
                        <head>A Corollary. 3.</head>
                        <p>It is euident thereby, a line being diuided by extreme and meane proportion, that the line whe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>by the greater segment excedeth the lesse, together with the lesse segment, do make a line diuided by extreme and meane proportion: whose lesse segment, is the sayd line of exceesse, or difference betwene the segments.</p>
                        <div type="proof">
                           <head>Iohn Dee.</head>
                           <p>¶ Two new wayes, to deuide any right line geuen by an extreme and meane proportion: demonstrated and added by M. Dee.</p>
                        </div>
                     </div>
                     <div type="problem">
                        <head>A Probleme.</head>
                        <p>
                           <hi>To deuide by an extreme and meane proportion, any right line geuen, in length and position.</hi>
                        </p>
                        <p>Suppose a line geuen in length and position<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> to be AB. I say that AB is to be deuided by an extreme, and meane proportion.<note place="margin">
                              <hi>Construction.</hi>
                           </note> Deuide AB into two equall parts as in the point C. Produce AB directly, from the point B, to the point D: making BD, equal to BC. To the line AD, and at the point D, let a line be drawen <note n="*" place="margin">
                              <hi>Though I say, perpe<g ref="char:cmbAbbrStroke">̄</g>dicular: yes you may perceue how in<g ref="char:EOLhyphen"/>finite other p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>
                                 <g ref="char:EOLhyphen"/>s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                                    <desc>••</desc>
                                 </gap>io<g ref="char:cmbAbbrStroke">̄</g>s will serue: so that DI and AD make an angle for a tri<g ref="char:EOLhyphen"/>angle to haue his sides propor<g ref="char:EOLhyphen"/>tionally cut. &amp;c.</hi>
                           </note>perpendicular: by the 11. of the first, which let be DF: (of what length you will). From DF and at the point D, cut of the sixth parte of
<figure/>
DF: by the 9. of the sixth. And let that sixth part, be the line DG. Vppon DF, as a diame<g ref="char:EOLhyphen"/>ter, describe a semicircle: which let be DH<g ref="char:EOLhyphen"/>F. From the point G, rere a line perpendicu<g ref="char:EOLhyphen"/>lar to DF, which suppose to be GH: and let it come to the circumference of DHF, in the point H. Draw right lines, HD, and HF. Pro<g ref="char:EOLhyphen"/>duce DH, from the point H, so long, till a line adioyned with DH, be equall to HF, which let be DI, equall to HF. From the point H, to the point B, (the one ende of our line geuen) let a right line be drawen: as H<g ref="char:EOLhyphen"/>B. From the point I, let a line be drawen, to the line AB: so that it be also parallel to the line HB. Which parallel line suppose to be I<g ref="char:EOLhyphen"/>K: cutting the line AB, at the point K. I say that AB, is deuided by an extreme &amp; meane proportion, in the point K. For the triangle DKI, hauing HB, parallel to IK, hath his sides DK and DI,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> cut proportionally, by the 2. of the sixth. Wherefore as IH is to HD: so is KB, to BD. And therfore compoundingly, (by the 18. of the fiueth) as DI, is to DH: so is DK to DB. But by construction DI is equall to HF: wherefore by the 7. of the fifth, DI is to DH, as HF is to DH. Wherefore by the 11. of the fifth, DK is to DB, as HF is to DH. Wherefore the square of DK is to the square of DB, as the square of HF, is to the square of DH: by the 22. of the sixth. But the square of HF, is to the square of DH: as the line GF is to the line GD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> by my corrollary vpon the 5, probleme of my additions to the second proposition of the twelfth. Wherefore by the 11. of the fifth, the square of DK is to the square of DB, as the line GF is to the line GD. But by construction, GF is quintuple to GD. Wherefore the square of DK is quintuple to the square of DB: and therefore, the double of DB, is deuided by an ex<g ref="char:EOLhyphen"/>treme and meane proportio<g ref="char:cmbAbbrStroke">̄</g>, and BK is the greater segment therof, by the 2. of this thirtenth. Where<g ref="char:EOLhyphen"/>fore seing AB is the double of DB by construction: the line AB is deuided by an extreme and meane
<pb n="394" facs="tcp:7062:423"/>
proportion: and his greater segment, is the line BK. Wherefore AB is deuided by an extreme and meane proportion, in the point K. We haue therefore deuided by extreme and meane proportion any line geuen in length and position. Which was requisite to be done.</p>
                        <div type="proof">
                           <head>The second way to execute this probleme.</head>
                           <p>Suppose the line geuen to be <hi>AB.</hi> Deuide <hi>A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>
                              </hi> into two equall parts: as suppose it to be done in the point <hi>C.</hi> Produce <hi>AB</hi> from the point <hi>B</hi>: adioyning a line equall to <hi>BC,</hi> which let be <hi>BD.</hi> To the right line <hi>AD,</hi> and at the point <hi>D,</hi> erect a perpendicular line equall to <hi>BD,</hi> let that be <hi>DE.</hi> Produce <hi>ED</hi> fro<g ref="char:cmbAbbrStroke">̄</g> the point <hi>D</hi> to the point <hi>F</hi>: making <hi>DF</hi> to contayne fiue such equall partes, as <hi>DE</hi> is one. Now vpon <hi>EF</hi> as a diameter, describe a semicircle which let <gap reason="illegible" resp="#PDCC" extent="1 word">
                                 <desc>〈◊〉</desc>
                              </gap> 
                              <hi>EKF,</hi> and let the
<figure/>
point where the circumference of <hi>EKF,</hi> doth cut the line <hi>AB,</hi> be the point <hi>K.</hi> I say that <hi>AB,</hi> is deuided in the point <hi>K,</hi> by an extreme and meane proportion. For by the 13. of the sixth <hi>ED, DK,</hi> &amp; <hi>DF,</hi> are three lines in continuall proportion, (<hi>DK</hi> being the middle proportionall)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> Wherefore by the corollary of the 20. of the sixth, as <hi>ED</hi> is to <hi>DF,</hi> so is the square of <hi>ED,</hi> to the square of <hi>DK,</hi> but by construction, <hi>ED,</hi> is sub<g ref="char:EOLhyphen"/>quintuple to <hi>DF.</hi> Wherefore the square of <hi>ED,</hi> is subquintuple to the square of <hi>DK.</hi> And therefore the square of <hi>DK,</hi> is quintuple to the square of <hi>ED.</hi> And <hi>ED</hi> is equall to <hi>ED,</hi> by construction, therefore the square of <hi>DK,</hi> is quintuple to the square of <hi>E</hi> D. Wherefore the dou<g ref="char:EOLhyphen"/>ble of <hi>BD,</hi> is deuided by an extreme and meane proportion: whose greater segment is <hi>BK</hi>
                              <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> by the second of this thirte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>th. But by con<g ref="char:EOLhyphen"/>struction, <hi>AB,</hi> is the double of <hi>
                                 <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                    <desc>•</desc>
                                 </gap>D</hi>
                              <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> Wherefore <hi>AB,</hi> is diuided by ex<g ref="char:EOLhyphen"/>treme and meane proportion, and his greater segment, is <hi>BK</hi>: and thereby, <hi>K</hi>
                              <gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap> the point of the diuision. We haue therefore deuided by extreme and meane proportion, any right line geuen, in length and po<g ref="char:EOLhyphen"/>sition. Which was to be done.</p>
                        </div>
                     </div>
                     <div type="note">
                        <head>Note<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </head>
                        <p>Ech of these wayes, may well be executed: But in the first, you haue this auantage: that the diame<g ref="char:EOLhyphen"/>ter is taken at pleasure. Which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>n the second way, is euer iust thrise so long, as the line geuen to be de<g ref="char:EOLhyphen"/>uided.</p>
                        <closer>
                           <signed>Iohn Dee.</signed>
                        </closer>
                     </div>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 4. Theoreme. The 4. Proposition. If a right line be deuided by an extreame and meane proportion: the squares made of the whole line and of the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t, are treble to the square made of the greater segment.</head>
               <p>SVppose that the right line AB, be deui<g ref="char:EOLhyphen"/>ded
<figure/>
by an extreame &amp; meane proportio<g ref="char:cmbAbbrStroke">̄</g>, in the point C. And let the greater seg<g ref="char:EOLhyphen"/>ment thereof be AC. Then I say, that the squares made of the lines AB, and BC, are treble to the square of the line AC. Describe (by the 46. of the first) vpon the line AB, a square ADEB. And make perfect the figure. Now forasmuch as the line AB,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> is deuided by an extreame and meane proportion, in the point C: and the greater segme<g ref="char:cmbAbbrStroke">̄</g>t thereof, is the line AC, therefore that which is contayned vnder the lines AB and BC is equall to the square of the line AC. But that which is contayned vnder the lines AB and CB is the pa<g ref="char:EOLhyphen"/>rallelogramme
<pb facs="tcp:7062:424"/>
AK, and the square of the line A<g ref="char:EOLhyphen"/>C
<figure/>
is the square FD. Wherefore the parallelo<g ref="char:EOLhyphen"/>gramme AK is equall to the square FD. And the parallelogramme AF is equall to the pa<g ref="char:EOLhyphen"/>rallelogramme FE, put the square CK common to them both wherfore the whole parallelogra<g ref="char:cmbAbbrStroke">̄</g>me AK is equall to the whole parallelogramme C<g ref="char:EOLhyphen"/>E. Wherefore the parallelogrammes CE and A<g ref="char:EOLhyphen"/>K are double to the parallelogramme AK. B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t the parallelogrammes AK and CE, are the gno<g ref="char:EOLhyphen"/>mon LMN, and the square CK. Wherefore the gnomon LMN and the square CK, are double to the parallelogramme AK. But it is proued that the parallelogramme AK is equal to the square DF. Wherefore the gnomon LMN and the square CK are double to the square DF. Wherefore the gnomon LMN and the squares CK and DF, are treble to the square DF. But the gno<g ref="char:EOLhyphen"/>mon LMN and the squares CK and DF, are the whole square AE together with the square CK, which are the squares of the lines AB and BC. And DF is the square of the line AC. Wherefore the squares of the lines AB and BC, are treble to the square of the line AC. If therefore a right line be deuided by an extreame and meane proportion, the squares made of the whole line and of the lesse segment, are treble to the square made of the greater segment: which was required to be proued.</p>
               <p>
                  <hi>Looke for an other demonstration of this proposition after the fifth proposition of this booke.</hi>
               </p>
            </div>
            <div type="part">
               <head>¶ The 5. Theoreme. The 5. Proposition. If a right line be deuided by an extreame and meane proportion, and vnto it be added a right <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ine,<note place="margin">I. Dee. This is most eui<g ref="char:EOLunhyphen"/>dent of my se<g ref="char:EOLhyphen"/>cond Theoreme, added to the third propositio<g ref="char:cmbAbbrStroke">̄</g>. For to adde to a whole line, a line equall to the greater seg<g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>t: &amp; to adde to the greater segment a line equall to the whole line, is all one thing, in the line produ<g ref="char:EOLhyphen"/>ced. By the whole line, I meane the line diuided by ex<g ref="char:EOLhyphen"/>treme and meane propor<g ref="char:EOLhyphen"/>tion.</note> equall to the greater segment, the whole right line is deuided by an extreame and meane proportion, and the greater segment thereof, is the right line geuen at the beginning.</head>
               <p>SVppose that the right line AB be deuided by an extreame and meane proporti<g ref="char:EOLhyphen"/>on in the point C, and let the greater segment thereof be AC. And vnto the line AB, adde the line AD equall to the line AC. Then I say that the line D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is deuided by an extreame and meane proportion in the point A: and the greater segment thereof, is the right line
<figure/>
p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t at the beginning, namely, AB. Describe (by the 46. of the first) vp<g ref="char:EOLhyphen"/>on on the line AB a square AE, and make perfect the figure. And foras<g ref="char:EOLhyphen"/>much as the line AB, is deuided by an extreame and meane proportion in the point C, therefore that which is contayned vnder the lines AB and BC is equall to the square of the line AC. But that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines AB and B<g ref="char:EOLhyphen"/>C
<pb n="395" facs="tcp:7062:424"/>
is the parallelogramme CE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the square of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ne AC is the square CH. Wherefore the parallelogramme CE is equall to the square CH. But vnto the square CH is equall the square DH, by the first of the sixth: and vnto the parallelogramme CE, is equall the pa<g ref="char:EOLhyphen"/>rallelogramme HE. Wherefore the parallelogramme DH is equall to the parallelogramme HE. Adde the parallelogramme HB, common to them both. Wherefore the whole parallelo<g ref="char:EOLhyphen"/>gramme DK is equall to the whole square AE. And the parallelogramme DK, is that which is contayned vnder the lines BD and DA, for the line AD is equall to the line DL, &amp; the square AE is the square of the line AB. Wherfore that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines AD and DB is equall to the square of the line AB. Wherefore as the line DB is to the line BA, so is the line BA to the line AD, by the 17. of the sixth. But the line DB is greater then the line BA. Wherefore the line BA is greater then the line AD. Wherefore the line BD is deuided by an extreame and meane proportion in the point A, and his greater segment is the line AB. If therefore a right line be deuided by an extreame and meane proportion, and vnto <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t be added a right line, equall to the greater segment: the whole <milestone type="tcpmilestone" unit="unspecified" n="1"/> right line is deuided by an extreame and meane proportion, and the greater segment therof, is the right line geuen at the beginning: which was required to be demonstrated.</p>
               <p>
                  <hi>This proposition is agayne afterward demonstrated.</hi>
               </p>
               <div type="section">
                  <head>A Corollary added by <hi>Campane.</hi>
                  </head>
                  <p>
                     <hi>Hereby it is <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> from the grea<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> of a line deuided by an extreame &amp; meane proportion,</hi>
                     <note place="margin">
                        <hi>This is before demonstrated most euidently and briefly by M. Dee, after the</hi> 3. <hi>proposi<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> 
                     <hi>be <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> away <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> segment: the sayd great a segment shall be deuided by an extreame and meane proportion, and the greater segment thereof shall be the line taken away.</hi>
                  </p>
                  <p>As let the line <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>, be deuided by an extreame and mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportion, in the point C. And le<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> line <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> A. D. I say that A<g ref="char:EOLhyphen"/>C is also deuided by an extreame and mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proportion in the point D, and that his greater portion is DC. For, by the definiti<g ref="char:EOLhyphen"/>no (of a line so deuided) AB, is to AC, as AC is to CB. But as AC is to CB, so is AC to DC, by the 7. of the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> is equall to CB) wherefore, by the 11. of the fifth, as AB is to AC, so is AC to CD: and therefore by the 19. of the fifth, as AB is to AC, so is the residue CB, to the residue AD. But CB is to AD, as DC is to AD (by the 7. of the fifth) for DC is by construction equall to C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. Wherefore, <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> so by the definition of A line de<g ref="char:EOLunhyphen"/>uided by an extreame and meane <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the point D, to be deuided, by an ex<g ref="char:EOLhyphen"/>treame and meane proportion: which was to be proued.</p>
                  <p>
                     <hi>Two Corollaries (added by M. Dee) following chiefely vpon the veritie, and demonstration of his Additions, vnto the 3. porposition annexed, and partely vpon this fifth, by Euclide demonstrated.</hi>
                  </p>
               </div>
               <div type="section">
                  <head>A Corollary. 1.</head>
                  <p>
                     <hi>As any line being deuided by an extreame and middle proportion, doth geue vs three right lines, <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <milestone type="tcpmilestone" unit="unspecified" n="1"/> 
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <milestone type="tcpmilestone" unit="unspecified" n="2"/> 
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <milestone type="tcpmilestone" unit="unspecified" n="3"/> 
                        <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>
                        <milestone type="tcpmilestone" unit="unspecified" n="4"/> these fower wayes, we haue two line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> deuided by an extreame and meane proportion; (it is to w<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> geuen, and the other made) and with all, in euery way, we haue fower <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> continuall proportion.</hi>
                     <note place="margin">
                        <hi>Note.</hi>
                     </note>
                  </p>
                  <p>Of these two lines, (by extreame and meane proportion deuided,) their demonstrations<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are after <hi>Euclides</hi> 3, proposition added: and here in this fifth by <hi>Euclide</hi> proued. But of the fower lines in continu<g ref="char:EOLhyphen"/>all proportion, seing, the demonstration is most easy for any man to frame. I will here, but note the lines
<pb facs="tcp:7062:425"/>
v<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>to you: as in euery of the fower places, t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e constructions haue them lettred and specified. As in my <milestone type="tcpmilestone" unit="unspecified" n="1"/> first way added after the third proposition,<note place="margin">
                        <hi>Note</hi> 4. <hi>Proportional lines.</hi>
                     </note> DC, AD, AC, and CB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are fower lines in continuall propor<g ref="char:EOLhyphen"/>tion.
<figure/>
                     <milestone type="tcpmilestone" unit="unspecified" n="2"/> And in the second way, AB, CQ (equall to AC) CB, and BQ: are fower lines in continuall pro<g ref="char:EOLhyphen"/>portion. <milestone type="tcpmilestone" unit="unspecified" n="3"/> And in the third way, AB, AC, C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, (equall to CB) and AR, are the fower lines in continuall <milestone type="tcpmilestone" unit="unspecified" n="4"/> proportion. And in the fourth way (by <hi>Euclide</hi> declared) DB, AB, AC (equall to AD) and CB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are fow<g ref="char:EOLhyphen"/>er <milestone type="tcpmilestone" unit="unspecified" n="1"/> lines in continuall proportion.<note place="margin">
                        <hi>Note two mid<g ref="char:EOLhyphen"/>dle propor<g ref="char:EOLhyphen"/>tionals.</hi>
                     </note> So, that in the first way you haue AD, and AC, middle proportionals <milestone type="tcpmilestone" unit="unspecified" n="2"/> betwene DC, and CB. In the second way, you haue CQ, and CB, betwene AB, and BQ. In the third <milestone type="tcpmilestone" unit="unspecified" n="3"/> way, you haue AC, CR, betwene AB and AR: and in the fourth way, you haue AB, and AC, betwene <milestone type="tcpmilestone" unit="unspecified" n="4"/> DB, and CB.</p>
               </div>
               <div type="section">
                  <head>A Corollary. 2.</head>
                  <p>It is also manifest, that you may by any of the fower wayes, here specified, proceede infinitely in the proportion of a line deuided by extreame &amp; middle proportion:<note place="margin">Note <hi>4.</hi> wayes of progres<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ion, in the proportion of a line deuided by extreme and middle propor<g ref="char:EOLhyphen"/>tion.</note> And in the first, and fourth wayes, en<g ref="char:EOLhyphen"/>creasing continually the quantities of the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s made: but in the second and third wayes, dim<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nishing continually the quantities of the sayd whole lines, made (and thereby their segmentes). And ye<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, ne<g ref="char:EOLhyphen"/>uerthelesse reteyning in euery line made (by <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>y of the wayes) and in his segments all, and the same properties, which the first line, and his segment <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> haue. After which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>are of Progression, as the termes in continuall proportion do encrease, and are m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e in number: So, likewise, do the middle proportionalls, (accordingly) become moe: But euer fower in number, by two, then the termes of the Progression are.</p>
                  <div type="subsection">
                     <head>¶What Resolution is.</head>
                     <p>Resolution, is the assumption or taking of the thing which is to be proued, as graunted, and by thinges which necessarily follow it,<note place="margin">What resolution and composition is, hath before bene taught in the beginning of the first booke.</note> to passe vnto s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>me truth graunted.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ What Composition is.</head>
                     <p>Composition, is an assumption or taking of a thing graunted, and by thinges which of necessity follow it, to passe vnto the finding out of the thing sought or to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>Resolution of the first Theoreme.</head>
                     <p>Suppose that a certaine right line AB, be diuided by an extreame &amp; meane proportion in the point C, &amp; let the greater segme<g ref="char:cmbAbbrStroke">̄</g>t therof be AC, vnto which adde a line equal to the halfe of the line AB, and let that line be AD. Then I say that the square of the line CD is quin<g ref="char:EOLhyphen"/>tuple to the square of AD. For forasmuch as the square of the line CD is quintuple to the square of AD: but the square of the line
<figure/>
CD is (by the 4. of the second) equall to that which is composed of the squares of the lines CA, &amp; AD, together with that
<pb n="396" facs="tcp:7062:425"/>
which is contayned vnder the lines CA, and AD twise. Wh<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>f<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>re that which is composed of the squares of the lines CA, &amp; AD, together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines C<g ref="char:EOLhyphen"/>A, &amp; AD twise is quintuple to the square of the line AD. Wherfore, that which is composed of the square of the line CA together with that, which is contayned vnder the lines CA, and AD twise is quadruple to the square of the line AD. But vnto that, which is contayned vnder the lines CA, and AD, twise, is equall that which is contayned vnder the lines CA, and AB, for the line AB is double to the line AD. And vnto the square of the line A<g ref="char:EOLhyphen"/>C, is equall that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AB &amp; BC, for the line AB, is by supposi<g ref="char:EOLhyphen"/>tion diuided by an extreme and meane proportion, in the point C. Wherefore, that, which is contayned vnder the lines AB, and AC, together with that which is contayned vnder the lines AB, and BC, is quadruple to the square of the line AD. But that, which is composed of that which is contayned vnder the lines AB, and AC, together with that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines AB, and BC, is the square of the line AB (by the 2. of the second.) Wher<g ref="char:EOLhyphen"/>fore the square of the line AB is quadruple to the square of the line AD. And so is it in deede: for the line AB is double to the line AD: as was at the first supposed.</p>
                  </div>
                  <div type="subsection">
                     <head>Composition of the first Theoreme.</head>
                     <p>Now forasmuch as the square of the line AB is quadruple to the square of the line AD, but the square of the line AB, is that which is contayned vnder the lines AB, and AC, toge<g ref="char:EOLhyphen"/>ther with that, which is contayned vnder the lines BA, and BC. Wherefore that which is contayned vnder the lines BA, and AC, together with that which is contayned vnder the lines BA, and BC is quadruple to the square of the line AD. But that which is contayned vnder the lines BA, and AC, is equall to that which is contayned vnder the lines DA, and AC twise (by the 1. of the sixth), and that which is contayned vnder the lines AB, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>C is equall to the square of the line AC, by the definition of a line diuided by extreme and meane proportion. Wherefore the square of the line AC, together with that, which is contay<g ref="char:EOLhyphen"/>ned vnder the lines DA, and AC twise, is quaduple to the square of the line DA. Wherfore that which is composed of the squares of the lines DA, and AC, together with that which is contayned vnder the lines DA, and AC, twise, is quintuple to the square of the line DA. But that which is composed of the squares of the lines DA, and AC, together with that which is contayned vnder the lines DA, and AC twise, is equall to the square of the line CD (by the 4. of the second). Wherefore the square of the line CD is quintuple to the square of the line AD: which was required to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>Resolution of the 2. Theoreme.</head>
                     <p>Suppose that a certayne right line, CD, be quintuple to a segme<g ref="char:cmbAbbrStroke">̄</g>t of the same line, namely, to DA: and let the double of the line DA, be AB. The<g ref="char:cmbAbbrStroke">̄</g> I say that the line AB is diuided by an extreme and meane proportion in the point C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and the greater segme<g ref="char:cmbAbbrStroke">̄</g>t therof is AC, which is the rest of the right line put at the beginning. For forasmuch as the line AB is diuided by an extreame and meane proportion in the poynt C, and the greater segment thereof is the line AC, therefore that which is contained vnder the lines A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, and BC, is equall to the square of the line AC. But that which is contay<g ref="char:EOLhyphen"/>ned
<figure/>
vnder the lines BA, and AC, is e<g ref="char:EOLhyphen"/>qual to that which is contayned vn<g ref="char:EOLhyphen"/>der the lines DA, and AC twise<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> for the line BA, is double to the line AD. Wherefore that which is contayned vnder the lines AB, and BC together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines BA, and AC, which is the square of the line AB (by the 2. of the second) is equall to
<pb facs="tcp:7062:426"/>
that which is contayned vnder y<hi rend="sup">e</hi> lines DA, &amp; AC, twise together with the square of the line AC. But the square of the line AB, is quadruple to the square of the line DA. Wherfore that which is contayned vnder the lines DA, and AC, twise, together with the square of the line AC, is quadruple to the square of the line AD. Wherefore the squares of the lines DA, and AC, together with that which is contayned vnder the lines DA, and AC, twise, which is the square of the line DC, are quintuple to the square of the line DA. And so are they in deede by supposition.</p>
                  </div>
                  <div type="subsection">
                     <head>Composition of the 2. Theoreme.</head>
                     <p>Now forasmuch as the square of the line <hi>CD,</hi> is quintuple to the square of the line <hi>DA</hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> But the square of the line <hi>CD,</hi> is that which is co<g ref="char:cmbAbbrStroke">̄</g>posed of the squares of y<hi rend="sup">e</hi> lines <hi>DA,</hi> &amp; <hi>AC,</hi> together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines <hi>DA,</hi> &amp; <hi>AC,</hi> twise: Wherfore y<hi rend="sup">e</hi> squares of the line <hi>DA,</hi> &amp; <hi>AC</hi> together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines <hi>DA,</hi> &amp; <hi>AC,</hi> twise, are quintuple to the square of the line <hi>DA.</hi> Wherfore, by diuisio<g ref="char:cmbAbbrStroke">̄</g>, that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines <hi>DA,</hi> and <hi>AC,</hi> twise together with the square of the line <hi>CA,</hi> is quadruple to the square of the line <hi>AD.</hi> And the square of the line <hi>AB,</hi> is quadruple to the square of the line <hi>AD.</hi> Wherefore that which is contayned vnder the lines <hi>DA,</hi> and <hi>AC</hi> twise, which is that, which is contayned vnder the lines <hi>BA,</hi> and <hi>AC</hi> once, together with the square of the line <hi>AC,</hi> is equall to the square of the line <hi>AB.</hi> But the square of <hi>BA,</hi> is that which is contayned vnder the lines <hi>BA,</hi> and <hi>AC,</hi> together with that which is contayned vnder the lines <hi>BA,</hi> and <hi>BC.</hi> Wherfore that which is contayned vnder the lines <hi>BA,</hi> &amp; A<g ref="char:EOLhyphen"/>C, together with that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines <hi>AB,</hi> &amp; <hi>BC,</hi> is equall to that which is contayned vnder the lines <hi>BA,</hi> and <hi>AC</hi> together with the square of the line <hi>AC.</hi> Now then taking away that which is common to them both, namely, that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the lines <hi>BA,</hi> and <hi>AC,</hi> the residue, namely, that which is contayned vnder the lines <hi>AB, BC</hi> is equall to the square of the line <hi>AC.</hi> Wherefore as the line <hi>BA,</hi> is to the line <hi>AC,</hi> so is the line <hi>AC</hi> to the line <hi>CB.</hi> But the line <hi>BA</hi> is greater then the line <hi>AC,</hi> wherefore the line <hi>AC</hi> also is greater then the line <hi>CB.</hi> Wherefore the line <hi>AB</hi> is diuided by an ex<g ref="char:EOLhyphen"/>treame, and meane proportion in the poynt <hi>C,</hi> and the greater segment thereof is the line <hi>AC,</hi> which was required to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>Resolution of the 3. Theoreme.</head>
                     <p>Suppose that a certayne right line <hi>AB,</hi> be diuided by an extreame, and meane propor<g ref="char:EOLhyphen"/>tion in the point <hi>C</hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and let the greater segment thereof be the line <hi>AC,</hi> and let the halfe of the line <hi>AC,</hi> be the line <hi>CD.</hi> Then I say that the square of the <hi>BD</hi> is quintuple to the square of the line CD. For forasmuch as the square of the line <hi>BD,</hi> is quintuple to the square of the line <hi>CD.</hi> But the square of the line <hi>DB,</hi> is that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi> together with the square of the line <hi>DC</hi> (by the 6. of the second). Wherefore that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi>
                        <figure/>
together with the square of the line <hi>DC,</hi> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> quintuple to the square of the line <hi>DC.</hi> Wherefore, that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi> is quadruple to the square of the line <hi>DC.</hi> But vnto that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi> is equall the square of the line <hi>AC</hi>: for the line <hi>AB,</hi> is diuided by an extreame and meane proportion in the point <hi>C.</hi> Wherefore the square of the line <hi>AC</hi> is quadruple to the square of the line <hi>DC</hi>: and so is it in deede, for the line <hi>AC</hi> is double to the line <hi>DC.</hi>
                     </p>
                  </div>
                  <div type="subsection">
                     <pb n="397" facs="tcp:7062:426"/>
                     <head>Composition of the 3. Theoreme.</head>
                     <p>Forasmuch as the line <hi>AC</hi> is double to the line <hi>DC,</hi> therefore the square of the line <hi>AC</hi> is quadruple to the square of the line <hi>DC</hi> (by the 20. of the sixth). But vnto the square of the line <hi>AC,</hi> is equall that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi> by supposi<g ref="char:EOLhyphen"/>tion: wherefore that which is contayned vnder the lines <hi>AB,</hi> and <hi>BC,</hi> is quadruple to the square of the line <hi>CD.</hi> Wherefore, that which is contayned vnder <hi>AB,</hi> and <hi>BC,</hi> to<g ref="char:EOLhyphen"/>ther with the square the line <hi>DC,</hi> which is the square of the line <hi>DB</hi> (by the 6. of the se<g ref="char:EOLhyphen"/>cond) is quintuple to the square of the line <hi>DC</hi>: which was required to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>Resolution of the 4. Theoreme.</head>
                     <p>Suppose that a certayne right line AB, be diuided by an extreme and meane proportion in the point C. And let the greater segment thereof be AC. Then I say that the squares of the lines AB, and BC, are treble to the square of the line AC. For forasmuch as the squares of the lines AB, and BC, are treble to the square of the line AC, but the squares of the lines AB, and BC, are that which is contayned vnder AB, and BC, twise together with the square of the line AC (by the 7. of the second). Wherefore that which is contayned vnder the lines AB, and BC, twise, together with the square
<figure/>
of the line AC, is treble to the square of the line AC. Wherefore, that which is contayned vnder the lines AB, &amp; BC, twise, is double to the square of the line AC. Wherefore that which is contayned vnder the lines AB, and BC, once, is equall to the square of the line AC. And so it is in deede. For the line AB is diuided by an extreme, and meane proportion in the point C.</p>
                  </div>
                  <div type="subsection">
                     <head>Composition of the 4. Theoreme.</head>
                     <p>Forasmuch therefore as the line AB, is diuided by an extreme and meane proportion in the poynt C, and the greater segment thereof is the line AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> therfore that which is contay<g ref="char:EOLhyphen"/>ned vnder the lines AB, and BC is equall to the square of the line AC. Wherfore that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AB, and BC twise is double to the square of AC. Wherfore that which is contayned vnder the lines AB, and BC, twise, together with the square of the line AC, is treble to the square of the line AC. But that which is contayned vnder the lines A<g ref="char:EOLhyphen"/>B, and BC, twise, together with the square of the line AC, is the squares of the lines AB, and BC (by the 7. of the second). Wherefore the squares of the lines AB, and BC, are treble to the square of the line AC: which was required to be demonstrated.</p>
                  </div>
                  <div type="subsection">
                     <head>Resolution of the 5. Theoreme.</head>
                     <p>Suppose that a certaine right line AB, be diuided by an extreme and meane proportion in the point C. And let the greater segment therof be the line AC. And vnto the line AB, adde a line equall to the line AC, and let the same be AD. The<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> I say that the line D<g ref="char:EOLhyphen"/>B, is diuided by an extreme and meane proportion in the point A. And the greater segment therof is the line AB. For forasmuch as the line DB is diuided by an extreme &amp; meane pro<g ref="char:EOLhyphen"/>portion in the point A, and the greater segment thereof is the line AB, therfore as the line DB, is to the line BA, so is the line B<g ref="char:EOLhyphen"/>A,
<figure/>
to the line AD: but the line AD, is equall to the line AC: wherefore as
<pb facs="tcp:7062:427"/>
the line DB, is to the line BA, so is the line BA to the line AC. Wherfore by conuersio<g ref="char:cmbAbbrStroke">̄</g> as the line BD is to the line DA, so is the line AB to the line BC (by the corollary of the 19. of the fifth): wherfore by diuision, by the 17. of the fifth, as the line BA, is to the line AD, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o is the line AC, to the line CB. But the line AD is equall to the line AC. Wherfore as the line BA, is to the line AC, so is the line AC to the line CB. And so it is indeede, for the line AB is, by supposition, diuided by an extreme and meane proportion in the point C.</p>
                  </div>
                  <div type="subsection">
                     <head>Composition of the 5. Theoreme.</head>
                     <p>Now forasmuch as the line AB, is diuided by an extreme and meane proportion in the point C: therefore as the line BA is to the line AC, so is the line AC to the line CB: but the line AC is equall to the line AD. Wherefore as the line BA is to the line AD, so is the line AC to the line CB. Wherfore by composition (by the 18. of the fifth) as the line BD is to the line DA, so is the line AB to the line BC. Wherefore by conuersion (by the corollary of the 19. of the fiueth) as the line DB is to the line BA, so is the line BA to the line AC: but the line AC is equall to the line AD. Wherefore as the line DB is to the line BA, so is the line BA to the line AC. Wherfore the line DB, is deuided by an extreme and meane proportion in the point A: and his greater segment is the line AB: which was required to be demon<g ref="char:EOLhyphen"/>strated.</p>
                  </div>
                  <div type="subsection">
                     <head>An Aduise, by Iohn Dee, added.</head>
                     <p>SEing, it is doubteles, that this parcel of <hi>Resolution and Composition,</hi> is not of <hi>Euclides</hi> doyng: it can not <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ustly be imputed to <hi>Euclide,</hi> that he hath, therby, eyther superfluitie or any part disproportioned in his whole <hi>Composition Elementall.</hi> And though, for one thing, one good demonstration well suffiseth: for stablishing of the veritie: yet, o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> one thing diuersly demonstrated: to the diligent examiner of the diuerse meanes, by which, that varietie ariseth, doth grow good occasions of inuenting demon<g ref="char:EOLhyphen"/>strations, where matter is more straunge, harde, and barren. Also, though resolution were not in all <hi>Euclide</hi> before vsed: yet thankes are to be geuen to the Greke Scholic writter, who did leaue both the definition, and also, so short and easy examples of a Method, so auncient, and so profitable. The antiquity of it, is aboue 2000. yeares: it is to we<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e, euer since Plato his time, and the profite, therof so great, that thus I finde in the Greeke recorded.<note n="*" place="margin">
                           <hi>Proclus in the Greeke: in the</hi> 58. <hi>page.</hi>
                        </note> 
                        <gap reason="foreign">
                           <desc>〈 in non-Latin alphabet 〉</desc>
                        </gap>. Proclus hauing spoken of some by nature, excellent in inuenting demonstrations, pithy and breif sayeth:
<q>Yet are there Methods geuen [for that purpose]. And in dede, that, the best, which, by Resolution, reduceth the thing inquired of, to an vndoubted principle. Which Method, Plato, taught Leodamas (as i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> re<g ref="char:EOLhyphen"/>ported)</q>
And he is registred, thereby, to haue bene the inuenter of many things in Geometry.</p>
                     <p>And, verely, in Problemes, it is the chief ayde for winning and ordring a demonstration: first by Supposition, of the thing inquired of, to be done: by due and orderly Resolution to bring it to a stay, at an vndoubted veritie. In which point of Art, great abundance of examples, are to be seen, in that excellent and mighty Mathematici<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n, Archimedes: &amp; in his expositor, Eutocius, in Menaechmus like<g ref="char:EOLhyphen"/>wise: and in Diocles booke, de Pyti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s: and in many other. And now, for as much as, our <hi>Euclide</hi> in the last six Propositions of this thirtenth booke propoundeth, and concludeth those Problemes, which were the ende, Scope, and principall purpose, to which all the premisses of the 12. bookes, and the rest of this thirtenth, are directed and ordered: It shall be artificially done, and to a great commodity, by Resolution, backward, from these 6. Problemes, to returne to the first definition of the first booke: I meane, to the definition of a point. Which, is nothing hard to do. And I do counsaile all such, as desire to attein<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, to the profound knowledge of Geometrie, Arithmeticke, or any braunche of the sciences Mathematicall, so by Resolution, (discreatly and aduisedly) to resolue, vnlose, vnioynt and disseauer euery part of any worke Mathematicall, that, therby<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> aswell, the due placing of euery verity, and his proofe: as also, what is either superfluous, or wanting, may euidently appeare. For so to inuent, &amp; there with to order their writings, was the custome of them, who in the old time, were most excellent. And I (for my part) in writing any Mathematicall conclusion, which requireth great discourse, at length haue found, (by experience) the commoditie of it, such: that to do other wayes, were to me a confu<g ref="char:EOLhyphen"/>sion, and an vnmethodicall heaping of matter together: besides the difficulty of inuenting the matter to be disposed and ordred. I haue occasion, thus to geue you friendely aduise, for your be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ofe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> because some, of late, haue inueyed against <hi>Euclide,</hi> or <hi>Theon</hi> in this place, otherwise than I would wish they had.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <pb n="398" facs="tcp:7062:427"/>
               <head>The 6. Theoreme. The 6. Proposition<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> If a rationall right line be diuided by an extreme and meane proportion: eyther of the segments, is an irrationall line of that kinde, which is called a residuall line.</head>
               <p>SVppose that AB, beyng a rationall line be deuidedly <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> extreme and meane proportion in the point C, and let the greater segment thereof be AC. Then I say that eyther of the lines AC, and CB, is an irrationall line of that kinde, which is called a residuall. Extend the line AB, to the point D:<note place="margin">Construction.</note> and let the line AD, be equall to halfe of the line AB.
<figure/>
Now forasmuch as the right line AB,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> is diuided by an extreme &amp; meane propor<g ref="char:EOLhyphen"/>tion in the point C, and vnto the greater segme<g ref="char:cmbAbbrStroke">̄</g>t AC is added a line AD, equall to the halfe of the right line AB: therfore (by the 1. of the thirtenth) the square of the line CD, is quintuple to the square of the line AD. Where<g ref="char:EOLhyphen"/>fore the square of the line CD, hath to the square of the line AD, that proportion that nu<g ref="char:cmbAbbrStroke">̄</g>ber hath to nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore the square of the line CD is comme<g ref="char:cmbAbbrStroke">̄</g>surable to the square of the line AD. But the square of the line DA is rationall, for the line DA is rationall; forasmuch as it is the halfe of the rationall line AB. Wherefore the square of the line CD, is rationall. Wherefore also the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CD, is rationall. And forasmuch as the square of the line CD; hath not to the square of the line AD, that proportion that a square number hath to a square number, therfore (by the 9. of the tenth) the line CD, is incommensurable in length, to the line AD. Wherefore the lines CD, and DA are rationall commensurable in power only. Wherfore the line AC is a residuall line, by the 73. of the tenth. Againe, forasmuch as the line AB, is deuided by an extreme and meane proportion, and the greater segment thereof is AC, therfore that which is co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the lines AB, &amp;. BC, is equall to the square of the line AC. Wherefore the square of the line AC, applyed to the rationall line AB, maketh the bredth BC. But the square of a residuall line, applied to a rationall line maketh the bredth a first residuall line (by the 97. of the tenth). Wherefore the line CB, is a first residuall line. And it is proued that the line AC, is also a residuall line. If therefore a rationall right line be diuided by an extreme and mean<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proportion, either of the segments, is an irrationall line of that kinde, which is called a residuall line, which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶A Corollary added by <hi>Campane.</hi>
                  </head>
                  <p>Hereby it is manifest, that if the greater segment be a rationall line: the lesse segment shalbe a resi<g ref="char:EOLhyphen"/>duall line.</p>
                  <p>For if the greater segment AC, of the right line ACB be diuided into two equall partes in the point D, the square of the line DB, shalbe quint<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>pl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the square of the line DC, (by the 3. of this booke.) And forasmuch as the line CD, (beyng the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>alfe of the
<figure/>
rationall line supposed AC) is rationall by the 6. diffini<g ref="char:EOLhyphen"/>tion of the tenth: And vnto the square of the line DC, the square of the line DB is commensurable, (for it is quintuple vnto it) wherfore the square of the line DB, is rationall. Wherfore also the line DB is rationall. And forasmuch as the squares of the lines DB &amp; DC, are not in proportion, as a square nu<g ref="char:cmbAbbrStroke">̄</g>ber is to a square number: therefore the lines DB and DC are incommensurable in length (by the 9. of the tenth.) Wherefore they are commensurable in power only: Wherefore by the 73. of the tenth, the line BC, which is the lesse segment, is a residuall line.</p>
               </div>
            </div>
            <div type="part">
               <head>The 7. Theoreme. The 7. Proposition. If an equilater Pe<g ref="char:cmbAbbrStroke">̄</g>tagon haue three of his angles, whether they follow in or<g ref="char:EOLhyphen"/>der,
<pb facs="tcp:7062:428"/>
or not in order, equall the one to the other: that Pentagon shalbe equi<g ref="char:EOLhyphen"/>angle.</head>
               <p>SVppose that ABCDE, be an equilater pentagon. And let the angles of the sayd Pentagon,<note place="margin">Two cases in this proposi<g ref="char:EOLhyphen"/>tion.</note> namely, first, three angles folowing in order, which are at the points A, B, C, be equal the one to the other. Then I say that the Pentagon ABC<g ref="char:EOLhyphen"/>DE is equiangle.<note place="margin">Construction.</note> Draw these right lines AC, BE, and FD. Now forasmuch as these two lines CB,<note place="margin">Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> first case.</note> and BA, are equall to these two lines BA, and AE, the one to the other, and the angle CBA is equall to the angle BAE:<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore (by the 4. of the first) the base A<g ref="char:EOLhyphen"/>C is equall to the base BE, and the triangle ABC is equall to the triangle ABE, and the rest of the angles are equal to the rest of the angles, vnder
<figure/>
which are subtended equall sides. Wherefore the an<g ref="char:EOLhyphen"/>gle BCA is equall to the angle BEA, and the angle ABE to the angle CAB. Wherefore also the side A<g ref="char:EOLhyphen"/>F, is equall to the side BF (by the 6. of the first). And it was proued that the whole line AC is equal to the whole line BE. Wherefore the residue CF is equall to the residue F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And the line CD, is equall to the line DE. Wherefore these two lines FC, and CD are e<g ref="char:EOLhyphen"/>quall to these two lines FE, &amp; ED, and the base F<g ref="char:EOLhyphen"/>D, is common to them both. Wherefore the angle F<g ref="char:EOLhyphen"/>CD, is equall to the angle FED (by the 8. of the first). And it is proued that the angle BCA, is equal to the angle AEB. Wherefore the whole angle BCD is equall to the whole angle AED. But the angle BCD, is supposed to be equall to the angles A, and B. Wherefore the angle AE<g ref="char:EOLhyphen"/>D, is equall to the angles A and B. In like sort also may we proue that the angle CDE, is equall to the angles A, and B. Wherefore the Pentagon ABCDE is equiangle.</p>
               <p>
                  <note place="margin">The second case.</note>But now supp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se that three angles, which folow not in order, be equall the one to the o<g ref="char:EOLhyphen"/>ther, namely, let the angles A, C, D, be equall. Then I say that in this case also the Pentagon ABCDE is equiangle. Draw a right line from the point B, to the point D. Now forasmuch as these two lines BA, and A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, are equall to these two lines BC, and CD, and they compre<g ref="char:EOLhyphen"/>hende equall angles, therefore (by the 4. of the first) the base BE, is equall to the base BD. And the triangle ABE, is equall to the triangle BDC, and the rest of the angles are equall to the rest of the angles, vnder which are subtended equall sides: wherefore the angle AEB, is equall to the angle CDB. And the angle BED, is equall to the angle BDE, (by the 5. of the first) for the side BE, is equall to the side BD. Wherefore the whole angle AED, is equal to the whole angle CDE. But the angle CDE, is supposed to be equall to the angles A, and C. Wherefore the AED, is equall to the angles A, and C. And by the same reason also the an<g ref="char:EOLhyphen"/>gle ABC, is equall to the angles A; C, and D. Wherefore the Pentagon ABCDE is equian<g ref="char:EOLhyphen"/>gle. If therefore an equilater Pentagon haue three of his angles, whither they follow in order, or not in order, equall the one to the other: that Pentagon shalbe equiangle: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <div type="part">
               <head>The 8. Probleme. The 8. Proposition.
If in an equilater &amp; equiangle Pe<g ref="char:cmbAbbrStroke">̄</g>tagon two right lines do subtend two of the angles following in order: those lines doo diuide the one the other by an extreme and meane proportion: and the greater segments of those lines are ech equall to the side of the Pentagon.</head>
               <p>
                  <pb n="399" facs="tcp:7062:428"/>SVppose that ABCDE be an equilater and equiangle Pentagon. And let two right lines AC, and BE, subtend the two angles A, and B, which follow in order. And let them cut the one the other in the point H. Then I say that either of those lines is diuided by an extreme &amp; meane proportio<g ref="char:cmbAbbrStroke">̄</g> in the point H: And that eche of the greater segments of those lines are equal to the side of the Pentagon Circumscribe (by the 14. of the fourth) about the Pentago<g ref="char:cmbAbbrStroke">̄</g>. ABCDE,<note place="margin">Construction.</note> a circle ABCDE. And forasmuch as these two right lines EA, and AB, are equall to these two right lines AB, and BC,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and they contayne equall angles: therefore (by the 4. of
<figure/>
the first) the base BE, is equal to the base AC: and the triangle ABE, is equall to the triangle ABC, and the angles remayning, are equall to the angles remay<g ref="char:EOLhyphen"/>nyng, the one to the other, vnder which are sub<g ref="char:EOLhyphen"/>tended equall sides. Wherefore the angle BAC, is equall to the angle ABE. Wherfore the angle AHE is double to the angle BAH, (by the 32. of the first) for it is an outward angle of the triangle ABH. And the angle EAC is double to the angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BAC (by the last of the sixth). For the circumference EDC is dou<g ref="char:EOLhyphen"/>ble to the circumference CB. Wherefore the angle H<g ref="char:EOLhyphen"/>AE is equall to the angle AHE. Wherefore also the right line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E, is (by the 6. of the first) equall to the right line EA, that is to the line AB. And forasmuch as the right line BA is equall to the right line AE, therefore the angle ABE is equall to the angle AEB. But it is proued that the angle ABE is equal to the angle BAH: wherefore also the angle BEA is equall to the angle BAH. And in the two triangles ABE, and ABH, the angle ABE is common to them both, wherefore the angle remayning, namely, BAE is equall to the angle remayning, namely, to AHB (by the corollary of the 32. of the first). Wherfore the triangle ABE, is e<g ref="char:EOLhyphen"/>quiangle to the triangle ABH. Wherefore proportionally as the line EB, is to the line BA, so is the line AB to the line BH (by the 4. of the sixth). But the line BA is equall to the line EH. Wherefore as the line B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the line EH, so is the line EH to the line H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. But the line BE is greater the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line BA<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore the line EH, also is greater then the line H<g ref="char:EOLhyphen"/>B. Wherefore the line BE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is diuided by an extreme and meane proportion in the point H (by the 3. diffinition of the sixth) and his greater segment EH is equall to the side of the Penta<g ref="char:EOLhyphen"/>gon. In like sort also may we proue that the line AC is diuided by an extreme and meane pro<g ref="char:EOLhyphen"/>portion in the point H, and that his greater segment CH, is equall to the side of the Penta<g ref="char:EOLhyphen"/>gon. (For the whole line AC is equall to the whole line BE, and it hath bene proued that the parts taken away BH, and AH are equall wherfore the residue CH is equall to the residue EH (by the 10. of the fifth). If therefore in an equilater and equiangle Pentagon two right lines <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>end two of the angles following in order: those lines doo diuide the one the o<g ref="char:EOLhyphen"/>ther by an extreme and meane proportion: and the greater segments of those lines are eche equal to the side of the Pentagon: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Theoreme. The 9. Proposition. If the side of an equilater hexagon, and the side of an equilater decagon or <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>gled figure, which both are inscribed in one &amp; the selfe same circle, be added together: the whole right line made of them is a line diuided by a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> extreame and meane proportion, and the greater segment of the same is
<pb facs="tcp:7062:429"/>
the side of the hexagon.</head>
               <p>SVppose that there be a circle ABC. And let the side of a decagon or tenangled figure inscribed in the circle ABC, be BC, and let the side of an hexagon or sixe angled figure inscribed in the same circle, be CD. And let the lines BC and CD be so ioyned together directly that they both make one right line, namely, BD. Then I say that the line BD is diuided by an extreame
<figure/>
and meane proportion in the point C: and that the greater segme<g ref="char:cmbAbbrStroke">̄</g>t therof is the line CD.<note place="margin">Construction.</note> Take (by the 1. of the third) the centre of the circle. And let it be the point E: and draw these right lines EB, EC, and ED. And extend the line BE to the point A. Now forasmuch as BC is the side of an equilater decagon,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> therefore the circumference or semicir<g ref="char:EOLhyphen"/>cle ACB is quintuple to the circumference CB. Where<g ref="char:EOLhyphen"/>fore the circumference AC is quadruple to the circumfe<g ref="char:EOLhyphen"/>rence CB. But as the circumference AC is to the circum<g ref="char:EOLhyphen"/>ference CB, so is the angle AEC to the angle CEB, by the last of the sixth. Wherefore the angle AEC is quadruple to the angle CEB. And forasmuch as the angle EBC is equall to the angle ECB (by the 5. of the first,) for the line EB is equall to the line EC, by the diffinition of a circle, therefore the angle AEC is double to the angle ECB, by the 32. of the first. And forasmuch as the right line EC is equall to the right line CD, by the corollary of the 15. of the fourth (for either of them is equall to the side of the hexagon inscribed in the circle ABC) therefore the angle CED is equall to the angle CDE wherefore the angle ECB is double to the angle EDC, by the 32. of the first. But it is proued that the angle AEC is double to the angle ECB, wherefore the angle AEC is qua<g ref="char:EOLhyphen"/>druple to the angle EDC. And it is proued that the angle AEC is quadruple to the angle BEC. Wherefore the angle EDC is equall to the angle BEC. And the angle EBD is com<g ref="char:EOLhyphen"/>mon to the two triangles BEC and BED: wherefore the angle remayning BED is equall to the angle remayning ECB, by the corollary of the 32. of the first. Wherefore the triangle EBD is equiangle to the triangle EBC. Wherfore, by the 4. of the sixt, proportionally, as the line BD is to the line BE, so is the line BE to the line BC. But the line EB is equall to the line CD. Wherefore as the line BD is to the DC, so is the line DC to the line CB. But the line BD is greater then the line DC: wherefore also the line DC is greater then the line C<g ref="char:EOLhyphen"/>B. Wherefore the right line BD is diuided by an extreame and meane proportion in the point C: and his greater segment is DC. If therefore the side of an equilater hexagon, and the side of an equilater decagon or tenangled figure, which both are inscribed in one and the selfe same circle, be added together<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the whole right line made of them, is a line diuided by an extreame and meane proportion, and the greater segment of the same is the side of the hexa<g ref="char:EOLhyphen"/>gon: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <note place="margin">
                        <hi>This Corollary is the</hi> 3. <hi>propo<g ref="char:EOLhyphen"/>sition of the</hi> 
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>4. <hi>booke af<g ref="char:EOLhyphen"/>ter Campane.</hi>
                     </note>Hereby it is manifest, that the side of an exagon inscribed in a circle being cut by an extreame and meane proportion, the greater segment thereof is the side of the decagon inscribed in the same circle. For if from the right line DC be cut of a right line equall to the line CB, we may thus reason, as the whole DB is to the whole DC, so is the part taken away DC to the part taken away CB: wherefore by the 19. of the fifth, the residue is to the residue as the whole is to the whole. Wherefore the line D<g ref="char:EOLhyphen"/>C is cutte like vnto the line DB: and therefore is cut by an extreame and meane proportion.</p>
               </div>
               <div type="section">
                  <pb n="400" facs="tcp:7062:429"/>
                  <head>
                     <hi>Campane</hi> putteth the conuerse of this proposition after this maner.</head>
                  <p>
                     <hi>If a line be diuided by an <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ane proportion, of <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> circle the greater segment is the side of an equilater Hexagon, of the same shall the lesse segment be the side of an equilater Decagon. And of what circle the lesse segment is the side of an equilater Decagon, of the same is the greater seg<g ref="char:EOLhyphen"/>ment the side of an equilater Hexagon.</hi>
                  </p>
                  <p>For the former figure remayning, suppose that the line BD be diuided by an extreme and meane proportion in the point C: and let the greater segment therof be DC. The<g ref="char:cmbAbbrStroke">̄</g> I say that of what circle the line DC is the side of an equilater Hexagon, of the same circle is the line CB the side of an equi<g ref="char:EOLhyphen"/>later decagon: and of what circle, the line BC is the side of an equilater Decagon, of the same is the line DC the side of an equilater Hexagon.<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</hi>
                     </note> For if the line
<figure/>
DC be the side of an Hexagon inscribed in the circle, then by the corollary of the 15. of the fourth, the line DC is e<g ref="char:EOLhyphen"/>qual to the line BE. And forasmuch as the proportio<g ref="char:cmbAbbrStroke">̄</g> of the line BD to the line DC is as the proportion of the line D<g ref="char:EOLhyphen"/>C to the line CB, by supposition: therfore (by the 7. of the fifth) the proportion of the line BD to the line BE, is as the proportion of the line BE to the line BC. Wherefore (by the 6. of the sixth) the two triangles DEB, EBC are equiangle (for the angle B is common to eche triangle). Wherefore the angle D is equall to the angle CEB: for they are subtended of sides of like proportion. And foras<g ref="char:EOLhyphen"/>much as the angle AEC is quadruple to the angle D (by the 32. of the first twise taken, and by the 5. of the same) therefore the same angle AEC is quadruple to the angle CEB. Wherefore (by the last of the sixth) the circumfe<g ref="char:EOLhyphen"/>rence AC is quadruple to to the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce CB. Wher<g ref="char:EOLhyphen"/>fore the line BC is the side of a decagon inscribed in the circle ACB.</p>
                  <p>But now if the line BC be the side of a decagon in<g ref="char:EOLhyphen"/>scribed in the circle ABC, the line CD shalbe the side of an Hexagon inscribed in the same circle.<note place="margin">
                        <hi>Demonstrati<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond part.</hi>
                     </note> For let DC be the side of an Hexagon inscribed in the circle H. Now by the first part of this proposition the line BC shalbe the side of a decagon inscribed in the same circle. Suppose that in the two circles AC<g ref="char:EOLhyphen"/>B and H be inscribed equilater decagons, all whose sides shalbe equall to the line CB. And forasmuch as euery equilater figure inscribed in a circle is also equiangle, therefore bothe the decagons are equi<g ref="char:EOLhyphen"/>angle. And forasmuch as al the angles of the one taken to<g ref="char:EOLhyphen"/>gether
<figure/>
are equall to al the angles of the other taken tog<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>ther, as it is easy to be proued by that which is added af<g ref="char:EOLhyphen"/>ter the 32. of the first, therefore one of these decagons is equiangle to the other: and therfore the one is like to the other by the diffinition of like superficieces. And for that if there be two like rectiline figures inscribed in two cir<g ref="char:EOLhyphen"/>cles, the proportion of the sides of like proportion of those figures, shalbe as the proportion of the Diameters of those circles, as it is easy to proue by the corollary of the 20. of the sixth, and first of this book<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>: but the sides of the like deca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ons inscribed in the two circles ABC and H are equall: therefore theyr Diameters also are equall. Wherefore also theyr semidiameters are equall. But the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>midi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>meters and the side of the Hex<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>gon are equall, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ollary of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the fourth. Wherefore the line DC is the side of an hexagon in the circle ABC, as also it is the side of an hexagon inscribed in the circle F, which is equall to the ci<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cle ABC: which was required to be proued.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 10. Theoreme. The 10. Proposition. If in a circle be described an equilater Pentagon, the side of the Pentagon containeth in power both the side of an hexagon and the side of a decagon,
<pb facs="tcp:7062:430"/>
being all described in one and the selfe same circle.</head>
               <p>SVppose that ABCDE be a circle. And in the circle ABCDE describe (by the 11. of the fourth) a pentagon figure ABCDE. Then I say, that the side of the pentagon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure ABCDE containeth in power both the side of an hexagon figure, and of a decagon figure, being described in the circle ABCDE.<note place="margin">Construction.</note> Take (by the 1. of the third) the centre of the circle, and let the same be F. And drawing a right line from the point A to the point F, extend it to the point G. And drawe a right line from the point F to the point B. And from the point F drawe (by the 12. of the first) vnto the line AB a perpendicular line FH: and extend it to the point K. And drawe the right lines AK and KB. And againe, from the point F draw (by the same) vnto the line AK, a perpendi<g ref="char:EOLhyphen"/>cular line FN: and extend FN to the point M, which line let cut the line AB in the point L. And draw a right line from the
<figure/>
point K to the point L. Now foras<g ref="char:EOLhyphen"/>much as the circumference ABCG is equall to the circumference AEDG,<note place="margin">Dem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nstra<g ref="char:EOLhyphen"/>tion.</note> of which the circumference ABC is equall to the circumference AED, therfore the rest of the circumference, namely, CG, is equall to the rest of the circumference, namely, to DG. But the circum<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>en<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e CD is subtended of the side of a p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>agon: Wherefore the circumference CG is subtended of the side of a decagon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure. And foras<g ref="char:EOLhyphen"/>much as the line BH is equall to the line HA (by the 3. of the third) and the line FH is common to them both, &amp; the angles at the point H are right angles. Wherefore (by the 4. of the first) the angle AFK is equall to the angle KFB. Wherefore also (by the 26. of the third) the circumference AK, is equall to the circumference KB. Wherefore the circumference AB is double to the circum<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>erence BK. Wherefore the right line AK is the side of a decagon fi<g ref="char:EOLhyphen"/>gure. And by the same reason also the circumference AK is double to the circumference KM. And forasmuch as the circumference AB is double to the circumference BK, but the cir<g ref="char:EOLhyphen"/>cumference CD is equall to the circumference AB: wherefore the circumference CD is double to the circumference BK. But the circumference CD is double to the circumference CG: wherefore the circumference CG is equall to the circumference BK. But the circum<g ref="char:EOLhyphen"/>ference BK is double to the circumference KM (for that the circumference KA is double thereunto) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore also the circumference CG is double to the circumference KM. But the circumference CB is also double to the circumference BK (for the circumference CB is equall to the circum<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>erence BA). Wherefore the whole circumference GB is double to the whole circumference BM, by the 12. of the fift. Wherefore also the angle GFB is double to the angle BFM, by the last of the sixt. But the angle GFB is double to the angle FAB (by the 32. of the first, or 20. of the third). For the angle FAB is equall to the angle ABF. Wherefore the angle BFL is equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the angle FAB, by the 15. of the fift. And the angle ABF is common to the two triangles ABF and BFL. Wherefore (by the Corollary of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. of the first) the angle remayning AFB is equall to the angle remayning BLF. Where<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the triangle ABF is equiangle to the triangle BFL. Wherefore (by the 4. of the sixt) proportionally as the right line AB is to the right line BF, so is the same right line FB to
<pb n="401" facs="tcp:7062:430"/>
the right line BL: wherefore that which is contained vnder the lines AB and BL is equall to the square of the line BF, by the 17. of the sixt. Againe forasmuch as the line AL is equall to the line LK [for by the last of the sixt, the angle KFL is equall to the angle AFL, which equall angles are contained vnder the lines FK, FL, and FA, FL, &amp; the line FK is equall to the line FA, and the line FL is common to them both. Wherefore, by the 4. of the first, the line AL is equall to the line LK] and the line LN is common to them both, &amp; maketh right angles at the point N, and (by the 3. of the third) the base AN is equall to the base KN. Wherfore also the angle LKN is equall to the angle LAN. But the angle LAN is equall to the angle KBL (by the 5. of the first). Wherefore the angle LKN is equall to the angle KBL. And the angle KAL is common to both the triangles AKB and AKL. Wherefore the angle remaining AKB is equall to the angle remayning ALK (by the Corol<g ref="char:EOLhyphen"/>lary of the 32. of the first). Wherefore the triangle KBA is equiangle to the triangle KLA. Wherefore (by the 4. of the sixt) proportionally as the right line BA is to the right line AK, so is the same right line KA to the right line AL. Wherefore that which is contained vnder the lines BA and AL is equall to the square of the line AK (by the 17. of the sixt). And it is proued that that which is contained vnder the lines AB and BL, is equall to the square of the line BF. Wherefore that which is contained vnder AB &amp; BL together with that which is con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ined vnder BA and AL (which by the 2. of the second, is the square of the line BA) is equall to the squares of the lines AF and AK. But the line BA is the side of the pentagon figure, and AF the side of the hexagon figure (by the Corollary of the 15. of the fourth), and AK the side of the decagon figure. Wherefore the side of a pentagon figure, containeth in power both the side of an hexagon figure, and of a decagon figure, being described all in one and the selfe same circle: which was required to be demonstrated.</p>
               <div type="section">
                  <head>¶A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>A perpendicular line from any angle drawen to the base of a pentagon, passeth by the centre.</hi>
                  </p>
                  <p>For if we drawe a right line from the poynt A to the poynt C, and an other from the poynt A to the point D: those right lines sh<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>l be equall, by the 4. of the first: and therefore in the triangle ACD the angles at the points C and D are, by the <hi>5.</hi> of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall. But the angles made at the point where the line AG cutteth the line CD, are by supposition right angles: wherefore, by the 26. of the first, the line CD is by the line AG diuided into two equall partes, and it is also diuided perpendicularly: wherefore by the corollary of the first of the third in the line AG is the centre of the circle: and there<g ref="char:EOLhyphen"/>fore the line AG passeth by the centre.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 11. Theoreme. The 11. Proposition. If in a circle hauing a rationall line to his diameter be inscribed an equila<g ref="char:EOLhyphen"/>ter pentagon: the side of the pentagon is an irrationall line, and is of that kinde which is called a lesse line.</head>
               <p>SVppose that in the circle ABCDE hauing a rationall line to his diameter be inscribed a pentagon figure ABCDE. Then I say that the side of the pentagon figure ABCDE, namely, the side AB, is an irrational line of that kinde which is called a lesse line. Take (by the 1. of the third) the centre of the circle, and let the same be the point F, and draw a right line from the point A to the point F,<note place="margin">Construstion.</note> and an other from the point F to the point B, and extend those lines to the pointes G and H. And draw a right line from the point A to the point C. And from the semidiameter FH take the fourth part (by the 9. of the sixt) and let the same be FK: But the line FH is rationall (for that it is the halfe of the diameter which is supposed to be rationall), wherefore also the
<pb facs="tcp:7062:431"/>
line FK is rationall. And the line or semidiameter BF is rationall. Wherefore the whole line BK is rationall.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as the circumference ACG is equall to the circumference ADG, of which the circumference ABC is equall to the circumference AED, wherefore the residue CG is equall to the residue GD. Now if we drawe a right line from the point A to the point D, it is manifest that the angles ALC and ALD are right angles. For foras<g ref="char:EOLhyphen"/>much as the circumference CG is equall to the circumference GD, therefore (by the last of the sixth) the angle CAG is equall to the angle DAG. And the line AC is equall to the line AD, for that the circumferences which they subtend are equall, and the line AL is c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>mmon to them both, therefore there are two lines AC and AL equall to two lines AD and AL, and the angle CAL is equall to the angle DAL. Wherefore (by the 4. of the first) the base CL is equall to the base LD, and the rest of the angles to the rest of the angles, and the line CD is double to the line CL. And by the same reason may it be proued, that the an<g ref="char:EOLhyphen"/>gles at the point M are right angles, and that the line AC is double to the line CM. Now for<g ref="char:EOLhyphen"/>asmuch as the angle ALC is equall to the angle AMF, for that they are both right angles, and the angle LAC is common to both the triangles ALC and AMF: wherefore the an<g ref="char:EOLhyphen"/>gle remayning, namely, ACL, is equal to the angle remayning AFM, by the corollary of the 32. of the first. Wherefore the triangle ACL is equiangle to the triangle AMF. Wherefore proportionally, by the 4. of the sixth, as the line LC is to the line CA, so is the line MF to the line FA. And in the same proportion also are the doubles of the antecedents LC and MF (by the 15. of the fifth). Wherefore as the double of the line LC is to the line CA, so is the double of the line MF to the line FA. But as the double of the line MF is to the line FA, so is the line MF to the halfe of the line FA, by the 15. of the fifth, wherefore as the double of the line LC is to the line CA, so is the line MF to the halfe of the line FA, by the 11. of the fifth. And in the same proportion, by the 15. of the fifth, are the halues of the consequents, namely, of CA and of the halue of the line AF. Wherefore as the double of the line LC is to the halfe of the line AC, so is the line MF to the fourth part of the line FA. But the double of the line LC is the line DC, and the halfe of the line CA is the line CM, as hath before bene proued, and the fourth part of the line FA is the line FK (for the line FK is the fourth part of the line FH by construction). Wherfore as the line DC is to the line CM, so is the line MF to the line FK. Wherfore by composition (by the 18. of the fifth) as both the lines DC and CM are to the line CM, so is the whole line MK to the line FK.
<figure/>
Wherefore also (by the 22. of the sixt) as the squares of the lines DC and CM are to the square of the line CM, so is the square of the line MK to the square of the line FK. And forasmuch as (by the 8. of the thirtenth) a line which is subten<g ref="char:EOLhyphen"/>ded vnder two sides of a pentagon figure, as is the line AC, being diuided by an extreame &amp; meane propor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ion, the greater segment is equall to the side of the pentagon figure, that is, vnto the line DC: and (by the 1. of the thirtenth) the greater segment hauing added vnto it the halfe of the whole, is in power quintuple to the square made of the halfe of the whole: and the halfe of the whole line AC is the line CM. Wherefore the square that is made of the lines DC and CM, that is, of the greater segment and of the halfe of the whole, as of one line, is quintuple to the square of the line CM, that is, of the halfe of the whole. But as the square made of the lines DC and CM, as of one line, is to the square of the line CM, so is it proued, that the square of the line MK is to the square of the line FK. Wherefore the square of the line M<g ref="char:EOLhyphen"/>K is quintuple to the square of the line FK. But the square of the line KF is rationall, as
<pb n="402" facs="tcp:7062:431"/>
hath before bene proued, wherefore also the square of the line MK is rationall, by the 9. diffi<g ref="char:EOLhyphen"/>nition of the tenth, for the square of the line MK hath to the square of the line KF that proportion that number, hath to number, namely, that 5. hath to 1. and therefore the sayd squares are commensurable, by the 6. of the tenth. Wherefore also the line MK is rationall. And forasmuch as the line BF is quadruple to the line FK (for the semidiameter BF is equal to the semidiameter FH), therfore the line BK is quintuple to the line FK. Wherefore the square of the line BK is 25. times so much as the square of the line KF, by the corollary of the 20 of the sixt. But the square of the line MK, is quintuple to the square of the FK, as is proued. Wherfore the square of the line BK is quintuple to the square of the line KM. Wher<g ref="char:EOLhyphen"/>fore y<hi rend="sup">e</hi> square of the line BK, hath not to y<hi rend="sup">e</hi> square of the line KM, that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square number, by the corollary of the 25. of the eight. Wherefore (by the 9. of the tenth) the line BK is incommensurable in length to the line KM, and either of the lines is rationall. Wherefore the lines BK and KM are rationall commensurable in power onely. But if fro<g ref="char:cmbAbbrStroke">̄</g> a rationall line be taken away a rationall line being commensurable in power onely to the whole, that which remayneth is irrationall, and is (by the 73. of the tenth) cal<g ref="char:EOLhyphen"/>led a residuall line. Wherefore the line MB is a residuall line. And the line conueniently ioy<g ref="char:EOLhyphen"/>ned vnto it, is the line MK. Now I say that the line BM is a fourth residuall line. Vnto the excesse of the square of the line BK aboue the square of the line KM, let the square of the line N be equall (which excesse how to finde out, is taught in the assumpt put after the 13. pro<g ref="char:EOLhyphen"/>position of the tenth). Wherefore the line BK is in power more then the line KM by the square of the line N. And forasmuch as the line KF is commensurable in length to the line FB, for it is the fourth part thereof, therefore (by the 16. of the tenth) the whole line KB is commensurable in length to the line FB. But the line FB is commensurable in length to the line BH, namely, the semidiameter<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the diameter: wherefore the line BK is commensura<g ref="char:EOLhyphen"/>ble in length to the line BH, by the 12. of the tenth. And forasmuch as the square of the line BK is quintuple to the square of the line KM, therefore the square of the line BK hath to the square of the line KM that proportion that fiue hath to one. Wherefore by conuersion of proportion (by the corollary of the 19. of the fifth) the square of BK hath to the square of the line N, that proportion that fiue hath to fower: &amp; therfore it hath not that proportio<g ref="char:cmbAbbrStroke">̄</g> that a square number hath to a square number, by the corollary of the 25. of the eight. Wherfore, the line BK is incommensurable in length to the line N (by the 9. of the tenth). Wherfore the line BK is in power more then the line KM, by the square of a line incommensurable in length to the line BK. Now then forasmuch as the whole line BK is in power more then the line conueniently ioyned, namely, then KM, by the square of a line inco<g ref="char:cmbAbbrStroke">̄</g>mensurable in length to the line BK, and the whole line BK is commensurable in length to the rationall line geuen BH: therefore the line MB is a fourth residuall line, by the diffinition of a fourth residuall line. But a rectangle parallelogramme contayned vnder a rationall line and a fourth residual line, is irrationall, and the line which contayneth in power the same parallelogramme is also irrationall, and is called a lesse line (by the 94. of the tenth). But the line AB contayneth in power the parallelogramme contayned vnder the lines HB and BM (for if we drawe a right line from the point A to the point H, the triangle ABH shall be like to the triangle ABM, by the 8. of the sixth. For from the right angle BAH is drawen to the base BH a perpendicu<g ref="char:EOLhyphen"/>lar line. And therefore as the line BH is to the line BA, so is the line AB to the line BM. this followeth also of the co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ollary of the sayd 8. of the sixth. Wherefore the line AB which is the side of the pentagon figure, is an irrationall line of that kinde which is called a lesse line. If therefore in a circle hauing a rationall line to his diameter be inscribed an equilater pen<g ref="char:EOLhyphen"/>tagon, the side of the pentagon is an irratio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all line, and is of that kinde which is called a lesse line: which was required to be demonstrated.</p>
            </div>
            <div type="part">
               <head>¶ The 12. Theoreme. The 12. Proposition. If in a circle be described an equilater triangle: the square made of the side of
<pb facs="tcp:7062:432"/>
the triangle, is treble to the square made of the line, which is drawen from the centre of the circle to the circumference.</head>
               <p>SVppose that ABC be a circle, and in it describe an equilater triangle ABC. Then I say that the square made of the side of the triangle ABC is treble to the square made of the line drawen from the center of the circle ABC to the circum<g ref="char:EOLhyphen"/>ference.<note place="margin">Constr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>yction.</note> Take (by the 1. of the third) the centre of the circle, and let the same be D. A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d draw a right line from the point A to the poynt D, and extend it to the point E. And draw a right line from the point B to the poynt E.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as the triangle ABC is equilater, t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>refore eche of these three circumferences AB,
<figure/>
AC, &amp; BEC is the third part of the whole circum<g ref="char:EOLhyphen"/>ference of the circle ABC wherefore the circumfe<g ref="char:EOLhyphen"/>rence BE is the sixth part of the circu<g ref="char:cmbAbbrStroke">̄</g>ference of the circle [for the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce of the semicircle ABE is equall to the circumference of the semicircle A<g ref="char:EOLhyphen"/>CE, from which taking away equal circumferences AB and AC, the circumference remayning BE shalbe equal to the circumference remayning EC]: wherefore the right line BE is the side of an equi<g ref="char:EOLhyphen"/>later hexagon figure described in the circle. Where<g ref="char:EOLhyphen"/>fore it is equall to the line drawen from the centre of the circle to the circumference, that is vnto the line DE (by the corollary of the 15. of the sixth). And forasmuch as the line AE is double to the line DE, therefore the square of the line AE is quadruple to the square of the line DE (by the 4. of the second): that is, to the square of the line BE. But the square of the line AE is equall to the squares of the lines AB, and BE (by the 47. of the first) for the angle ABE is (by the 31. of the third) a right angle. Wherfore the squares of the line AB &amp; BE are qua<g ref="char:EOLhyphen"/>druple to the square of the line BE. Wherefore taking away the square of the line BE, the squa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e of the line AB sh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lbe treble to the square of BE: but the line BE is equall to the line DE. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore the square of the line AB is treble to the square of the line DE. Wherefore the square made of the side of the triangle, is treble to the square made of the line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre of the circle to the circumference: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary added by <hi>Campane.</hi>
                  </head>
                  <p>Hereby it is manifest, that the line BC, which is the side of the equilater triangle, diuideth the semi<g ref="char:EOLhyphen"/>diameter DE into two equall parts. For let the poynt of the diuision be F. And suppose a line to be drawen from the poynt D to the B, and an other from the poynt D to the poynt C. Now it is manifest (by the 4. of the first) that the line BF is equall to the line FC, and therefore (by the 3. of the third) all the angles at the poynt F are <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ight angles. Wherefore (by the 47. of the first) the square of the line BD is equall to the squares o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>F and FD, and by the same the square of the line BE is equall to the squares of the lines BF, and FE: b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t the line BD is equall to the line BE (as hath before bene proued). Wherefore by the common sentence the twoo squares of the two lines BF and FD are equall to the two squares of the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> BF, and FE. Wherefore taking away the square of the line B<g ref="char:EOLhyphen"/>F which is co<g ref="char:cmbAbbrStroke">̄</g>mon to them both: the residue, namely, the square of the line DF shalbe equall to the re<g ref="char:EOLhyphen"/>sidue, namely, to the square of the line FE. Wherfore also the line FD is equal to the line FE. Wher<g ref="char:EOLhyphen"/>fore hereby it is manifest that a <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>erpendicular line drawen from the centre of a circle to the side of an equilater triangle inscribed in it, is equall to halfe of the line drawen from the centre of the same circle, to the circumference thereof.</p>
               </div>
               <div type="section">
                  <head>A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>The side of an equilater triangle is in power sesquitertia to the perpendicular line which is drawe<g ref="char:cmbAbbrStroke">̄</g> from one of the angles to the opposite side. For of what parts the line AB contayneth in power 12. of
<pb n="403" facs="tcp:7062:432"/>
such parts the line BF which is the halfe of AB contayned in power 3.<note place="margin">
                        <hi>This Corollary is the</hi> 11. <hi>prop<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>sition of the</hi> 14. <hi>booke after Campane.</hi>
                     </note> Wherefore the residue, namely, the perpendicular line AF contayneth in power of such parts 9. (for the squares of the lines AF, and BF are by the 47. of the first equall to the square of the line AB). Now 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. to 9. is sesquitertia<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wherfore the power of the line AB is to the power of the line AF in sesquitertia proportion.</p>
                  <p>Moreouer the side of the triangle is the meane proportionall betwene the diameter and the per<g ref="char:EOLhyphen"/>pendicular line: For (by the Corollary of the 8. of the sixth) the line AE is to the line AB as the line AB is to the line AF.</p>
                  <p>Farther the perpendicular line drawen from the angle diuideth the base into two equal parts and passeth by the center.<note place="margin">
                        <hi>This Corollary is the</hi> 3. <hi>Corolla<g ref="char:EOLhyphen"/>ry after the</hi> 17. <hi>proposition of the</hi> 14 <hi>booke af<g ref="char:EOLhyphen"/>ter Campane.</hi>
                     </note> For if there should be drawen any other right line fro<g ref="char:cmbAbbrStroke">̄</g> the point A to the poynt F, the<g ref="char:cmbAbbrStroke">̄</g> that which is drawen by the point D, two right lines should include a superficies, which is impossi<g ref="char:EOLhyphen"/>ble. Wherefore the contrary followeth, namely, that the line, which being drawen from the angle pas<g ref="char:EOLhyphen"/>seth by the center, is a perpendicular line to the base (by the 3. of the third).</p>
               </div>
            </div>
            <div type="part">
               <head>The 1. Probleme. The 13. Proposition. To make a <note n="*" place="margin">By the name o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> a Pyramis both here &amp; i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> this booke fol<g ref="char:EOLhyphen"/>lowing, vnder<g ref="char:EOLhyphen"/>stand a Tetra<g ref="char:EOLhyphen"/>hedron.</note> Pyramis, and to comprehend it in a sphere geuen: and to proue that the diameter of the sphere is in power sesquialtera to the side of the Pyramis.</head>
               <p>SVppose that the diameter of the sphere geuen be AB, and diuide AB in the poynt C, so that the line AC be double to the line BC (by the 9. of the sixth).<note place="margin">First part of the construction.</note> And vpon the line AB, making the center the point N, describe a semicircle ADB. And (by the 11. of the first) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rom the point C rayse vp vnto the line AB a perpendicular line CD. And drawe a right line from the point D to the point A. And describe a circle EFG hauing his semidiameter equall to the line CD. And describe in the circle EFG, an equilater trian<g ref="char:EOLhyphen"/>gle EFG (by t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e 2. of the fourth). And (by the
<figure/>
1. of the third) take the ce<g ref="char:cmbAbbrStroke">̄</g>tre of the circle, and let the same be the point H. And draw these right lines FH, HF, &amp; HG. And (by the 12. of the eleuenth) fro<g ref="char:cmbAbbrStroke">̄</g> the point H rayse vp vnto the playne superficies of the circle EFG a per<g ref="char:EOLhyphen"/>pe<g ref="char:cmbAbbrStroke">̄</g>dicular line HK, &amp; let the line HK be equal to the right line AC. And draw these right lines KE, KF, &amp; KG. Now forasmuch as the line HK is erected perpe<g ref="char:cmbAbbrStroke">̄</g>dicularly to the plaine super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies of the circle EFG,<note place="margin">First part of the demonstration.</note> therfore (by the 2. definition of the eleuenth) it maketh right angles with all the right lines that touch it, and which are in the selfe same superficies of the circle EFG. But it toucheth euery one of these right lines, HE, HF and HG: Where<g ref="char:EOLhyphen"/>fore the right line HK is erected perpendicu<g ref="char:EOLhyphen"/>larly to euery one of these lines HE, HF, and HG. And forasmuch as the line AC is equal to the line HK, and the line CD to the line HE, and they comprehende right angles, ther<g ref="char:EOLhyphen"/>fore the base DA is equall to the base KE (by the 4 of the first). And by the same reason ech of the lines KF, and KG is equall to the line DA. Wherefore the three lines KE, KF, and KG are equall the one to the other. And forasmuch as the line AC is double to the line CB, (by cons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ruction): therefore the line AB is treble to the line BC: but as the line AB is to the
<pb facs="tcp:7062:433"/>
line BC, so is the square of the line AD to the square of the line DC.
<q>[Whiche may thus be proued.<note place="margin">This Assumpt is againe at the end of the demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tio<g ref="char:cmbAbbrStroke">̄</g> of the pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>osi<g ref="char:EOLhyphen"/>tion proued.</note> It is manifest (by the Corollary of the 8. of sixth) that the line CD is the meane pro<g ref="char:EOLhyphen"/>portionall betwene the lines AC and CB. Wherfore (by the corollary of the 20. of the same) the square of the line AC is to the square of the line CD, as the line AC is to the line CB: wherefore by composition (by the 18. of the fiueth) the squares of the lines AC and CD are to the square of the line CD, as the line AB is to the line BC. But the squares of the lines AC and CD are (by the 47. of the first) equall to the square of the line AD: wherefore (by the 7. of the fiueth) the square of the line AD is to the square of the line DC, as the line AB is to the line BC.]</q>
Wherefore the square of the line AD is treble to the square of the line DC. And forasmuch as the line KE is equall to the line AD (as it hath ben proued) and the line HE is put equall to the line CD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore the square of the line KE is triple to the square of the line HE. But vnto the square of the same line HE is (by the 12. of the thir<g ref="char:EOLhyphen"/>tenth) the square of the line FE treble wherefore the line EF is equall to the line KE. Now the lines KE, KF, and KG are equall the one to the other, as it hath before ben proued, and so also are the lines EF, FG and GE, for that they are the sides of an equilater triangle. Wherefore euery one of these lines EF, FG, and GE, is equall to euery one of the lines KE, KF and KG. Wherefore these fower triangles EFG, KEF, KFG and KGE are equila<g ref="char:EOLhyphen"/>ter. Wherefore there is made a Pyramis consisting of fower equall and equilater triangles, whose base is the triangle EFG and toppe the poynt K.</p>
               <p>
                  <note place="margin">Second part of the Constructio<g ref="char:cmbAbbrStroke">̄</g>.</note>Now it is required to comprehende the
<figure/>
same Pyramis in the sphere geue<g ref="char:cmbAbbrStroke">̄</g>, and to proue that the diameter of the sphere is in power ses<g ref="char:EOLhyphen"/>quialtera to the side of the pyramis.<note place="margin">Second part of the demostratio<g ref="char:cmbAbbrStroke">̄</g>.</note> Adde vn<g ref="char:EOLhyphen"/>to the right line HK a right line directly, na<g ref="char:EOLhyphen"/>mely HL, and let the line HK be equall to the the line BC.<note n="*" place="margin">Looke at the end of this de<g ref="char:EOLhyphen"/>monstration for an other con<g ref="char:EOLhyphen"/>struction and demonst<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ation of this second part after Flus<g ref="char:EOLhyphen"/>sa<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> Now for that as the line AC is to the line CD, so is the line CD to the line CB (by the corollary of the 18. of the sixth) but the line AC is equal to the line KH, &amp; the line CD to the line HE, &amp; the line CB to the line HL. Therfore as the line KH is to the line HE, so is the line HE to the line HL. Where<g ref="char:EOLhyphen"/>fore that which is contayned vnder the lines HK and HL is equall to the square of the line EH. And ether of angles the HKE, &amp; EHL is a right angle, wherefore a semicircle descri<g ref="char:EOLhyphen"/>bed vpon the line KL shall passe by the poynt E.<note n="*" place="margin">Reade the two assumpts added by Campane (which are set at the end of the demonstration of this propositi<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n) for the bet<g ref="char:EOLhyphen"/>ter <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nderstan<g ref="char:EOLhyphen"/>ding of this rea<g ref="char:EOLhyphen"/>son.</note> For if we draw a right line from the point E to the poynt L, the angle LEK shalbe a right angle, for that the tria<g ref="char:cmbAbbrStroke">̄</g>gle ELK is equi<g ref="char:EOLhyphen"/>angle to either of the triangles ELH and E<g ref="char:EOLhyphen"/>HK (by the 8. of the sixth.) Now then if the the diameter KL abiding fixed, the semicircle be turned round about, vntil it returne vn<g ref="char:EOLhyphen"/>to the selfe same place from whense it began to be moued, it shall also passe by the pointes F and G. For drawing a right line from the ponit F to the point L, and an other from the poynt L to the point G, which also maketh at the points F and G right angles, the pyramis shall be contayned in the sphere geuen. For the line KL being the diameter of the sphere is equall to the line AB, which is the diameter of the sphere geuen, for the line KH is put equall to the line AC, and the line HL to the line CB.</p>
               <p>
                  <pb n="404" facs="tcp:7062:433"/>
                  <hi>[In the semicircle ADB of the former figure drawe the line DN. And diuide the line KG into two equall parts in the point M.</hi>
                  <note place="margin">An other con<g ref="char:EOLhyphen"/>struction and demonstration of the second part after F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>us<g ref="char:EOLhyphen"/>sas.</note> 
                  <hi>And draw a line from M to G. And forasmuch as by construction the line KH is equall to the line AC, and the line HL to the line CB: therefore the whole line AB is equal to the whole line KL. Wherefore also the halfe of the line KL, namely, the line LM, is equal to the se<g ref="char:EOLhyphen"/>midiameter BN: wherefore taking away from those equall lines, equall parts BC and LH, the residues NC, and MH shalbe equall. Wherefore in the two triangles MHG and NCD, the two sides about the equall right angles DCN and GHM, namely, the sides DC, CN, and GH, HM, are equall: wher<g ref="char:EOLhyphen"/>fore the bases MG and ND are equall (by the 4. of the first.) And by the same reason may it be proued that right lines drawen from the poynt M to the points E and F are equal to the line ND. But the right line ND is equall to the line AN, which is drawen from the centre to the circumference: wherefore the line MG is equall to the line MK, &amp; also to the lines ME, MF and ML. Wherfore making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the poynt M, and the space MK or MG describe a semicircle KGL: and the diameter KL abiding fixed let the sayd semicircle KGL be moued rounde about vntill it returne to the same place from whence it began to be moued: and there shalbe described a sphere about the centre M (by the 12. diffi<g ref="char:EOLhyphen"/>nition of the eleuenth) touching euery one of the angles of the Pyramis which are at the points K, E, F, G: for those angles are equally dista<g ref="char:cmbAbbrStroke">̄</g>t from the centre of the sphere, namely, by the semidiameter of the sayd sphere, as hath before bene proued. Wherefore in the sphere geuen whose diameter is the line K<g ref="char:EOLhyphen"/>L, or the line AB, is inscribed a Tetrahedron EFGK.]</hi>
               </p>
               <p>Now I say, that the diameter of the sphere is in power sesquialtera to the side of the Py<g ref="char:EOLhyphen"/>ramis.<note place="margin">Third part of the demonstra<g ref="char:EOLhyphen"/>tion.</note> For forasmuch as the line AC is double to the line CB (by co<g ref="char:cmbAbbrStroke">̄</g>structio<g ref="char:cmbAbbrStroke">̄</g>) therfore the line AB is treble to the line BC. Wherfore by conuersion) by the corollary of the 19. of the fiueth) the line AB is sesquialtera to the line AC. But as the line BA is to the line AC, so is the square of the line BA, to the square of the line AD. For if we draw a right line fro<g ref="char:cmbAbbrStroke">̄</g> the point B, to the point D, as the line BD is to the line AD, so is the same AD to the line AC, by reaso<g ref="char:cmbAbbrStroke">̄</g> of the likenes of the ttriangles DAB, &amp; DAC (by the 8. of the sixth): &amp; by reason also that as the first is to th third so is the square of the first to the square of the second (by the corollary of the 20. of the sixth). Wherfore the square of the line BA, is sesquialter to the square of the line AD. But the line BA is equal to the diameter of the sphere geue<g ref="char:cmbAbbrStroke">̄</g>, namely, to the line KL, as hath bene proued, &amp; the line AD is equal to the side of the pyramis inscribed in the sphere. Wherfore y<hi rend="sup">e</hi> diameter of the sphere is in power sesquialter to the side of the pyramis. Wherfore there is made a pyramis comprehended in a sphere geuen, and the diameter of the sphere is sesquialtera to the side of the pyramis: which was required to be done and proued.</p>
               <div type="section">
                  <head>¶An other demonstration to proue that as the line AB is to the line BC, so is the square of the line AD to the square of the line DC.</head>
                  <p>Let the description of the semicircle ADB be as in the first description. And vpon the line AC describe (by the 46. of the first) a square EC, and make perfecte the parallelogra<g ref="char:cmbAbbrStroke">̄</g>me FB. Now forasmuch as the triangle DAB is equiangle to the triangle DAC (by the 32. of the sixt: therfore as the line BA is to the line
<figure/>
AD, so is the line DA to the line AC, by the 4. of the sixt. Wherefore that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines BA and AC, is equall to the square of the line AD, by the 17. of the sixt. And for that as the line AB is to the line BC, so is the parallelogramme EB to the pa<g ref="char:EOLhyphen"/>rallelogramme FB, by the 1. of the sixt: and the parallelogramme EB is that which is con<g ref="char:EOLhyphen"/>tained vnder the lines BA and AC (for the line EA is equall to the line AC): and the parallelogramme BF is that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines AC and BC. Wherefore as the line AB is to the line BC, so is that which is contained vnder the lines BA and
<pb facs="tcp:7062:434"/>
AC, to that which is contained vnder the lines AC and CB. But that which is contai<g ref="char:EOLhyphen"/>ned vnder the lines BA and AC, is equall to the square of the line AD, by the Corollary of the 8. of the sixt, and that which is contained vnder the lines AC and CB, is equall to the square of the line CD, for the perpendicular line DC is the meane proportionall betwene the segmentes of the base, namely, AC and CB, by the former Corollary of the 8. of the sixt, for that the angle ADB is a right angle. Wherefore as the line AB is to the line BC, so is the square of the line AD to the square of the line DC, by the 11. of the fift: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
               </div>
               <div type="section">
                  <head>¶Two Assumptes added by <hi>Campane.</hi>
                  </head>
                  <div type="subsection">
                     <head>First Assumpt.</head>
                     <p>Suppose that vpon the line AB be erected perpendicularly the line DC, which line DC let be the meane proportionall betwene the partes of the line AB, namely, AC &amp; CB: so that as the line AC is to the line CD, so let the line CD be to the line CB. And vpon the line AB describe a semicircle. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the circumference of that semicircle shall passe by the point D, which is the end of the perpe<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>dicular line. But if not, then it shall either cut the
<figure/>
line CD, or it shall passe aboue it, and include it not touching it. First let it cut it in the point E. And drawe these right lines EB and EA. Wher<g ref="char:EOLhyphen"/>fore by the 31. of the third, the whole angle AE<g ref="char:EOLhyphen"/>B is a right angle. Wherefore by the first part of the Corollary of the 8. of the sixt, the line AC is to the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>C, as the line EC is to the line CB. But by the 8. of the fift, the proportion of the line AC to the line EC, is greater then the proporti<g ref="char:EOLhyphen"/>on of the same line AC to the line CD (for the line CE is lesse then the line CD). Now for that the line CE is to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> line CB, as the line AC is to the line CE, and the line CD is to the line CB, as the line AC is to the line CD, therefore by the 13. of the fift, the proportion of the line EC to the line CB, is greater then the proportion of the line CD to the line CB. Wherefore by the 10. of the fift, the line EC is greater then the line DC, namely, the part greater then the whole: which is impossible. Wherefore the circumference shall not cut the line CD. Now I say, that it shall not passe aboue the line CD, and not touch it in the point D. For if it be possible, let it passe aboue it, and extend the line CD to the circu<g ref="char:cmbAbbrStroke">̄</g>ference, and let it cut it in the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. And draw the lines FB and FA, and it shall followe as before that the line CD is greater then the line CF: which is impossible. Wherefore that is manifest which was required to be proued.</p>
                  </div>
                  <div type="subsection">
                     <head>¶Second Assumpt.</head>
                     <p>
                        <hi>If there be a right angle vnto which a base is subtended, and if vpon the same be described a se<g ref="char:EOLhyphen"/>micircle: the circumference thereof shall passe by the point of the right angle.</hi>
                     </p>
                     <p>For suppose that there be a right angle ABC, vnto which subtend the base AC, and vpon the line AC describe a semicircle. Then I say, that the circumference thereof shall passe by the point B. For if not, then it shall pa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>e either aboue the point B, or vnder the point B. First let it passe vnder the point B, and l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> the circumference be AEC. And (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. of the first) from the point B drawe vnto the line AC a perpendicular line BD, which let cut the
<figure/>
circumference of the semicircle in the point E. And drawe th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>se right lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>A and EC. Now it is ma<g ref="char:EOLhyphen"/>nifest, by the 31. of the third, that the angle AEC is a right angle. But (by the 21. of the first) the an<g ref="char:EOLhyphen"/>gle AEC is greater then the angle ABC: which is impossible, by the 10. common sentence. Where<g ref="char:EOLhyphen"/>fore the circumfere<g ref="char:cmbAbbrStroke">̄</g>ce of the semicircle passeth not vnder the angle B. Now I say, that it passeth not a<g ref="char:EOLhyphen"/>boue the angle B. For if it be possible, let it passe a<g ref="char:EOLhyphen"/>boue the point B, and let the circumference be A<g ref="char:EOLhyphen"/>FC: and produce the perpendicular line BD till it cut the circumference AFC in the point F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. And draw these right lines AF and FC. Wherefore a<g ref="char:EOLhyphen"/>gaine,
<pb n="405" facs="tcp:7062:434"/>
by the 31. of the third, the angle AFC is a right angle. But by supposition, the angle ABC is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> right angle, and is, by the 21. of the first, greater then the right angle AFC: which againe by the fore<g ref="char:EOLhyphen"/>sayd common sentence, is impossible. Wherfore a semicircle described vpon the base AC, passeth nei<g ref="char:EOLhyphen"/>ther vnder the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, nor aboue it. Wherefore it passeth by it: which was required to be proued.</p>
                     <p>The conuerse of this was added after the demonstration of the 31. of the third, out of Pelitarius. And these two Assumptes of Campane are necessary, for the better vnderstanding of the demonstration of the seco<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>d part of this 13. Proposition, wherein is proued that the pyramis is contained in the Sphere geuen.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶ Certaine Corollaryes added by <hi>Flussas.</hi>
                  </head>
                  <div type="subsection">
                     <head>First Corollary.</head>
                     <p>
                        <hi>The diameter of the Sphere is in power quadruple sesquialtera to the line which is drawen from the centre to the circumference of the circle which containeth the base of the pyramis.</hi>
                     </p>
                     <p>For forasmuch as it hath bene proued, that the diameter KL is in power sesquialter to the side EF: and it is proued also, by the 12. of this booke, that the side EF is in power triple to the line EH (which is drawen from the centre of the circle contayning the triangle EFG). But the proportion of the ex<g ref="char:EOLhyphen"/>tremes, namely, of the diameter to the line EH, consisteth of the proportions of the meanes, namely, of the proportion of the diameter to the line EF, and of the proportion of the line EF to the line EH, by the 5. definition of the sixt: which proportions, namely, triple, and sesquialter, added together, make quadruple sesquialter (as it is easie to proue by that which was taught in the declaration of the 5. defini<g ref="char:EOLhyphen"/>tion of the sixt booke). Wherefore the Corollary is manifest.</p>
                  </div>
                  <div type="subsection">
                     <head>¶ Second Corollary.</head>
                     <p>
                        <hi>Onely the line which is drawen from the angle of the pyramis to the base opposite vnto it,</hi>
                        <note place="margin">
                           <hi>This Corolla<g ref="char:EOLhyphen"/>ry is the</hi> 15. <hi>proposition of the</hi> 14. <hi>booke after Cam<g ref="char:EOLhyphen"/>pane.</hi>
                        </note> 
                        <hi>&amp; passing by the center of the Sphere, is perpendicular to the base, and falleth vpon the centre of the circle which containeth the base.</hi>
                     </p>
                     <p>For if any other line (then the line KMH which is drawen by the centre of the Sphere to the centre of the circle) should fall perpendicularly vpon the plaine of the base, then, from one and the selfe same point should be drawen to one and the selfe same plaine two perpendicular lines, contrary to the 13. of the eleuenth: which is impossible. Farther if from the toppe K should be drawen to the center of the base, namely, to the point H, any other right line not passing by the centre M, two right liues shoulde include a superficies contrary to the last common sentence: which were absurde. Wherefore onely the line which is drawen by the center of the Sphere to the centre of the base, is perpendicular to the sayd base. And the line which is drawen from the angle perpendicularly to the base, shall passe by the centre of the Sphere.</p>
                  </div>
                  <div type="subsection">
                     <head>Third Corollary.</head>
                     <p>
                        <hi>The perpendicular line which is drawen from the centre of the Sphere to the base of the pyramis,</hi>
                        <note place="margin">
                           <hi>This Corolla<g ref="char:EOLhyphen"/>ry Campane putteth as a Corollary af<g ref="char:EOLhyphen"/>ter the</hi> 17. <hi>pro<g ref="char:EOLhyphen"/>position of the</hi> 14. <hi>booke.</hi>
                        </note> 
                        <hi>is equall to the sixt part of the diameter of the Sphere.</hi>
                     </p>
                     <p>For it is before proued, that the line MH (which is drawen from the centre of the Sphere to the cen<g ref="char:EOLhyphen"/>tre of the base) is equall to the line NC: which line NC is the sixt part of the diameter AB, and ther<g ref="char:EOLhyphen"/>fore the line MH is the sixt part of the diameter of the Sphere. For the diameter AB is equall to the di<g ref="char:EOLhyphen"/>ameter of the Sphere, as hath also before bene proued.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 2. Probleme. The 14. Proposition. To make an octohedron, and to co<g ref="char:cmbAbbrStroke">̄</g>prehend it in the sphere geuen, namely, that wherein the pyramis was comprehended: and to proue that the diameter of the sphere is in power double to the side of the octohedron.</head>
               <p>
                  <pb facs="tcp:7062:435"/>
                  <note place="margin">Construction.</note>TAke the diameter of the former sphere geuen, which let be the line AB: and diuide it (by the 10. of the first into two equall partes in the point C. And de<g ref="char:EOLhyphen"/>scribe vpon the line AB a semicircle ADB. And (by the 11. of the first) fro<g ref="char:cmbAbbrStroke">̄</g> the point C rayse vp vnto the line AB a perpendicular line CD. And draw a right line from the point D to the point B. And describe a square EFGH hauing euery one of his sides equall to the line BD.
<figure/>
and draw the diagonal lines FH &amp; EG, cutting the one the other in the point K. And (by the 12. of the eleuenth) from the point K (namely, the point where the lines FH and EG cut the one the other) rayse vp to the playne superficies, wherein the square EF<g ref="char:EOLhyphen"/>GH is, a perpendicular line KL, and extend the line KL on the other side of the playne superficies to the point M. And let eche of the lines KL and KM be put equall to one of these lines KE, KF, K<g ref="char:EOLhyphen"/>H or KG. And draw these right lines LE, LF, L<g ref="char:EOLhyphen"/>G, LH, ME, MF, MG, and MH.<note place="margin">
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rist part of the demonstration.</note> Now forasmuch as the line KE is (by the corollary of the 34. of the first) equall to the line KH, and the angle<note n="*" place="margin">For the <hi>4.</hi> an<g ref="char:EOLhyphen"/>gles at the point K are equall to fower right an<g ref="char:EOLhyphen"/>gles by the Co<g ref="char:EOLhyphen"/>rollary of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e 15. of the first: and those <hi>4.</hi> an<g ref="char:EOLhyphen"/>gle, are equall the one to the o<g ref="char:EOLhyphen"/>ther by the <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>.</hi> of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>irst: and the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>efore ech is a right angle.</note> EKH is a right angle, therefore the square of HE is double to the square of EK, by the 47. of the first: Agayne forasmuch as the line LK is equall to the line KE by position, and the angle LKE is, by the second diffinition of the eleuenth, a right angle: therefore the square of the line EL is double to the square of the line EK: and it is proued that the square of the line HE is double to the square of the line EK. Wherefore the square of the line LE is equall to the square of the line EH. Wherefore also the line LE is equall to the line EH. And by the same rea<g ref="char:EOLhyphen"/>son the line LH is also equall to the line HE. Wherefore the triangle LHE is equilater. In like sort may we proue that euery one of the rest of the triangles whose bases are the sides of the square EFGH, and toppes the pointes L and M, are equilater. The sayd eight triangles also are equall the one to the other, for euery side of eche is equall to the side of the square E<g ref="char:EOLhyphen"/>FGH. Wherfore there is made an octohedron co<g ref="char:cmbAbbrStroke">̄</g>tained vnder eight triangles whose sides are equall. Now it is required to comprehend it in the sphere geuen, and to proue that the diame<g ref="char:EOLhyphen"/>ter of the sphere is in power double to the side of the octohedron.</p>
               <p>
                  <note place="margin">Second part of the demo<g ref="char:cmbAbbrStroke">̄</g>stratio<g ref="char:cmbAbbrStroke">̄</g>.</note>Forasmuch as these three lines LK, KM, and KE are equall the one to the other, ther<g ref="char:EOLhyphen"/>fore a semicircle described vpon the line LM shall passe also by the point E. And by the same reason, if the semicircle be turned round about, vntill it returne vnto the selfe same place from whence first it began to be moued, it shal passe by the pointes F, G, H, and the octohedron shall be comprehended in a sphere. I say also that it is comprehended in the sphere geuen. For forasmuch as the line LK is equal to the line KM, by position, and the line KE is common to them both and they contayne right angles by the 3. diffinition of the eleuenth, therefore (by the 4. of the first) the base LE is equall to the base EM. And forasmuch as the angle LEM is a right angle, by the 31. of the third, for it is in a semicircle, as hath bene proued, therefore the square of the line LM is double to the square of the line LE by the 47. of the first. Againe forasmuch as the line AC is equall to the line BC, therefore the line AB is double to the line BC, by the di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>finition of a circle. But as the line AB is to the line BC, so is the square of the
<pb n="406" facs="tcp:7062:435"/>
line AB to the square of the line BD, by the corollaries of the 8. and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. of the sixt. Where<g ref="char:EOLhyphen"/>fore the square of the line AB is double to the square of the line BD. And it is proued that the square of the line LM is double to the square of the line LE. Wherefore the square of the line BD is equall to the square of the line LE. For the line EH which is equall to the line LF, is put to be equall to the line DB. Wherefore the square of the line AB is equall to the square of the line LM. Wherefore the line AB is equall to the line LM. And the line AB is the diameter of the sphere geue<g ref="char:cmbAbbrStroke">̄</g>, wherefore the line LM is equall to the diameter of the sphere geuen. Wherefore the octoedron is contayned in the sphere geuen: <note n="*" place="margin">For the square of the line AB, which is proued equall to the square of the line LM, is dou<g ref="char:EOLhyphen"/>ble to the square of the line BD which is also e<g ref="char:EOLhyphen"/>quall to the square of the line LE.</note> and it is also proued that the diameter of the sphere is in power double to the side of the octohedron. Wherefore there is made an octohedron, and it is comprehended in the sphere geuen, wherein was comprehended the Pyramis: and it is proued that the diameter of the sphere is in power double to the side of the octohedrn: which was required to be doone, and to be proued.</p>
               <div type="section">
                  <head>Certayne Corollaries added by <hi>Flussas.</hi>
                  </head>
                  <div type="subsection">
                     <head>First Corollary.</head>
                     <p>
                        <hi>The side of a Pyramis is in power sesquitertia to the side of an oct<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hedron inscribed in the same Sphere.</hi>
                     </p>
                     <p>For forasmch as the diameter is in power double to the side of the octohedron, therefore of what partes the diameter contayneth in power 6. of the same, the side of the octohedron co<g ref="char:cmbAbbrStroke">̄</g>tayneth in power 3. but of what partes the diameter contayneth 6. of the same, the side of the pyramis contayneth 4. by the 13. of this booke. Wherefore of what partes the side of the pyramis contayneth 4. of the same the side of the octohedron contayneth 3.</p>
                  </div>
                  <div type="subsection">
                     <head>Second Corollary.</head>
                     <p>
                        <hi>An octohedron is deuided into two equall and like Pyramids.</hi>
                     </p>
                     <p>The common bases of these Pyramids are set vpon euery square contayned of the sides of the oc<g ref="char:EOLhyphen"/>tohedron, vpon which square are set the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> triangles of the octohedron:<note place="margin">
                           <hi>This Corollary is the</hi> 16. <hi>propo<g ref="char:EOLhyphen"/>sition of the</hi> 14. <hi>booke after Campane.</hi>
                        </note> which pyramids are by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. diffinition of the eleuenth equall and like. And the foresayd square common to those Pyramids, is the halfe of the square of the diameter of the sphere, for it is the square of the side of the octohedron.</p>
                  </div>
                  <div type="subsection">
                     <head>Third Corollary.</head>
                     <p>
                        <hi>The three diameters of the octohedron, do cutte the one the other perpendicularly into two equall parts, in the center of the sphere which contayneth the sayd octohedron.</hi>
                     </p>
                     <p>As it is manifest by the three diameters EG, FH, and LM which cutte the one the other in the center K equally and perpendicularly.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 3. Probleme. The 15. Proposition. To make a solide called a cube, and to comprehend it in the sphere geuen, namely, that Sphere wherein the former two solides were comprehend<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and to proue that the diameter of the sphere, is in power treble to the side of the cube.</head>
               <p>TAke the diameter of the sphere geuen, namely, AB, and diuide it in the point C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> So that let the line AC be double to the line BC by the 9. of the sixt. And vpon the
<pb facs="tcp:7062:436"/>
line AB describe a semicircle ADB. And (by the 11. of the first) from the p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ynt C r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>yse vp vnto the line AB a perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line CD. And draw a right lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> DB. And describe a squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> EFGH, hauing euery one of his sides equall
<figure/>
to the line DB, And from the pointes E, F, G, H, rayse vp (by the 12. of the eleuenth) vnto the playne superficies of the square EFGH perpendicular lines EK, FL, GM, and HN: and let euery one of the lines EK, FL, GM, and HN be put equall to one of the lines EF, FG, GH, or HE, which are the sides of the square, and draw these right lines KL, LM, MN, and NK.<note place="margin">First part of the demonstration.</note> Wherfore there is made a cube namely FN which is contayned vnder six e<g ref="char:EOLhyphen"/>quall squares.<note place="margin">Second part of the Constructio<g ref="char:cmbAbbrStroke">̄</g>.</note> Now it is required to compre<g ref="char:EOLhyphen"/>hend the same cube in the sphere geuen, and to proue that y<hi rend="sup">e</hi> diameter of the sphere is in power ble to the side of the cube.<note place="margin">Second part of the demonstra<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ion.</note> Draw these right lines KG and EG. And forasmuch as the an<g ref="char:EOLhyphen"/>gle KEG is a right angle, for that the line K<g ref="char:EOLhyphen"/>E is erected perpendicularly to the playne su<g ref="char:EOLhyphen"/>perficies E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and therefore also to the right line EG, by the 2. diffinitio<g ref="char:cmbAbbrStroke">̄</g> of the eleuenth, wherefore a semicircle described vpon the line KG shall <note n="*" place="margin">By the 2. As<g ref="char:EOLhyphen"/>sumpt of the <hi>13.</hi> of this booke.</note> passe by the poynt E. Agayne for<g ref="char:EOLhyphen"/>asmuch as the line FG is erected perpendicu<g ref="char:EOLhyphen"/>larly to either of these lines FL and FE, by the diffinition of a square, &amp; by the 2. diffini<g ref="char:EOLhyphen"/>tion of the eleuenth, therefore the line FG is erected perpendicularly to the playne superficies FK, by the 4. of the eleuenth. Wherefore if we draw a right line from the point F to the point K, the line GF shall be erected perpendicularly to the line KF, by the 2. diffinition of the ele<g ref="char:EOLhyphen"/>uenth. And by the same reason agayne a semicircle described vpon the line GK shall passe also by the point F. And likewise shall it passe by the rest of the pointes of the angles of that cube. If now the diameter KG abiding fixed the semicircle be turned round about vntill it returne into the selfe same place from whence it began first to be moued, the cube shalbe comprehe<g ref="char:cmbAbbrStroke">̄</g>ded in a sphere. I say also that it is comprehended in the sphere geuen.</p>
               <p>
                  <note place="margin">Third part of the demonstra<g ref="char:EOLhyphen"/>tion.</note>For forasmuch as the line GF is equall to the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E and the angle F is a right an<g ref="char:EOLhyphen"/>gle, therefore the square of the line EG is by the 47. of the first double to the square of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>F. But the line EF is equall to the line EK. Wherefore the square of the line EG is double to the square of the line EK. Wherfore the squares of EG and EK that is the square of the line GK, by the 47. of the first, are treble to the square of the line EK. And forasmuch as the line AB is treble to the line BC, but as the line AB is to the line BC, so is the square of the line AB to the square of the line BD, by the corollaries of the 8. and 20. of the sixt. Wherefore the square of the line AB is treble to the square of the line BD. And it is proued that the square of the line GK is treble to the square of the line KE, and the line KE is put equall to the line BD. Wherefore the line KG is also equall to the line AB. And the line AB is the diameter of the sphere geuen. Wherefore the line KG is equall to the diameter of the sphere geuen. Wherfore the cube is co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>ded in the sphere geuen: and it is also proued that the diameter of the Sphere is in power treble to the side of the cube: which was required t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e doone, and to be proued.</p>
               <div type="section">
                  <pb n="407" facs="tcp:7062:436"/>
                  <head>An other demonstration after <hi>Flussas.</hi>
                  </head>
                  <p>Suppose that the diameter of the Sphere geuen in the former Propositions, be the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. And let the center be the point C, vpon which describe a semicircle ADB. And from the diameter AB cut of a third part BG, by the 9. of the sixt. And from the point G raise vp vnto the line AB a perpendi<g ref="char:EOLhyphen"/>cular line DG, by the 11. of the first. And draw these right lines DA, DC, and DB. And vnto the right line DB put an equall right line ZI:
<figure/>
and vpon the line ZI describe a square EZIT. And fro<g ref="char:cmbAbbrStroke">̄</g> the pointes E, Z, I, T, erecte vnto the superficies EZIT per<g ref="char:EOLhyphen"/>pendicular lines EK, ZH, IM, TN (by the 12. of the eleuenth): and put euery one of those perpendicular lines equall to the line ZI. And drawe these right lines KH, HM, MN, and NK, ech of which shall be equall and parallels to the line ZI, and to the rest of the lines of the square, by the 33. of the first. And moreouer they shall containe equall angles (by the 10. of the eleuenth): and therefore the angles are right angles, for that EZIT is a square: wherfore the rest of the bases shall be squares. Wherfore the solide EZITKHMN being co<g ref="char:cmbAbbrStroke">̄</g>tained vnder 6. equall squares, is a cube, by the 21. definition of the eleuenth. Extend by the opposite sides KE and MI of the cube, a plaine KEI<g ref="char:EOLhyphen"/>M: and againe by the other opposite sides NT and HZ, extend an other plaine HZTN. Now forasmuch as ech of these plaines deuide the solide into two equall partes, namely, into two Prismes equall and like (by the 8. defi<g ref="char:EOLhyphen"/>nition of the eleuenth): therfore those plaines shall cut the cube by the cen<g ref="char:EOLhyphen"/>tre, by the Corollary of the 39. of the eleuenth. Wherefore the co<g ref="char:cmbAbbrStroke">̄</g>mon secti<g ref="char:EOLhyphen"/>on of those plaines shall passe by the centre. Let that common section be the line LF. And forasmuch as the sides HN and KM of the superficieces KE<g ref="char:EOLhyphen"/>IM and HZTN do diuide the one the other into two equall partes, by the Corollary of the 34. of the first, and so likewise do the sides ZT and EI: therefore the common section LF is drawen by these sections, and diuideth the plaines KEIM and HZTN into two equall partes, by the first of the sixt: for their b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ses are equall, and the altitude is one and the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ame, namely, the altitude of the cube. Wherefore the line LF shall diuide into two equall partes the diameters of his plaines, namely, the right lines KI, EM, ZN, and NT, which are the diameters of the cube. Wher<g ref="char:EOLhyphen"/>fore those diameters shall concurre and cut one the other in one and the selfe same poynt, let the same be O. Wherfore the right lines OK, OE, OI, OM, OH, OZ, OT, and ON, shall be equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ll the on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> to the other, for that they are the halfes of the diameters of equall and like rectangle parallelogra<g ref="char:cmbAbbrStroke">̄</g>mes. Wherefore making the centre the point O, and the space any of these lines OE, or OK. &amp;c. a Sphere described, shall passe by euery one of the angles of the cube, namely, which are at the pointes E, Z, I, T, K, H, M, N, by the 12. definition of the eleuenth, for that all the lines drawen from the point O to the an<g ref="char:EOLhyphen"/>gles of the cube are equall. But the right line EI containeth in power the two equall right lines EZ, and ZI, by the 47. of the first. Wherefore the square of the line EI is double to the square of the line ZI. And forasmuch as the right line KI subtendeth the right angle KEI (for that the right line KE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is erected perpendicularly to the plai<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uperficies of the right lines EZ and ZT (by the 4. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore the square of the line KI is equall to the squares of the lines EI and EK, but the square of the line EI is double to the square of the line EK (for it is double to the square of the line ZI, as hath bene proued, and the bases of the cube are equall squares). Wherefore the square of the line KI is triple to the square of the line KE, that is, to the square of the line ZI. But the right line ZI is equall to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> right line DB, by position, vnto whose square the square of the di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>meter AB is triple, by that which was demonstrated in the 13. Proposition of this booke. Wherefore the diameters KI &amp; DB are equall. Wherefore there is described a cube KI, and it is comprehended in the Sphere geuen wherin the other
<pb facs="tcp:7062:437"/>
solides were contained, the diameter of which Sphere is the line AB. And the diameter KI or AB of the same Sphere, is proued to be in power triple to the side of the cube, namely, to the line DB, or ZI.</p>
               </div>
               <div type="section">
                  <head>¶ Corollaryes added by <hi>Flussas.</hi>
                  </head>
                  <div type="subsection">
                     <head>First Corollary.</head>
                     <p>
                        <hi>Hereby it is manifest, that the diameter of a Sphere containeth in power the sides both of a pyra<g ref="char:EOLhyphen"/>mis and of a cube inscribed in it.</hi>
                     </p>
                     <p>For the power of the side of the pyramis is two thirdes of the power of the diameter (by the 13. of this booke). And the power of the side of the cube is, by this Proposition, one third of the power of the sayd diameter. Wherefore the diameter of the Sphere contayneth in power the sides of the pyra<g ref="char:EOLhyphen"/>mis and of the cube..</p>
                  </div>
                  <div type="subsection">
                     <head>¶ Second Corollary.</head>
                     <p>
                        <hi>All the diameters of a cube cut the one the other into two equall partes in the centre of the sphere which containeth the cube. And moreouer those diameters do in the selfe same point cut into two e<g ref="char:EOLhyphen"/>quall partes the right lines which ioyne together the centres of the opposite bases.</hi>
                     </p>
                     <p>As it is manifest to see by the right line LOF. For the angles LKO, and FIO, are equall, by the 29. of the first: and it is proued, that they are contained vnder equall sides: Wherefore (by the 4. of the first) the bases LO and FO are equall. In like sort may be proued, that the rest of the right lines which ioyne together the centres of the opposite bases do cut the one the other into two equall partes in the centre O.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶ The 4. Probleme. The 16. Proposition. To make an Icosahedron, and to comprehend it in the Sphere geuen, wher<g ref="char:EOLhyphen"/>in were contained the former solides, and to proue that the side of the Ico<g ref="char:EOLhyphen"/>sahedron is an irrationall line of that kinde which is called a lesse line.</head>
               <p>
                  <note place="margin">First part of the construction.</note>TAke the diameter of the Sphere, namely, the line AB: and deuide it in the point C, so that let the line AC be quadruple to the line CB, by the 9. of the sixt. And describe vpo<g ref="char:cmbAbbrStroke">̄</g> the line AB a semicircle ADB. And (by the 11. of the first) from the point Craise vp vnto the line AB a perpendicular line CD. And draw a right line from the point D to the point B. And describe a circle EFGHK whose line from the centre (which let be the point Z) to the circumference, let be equall to the line DB. And in the circle EFGHK describe (by the 11. of the fourth) an equilater and equiangle Pentagon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure EFGHK. And deuide the circumferences EF, FG, GH, HK, and KE, into two equall partes in the pointes L, M, N, X, O. Draw also these right lines LM, MN, NX, XO, and OL: and moreouer these lines OE, EL, LF, FM, MG, GN, NH, HX, XK, and FO, and they shall be the sides of an equilater decagon in<g ref="char:EOLhyphen"/>scribed in the circle EFGHK, by the 29. of the third. Wherefore the figure LMNXO is an equilater pentagon, by the 29. of the third, and the right line EO is the side of a decagon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r ten angled figure. Raise vp (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the eleuenth) from the pointes E, F, G, H, K, and the centre Z, vnto the plaine superficies of the circle, perpendicular lines EP, FR, GS, HT, KV, and ZW, and let ech of them be put equall to the line drawen from the centre of the circle EFGHK, to the circumference, namely, to the line ZE. Wherefore right lines drawen from W to P, from W to V, from W to T, from W to S, from W to R, shall be equall and parallels to right lines drawen from Z to E, from Z to K, from Z to H, from Z to G, and from Z to F, by the 6. and 7. of the eleuenth, and 33. of the first. Wherefore the plaine superficieces EFGHK, and PRSTV, which are extended by those parallel lines, are pa<g ref="char:EOLhyphen"/>rallel
<pb n="408" facs="tcp:7062:437"/>
superficieces, by the 15. of the eleuenth. Wherefore making the centre the point W, and the space WP, or WV, describe a circle, and it shall passe, by the pointes T, S, F, and shall be equall to the circle EFGHK. For the semidiameters of eche are equall. And drawe these right lines PR, RS, ST, TV, VP, and they shall make a pentagon, whose sides shall be e<g ref="char:EOLhyphen"/>quall to the sides of the Pentagon OLMNX, by the 29. of the first. For ech of them doth subtend two sides of the decagon, or
<figure/>
the fift part of equall circles. From the vpper pointes P, R, S, T, V, draw these lines PO, PL, RL, RM, SM, SN, TN, TX, VX, VO: which shall subtend right angles co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the sides of the decagon EL<g ref="char:EOLhyphen"/>FMGNHXKO, and the per<g ref="char:EOLhyphen"/>pendicular lines PE, RF, SG, TH, VK.</p>
               <p>Now forasmuch as the perpendicular lines PE, RF, SG, TH, and VK,<note place="margin">First part of the demonstration.</note> are put e<g ref="char:EOLhyphen"/>quall to the line ZE drawen from the centre, therefore they are equall to the side of an equi<g ref="char:EOLhyphen"/>later hexagon inscribed in the same circle (by the Corollary of the 15. of the fourth). Where<g ref="char:EOLhyphen"/>fore the right lines PO, PL, VO, and VX (which subtend the right angles contained vn<g ref="char:EOLhyphen"/>der those perpendicular lines and the sides of the decagon) containe them in power, by the 47. of the first. But the side of a pentagon (namely, the side LO or PV) containeth in power the sides of an hexagon and of a decagon inscribed in one and the selfe same circle, by the 10. of
<pb facs="tcp:7062:438"/>
this booke. Wherefore the subtending lines PO, PL, VO, VX, TX, TN, SN, SM, RM, RL, containe in power the selfe same square that the sides of the pentagon OLMNX con<g ref="char:EOLhyphen"/>taine, or that the sides of the pentagon PRSTV containe: and therefore those subtending lines are equall to the sides of the foresaid pentagons. Wherefore the triangles contained of those subtending lines and of the sides of the pentagons, and which are ten in number, name<g ref="char:EOLhyphen"/>ly, PLO, OVP, VOX, VXT, TXN, TNS, SNM, SMR, RML, and RLP, are equilater. Againe produce the right line ZW on either side to the points Q &amp; Y and vn<g ref="char:EOLhyphen"/>to the side of the decagon, namely, to the line OE, put the lines ZY and WQ equal. And for<g ref="char:EOLhyphen"/>asmuch as the right line QY is ere<g ref="char:EOLhyphen"/>cted
<figure/>
perpendicularly to the plaine su<g ref="char:EOLhyphen"/>perficies OLMNX, therefore it is also erected perpendicularly to the o<g ref="char:EOLhyphen"/>ther plaine superficies PRSTV,<note place="margin">Second part of the co<g ref="char:cmbAbbrStroke">̄</g>struction.</note> by the Corollay of the 14. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th. And drawe these right lines QP, QV, QT, QS, &amp; QR: and these lines also YL, YM, YN, YX, and YO.</p>
               <p>
                  <note place="margin">Second part of the demo<g ref="char:cmbAbbrStroke">̄</g>stratio<g ref="char:cmbAbbrStroke">̄</g>.</note>Now forasmuch as the lines QP, QV, QT, QS, and QR do eche subtend right an<g ref="char:EOLhyphen"/>gles contayned vnder the sides of an equilater hexagon &amp; of an equilater decagon inscribed in the circle PRSTV or in the circle EFGHK (which two circles are equall) therfore the sayd lines are eche equal to the side of the pentagon inscribed in the foresayd circle by the 10.
<pb n="409" facs="tcp:7062:438"/>
of this booke, and are equall the one to the other, by the 4. of the first, (for all the angles at the poynt W which they subtend are right angles). Wherefore the fiue triangles QPV, QPR, QRS, QST, and QTV, which are contayned vnder the sayd lines QV, QP, QR, QS, QT, and vnder the sides of the pentagon VPRST, are equilater, and equal to the ten for<g ref="char:EOLhyphen"/>mer triangles. And by the same reason the fiue triangles opposite vnto them, namely, the tri<g ref="char:EOLhyphen"/>angles YML, YMN, YNX, YXO, and YOL, are equilater and equal to the said ten triangles. For the lines YL, YM, YN, YX, and YO do subtend right angles co<g ref="char:cmbAbbrStroke">̄</g>tayned vnder the sides. of an equilater hexagon and of an equilater decago<g ref="char:cmbAbbrStroke">̄</g> inscribed in the circle EFGHK, which is equall to the circle PRSTV. Wherefore there is described a solide contayned vnder 20. equilater triangles. Wherefore by the last diffinition of the eleuenth there is described an Icosahedron.</p>
               <p>Now it is required to comprehend it in the sphere geuen, and to proue that the side of the Icosahedron is an irrationall line of that kinde which is called a lesse line. Forasmuch as the line ZW is the side of an hexagon, &amp; the line WQ is the side of a decagon, therfore the line ZQ is diuided by an extreme and meane proportion in the point W, and his greater segme<g ref="char:cmbAbbrStroke">̄</g>t is ZW (by the 9. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th). Wherfore as the line QZ is to the line ZW, so is the line Z<g ref="char:EOLhyphen"/>W to the line WQ. But the ZW is equall to the line ZL by construction, and the line WQ to the line ZY by construction also: Wherefore as the line QZ is to the line ZL, so is the line ZL to the line ZY, and the angles QZL<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and LZY are right angles (by the 2. diffinition of the eleuenth): If therfore we draw a right line from the poynt L to the poynt Q, the angle YLQ shalbe a right angle, by reaso<g ref="char:cmbAbbrStroke">̄</g> of the likenes of the triangles YLQ and ZLQ (by the 8. of the sixth). Wherfore a semicircle described vpo<g ref="char:cmbAbbrStroke">̄</g> the line QY, shal passe also by the point L (by the assumpts added by Campane after the 13. of this booke). And by the same reaso<g ref="char:cmbAbbrStroke">̄</g> al<g ref="char:EOLhyphen"/>so, for that as the line QZ is the line ZW, so is the line ZW to the line WQ, <note place="margin">
                     <p>For the line Q<g ref="char:EOLhyphen"/>W, is equall to the line IZ, &amp; the line ZW, is commo<g ref="char:cmbAbbrStroke">̄</g> to them both.</p>
                     <p>This part is a<g ref="char:EOLhyphen"/>gaine after<g ref="char:EOLhyphen"/>ward demon<g ref="char:EOLhyphen"/>strated by Flus<g ref="char:EOLhyphen"/>sas.</p>
                  </note> but the line ZQ is equall to the line YW, and the line ZW to the line PW: wherefore as the line YW is to the line WP, so is the line PW to the line WQ. And therefore agayne if we draw a right line from the poynt P to the point Y, the angle YPQ shalbe a right angle. Wherfore a semi<g ref="char:EOLhyphen"/>circle described vpon the line QY shal passe also by the point P, by the former assumpts: &amp; if the diameter QY abiding fixed the semicircle be turned round about, vntil it come to the selfe same place from whence it began first to be moued, it shall passe both by the point P, and also by the rest of the pointes of the angles of the Icosahedron, and the Icosahedron shalbe comprehended in a sphere. I say also that it is contayned in the sphere geuen.</p>
               <p>Diuide (by the 10. of the first) the line ZW into two equall parts in the point a. And for<g ref="char:EOLhyphen"/>asmuch as the right line ZQ is diuided by an extreme and meane proportion in the point W, and his lesse segment is QW, therefore the segment QW hauing added vnto it the halfe of the greater segment, namely, the line Wa, is (by the 3. of this booke) in power quintuple to the square made of the halfe of the greater segment: wherefore the square of the line Qa is quintuple to the square of the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>W. But vnto the square of the Qa, the square of the line QY is quadruple (by the corollary of the 20. of the sixth) for the line QY is double to the line Qa: and by the same reason vnto the square of the WA the square of the line ZW is quadruple: Wherefore the square of the line QY is quintuple to the square of the line ZW (by the 15. of the fiueth). And forasmuch as the line AC, is quadruple to the line CB, there<g ref="char:EOLhyphen"/>fore the line AB is quintuple to the line CB. But as the line AB is to the line BC, so is the square of the line AB to the square of the line BD (by the 8 of the sixth, and corollary of the 20. of the same). Wherfore the square of the line AB is quintuple to the square of the line B<g ref="char:EOLhyphen"/>D. And it is is proued that the square of the line QY is quintuple to the square of the line ZW. And the line BD is equall to the line ZW, for either of them is by position equall to the line which is drawen from the centre of the circle EFGHK to the circumference. Where<g ref="char:EOLhyphen"/>fore the line AB is equall to the YQ. But the line AB is the diameter of the sphere geuen: Wherefore the line YQ, which is proued to be the diameter of the sphere contayning the Ico<g ref="char:EOLhyphen"/>sahedron,
<pb facs="tcp:7062:439"/>
is equall to the diameter of the sphere geuen. Wherefore the Icosahedron is contay<g ref="char:EOLhyphen"/>ned in the sphere geue<g ref="char:cmbAbbrStroke">̄</g>. Now I say that the side of the Icosahedron is an irrationall line of that kinde which is called a lesse line. For forasmuch as the diameter of the sphere is rational, and is in power quintuple to the square of the line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre of the circle OLMNX: wherefore also the line which is drawen from the centre of the circle OLMNX is rationall: wherefore the diameter also being co<g ref="char:cmbAbbrStroke">̄</g>mensurable to the same line (by the 6. of the tenth) is ra<g ref="char:EOLhyphen"/>tionall. But if in a circle hauing a rationall line to his diameter be described an equilater pentagon, the side of the pentagon is (by the 11. of this booke) an irrationall line, of that kinde which is called a lesse line. But the side of the pentagon OLMNX is also the side of the Ico<g ref="char:EOLhyphen"/>sahedron described, as hath before ben proued. Wherfore the side of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> is an irra<g ref="char:EOLhyphen"/>tionall line of that kinde which is called a lesse line. Wherefore there is described an Icosahe<g ref="char:EOLhyphen"/>dron and it is contayned in the sphere geuen, and it is proued that the side of the', Icosa<g ref="char:EOLhyphen"/>hedron is an irrationall line of that kind which is called a lesse line. Which was required to be done, and to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest that the diameter of the sphere, is in power quintuple to the line which is drawen from the centre of the circle to the circumference, on which the Icosahedron is described. And that the diameter of the sphere: is com<g ref="char:EOLhyphen"/>posed of the side of an hexagon, and of two sides of a decagon described in one and the selfe same circle.</hi>
                  </p>
                  <p>
                     <hi>Flussas</hi> proueth the Icosahedron described, to be co<g ref="char:cmbAbbrStroke">̄</g>tayned in a sphere, by drawing right lines from the poynt, a, to the poyntes P and G after this maner.</p>
                  <p>Forasmuch as the lines ZW, WP are put equal to the line drawen from the centre to the circum<g ref="char:EOLhyphen"/>ference<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and the line drawen from the centre to the circumference is double to the line aW, by con<g ref="char:EOLhyphen"/>struction: therefore the line WP is also double to the same line aW. Wherefore the square of the line WP is quadruple to the square of the line aW (by the corollary of the 20. of the sixth). And those lines PW and Wa contayning a right angle PWa (as hath before bene proued) are subtended of the right line aP. Wherefore (by the 47. of the first) the line aP contayneth in power the lines PW, and Wa. Wherefore the right line aP is in power quintuple to the line Wa. Wherefore the right lines aP, and aQ being quintuple to one and the same line Wa, are (by the 9. of the flueth) equall. In like sorte also may we proue that vnto those lines aP and aQ, are equall the rest of the lines drawen from the poynt a to the rest of the angles R, S, T, V. For they subtend right angles contayned of the line Wa, and of the lines drawen from the centre to the circumference. And forasmuch as vnto the line Wa is equall the line Va, which is likewise erected perpendicularly vnto the other plaine superficies OLMNX: there<g ref="char:EOLhyphen"/>fore lines drawen from the point a to the angles O, L, M, N, X, and subtending right angles at the point Z contayned vnder lines drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre to the circumference, and vnder the line aZ, are equal not onely the one to the other, but also to the lines, drawen fro<g ref="char:cmbAbbrStroke">̄</g> the sayde poynt a, to the former angles at the poynts, P, R, S, T, V<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. For the lines drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre to the circumference of ech circle are equall, &amp; the line aW is equall to the line aZ. But the line aP is proued equal to the line aQ, which is the halfe of the whole QY. Therefore the residue aY is equall to the foresayd lines aP, aQ &amp;c. Wherefore making the centre the poynt a, and the space one of those lines aQ, aP, &amp;c. extende the su<g ref="char:EOLhyphen"/>perficies of a sphere, &amp; it shal touch the 12. angles of the Icosahedron, which are at the pointes O, L, M. N, X, P, K, S, T, V, Q, Y: which sphere is described, if vpo<g ref="char:cmbAbbrStroke">̄</g> the diameter QY, be drawen a semicircle, and the sayd semicircle be moued about, till it returne vnto the same place from whense it began first to be moued.</p>
               </div>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>The opposite sides of an Icosahedron are parallels. For the diameters of the sphere do fall vpon the opposite angles of the Icosahedron: as it was manifest by the right line QY. If therefore there be ima<g ref="char:EOLhyphen"/>gined to be drawen the two diameters PN, and OM they shall concurre in the point F: wherefore the right lines which ioyne them together, PV, and LN, are in one and the selfe same playne superficies, by
<pb n="413" facs="tcp:7062:439"/>
the 2. of the eleuenth. And forasmuch as the alternate angles at the endes of the diameters are equall (by the 8. of the first): for the triangles contayned vnder equall semidiamete<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s and the side of the Ico<g ref="char:EOLhyphen"/>sahedron are equiangle: therefore (by the 28. of the first) the lines PV and LN are paralles.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 5. Probleme. The 17. Proposition. To make a Dodecahedron, and to comprehend it in the sphere geuen, wherin were comprehended the foresayd solides: and to proue that the side of the dodecahedron is an irrationall line of that kind which is called a re<g ref="char:EOLhyphen"/>siduall line.</head>
               <p>TAke two playne superficieces or bases of the foresayde cube,<note place="margin">First part of the construc<g ref="char:EOLhyphen"/>tion.</note> which let be the two squares ABCD and CBEF, cutting the one the other in the line BC per<g ref="char:EOLhyphen"/>pendicularly according to the nature of a cube. And (by the 10. of the first) di<g ref="char:EOLhyphen"/>uide euery one of the lines AB, BC, CD, DA, EF, EB, and FC into two equall partes in the poyntes G, H, K, L, M, N, X. And drawe these right lines GK and HL, cutting the one the other in the point P, and likewise draw the right lines MH and NX cutting the one the other in the poynt O. And diuide euery one of these right lines NO, OX, HP, and LP, by an extreme and meane proportion in the
<figure/>
points R, S, T, ct, and let their greater segments be RO, OS, TP, and Pct. And (by the 12. of the eleuenth) fro<g ref="char:cmbAbbrStroke">̄</g> the poynts R, O, S, rayse vp to the outward part of the playne superficies E<g ref="char:EOLhyphen"/>BCF of the foresayd cube, perpendicular lines RV, &amp; OYSZ: and let eche of those perpe<g ref="char:cmbAbbrStroke">̄</g>di<g ref="char:EOLhyphen"/>cular lines be equall to one of these lines RO, OS or TP, which perpendiculars shalbe paral<g ref="char:EOLhyphen"/>lels (by the 6. of the eleuenth), and likewise from the pointes T, P, ct, rayse vp vnto the out<g ref="char:EOLhyphen"/>ward part of the playne superficies ABCD of the sayd cube, these perpendicular lines TW, Pst, and ctl, eche of which perpendicular lines put equall also to the line OS, or OR or TP<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the sayd perpendiculars shalbe parallels (by the foresayd 6. of the eleuenth)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And draw these right lines YH, HW, BW, WC, CZ and CB.<note place="margin">The pentagon VBWCZ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> proued equi<g ref="char:EOLhyphen"/>later.</note> Now I say that the pentagon figure VBWCZ is equilater and in one and the self <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame plaine superficies, and moreouer is equi<g ref="char:EOLhyphen"/>angle. Draw these right lines TB, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>B, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>B. And forasmuch as the right line NO is diuided by an extreme and meane proportion in the poynt R, and his greater segment is the line RO, therefore the square of the lines NO and NR are treble to the square of the line RO (by the 4. of this booke). But the line ON is equall to the line NB, and the line OR to the line RV. Wherefore the squares of the lines BN and RN are treble to the square of the line RV. But vnto the squares of the lines BN and NR is equall the square of the line BR (by the 47. of the first). Wherefore the square of the line BR is treble to the square of the line RV. Wherefore the squares of the lines BR and RV are quadruple to the square of the line RV. But vnto the squares of the lines BR and RV is equall the square of the line
<pb facs="tcp:7062:440"/>
BV (by the 47. of the first): for the angle BRV is a right angle (by the 2. diffinition of the eleuenth). Wherefore the square of the line BV is quadruple to the square of the line VR. Wherefore the line BV is double to the line RV (by the Corollary of the 20. of the sixth). And the line ZV is also double to the line RV (for<note n="*" place="margin">The lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> SR is equal &amp; pa<g ref="char:EOLhyphen"/>rallel to the line ZV (by the <hi>33.</hi> of the first) for they ioyne toge<g ref="char:EOLhyphen"/>ther equall and parallel lines, and are in the same su<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ficies with them, by the <hi>7.</hi> of the ele<g ref="char:EOLhyphen"/>uenth.</note> that the line SR is double to the line O<g ref="char:EOLhyphen"/>R, that is, to the line RV which is equall to the line OS). Wherfore the line BV is equall to the line VZ. And forasmuch as the two lines BN
<figure/>
and NR are equall to the two lines BH and H<g ref="char:EOLhyphen"/>T, namely, the wholes and the lesse segme<g ref="char:cmbAbbrStroke">̄</g>ts, and they comprehend right angles, namely, of the squares, BO, and BP, therefore (by the 4. of the first) the bases BR and BT are equall. And for<g ref="char:EOLhyphen"/>asmuch as the lines BR, and BT are equall, and the two lines RV, and TW are also by constru<g ref="char:EOLhyphen"/>ction equall, and the angles BRV, and BTW are by supposition right angles: therefore againe (by the 4. of the first) the bases BV and BW are equall: but the line BV, is proued equall to the line VZ. Wherfore the line BW is also equall to the line VZ. In like sort also may we proue that either of these lines WC, CZ is equal to the same line VZ. Wherefore the pentagon figure BVZ<g ref="char:EOLhyphen"/>CW is equilater.</p>
               <p>
                  <note place="margin">The pentagon VBWCR, proued to be in one and the selfe same playne su<g ref="char:EOLhyphen"/>perficies.</note>Now I say that it is in one and the self same playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies. Forasmuch as the line ZV is a parallell to the line SR (as was before proued) but vnto the same line SR, is the line CB a parallell (by the 28. of the first). Wherfore (by the 9. of the eleuenth) the line VZ is a parallell to the line CB. Wherefore, by the seue<g ref="char:cmbAbbrStroke">̄</g>th of the e<g ref="char:EOLhyphen"/>leuenth, the right lines which ioyne the<g ref="char:cmbAbbrStroke">̄</g> together are in the selfe same playne wherein are the parallell lines. Wherefore the Trapesium BVZ<g ref="char:EOLhyphen"/>C is in one playne. And the triangle BWC is in one playne (by the 2. of the eleuenth). Now to proue that the Trapesium BVZC &amp; the triangle BWC are in one and the self same plaine, we must proue that the right lines YH, and HW are made directly one right line: which thing is thus proued. Forasmuch as the line HP is diuided by an extreme and meane pro<g ref="char:EOLhyphen"/>portion in the point T, and his greater segment is the line PT, therefore as the line HP is to the line PT, so is the line PT to the line TH. But the line HP is equall to the line HO, and the line PT to either of these lines TW and OY. Wherefore as the line HO is to the line O<g ref="char:EOLhyphen"/>Y, so is the line WT to the line TH. But the lines HO and TW being sides of like proportion are parallels (by the 6. of the eleuenth): (For either of them is erected perpendicularly to the plaine superficies BD)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the lines TH and OY are parallels, which are also sides of like proportion, by the same 6. of the eleuenth, (For either of them is also erected perpendicu<g ref="char:EOLhyphen"/>larly to the playne superficies BF.) But when there are two triangles, hauing two sides pro<g ref="char:EOLhyphen"/>portionall to two sides, so set vpon one angle, that their sides of like proportio<g ref="char:cmbAbbrStroke">̄</g> are also parallels (as the triangles YOH and HTW are)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> whose two sides, OH &amp; HT, being in the two ba<g ref="char:EOLhyphen"/>ses of the cube making an angle at the point H, the sides remayning of those triangles shal (by the 32. of the sixth) be in one right line. Wherfore the lines YH &amp; HW make both one right line. But euery right line is (by the 3. of the eleuenth) in one &amp; the self same plaine superficies. Wherefore if ye draw a right line from B to Y, there shalbe made a triangle BWY, which shalbe in one and the selfe same plaine (by the 2. of the eleuenth). And therefore the whole
<pb n="411" facs="tcp:7062:440"/>
pentagon figure VBWCZ is in one and the selfe same playne superficies.</p>
               <p>Now also I say that it is equiangle.<note place="margin">The pentagon VBWCZ, it proued equi<g ref="char:EOLhyphen"/>angle.</note> For forasmuch as the right line NO is diuided by an extreame and meane proportion in the point R, and his greater segment is OR, therefore as both the lines NO and OR added together is to the line ON, so (by the 5. of this booke) is the line ON to the line OR. But the line OR is equall to the line OS. Wherefore as the line SN is to the line NO, so is the line NO to the line OS. Wherfore the line SN is diuided by an extreme and meane proportion in the point O, and his greater segment is the line NO. Wherefore the squares of the lines NS and SO are treble to the square of the line NO (by the 4. of this booke). But the line NO is equall to the NB, and the line SO to the line SZ: wherfore the squares of the lines NS and ZS are treble to the square of the line NB: wher<g ref="char:EOLhyphen"/>fore the squares of the lines ZS, SN and NB, are quadruple to the square of the line NB. But vnto the squares of the lines SN &amp; NB (by the 47. of the first) is equal the square of the line SB: wherefore the squares of the lines BS and SZ, that is, the square of the line BZ, by the 47. of the first, (for the angle ZSB is a right angle by positio<g ref="char:cmbAbbrStroke">̄</g>) is quadruple to the square of the line NB. Wherfore the line BZ is double to the line BN (by the Corollary of the 20. of the sixth). But the line BC is also double to the line BN. Wherefore the line BZ is equall to the line BC. Now forasmuch as these two lines BV and VZ are equall to these two lines B<g ref="char:EOLhyphen"/>W and WC, and the base BZ is equall to the base BC, therefore (by the 8. of the first) the an<g ref="char:EOLhyphen"/>gle BVZ is equall to the angle BWC. And in like sort (by the 8. of the first) may we proue that the angle VZC is equall to the angle BWC (prouing first that the lines CB and CV are equal: which are proued equal by this, that the line NS is equal to the line XR, and ther<g ref="char:EOLhyphen"/>fore the line CR is equal to the line BS, by the 47. of the first: wherfore also by the same y<hi rend="sup">e</hi> line CV is equal to the line BZ, that is, to the line BC (for the lines BC &amp; BZ are proued equal.) Wherefore the three angles BWC, BVZ, and VZC are equall the one to the other. But if in an equilater pentagon figure there be thre angles equall the one to the other, the pentagon is (by the 7. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th) equiangle: wherfore the pentagon BVZCW is equiangle. And it is also proued that it is equilater. Wherfore the pentagon BVZCW is both equilater &amp; e<g ref="char:EOLhyphen"/>quiangle. And it is made vpon one of the sides of the cube, namely, vpon BC. <note n="*" place="margin">Looke for a farther con<g ref="char:EOLhyphen"/>struction after Flussas at the ende of the de<g ref="char:EOLhyphen"/>monstration.</note> If therefore vpon euery one of the twelve sides of the cube be vsed the like construction, there shal then be made a dodecahedron contayned vnder twelue pentagons equilater and equiangle.</p>
               <p>Now it is required to comprehend it in the sphere geuen, and to proue that the side of the dodecahedron is an irrationall line of that kinde which is called a residuall line. Extend the line YO, and let the line extended be YQ: now then the line YQ shall light vppon the diameter of the cube, and shall diuide the one the other into two equall parts.<note place="margin">That the do<g ref="char:EOLhyphen"/>decahedron is contayned in the sphere geuen.</note> For this is manifest to se by the 39. of the eleuenth. (For if by the two lines NX and MH be drawen two playnes perpendicularly to the bases, and cutting the cube, the common section of those playnes shalbe the line YO produced: for their common section is from the poynt O erected perpendicularly to the plaine EBCF, by the 19. of the eleuenth). Let them cut the one the o<g ref="char:EOLhyphen"/>ther in the point Q: wherefore Q is the centre of the sphere which comprehendeth the cube, and YQ is the halfe diameter of the sphere by that which was demo<g ref="char:cmbAbbrStroke">̄</g>strated in the 15. of this booke: wherefore the right lines drawen from the centre Q to all the angles of the cube shalbe equall. And draw a right line from the point V to the point Q. Now forasmuch as the right line NS i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> diuided by an extreme and meane proportion in the point O, and his greater seg<g ref="char:EOLhyphen"/>ment is the line NO, as hath before ben proued, therefore the squares of the lines NS and SO are treble to the square of the line NO, by the 4. of this booke. But the line NS is equal to the line YQ (for the line NO is equal to the line OQ as hath before ben proued, &amp; the line YO to the line OS) being both lesse segmentes: but the line OS is equall to the line TV, for the line RO is equall thereunto: wherefore the squares of the lines QY and YV are treble to the square of the line NO. But vnto the squares of the lines QY &amp; YV the square of the line VQ is equall (by the 47. of the first): wherefore the square of the line VQ is treble to the
<pb facs="tcp:7062:441"/>
square of the line NO. But the semidiameter
<figure/>
of the sphere co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>ding the said cube is in po<g ref="char:EOLhyphen"/>wer treble to the half of the side of the cube. For we haue before (in the 15. of this booke) taught how to make a cube, and to comprehende it in a sphere, and haue proued that the diameter of the sphere is in power treble to the side of the cube. Now in what proportio<g ref="char:cmbAbbrStroke">̄</g> y<hi rend="sup">e</hi> whole is to the whole, in the same is the halfe to the halfe (by the 15. of the fifth). But the line NO is the half of the side of the cube. Wherefore the line VQ is equall to the semidiameter of the sphere co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>ding the cube. But the point Q is the centre of the sphere co<g ref="char:cmbAbbrStroke">̄</g>prehending the cube. Wherefore the point V, which is one of the angles of the dodecahedron, toucheth the superficies of the sphere geuen. In like sort also may we proue, that euery one of the rest of the angles of the dodecahedron toucheth the superficies of the sphere. Wherefore the dodecahedron is comprehended in the sphere geuen.</p>
               <p>Now I say, that the side of the dodecahedron is an irrationall line of that kinde which is called a residuall line.<note place="margin">That the side of the dodeca<g ref="char:EOLhyphen"/>hedron is a residuall line.</note> For forasmuch as the line NO is diuided by an extreme and meane proportion in the point R, and his greater segment is the line OR, and the line OX is also di<g ref="char:EOLhyphen"/>uided by an extreme and meane proportion in the point S, and his greater segment is the line OS. Wherefore the whole line NX is diuided by an extreme and meane proportion, and his greater segment is the line RS. (For for that as the line ON is to the line OR, so is the line OR to the line NR, and in the same proportion also are their doubles (for the partes of eque<g ref="char:EOLhyphen"/>multiplices haue one and the selfe same proportion with the whole, by the 15. of the fifth). Wherefore as the line NX is to the line RS, so is the line RS to both the lines NR and SX added together. But the line NX is greater then the line RS, by both the lines NR and SX added together. Wherefore the line NX is diuided by an extreme and meane proportion, and his greater segment is the line RS. But the line RS is equall to the line VZ, as hath before bene proued. Wherefore the line NX is diuided by an extreme and meane proportion, and his greater segment is the line VZ. And forasmuch as the diameter of the Sphere is ratio<g ref="char:EOLhyphen"/>nall, and is in power treble to the side of the cube, by the 15. of this booke, therefore the line NX, being the side of the cube, is rationall. But if a rationall line be diuided by an extreme and meane proportion, either of the segmentes is (by the 6. of this booke) an irrationall line of that kinde which is called a residuall line. Wherefore the line VZ being the side of the do<g ref="char:EOLhyphen"/>decahedron, is an irrational line of that kinde which is called a residuall line. Wherfore there is made a dodecahedron, and it is co<g ref="char:cmbAbbrStroke">̄</g>prehended in the Sphere geuen, wherein the other solides were contained, and it is proued that the side of the dodecahedron is a residuall line: which was required to be done, and also to be proued.</p>
               <div type="section">
                  <head>¶ Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest, that the side of a cube being diuided by an extreme
<pb n="412" facs="tcp:7062:441"/>
and meane proportion, the greater segment thereof is the side of the dode<g ref="char:EOLhyphen"/>cahedron.</hi> As it was manifest by the line VZ which was proued to be the greater segment of the right line NX, namely, of the side of the cube.</p>
                  <div type="subsection">
                     <head>A further construction of the dodecahedron after <hi>Flussas.</hi>
                     </head>
                     <p>Forasmuch as it hath bene proued that the pentagon BVZCW is equilater and equiangle and toucheth one of the sides of the cube.<note place="margin">
                           <hi>Draw in the former figure these lines, ctA, ctL, c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>D.</hi>
                        </note> Let vs show also by what meanes vpon eche of the 12. sides of the cube may in like sort be applyed pentagons ioyning one to the other, and composing the 12. bases of the dodecahedron. Draw in the former figure these right lines AI, ID, IL, <hi>ct</hi>K. Now forasmuch as the line PL was in the point ct diuided like vnto the lines PH, ON. or OX, and vpon the pointes T, P, ct, were erected perpendicular lines equall vnto the line OY, and the rest: namely, vnto the greater segme<g ref="char:cmbAbbrStroke">̄</g>t: and the lines T W and ct I were proued parallels, therefore the lines WI and Tct are parallels, by the 7. of the eleuenth, and 33. of the first. Wherefore also, by the 9. of the eleuenth, the lines WI and DC are parallels. Wherefore by the 7. of the eleuenth CWID is a playne superficies. And the triangle AID is a playne superficies, by the 2. of the eleuenth. Now it is manifest that the right lines ID, &amp; IA are equall to the right line WC. For the right lines AL &amp; <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ct (which are equall to the right lines BH, &amp; HT) do make the subte<g ref="char:cmbAbbrStroke">̄</g>ded lines A<hi>ct</hi> and BT equall by the 4. of the first. And agayne forasmuch as the lines BT and TW contayne a right angle BTW, as also doo the right lines Act and ctI contayne the right an<g ref="char:EOLhyphen"/>gle ActI (for the right lines WT, and Ict are erected perpendicularly vnto one and the selfe same playne ABCD by supposition). And the squares of the lines BT and TW are equall to the squares of the lines Act, and ctI (for it is proued that the line BT is equall to the line Act, and the line TW to the line ctI). And vnto the squares of the lines BT and TW is equall the square of the line BW, by the 47. of the first: likewise by the same vnto the squares of the lines Act and ctI is equall the square of the line AI. Wherefore the square of the line BW is equall to the square of the line AI, wherefore also the line BT is equall to the line AI. And by the same reason are the lines ID and WC equall to the same lines. Now forasmuch as the lines AI and ID, and the lines AL and LD are equall, and the base IL is common to them both, the angles ALI and DLI shalbe equall, by the 8. of the first: and therefore they are right angles, by the 10. diffinition of the first. And by the same reason are the angles WHB, and W<g ref="char:EOLhyphen"/>HC right angles. And forasmuch as the two lines HT and TW are equall to the two lines Lct and ct<g ref="char:EOLhyphen"/>I, and they contayne equall angles, that is, right angles by supposition, therefore the angles WH<g ref="char:EOLhyphen"/>T, and ILct, are equall by the 4. of the first. Wherefore the playne superficies AID is in like sort incli<g ref="char:EOLhyphen"/>ned to the playne superficies ABCD, as the playne superficies BWC is inclined to the same playne A<g ref="char:EOLhyphen"/>BCD, by the 4. diffinition of the eleuenth. In like sort may we proue that the playne WCDI is in like sort inclined to the playne ABCD, as the playne BVZC is to the playne EBCF. For that in the trian<g ref="char:EOLhyphen"/>gles YOH and <hi>ct</hi>PK which consist of equall sides (eche to his correspondent side), the angles YHO, and <hi>ct</hi>KP, which are the angles of the inclination, are equall. And now if the right line <hi>ct</hi>K be extended to the point a, and the pentagon CWIDa be made perfect, we may, by the same reason, proue that that playne is equiangle and equilater, that we proued the pentagon BVZCW to be equaliter and e<g ref="char:EOLhyphen"/>quiangle. And likewise if the other playnes BWIA and AID be made perfect, they may be proued to be equall and like pentagons and in like sort situate, and they are set vpon these common right lines B<g ref="char:EOLhyphen"/>W, WC, WI, AI, and ID. And obseruing this methode, there shall vpon euery one of the 12. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>id<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s of the cube be set euery one of the 12. pentagons which compose the dodecahedron.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶ Certayne Corollaryes added by <hi>Flussas.</hi>
                  </head>
                  <div type="subsection">
                     <head>First Corollary.</head>
                     <p>The side of a cube, is equall to the right line which subtendeth the angle of the pentagon of a dode<g ref="char:EOLhyphen"/>cahedron contayned in one and the selfe same sphere with the cube.</p>
                     <p>
                        <hi>For the angles BWC and AID, are subtended of the lines BC and AD. Which are sides of the Cube<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </p>
                  </div>
                  <div type="subsection">
                     <head>¶Second Corollary.</head>
                     <p>
                        <hi>In a dodecahedron there are sixe sides euery two of which are parallels and opposite, whose secti<g ref="char:EOLhyphen"/>ons into two equall partes, are coupled by three right lines, which in the center of the sphere which contayneth the dodecahedron, deuide into two equall partes and perpendicularly both them selues and also the sides.</hi>
                     </p>
                     <p>
                        <pb facs="tcp:7062:442"/>For vpon the sixe bases of the cube are set sixe sides of the dodecahedron, as it hath bene pro<g ref="char:EOLhyphen"/>ued (by the lines ZV, WI &amp;c.) which are cutte into two equall partes by right lines, which ioyne together the centers of the bases of the cube, as the line YO produced, and the other like. Which lines coupling together the centers of the bases are three in number, cutting the one the other perpendicu<g ref="char:EOLhyphen"/>larly (for they are parallels to the sides of the cube) and they cutte the one the other into two equall partes in the center of the sphere which contayneth the cube (by that which was demonstrated in the 15. of this booke). And vnto these equall lines, ioyning together the centers of the bases of the cube, are without the bases added equall partes OY, P<hi>ct,</hi> and the other like, which by supposition are equall to halfe of the side of the dodecahedron. Wherefore the whole lines, which ioyne together the sectoins of the opposite sides of the dodecahedron, are equall, and they cut those sides into two equall partes and perpendicularly.</p>
                  </div>
                  <div type="subsection">
                     <head>Third Corollary.</head>
                     <p>
                        <hi>A right line ioyning together the poynts of the sections of the opposite sides of the dodecahedron in<g ref="char:EOLhyphen"/>to two equall partes, being diuided by an extreame and meane proportion: the greater segment thereof shalbe the side of the cube, and the lesse segment the side of the dodecahedron contayned in the selfe same sphere.</hi>
                     </p>
                     <p>For it was proued that the right line YQ is diuided by an extreame and meane proportion in the poynt O, and that his greater segment OQ is halfe the side of the cube, and his lesse segment OY is halfe of the side VZ (which is the side of the dodecahedron). Wherefore it followeth (by the 15. of the fifth) that their doubles are in the same proportion. Wherefore the double of the line YQ which ioy<g ref="char:EOLhyphen"/>neth the poynt opposite vnto the line Y, is the whole: and the greater segment is the double of the line OQ which is the side of the cube: &amp; the lesse segment is the double of the line YO, which is equall to the side of the dodecahedron, namely, to the side VZ.</p>
                  </div>
               </div>
            </div>
            <div type="part">
               <head>¶The 6. Probleme. The 18. Proposition. To finde out the sides of the foresayd fiue bodies, and to compare them together.</head>
               <p>TAke the diameter of the Sphere geuen, and let the same be AB, and diuide it in the point C, so that let the line AC be equall to CB, by the 10. of the first: and in the point D, so that let AD be double to DB, by the 9. of the sixt. And vpon the line AB describe a semicircle AEB. And from the pointes C and D, raise vp (by the 11. of the first) vnto the line AB perpendicular lines CE and DF. And draw these right lines AF, FB, and BE. Now forasmuch as the line AD is double to the line DB, therefore the line AB is treble to the line DB. Wherefore the line BA is sesquialter to the line AD (for it is as 3. to 2.). But as the line BA is to the line AD, so is the square of the line BA to the square of the line AF (by the 6. of the sixt, or by the Co<g ref="char:EOLhyphen"/>rollary of the same, and by the Corollary of the 20. of the same): for the triangle AFB is equiangle to the triangle AFD. Wherefore the square of the line BA is sesquialter to the square of the line AF. But the diameter of a sphere is in power sesquialter to the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the pyramis,<note place="margin">The side of a pyramis.</note> by the 13. of this booke, and the line AB is the diameter of the sphere. Wherefore the line AF is equall to the side of the pyramis.</p>
               <p>Againe forasmuch as the line AB is treble to the line BD: but as the line AB is to the line BD, so is the square of the line AB to the square of the line FB, by the Corollaries of the 8. and 20. of the sixt. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the square of the line AB is treble to the square of the line FB. But the diameter o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a sphere is in power treble to the side of the cube (by the 15. of this booke) and the diameter of the sphere is the line AB. Wherefore the line BF is the side of the cube.<note place="margin">The side of a cube.</note>
               </p>
               <p>And forasmuch as the line AC is equall to the line CB, therefore the line AB is double to the line CB. But as the line AB is to the line CB, so is the square of the line AB to the square of the line BE (by the former Corollaries): Therefore the square of the line AB is double to
<pb n="413" facs="tcp:7062:442"/>
the square of the line BE. But the diameter of the sphere is in power double to the side of the octohedron,<note place="margin">The side of <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> octohedron.</note> and AB is the diameter of the sphere geuen: wherefore the line BE is the side of the octohedron. Raise vp (by the 11. of the first) from the point A vnto the right line AB a perpendicular line AG. And put the line AG equall to the line AB. And drawe a right line from the point G to the point C. And let the line GC cut the circumference in the point H. And (by the 12. of the first) from the point H drawe vnto the line AB a perpendicular line HK. Now forasmuch as the line GA is double to the line AC (for GA is equall to AB). But as GA is to AC, so is HK to KC (by the 28. of the first, and Corollary of the 2. of the sixt): wherefore the line HK is double to the line KC. Wherefore the square of the line HK is quadruple to the square of the line KC, by the Corollary of the 20. of the sixt. Wherefore the squares of the lines HK and KC, which are all one with the square of the line HC, by the 47. of the first, is quintuple to the square of the line KC. But the line HC is e<g ref="char:EOLhyphen"/>quall to the line CB, by the definition of a circle. Wherfore the square of the line BC is quin<g ref="char:EOLhyphen"/>tuple to the square of the line
<figure/>
CK. And forasmuch as the line AB is double to the line BC, of which the line AD is double to the line DB. Wherfore the resi<g ref="char:EOLhyphen"/>due, namely, BD, is double to the residue, namely, to DC (by the 19. of the fift). VVherefore the line BC is treble to the line CD. Wherefore the square of the line BC is nonecuple to the square of the line CD, by the Corollary of the 20. of the sixt. But the square of BC is onely quintuple to the square of CK. Wherefore the square of CK is greater the<g ref="char:cmbAbbrStroke">̄</g> the square of CD, by the 10. of the fift. Wherefore also the line CK is greater then the line DC. Vn<g ref="char:EOLhyphen"/>to the line CK put (by the 2. of the first) an equall line CL. And from the point L raise vp vnto the line AB a perpendicular line LM. And drawe a right line from the point M to the point B. Now forasmuch as the square of the line CB is quintuple to the square of the line CK, and the line AB is double to the line BC, and the line KL is double to the line CK: therefore the square of the line AB is quintuple to the square of the line KL, by the 15. of the fift. But the diameter of a sphere is in power quintuple to the line which is drawen from the centre of the circle to the circumference on which the Icosahedron is described, by the Co<g ref="char:EOLhyphen"/>rollary of the 16. of the sixt. And the line AB is the diameter of the sphere: wherefore the line KL is the semidiameter of the circle on which the Icosahedron is described. Wherefore the line KL is the side of an hexagon figure described in the same circle, by the Corollary of the 15. of the fourth. And forasmuch as the diameter of the sphere is made of the side an hex<g ref="char:EOLhyphen"/>agon figure, and of two sides of a decagon being <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of them described in one and the selfe same circle (by the Corollary of the 16. of this booke): and the line AB is the diameter of the sphere, and the line KL is the side of the hexagon, and the line AK is equall to the line LB: wherefore either of the lines AK and LB is the side of a decagon described in the circle on which the Icosahedron is described (that is, in the circle whose semidiameter is the line KL). And forasmuch as the line LB is the side of a decagon, and the line ML of an hexagon (for ML is equall to KL, for that it is equall to KH, by the 14. of the third, for they are equally
<pb facs="tcp:7062:443"/>
distant from the centre, and ech of the lines HK and KL is double to KC). Wherefore the line M B is the side of a pentagon, by the 10. of this booke. But the side of the pentagon is also the side of the Icosahedron,<note place="margin">The side of an Icosahedron.</note> by that which was demonstrated in the 10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of this booke. Where<g ref="char:EOLhyphen"/>fore the line MB is the side of an Icosahedron.</p>
               <p>And forasmuch as the line FB is the side of a cube, let it be diuided by an extreme and meane proportion in the point N, and let the greater segment therof be NB. Wherefore the line NB is the side of a Dodecahedron,<note place="margin">The sides of a dodecahedron.</note> by the Corollary of the 17. of this booke.</p>
               <p>
                  <note place="margin">Comparison of the fiue sides of the foresayd bodies.</note>And forasmuch as it hath bene proued, by the 13. of this booke, that the diameter of the sphere is in power sesquialter to AF the side of the pyramis, and is in power double to BE the side of the octohedron, by the 14. of the same, and is in power treble to FB the side of the cube, by the 15. of the same. Wherefore it followeth, that of what partes the diameter of the sphere containeth sixe, of such partes the side of the pyramis containeth fower: and the side of the octohedron three: and the side of the cube two. Wherefore the side of the pyramis is in power to the side of the octohedron in sesquitertia proportion: and is in power to the side of the cube in double proportion. And the side of the octohedron is in power to the side of the cube in ses<g ref="char:EOLhyphen"/>quialtera proportion. Where<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oresaid sides of the three figures, that is, of the pyra<g ref="char:EOLhyphen"/>mis, of the octohedron, and o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the cube, are the one to the other in rationall proportions. Wher<g ref="char:EOLhyphen"/>fore they are rationall. But the other two sides, namely, the sides of the Icosahedron and of the Dodecahedron, are neither the one to the other, nor also to the foresaid sides, in rationall pro<g ref="char:EOLhyphen"/>portions: <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or they are irrationall lines, namely, a lesse line, and a residuall line.</p>
               <p>
                  <note place="margin">The side of the Ico<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>edron proued <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>reater then the side o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the dodeca<g ref="char:EOLhyphen"/>hedron.</note>But that M<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the side of the
<figure/>
I<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>sah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron is greater then NB the side of the Dodecahedron may thus be proued. Forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the triangle FDB is equian<g ref="char:EOLhyphen"/>gle to the triangle FAB, by the 8. of the sixt, therefore proporti<g ref="char:EOLhyphen"/>onally, as the line BD is to the line B F, so is the line BF to the line BA. And forasmuch as there are three right lines pro<g ref="char:EOLhyphen"/>portionall, there<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ore as the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst is to the third, so is the square made of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst to the square made of the second, by the Corol<g ref="char:EOLhyphen"/>lary of the 20. of the sixt. Wher<g ref="char:EOLhyphen"/>fore as the line DB is to the line BA, so is the square of the line DB to the square of the line BF. Wherefore (by conuersion, by the Corollary of the 4. of the fiueth) as the line AB is to the line BD, so is the square of the line FB to the square of the line B D. But the line AB is treble to the line BD, as hath before bene proued. Wherefore the square of the line FB is treble to the square of the line BD. But the square of the line AD is qua<g ref="char:EOLhyphen"/>druple to the square of the line DB, by the Corollary of the 20. of the sixt, for the line AD is double to the line DB. Wherefore the square of the line AD is greater then the square of the line FB, by the 10. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift. Wherefore also the line AD is greater then the line FB. Wherefore the line AL is much greater then the line FB. And the line AL being diuided by an extreme &amp; meane proportion, his greater segment is the line KL, by the 9. of this boke: (for the line LK is the side of an hexagon, and the line KA is the side of a decagon inscri<g ref="char:EOLhyphen"/>bed in one and the same circle, as hath before bene proued): and the line FB being diuided
<pb n="414" facs="tcp:7062:443"/>
by an extreme and meane proportion, his greater segment is NB. Wherefore the line KL is greater then the line NB. (<note n="*" place="margin">This Assumpt is afterward proued in the <hi>14.</hi> book<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> and <hi>4.</hi> proposition.</note> For two lines diuided by an extreme and meane proportion, are euery way like proportionall). But the line KL is equall to the line LM. Wherefore the line LM is greater then the line NB. But the line MB is greater then the line LM. Wher<g ref="char:EOLhyphen"/>fore the line MB being the side of the Icosahedron, is much greater then the line NB be<g ref="char:EOLhyphen"/>ing the side of the Dodecahedron: which was required to be done, and to be proued.</p>
               <p>An other way to proue that the line MB is greater then the line NB.<note place="margin">An other demo<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>monstration to proue that the side of the Ico<g ref="char:EOLhyphen"/>sahedro<g ref="char:cmbAbbrStroke">̄</g> is grea<g ref="char:EOLhyphen"/>ter then the side of the dodecahe<g ref="char:EOLhyphen"/>dron.</note> Forasmuch as the line AD is double to the line DB, therefore the line AB is treble to the line DB. But as AB is to BD, so is the square of the line AB to the square of the line BF, by the 8. of the sixt (for the triangle FAB is equiangle to the triangle FDB). Wherefore the square of the line AB is treble to the square of the line BF. And it is before proued, that the square of the line AB is quintuple to the square of the line KL. Wherefore fiue squares made of the line KL, are equall to three squares made of the line FB. But three squares made of the line FB, are greater then sixe squares made of the line NB, as is straight way proued. Wherfore fiue squares made of the line KL, are greater then sixe squares made of the line NB. Wher<g ref="char:EOLhyphen"/>fore also one square made of the line KL, is greater then one square made of the line NB. Wherefore the line KL is greater then the line NB. But the line KL is equall to the line LM. Wherefore the line LM is greater then the line NB. Wherefore the line MB is much greater then the line NB: which was required to be proued.</p>
               <p>But now let vs proue that three squares made of the line FB,<note place="margin">That <hi>3.</hi> squares of the line FB are greater the<g ref="char:cmbAbbrStroke">̄</g> 
                     <hi>6.</hi> squares of the line NB.</note> are greater then sixe squares made of the line NB. Forasmuch as the line BN is greater then the line NF, for it is the greater segment of the line BF diuided by an extreme and meane proportion, therefore that which is contained vnder the lines BF, and BN, is greater then that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vn<g ref="char:EOLhyphen"/>der the lines BF and FN, by the 1. of the sixt. Wherefore that which is contained vnder the lines BF and BN, together with that which is contained vnder the lines BF and FN, is greater then that which is contained vnder the lines BF and FN twise. But that which is contained vnder the lines BF and FN; together with that which is contained vnder the lines BF and BN, is the square of the line BF, by the 2. of the second, and that which is con<g ref="char:EOLhyphen"/>tained vnder the lines BF and FN once, is equall to the square of NB. For the line FB is diuided by an extreme and meane proportion in the point N: (and (by the 17. of the sixt) that which is contained vnder the extremes, is equall to the square made of the midle line). Wherefore the square of the line FB, is greater then the double of the square of the line BN. Wherefore one of the squares made of the line BF, is greater then two squares made of the line BN. Wherefore also three squares made of the line FB, are greater then sixe squares made of the line BN: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>Now also I say that besides the fiue foresayd solides there can not be described any other solide co<g ref="char:cmbAbbrStroke">̄</g>prehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder figures equilater &amp; equiangle the one to the other.<note place="margin">That there can be no other so<g ref="char:EOLhyphen"/>lide besids these fiue, contayned vnder equilater and equiangle bases.</note> For of two trian<g ref="char:EOLhyphen"/>gles, or of any two other playne superficieces can not be made a solide angle (for, that is co<g ref="char:cmbAbbrStroke">̄</g>trary to the diffinition of a solide angle). Vnder three triangles is contayned the solide angle of a pyramis: vnder fower, the solide angle of an octohedro<g ref="char:cmbAbbrStroke">̄</g>: vnder fiue, the solide angle of an Ico<g ref="char:EOLhyphen"/>sahedro<g ref="char:cmbAbbrStroke">̄</g>: of sixe, equilater &amp; equiangle triangles set to one point can not be made a solide an<g ref="char:EOLhyphen"/>gle. For forasmuch as the angle of an equilater triangle is two third partes of a right angle, the sixe angles of the solide shalbe equall to fower right angles, which is impossible. For euery solide angle is (by the 21. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) contayned vnder playne angles lesse the<g ref="char:cmbAbbrStroke">̄</g> fower right angles. And by the same reason can not be made a solide angle contained vnder more the<g ref="char:cmbAbbrStroke">̄</g> sixe playne superficiall angles of equilater triangles. Vnder three squares is contained the angle of a cube. Vnder fower squares it is impossible that a solide angle should be contayned: for then
<pb facs="tcp:7062:444"/>
agayne it should be contayned vnder fower right angles. Wherefore much lesse can any solide angle be contayned vnder more squares then fower. Vnder three equilater and equiangle pen<g ref="char:EOLhyphen"/>tagons is contayned the solide angle of a dodecahedron. But vnder fower it is impossible. For forasmuch as the angle of a pentagon is a right angle and the fift part more of a right angle, the fower angles shalbe greater then fower right angles: which is impossible. And ther<g ref="char:EOLhyphen"/>fore much lesse can a solide angle be composed of more pentagons then fower. Neither can a solide angle be contayned vnder any other equilater and equiangle figures of many angles, for that that also should be absurd. For the more the sides increase, the greater are the angles which they contayne, and therfore the farther of are the superficiall angles contayned of those sides from composing of a solide angle. Wherefore besides the foresayd fiue figures there can not be made any solide figure contayned vnder equall sides and equall angles: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
               </div>
               <div type="section">
                  <head>An Assumpt.</head>
                  <p>But now that the angle of an equilater and equiangle pentagon is a right angle and a fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> more of a right angle, may thus be proued. Suppose that ABCDE be an equilater and equiangle pentagon.<note place="margin">That the angle of an equilater and equiangle Pentagon is one right angle and a <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> part <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> which thing was also before proued in the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> of the <hi>32.</hi> of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>irst.</note> And (by the 14. of the fourth) describe about it a circle ABCDE. And take (by the 1. of the third) the center thereof,
<figure/>
and let the same be F. And draw these right lines FA, FB, FC, FD, FE. Wherefore those lines do diuide the angles of the pentagon into two equall partes in the poyntes A, B, C, D, E, by the 4. of the first. And <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>orasmuch as the fiue angles that are at the poynt F a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e equall to fower right angles, by the corollary of the 15. of the first, and they are equall the one to the other by the 8. of the first: therfore one of those angles, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>or example sake, the angle AFB is a fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>th part lesse then a right angle. Wherfore the angles remayning, namely, FAB, &amp; ABF, are one right angle and a fifth part ouer. But the angle F<g ref="char:EOLhyphen"/>AB is equall to the angle FBC. Wherefore the whole angle ABC being one of the an<g ref="char:EOLhyphen"/>gles of the pentagon is a right angle and a fifth part more then a right angle: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ A Corollary added by <hi>Flussas.</hi>
                  </head>
                  <p>
                     <hi>Now, let vs teach, how those fiue solides, haue eche like inclinations of theyr bases.</hi>
                  </p>
                  <p>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>st let vs take a Pyramis, and diuide one of the sides thereof into two equall parts: and from the two angles opposite vnto that side, d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>aw perpe<g ref="char:cmbAbbrStroke">̄</g>diculars, which shall fall vpon the section, by the co<g ref="char:EOLhyphen"/>rollary of the 12. of the thirtenth,<note place="margin">
                        <hi>The sides of the angle of the in<g ref="char:EOLhyphen"/>cl<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap>tion of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> are <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> rati<g ref="char:EOLhyphen"/>onall.</hi>
                     </note> and at the sayd poynt of diuision (as may easily be proued). Wherfore they shal containe the ang<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e of the inclination of the plaines, by the 4. diffinition of the eleuenth, which angle is subtended of the opposite side of the pyramis. Now forasmuch as the rest of the angles of the inclination of the playnes of the Pyramis, are contayned vnder two perpe<g ref="char:cmbAbbrStroke">̄</g>dicular lines of the triangles, and are subtended of the side of the Pyramis, it foloweth, by the 8. of the fir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t, that those angles are equall. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e (by the 5. diffinition of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) the superficieces are in like sort inclined the one to the other.</p>
                  <p>
                     <note place="margin">
                        <hi>The sides of the angle of the in<g ref="char:EOLhyphen"/>clination of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>f t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap>.</hi>
                     </note>One of the sides of a Cube being diuided into two equall parts, if from the sayd section be drawen in two of the bases thereof, two perpendicular lines, they shalbe parallels and equall to the sides of the square which co<g ref="char:cmbAbbrStroke">̄</g>tayne a right angle. And forasmuch as all the angles of the bases of the Cube are right angles: therefore those perpendiculars falling vpon the section of the side common to the two bases, shall cont<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>yne a right angle (by the 10. of the eleuenth): which selfe angle is the angle of inclination (by the 4. diffinition of the eleuenth) and is subtended of the diameter of the base of the Cube. And by
<pb n="415" facs="tcp:7062:444"/>
the same reason may we proue that the rest of the angles of the inclination of the bases of the cube are right angles. Wherefore the inclinations of the superficieces of the cube the one to the other are equal (by the 5. diffinition of the eleuenth.</p>
                  <p>In an Octohedron take the diameter which coupleth the two opposite angles.<note place="margin">
                        <hi>That the plai<g ref="char:EOLhyphen"/>nes of an octo<g ref="char:EOLhyphen"/>hedron are in li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e sort incli<g ref="char:EOLhyphen"/>ned.</hi>
                     </note> And from those opposite angles draw to one and the sel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e same side of the Octohedron, in two bases thereof, two per<g ref="char:EOLhyphen"/>pendicular lines, which shall diuide that side into two equall parts and perpendicularly (by the Corol<g ref="char:EOLhyphen"/>lary of the 12. of the thirtenth). Wherefore those perpendiculars shall contayne the angle of the incli<g ref="char:EOLhyphen"/>nation of the bases (by the 4. diffinition of the eleue<g ref="char:cmbAbbrStroke">̄</g>th): and the same angle is subtended of the diame<g ref="char:EOLhyphen"/>ter of the Octohedro<g ref="char:cmbAbbrStroke">̄</g>. Wherfore the rest of the angles after the same maner described in the rest of the bases, being comprehended and subtended of equall sides, shall (by the 8. of the first) be equall the one to the other. And therefore the inclinations of the playnes in the Octohedron, shal (by the 5. diffinition of the eleuenth) be equall.</p>
                  <p>In an Icosahedron let there be drawen from the angles of two of the bases,<note place="margin">
                        <hi>That the plai<g ref="char:EOLhyphen"/>nes of an Icosa<g ref="char:EOLhyphen"/>hedron are in like sort incli<g ref="char:EOLhyphen"/>ned.</hi>
                     </note> to one side common to both the sayd bases perpendiculars, which shall contayne the angle of the inclination of the bases (by the 4. diffinition of the eleuenth): which angle is subtended of the right line which subtendeth the angle of the pentagon which contayneth fiue sides of the Icosahedron, by the 16. of this booke: for it coupleth the twoo opposite angles of the triangles which are ioyned together. Wherefore the rest of the angles of the inclination of the bases being after the same maner found out, they shalbe contayned vnder equall sides, and subtended of equall bases, and therefore (by the 8. of the fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>st) those angles shal<g ref="char:EOLhyphen"/>be equall. Wherfore also al the inclinations of the bases of the Icosahedron the one to the other shalb<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> equall, by the 5. diffinition of the eleuenth.</p>
                  <p>In a Dodecahedron,<note place="margin">
                        <hi>That the plai<g ref="char:EOLhyphen"/>nes of a D<gap reason="illegible" resp="#PDCC" extent="5 letters">
                              <desc>•••••</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>hedron are <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> like sort incli<g ref="char:EOLhyphen"/>ned.</hi>
                     </note> from the two opposite angles of two next pentagons draw to theyr common side perpendicular lines, passing by the centres of the sayd pentagons, which shal, where they fal, diuide the side into two equall parts by the 3. of the third. (For the bases of a Dodecahedron are contayned in a circle) And the angle contayn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d vnder those perpendicular lines is the inclination of those bases (by the 4. diffinition of the eleuenth). And the foresayd opposite angles are coupled by a right line equal to the right line which coupleth the opposite sections into two equall parts of the sides of the dodecahe<g ref="char:EOLhyphen"/>dro<g ref="char:cmbAbbrStroke">̄</g> (by the 33. of the first). For they couple together the halfe sids of the dodecahedro<g ref="char:cmbAbbrStroke">̄</g>, which halfes are parallels and equall, by the 3. corollary of the 17. of this booke: which coupling lines also are equall, by the same corollary. Wherefore the angles being contayned of equal perpendicular lines, and subtended of equall coupling lines, shall (by the 8. of the first) be equal. And they are the angles of the inclinations. Wherefore the bases of the dodecahedron are in like sort inclined the one to the other (by the 5. diffi<g ref="char:EOLhyphen"/>nition of the eleuenth).</p>
                  <p>
                     <hi>Flussas</hi> after this teacheth how to know the rationality or irrationality of the sides of the triangles, which contayne the angles of the inclinations of the superficieces of the foresayd bodies.</p>
                  <p>In a Pyramis the angle of the inclinatio<g ref="char:cmbAbbrStroke">̄</g> is contayned vnder two perp<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>dicular lines of the triangles,<note place="margin">
                        <hi>The sides of the angle of the in<g ref="char:EOLhyphen"/>clination of the supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ficieces of the Tetrahedro<g ref="char:cmbAbbrStroke">̄</g> are proued rati<g ref="char:EOLhyphen"/>onall.</hi>
                     </note> and is subtended of the side of the Pyramis Now the side of the pyramis is in power sesquitertia to the perpendicular line, by the corollary of the 12. of this booke: and therfore the triangle co<g ref="char:cmbAbbrStroke">̄</g>tained of those perpe<g ref="char:cmbAbbrStroke">̄</g>dicular lines and the side of pyramis, hath his sides rational &amp; commensurable in power the one to the other.</p>
                  <p>Forasmuch as the twoo sides of a Cube (or right lines equall to them) subtended vnder the dia<g ref="char:EOLhyphen"/>meter of one of the bases,<note place="margin">
                        <hi>The sides of the angle of the in<g ref="char:EOLhyphen"/>clination of the superficieces of the cube proued rationall.</hi>
                     </note> doo make the angle of the inclination: and the diameter of the cube is in power sesquialter to the diameter of the base, which diameter of the base is in power double to the side (by the 47. of the first): therefore those lines are rationall and commensurable in power.</p>
                  <p>In an Octohedron,<note place="margin">
                        <hi>The sides of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> angle &amp;c. of the octohedron proued ratio<g ref="char:EOLhyphen"/>nall.</hi>
                     </note> whose two perpendiculars of the bases contayne the angle of the inclination of the Octohedron, which angle also is subtended of the diameter of the Octohedron, the diameter is in power double to the side of the Octohedron, &amp; the side is in power sequitertia to the perpe<g ref="char:cmbAbbrStroke">̄</g>diclar line, by the 12. of this booke: wherfore the diameter thereof is in power duple superbipartiens tertias to the perpendicular line. Wherfore also the diameter and the perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line are rationall and commensu<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>able (by the 6. of the tenth.)</p>
                  <p>As touching an Icosahedron, it was proued in the 16. of this booke, that the side thereof is a lesse line, when the diameter of the sphere is rationall. And forasmuch as the angle of the inclination of the bases thereof, is contayned of the perpendicular lines of the triangles, and subtended of the right line which subtendeth the angle of the Pentagon which contayneth fiue sides of the Icosahedron: and vn<g ref="char:EOLhyphen"/>to the perpendicular lines the side is commensurable (namely, is in power sesquitertia vnto them, by the Corollary of the 12. of this booke): therefore the perpendicular lines which contayne the angles are irrationall lines, namely, lesse lines (by the 105. of the tenth booke.) And forasmuch as the diameter contayneth in power both the side of the Icosahedron, and the line which subtendeth the foresayd an<g ref="char:EOLhyphen"/>gle, if from the power of the diameter which is rationall, be taken away the power of the side of the Icosahedron which is irrationall, it is manifest that the residue which is the power of the subtending line shalbe irrationall. For if it shoulde be rationall, the number which measureth the whole power of the diameter, and the part taken away of the subtending line, should also, by the 4. common sentence of
<pb facs="tcp:7062:445"/>
the seuenth measure the residue, namely, the power of the side:<note place="margin">
                        <hi>The sides of the angle &amp;c. of the Icosahedron proued irrati<g ref="char:EOLhyphen"/>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>all.</hi>
                     </note> which is irrationall for that it is a lesse line, which were absurd. Wherefore it is manifest that the right lines which compose the angle of the inclination of the bases of the Icosahedron are Irrationall lines. For the subtending line hath to the line contayninge, a greater proportion, then the whole hath to the greater segment.</p>
                  <p>The angle of the inclination of the bases of a dodecahedron, is contayned vnder two perpendicu<g ref="char:EOLhyphen"/>lars of the bases of the dodecahedron, and is subtended of that right line, whose greater segment is the side of a Cube inscribed in the dodecahedron, which right line is equall to the line which coupleth the sections, into two equal parts, of the opposite sides of the dodecahedron. And this coupling line we say is an irrationall line, for that the diameter of the sphere contayneth in power both the coupling line, and the side of the dodecahedron: but the side of the dodecahedron is an irrationall line, namely, a re<g ref="char:EOLhyphen"/>siduall line (by the 17. of this booke). Wherefore the residue namely, the coupling line is an irrationall line, as it is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>asy to proue by the 4. co<g ref="char:cmbAbbrStroke">̄</g>mon sentence of the seue<g ref="char:cmbAbbrStroke">̄</g>th. And that the perpe<g ref="char:cmbAbbrStroke">̄</g>dicular lines which contayne the angle of the inclination are irrationall, is thus proued.</p>
                  <p>Suppose that there be a Pentagon ABCDE, and draw in it the perpendicular line, AG, and let the line subtending the angle of the pentagon be AC. Now forasmuch as the right line AC is the side of the cube, and CD the side the Dodecahedron inscribed in one and the selfe same sphere, by the 2. Corollary of the 17. of this booke: but the line AC is commensurable to the diameter of the sphere, by the 15. of this booke, and is therefore rationall, by the 6.
<figure/>
diffinitio<g ref="char:cmbAbbrStroke">̄</g> of the tenth: &amp; the right line CD which is the side of the dodecahedron is irrationall (by the 17. of this booke). Wherfore the line CG which is the halfe of the line CD is irrationall by the 103. of the 10. boke. And the right line AC contayneth in power the two right lines AG and GC (by the 47. of the first). If therfore from the power of the right line AC being rationall, be take<g ref="char:cmbAbbrStroke">̄</g> away the power of the line CG being irrationall, the power remayning, namely, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he power of the line AG, shall of necessitie be irrationall. For if the power of the line AG taken away should be rationall, and the whole power of the line AC is rationall, the residue, namely, the power of the line CG should be also rationall, and should be measured by the selfe same numbers, by the 4. common sentence of the seuenth. But it is proued that the line C<g ref="char:EOLhyphen"/>G is irrationall, for it is the halfe of the whole residuall line CD (by the 17. of the thirtenth): which is impossible. Wherefore the perpendicular line AG is ir<g ref="char:EOLhyphen"/>rationall. Wherfore the angle of the inclination of the dodecahedron, which is contayned vnder two perpendicular lines of the pentagon, and is subtended of a right line,<note place="margin">
                        <hi>The sides of the angle &amp;c. of the dodecahedro<g ref="char:cmbAbbrStroke">̄</g> proued irratio<g ref="char:EOLhyphen"/>nall.</hi>
                     </note> which coupleth the sections into two equall parts of the opposite sides of the dodecahedron (by the 2. corollary of the 18. of this booke) which line we haue proued to be irrationall (for that it is equall to the two lines AC, and CD by the 4. corollary of the 17. of this booke) is contayned vnder irrationall right lines.</p>
                  <p>By the proportion of the subtending line (of the foresayd angles of inclination) to the lines which containe the angle, is found out the obliquitie of the angle.<note place="margin">
                        <hi>How to know whether the an<g ref="char:EOLhyphen"/>gle of the incli<g ref="char:EOLhyphen"/>natio<g ref="char:cmbAbbrStroke">̄</g> be a right angle, an acute angle, or an ob<g ref="char:EOLhyphen"/>lique angle.</hi>
                     </note> For if the subtending line be in power dou<g ref="char:EOLhyphen"/>ble to the line which contayneth the angle, then is the angle a right angle (by the 48. of the first,) But if it be in power lesse then the double it is an acute angle (by the 23. of the second). But if it be in power more then the double, or haue a greater proportion then the whole hath to the greater segme<g ref="char:cmbAbbrStroke">̄</g>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the angle shalbe an obtuse angle (by the 12. of the second and 4. of the thirtenth). By which may be proued that the square of the whole is greater then the double of the square of the greater segment.</p>
                  <p>This is to be noted that that which <hi>Flussas</hi> hath here taught touching the inclinations of the bases of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iue regular bodies, Hypsicles teacheth after the 5 proposition of the 15. booke. Where he confes<g ref="char:EOLhyphen"/>seth, that he receiued it of one Isidorus, and seking to make the mater more cleare, he endeuored himselfe to declare, that the angles of the inclination of the solides are geuen, and that they are either acute or obtuse, according to the nature of the solide: although <hi>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uclid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> in all his 15. bookes hath not yet shewed, what a thing geuen is. Wherefore <hi>Flussas</hi> framing his demo<g ref="char:cmbAbbrStroke">̄</g>stration vpon an other ground procedeth after an other maner, which semeth more playne, and more aptly here<g ref="char:EOLhyphen"/>to be placed then there. Albeit the reader in that place shal not be frustrate of his also.</p>
               </div>
            </div>
            <trailer>The ende of the thirtenth Booke of Euclides Elementes.</trailer>
         </div>
         <div n="14" type="book">
            <pb n="416" facs="tcp:7062:445"/>
            <head>¶The fourtenth booke of Euclides Elementes.</head>
            <p>IN this booke, which is commonly accompted the 14. booke of <hi>Euclide</hi> is more at large intreated of our prin<g ref="char:EOLhyphen"/>cipal purpose:<note place="margin">
                  <hi>The argument of the fourtenth booke.</hi>
               </note> namely, of the comparison and propor<g ref="char:EOLhyphen"/>tion of the fiue regular bodies (customably called the 5. figures or formes of <hi>Pythagoras</hi>) the one to the other, and also of their sides together, eche to other: which thinges are of most secret vse, and inestimable pleasure, and commoditie to such as diligently search for them, and attayne vnto them. Which thinges also vndoub<g ref="char:EOLhyphen"/>tedly for the woorthines and hardnes thereof (for thinges of most price are most hardest) were first sear<g ref="char:EOLhyphen"/>ched, and found out of Philosophers, not of the inferi<g ref="char:EOLhyphen"/>or or meane sort, but of the depest and most grounded Philosophers, and best exercised in Geometry. And albeit this booke with the booke following, namely, the 15. booke, hath bene hetherto of all men for the most part, and is also at this day numbred and accompted amo<g ref="char:cmbAbbrStroke">̄</g>gst <hi>Euclides</hi> bookes, and supposed to be two of his, namely, the 14. and 15. in order: as all exemplars (not onely new and lately set abroade, but also old monumentes written by hand) doo manifestly witnes: yet it is thought by the best learned in these dayes, that these two bookes are none of <hi>Eu<g ref="char:EOLhyphen"/>clides,</hi> but of some other author, no lesse worthy, nor of lesse estimation and authoritie, notwithstanding, then <hi>Euclide. Apollonius</hi> a man of deepe knowledge a great Philosopher and in Geometrie maruelous (whose wo<g ref="char:cmbAbbrStroke">̄</g>derful bookes writte<g ref="char:cmbAbbrStroke">̄</g> of the sections of cones, which exercise &amp; occupy thewittes of the wisest and best learned, are yet remayning) is thought, and that not without iust cause, to be the author of them, or as some thinke <hi>Hypsicles</hi> him selfe. For what can be more playnely, then that which he him selfe witnes<g ref="char:EOLhyphen"/>seth in the preface of this booke. <hi>Basilides of Tire</hi> (sayth Hypsicles) <hi>and my father together, scanning, and peysing a writing or books of Apollonius, which was of the comparison of a dodecahe<g ref="char:EOLhyphen"/>dron to an Icosahedron inscribed in one and the selfe same sphere, and what proportion these figures had the one to the other, found that Apollonius had fayled in this matter. But afterward</hi> (sayth he) <hi>I found an other copy or booke of Apollonius, wherein the demonstration of that matter was full and perfect, and shewed it vnto them, whereat they much reioysed.</hi> By which wordes it semeth to be manifest that <hi>Apollonius</hi> was the first author of this booke, which was afterward set forth by <hi>Hypsicles.</hi> For so his owne wordes after in the same preface seme to import.</p>
            <div type="part">
               <head>The Preface of <hi>Hypsicles</hi> before the fourtenth booke.</head>
               <p>FRend <hi>Protarchus,</hi> whe<g ref="char:cmbAbbrStroke">̄</g> that <hi>Basilides</hi> of Tire came into <hi>Alexandria,</hi> hauing familiar frendship with my father by reason of his knowledge in the mathe<g ref="char:EOLhyphen"/>maticall sciences, he remayned with him a long time, yea euen all the time of the pestilence. And sometime reasoning betwene themselues of that which <hi>A<g ref="char:EOLhyphen"/>pollonius</hi> had written touching the comparison of a dodecahedron and of an Icosahedron inscribed in one and the selfe same sphere, what proportion such bodies haue the one to the other, they iudged that <hi>Apollonius</hi> had somewhat erred therein. Wherefore they (as my father declared vnto me) diligently weighing it, wrote it perfectly. Howbeit after<g ref="char:EOLhyphen"/>ward I happened to finde an other booke written of <hi>Apollonius,</hi> which contayned in it the
<pb facs="tcp:7062:446"/>
right demonstration of that which they sought for: which when they saw, they much reioy<g ref="char:EOLhyphen"/>sed. As for that which <hi>Apollonius</hi> wrote, may be sene of all men, for that it is in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uery mans hand. And that which was of vs more diligently afterward written agayne, I thought good to send and dedicate vnto you, as to one whome I thought worthy commendation, both for that deepe knowledge which I know you haue in all kindes of learning, and chiefely in Geome<g ref="char:EOLhyphen"/>trie, so that you are able redily to iudge of those thinges which are spoken, and also for the greate loue and good will which you beare towardes my father and me. Wherfore vouchsafe gently to accept this, which I send vnto you. But now is it time to end our preface, and to be<g ref="char:EOLhyphen"/>gin the matter.</p>
               <div type="section">
                  <head>¶ The 1. Theoreme. The 1. Proposition. <note place="margin">First proposi<g ref="char:EOLhyphen"/>tion after Flussas.</note>A perpendicular line drawen from the centre of a circle to the side of a Pentagon described in the same circle: is the halfe of these two lines, name<g ref="char:EOLhyphen"/>ly, of the side of an hexagon figure, and of the side of a decagon figure be<g ref="char:EOLhyphen"/>ing both described in the selfe same circle.</head>
                  <p>SVppose that the circle be ABC.<note place="margin">Construction.</note> And let the side of an equilater Pentagon described in the circle ABC, be BC. And (by the 1. of the third) take the centre of the circle, and let the same be D. And (by the 12. of the first) from the point D draw vnto the line BC a perpendicular line DE. And extend the right line DE direct<g ref="char:EOLhyphen"/>ly to the point F. Then I say, that the line DE (which is drawen from the centre to BC the side of the pentagon) is the halfe of the sides of an hexagon and of a decagon taken together and descri<g ref="char:EOLhyphen"/>bed in the same circle. Draw these right lines DC and CF. And vnto the line EF put an equall line GE. And draw a right
<figure/>
line from the point G to the point C.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as the circumference of the whole circle is quintuple to the circu<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>ference BFC (which is subtended of the side of the penta<g ref="char:EOLhyphen"/>gon) and the circumference ACF is the halfe of the cir<g ref="char:EOLhyphen"/>cumference of the whole circle, and the circumference CF (which is subtended of the side of the decagon) is the halfe of the circumference BCF: therefore the circumference ACF is quintuple to the circumference CF (by the 15. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>t). Wherefore the circumference AC is qradruple to the circumference FC. But as the circumference AC is to the circumference FC, so is the angle ADC to the angle FDC, by the last of the sixt. Wherefore the angle ADC is quadruple to the angle FDC. But the angle ADC is double to the angle EFC, by the 20. of the third: Wherefore the angle EFC is double to the angle GDC. But the angle EFC is equall to the angle EGC, by the 4. of the first. Wherfore the angle EGC is double to the an<g ref="char:EOLhyphen"/>gle EDC. Wherefore the line DG is equall to the line GC (by the 32. and 6. of the first). But the line GC is equall to the line CF, by the 4. of the first. Wherfore the line DG is equall to the line CF. And the line GE is equall to the line EF (by construction). Wherefore the line DE is equall to the lines EF and FC added together. Vnto the lines EF and FC adde the line DE. Wherefore the lines DF and FC added together, are double to the line DE. But the line DF is equall to the side of the hexagon: and FC to the side of the decagon.
<pb n="417" facs="tcp:7062:446"/>
Wherefore the line DE is the halfe of the side of the hexagon, and of the side of the decagon being both added together and described in one and the selfe same circle.</p>
                  <p>It is manifest <note n="*" place="margin">This is mani<g ref="char:EOLhyphen"/>fest by the <hi>12.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g> of the thirtenh booke as Campane well gathereth in a Corollary of the same.</note> by the Propositions of the thirtenth booke, that a perpendicular line drawen from the centre of a circle to the side of an equilater triangle described in the same circle; is halfe of the semidiameter of the circle. Wherefore by this Proposition, a perpendicular dra<g ref="char:EOLhyphen"/>wen from the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ntre of a circle to the side of a Pentag<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>n, is equall to the perpendicular drawen from the centre to the side of the triangle, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nd to halfe of the side of the decagon described in the same circle.</p>
               </div>
               <div type="section">
                  <head>¶ The 2. Theoreme. The 2. Proposition. One and the selfe same circle comprehendeth both the Pentagon of a Do<g ref="char:EOLhyphen"/>decahedron, and the triangle of an Icosahedron,<note place="margin">The <hi>4.</hi> p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>pos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>tio<g ref="char:cmbAbbrStroke">̄</g> after Flussas.</note> described in one and the selfe same Sphere.</head>
                  <p>THis Theoreme is described of <hi>Aristeus</hi> in that booke whose title is, <hi>The comparison of the fiue figures,</hi> and is described of <hi>Apollonius</hi> in his se<g ref="char:EOLhyphen"/>cond edition of the co<g ref="char:cmbAbbrStroke">̄</g>parison of a Dodecahedron to an Icosahedron, which is, <note n="*" place="margin">This is after<g ref="char:EOLhyphen"/>ward proued in the <hi>4.</hi> proposi<g ref="char:EOLhyphen"/>tion.</note> that as the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>icies of a Dodecahedro<g ref="char:cmbAbbrStroke">̄</g> is to the superficies of an Icosahedron, so is the Dode<g ref="char:EOLhyphen"/>cahedron to an Icosahedron, for that a perpendicular line drawen from the centre of a sphere to the pentagon of a dodecahedron and to the triangle of an Icosahedron is one and the selfe same. Now must we also proue that one and the selfe same circle comprehendeth both the pentagon of a Dodecahedron, and also the triangle of an Icosahedron described in one and the selfe same sphere, first this being proued:<note place="margin">This Assumpt is the <hi>3.</hi> proposi<g ref="char:EOLhyphen"/>tion after Flussas.</note>
                     <figure/>
If in a circle be described an equilater pentagon, the squares which are made of the side of the pentagon, and of that right line which is sub<g ref="char:EOLhyphen"/>tended vnder two sides of the pentagon, are quintuple to the square of the semidiameter o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the circle. Suppose that ABC be a circle.<note place="margin">Construction of the Assumpt.</note> And let the side of a pentagon in the circle ABC, be AC. And take (by the 1. of the third) the centre of the circle and let the same be D. And (by the 12. of the first) from the point D draw vnto the line AC a perpendicular line DF. And extend the line DF on either side to the pointes B and E. And draw a right line from the point A to the point B. Now I say, that the squares of the lines BA and AC are quint<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ple to the square of the line DE. Drawe a right line from the point A to the point E. Wherefore the line AE is the side of a decagon figure. And forasmuch as the line BE is double to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> line DE:<note place="margin">Demonstration of the Assumpt.</note> therefore the square of the line BE is quadruple to the square of DE (by the 20. of the sixt). But vnto the square of the line BE, are equall the squares of the lines BA and AE (by the 47. of the first, for the angle BAE is a right angle, by the 31. of the third). Wherefore the squares of the lines BA and AE, are quadruple to the square of the line DE. Wherfore the squares of the lines AB, AE, and DE, are quintuple to the square of the line DE. But the squares of the lines DE and AE, are equall to the square of the line AC (by the 10. of the thirtenth). Wherefore the squares of the lines BA and AC, are quintuple to the square of the line DE.</p>
                  <p>This being thus proued, now is to be demonstrated that one and the selfe same circle co<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>prehendeth both the pentagon of a dodecahedron, &amp; the triangle of an Icosahedron described in one &amp; the self same cir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>le.<note place="margin">Construction of the proposition.</note> Take the diameter of the sphere, &amp;<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> let the same be AB. And in the same sphere describe a dodecahedron, &amp; also an Icosahedron. And let one of the pe<g ref="char:cmbAbbrStroke">̄</g>tagons of the dodecahedron be CDEFG, &amp; let one of the triangles of the Icosahedron be KLH. Now I say that the semidiameters of the circles which are described about them are equall,
<pb facs="tcp:7062:447"/>
that is, that one and the selfe same circle contayneth both the pentagon CDEFG, and the triangle KLH. Draw a right line from the point D to the point G. Wherfore the line DG is the side of a cube (by the corollary of the 17. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th). Take a certayne right line MN. And let the square of the line AB be quintuple to the square of the line MN, by the assumpt put after the 6. propositio<g ref="char:cmbAbbrStroke">̄</g> of the te<g ref="char:cmbAbbrStroke">̄</g>th. But the diameter of a sphere is in power quintuple to the square of the semidiameter of the circle, on which is described the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> (by the corolla<g ref="char:EOLhyphen"/>ry of the 16. of the thirtenth). Wherefore the line MN is the semidiameter of the circle on which is described the Icosahedron. Diuide (by the 30. of the sixth) the line MN by an extreame and meane proportion in the poynt X. And
<figure/>
let the greater segment thereof be MX. Wherefore the line MX is the side of a decagon described in the same circle (by the corollary of the 9. of the thirtenth).<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>op<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>sition.</note> And forasmuch as the square of the line AB is quintuple to the square of the line MN: But the square of the line BA is treble to the square of the line DG (by the co<g ref="char:EOLhyphen"/>rollary of the 15, of the thirtenth). Wherfore three squares of the line DG are equall to fiue squares of the line MN.<note n="*" place="margin">
                        <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                           <desc>〈1 paragraph〉</desc>
                        </gap>
                     </note> But as thre squares of the line DG are to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iue squares of the line MN, so are three squares of the line CG to fiue squares of the line MX. Wherfore three squares of y<hi rend="sup">e</hi> line CG are equll to fiue squares of the line MX. But fiue squares of the line CG are equal to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>iue squares of the line MN &amp; to fiue squares of the MX. For (by the 10. of the thirtenth) one square of the line CG is equall to one square of the line MN &amp; to one square of the line MX. Wherfore fiue squares of the line CG are equall to thre squares of the line DG and to three squares of the line CG (as it is not hard to proue, marking what hath before bene proued). But three squares of the line DG, together with three squares of the line CG, are equall to fiftene squares of the semidiameter of the circle described about the pentagon CDEFG (for it was before proued in the assumpt put in this proposition) that the square<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of DG and G C taken once, are quintuple to the square of the semidiameter of the circle d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>scribed about the pentagon CDEFG). And fiue squares of the line KL are equall to fiftene squares of the semidiameter of the circle described about the triangle KLH. (For by the 12. of the thirtenth, one square of the line LK is triple to one square of the line drawne from the centre to the circumference). Wherefore fiftene squares of the line drawne from the centre to the circumference (of the circle which contayneth the pentagon CDEFG) are equall to fiftene squares of the line drawne from the centre to the circumference of the circle which contayneth the triangle KLH): wherefore one of the squares which is drawne from the centre to the circumference of the one circle, is equall to one of the squares which is drawne from the centre to the circumference of the other circle. Wherefore the diameter is equall to the diameter, wherefore one and the selfe same circle comprehendeth both the pen<g ref="char:EOLhyphen"/>tagon of a dodecahedron and the triangle of an Icosahedron described in one and the selfe same circle: which was required to be proued.</p>
               </div>
               <div type="section">
                  <head>¶ The 3. Theoreme. The .3 Proposition. <note place="margin">Th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 5. propo<g ref="char:EOLhyphen"/>sition a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>.</note>If there be an equilater and equiangle pe<g ref="char:cmbAbbrStroke">̄</g>tagon, aud about it be described a circle, and from the centre to one of the sides be drawne a perpendicular line, that which is contayned vnder one of the sides and the perpendicular
<pb n="418" facs="tcp:7062:447"/>
line thirty times, is equall to the superficies of the dodecahedron.</head>
                  <p>SVppose that ABCD be an equilater and equiangle pentagon. And about the same pentagon, describe (by the 14. of the fourth) a circle.<note place="margin">Construction.</note> And let the centre ther<g ref="char:EOLhyphen"/>of be the poynt F. And from the poynt F draw (by the 12. of the first) vnto the line CD a perpendicular line FG. Now I say that that which is contayned vn<g ref="char:EOLhyphen"/>der the lines CD and GF thirty times, is equall to 12. pentagons of the same quantitie that the pentagon ABCD is. Draw these right lines CF and FD.<note place="margin">Demo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>stra<g ref="char:EOLhyphen"/>tion.</note> Now forasmuch as that which is contayned vnder the lines CD and FG is double to
<figure/>
the triangle CDF (by the 41. of the first) therefore that which is contayned vnder the lines CD and FG fiue times is equall to ten of those triangles. But ten of those triangles are two pentago<g ref="char:cmbAbbrStroke">̄</g>s, and sixe times ten of those triangles are all the pentagons. Wherefore that which is contayned vnder the lines CD and FG thirty times is equall to 12. pentagons But 12. pentagons are the superficies of dodecahedron. Wher<g ref="char:EOLunhyphen"/>fore that which is contayned vnder the lines CD and FG thirty times is equall to the superficies of the dodecahedron.</p>
                  <p>In like sort also may we proue that if there be an equilater triangle, as for example,<note place="margin">The <hi>5.</hi> propo<g ref="char:EOLhyphen"/>sition a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ter F<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ussas.</note> the triangle ABC, and about it be described a circle, and the centre of the circle be the point D, and the perpendicular line be the line DE: that which is contay<g ref="char:EOLhyphen"/>ned
<figure/>
vnder the lines BC and DE thirty times, is equall to the su<g ref="char:EOLhyphen"/>perficies of the Icosahedron.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> For agayne forasmuch as that which is contayned vnder the lines DE and BC is double to the trian<g ref="char:EOLhyphen"/>gle DBC (by the 41. of the first): therefore two triangles are e<g ref="char:EOLhyphen"/>quall to that which is contayned vnder the lines DE and BC, and three of those triangles contayne the whole triangle. Where<g ref="char:EOLhyphen"/>fore sixe such triangles as DBC is, are equall to that which is contayned vnder the lines DE and BC thrise. But sixe s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ch tri<g ref="char:EOLhyphen"/>angles as DBC is, are equall to two such triangles as ABC is. Wherefore that which is contained vnder the lines DE and BC thrise, is equall to two such triangles as ABC is. But two of those triangles take<g ref="char:cmbAbbrStroke">̄</g> ten times contayneth the whole Icosahe<g ref="char:EOLhyphen"/>dron. Wherfore that which is contayned vnder the lines DE &amp; BC thirty times, is equall to twenty such triangles as the triangle ABC is, that is, to y<hi rend="sup">e</hi> whole superficies of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>. <note n="*" place="margin">This is the reason of the Corollary following.</note> Wherefore as the superficies of the dodecahedron is to the supe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ficies of the Icosahedron, so is that which is con<g ref="char:EOLhyphen"/>tayned vnder the lines CD and FG to that which is contayned vnder the lines BC and DE.</p>
                  <div type="subsection">
                     <head>¶ Corollary.</head>
                     <p>By this it is manifest, that as the superficies of the Dodecahedron is to the superficies of the Icosahedron,<note place="margin">A Corollary which also Flussas put<g ref="char:EOLhyphen"/>teth as a Co<g ref="char:EOLhyphen"/>rollary after the <hi>5.</hi> proposi<g ref="char:EOLhyphen"/>tion in his order.</note> so is that which is contained vnder the side of the Pentagon, and the perpe<g ref="char:cmbAbbrStroke">̄</g>dicular line which is drawen from the cen<g ref="char:EOLhyphen"/>tre of the circle described about the Pentagon to the same side, to that which is contained vnder the side of the Icosahedron and the perpendicular line which is drawen from the centre of the circle described about the tri<g ref="char:EOLhyphen"/>angle to the same side: so that the Icosahedron and Dodecahedron be both described in one and the selfe same Sphere.</p>
                  </div>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:448"/>
                  <head>¶ The 4. Theoreme. The 4. Proposition. <note place="margin">The <hi>6.</hi> p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>positio<g ref="char:cmbAbbrStroke">̄</g> 
                        <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>ter Flussas.</note>This being done, now is to be proued, that as the superficies of the Do<g ref="char:EOLhyphen"/>decahedron is to the superficies of the Icosahedron, so is the side of the cube to the side of the Icosahedron.</head>
                  <p>
                     <note place="margin">Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>struction.</note>TAke (by the 2. Theoreme of this booke) a circle containing both the penta<g ref="char:EOLhyphen"/>gon of a Dodecahedron, and the triangle of an Icosahedron, being both de<g ref="char:EOLhyphen"/>scribed in one and the selfe same sphere, and let the same circle be DBC. And in the circle DBC describe the side of an equilater triangle, namely, CD, and the side of an equilater pentagon, namely, AC. And take (by the 1. of the third) the centre of the circle, and let the same be E. And from the point E drawe vnto the lines DC and AC, perpendicular lines EF and EG. And extend the line EG directly to the point B. And drawe a right line from the point B to the point C. And let the side of the cube be the line H. Now I say, that as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the line H to the line CD.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Forasmuch as the line made of the lines EB and BC added together (namely, of the side of the hexagon, and of the side of a decagon) is (by the 9. of the thirtenth) diuided by an extreme and meane proportion, and his greater segment is the line BE: and the line EG is also (by the 1. of the fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>retenth) the halfe of the same line, and the line EF is the halfe of the line BE (by the Corollary of the 12. of the thirtenth). Wherefore the line EG being diuided by
<figure/>
an extreme and meane proportion,<note n="*" place="margin">This is not hard to proue by the <hi>15. 16.</hi> and <hi>19.</hi> of the <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>eth.</note> his greater seg<g ref="char:EOLhyphen"/>ment shall be the line EF. And the line H also being diuided by an extreme &amp; meane proportion, his grea<g ref="char:EOLhyphen"/>ter segment is the line CA, as it was proued<note n="†" place="margin">In the Corol<g ref="char:EOLhyphen"/>lary of the <hi>17.</hi> of the t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>irte<g ref="char:cmbAbbrStroke">̄</g>th.</note> in the Dodecahedron. <note n="*" place="margin">
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> againe is required the Assumpt which is afterward proued in this <hi>4</hi> proposition.</note> Wherefore as the line H is to the line CA, so is the line EG to the line EF. Wherefore (by the 16. of the sixt) that which is contained vnder the lines H and EF, is equall to that which is contained vnder the lines CA and EG. And for that as the line H is to the line CD, so is that which is contained vn<g ref="char:EOLhyphen"/>der the lines H and EF, to that which is contained vnder the lines CD and EF (by the 1. of the sixt). But vnto that which is contained vnder the lines H and EF, is equall that which is contained vnder the lines CA and EG. Wherefore (by the 11. of the fift) as the line H is to the line CD, so is that which is contained vnder the lines CA and EG, to that which is contained vnder the lines CD and EF, that is (by the Corollary next going before) as the superficies of the Dodeca<g ref="char:EOLhyphen"/>hedron is to the superficies of the Icosahedron, so is the line H to the line CD.</p>
                  <div type="subsection">
                     <head>An other demonstration to proue that as the superficies of the Dodecahe<g ref="char:EOLhyphen"/>dron is to the superficies of the Icosahedron, so is the side of the cube to the side of the Icosahedron. <note n="†" place="margin">But first the Assumpt follow<g ref="char:EOLhyphen"/>ing, the con<g ref="char:EOLhyphen"/>struction wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>f here beginne<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>h, is to be proued.</note>
                     </head>
                     <p>LEt there be a circle ABC. And in it describe two sides of an equilater pen<g ref="char:EOLhyphen"/>tagon (by the 11. of the fift) namely, AB and AC: and draw a right line from the point B to the point C. And (by the 1. of the third) take the centre of the circle, and let the same be D. And draw a right line from the point A to the point D, and extend it directly to the point E, and let it cut the line BC in the point G. And let the line DF be halfe to the line DA, and let
<pb n="419" facs="tcp:7062:448"/>
the line GC be treble to the line HC, by the 9. of the sixt.<note place="margin">The Assumpt, which also Flus<g ref="char:EOLhyphen"/>sas putteth as an Assumpt a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ter the <hi>6.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> Now I say, that that which is contained vnder the lines AF and BH, is equall to the pentagon inscribed in the circle ABC. Draw a right line from the point B to the point D. Now forasmuch as the line AD is double to the line DF, there<g ref="char:EOLhyphen"/>fore the line AF is sesquialter to the line AD.
<figure/>
Againe,<note place="margin">Demonstration of the Assumpt.</note> forasmuch as the line GC is treble to the line CH, therefore the line GH is double to the line CH. Wherefore the line GC is sesquialter to the line HG. Wherefore as the line FA is to the line AD, so is the line GC to the line GH. Wherefore (by the 16. of the sixt) that which is contained vnder the lines AF &amp; HG, is equall to that which is contained vnder the lines DA and GC. But the line GC is equall to the line BG (by the 3. of the third). Wherfore that which is contained vnder the lines AD and BG, is e<g ref="char:EOLhyphen"/>quall to that which is contained vnder the lines AF and GH. But that which is contained vn<g ref="char:EOLhyphen"/>der the lines AD and BG, is equall to two such triangles as the triangle ABD is (by the 41. of the first). Wherefore that which is contained vnder the lines AF and GH, is equall to two such triangles as the triangle ABD is. Wherefore that which is contained vnder the lines AF and GH <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>iue times, is equall to ten triangles. But ten triangles are two pentagons. Wherefore that which is contained vnder the lines AF and GH fiue times, is equall to two pentagons. And forasmuch as the line GH is double to the line HC, therefore that which is contained vnder the lines AF and GH, is double to that which is contained vnder the lines AF and HC (by the 1. of the sixt). Wherefore that which is contained vnder the lines AF and CH twise, is equall to that which is contained vnder the lines AF and GH once. Take eche of those parallelogrammes fiue times. Wherefore that which is contained vnder the lines AF and HC ten times, is equall to that which is contained vnder the lines AF &amp; GH fiue times, that is, to two pentagons. Wherefore that which is contained vnder the lines AF and HC fiue times, is equall to one pentagon. But that which is contained vnder the lines AF and HC fiue times, is equall (by the 1. of the sixt) to that which is contained vnder the lines AF and HB, for the line HB is quintuple to the line HC (as it is easie to see by the con<g ref="char:EOLhyphen"/>struction) and they are both vnder one &amp; the selfe same altitude, namely, vnder AF. Wher<g ref="char:EOLhyphen"/>fore that which is contained vnder the lines AF and BH, is equall to one pentagon.</p>
                  </div>
                  <div type="subsection">
                     <head>This being proued, now let there be drawne a Circle comprehending both the Pentagon of a Dodecahedron, and the triangle of an Icosahedron, being both described in one and the selfe same Sphere.</head>
                     <p>
                        <note place="margin">Construction pertaining to the second d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                           <g ref="char:EOLhyphen"/>monstration of the <hi>4.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g>.</note>LEt the circle be ABC. And in it describe as before, two sides of an equilater pentagon, namely BA and AC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and draw a right line from the point B to the point C: and take the centre of the circle and let the same be E. And from the point A to the point E draw a right line AE: and extend the line AE to the point F. And let it cut the line BC in the point K. And let the line AE be do<g ref="char:EOLhyphen"/>ble to the line EG, &amp; let the line CK be treble to the line CH, by the .9. of the sixth. And fro<g ref="char:cmbAbbrStroke">̄</g> the point G raise vp (by the .11. of the first) vnto the line AF a perpendicular line GM: and extend the line GM directly to the point D. Wherfore the line MD is the side of an equi<g ref="char:EOLhyphen"/>liter tria<g ref="char:cmbAbbrStroke">̄</g>gle, by the corollary of the .1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. of the thirtenth: draw these right lines AD and AM. Wherfore ADM is an equilater triangle.<note place="margin">Second demon<g ref="char:EOLhyphen"/>stration o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> the <hi>4.</hi> proposition.</note> And for as much as that which is contained vn<g ref="char:EOLhyphen"/>der the lines AG and BH is equal to the pentagon (by the former assump<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>) and that which
<pb facs="tcp:7062:449"/>
is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines AG and GD is equal
<figure/>
to the triangle ADM: therefore as that which is contained vnder the lines AG and HB is to that which is contained vnder the lines AG and GD, so is the pentagon to the tria<g ref="char:cmbAbbrStroke">̄</g>gle. But as that which is contained vnder the lines BH &amp; AG is to that which is contained vnder the lines AG and GD, so is the line BH to the line DG (by the .1. of the sixth) wherefore (by the .15. of the fifth) as 12. such lines as BH is, are to .20 such lines as DG is, so are 12. pentagons to 20. triangles, that is the su<g ref="char:EOLhyphen"/>perficies of the Dodecahedron, to the superficies of the Icosahedron. And 12. suche lines as BH is, are equall to tenne suche lines as BC is (for the line HB is quintuple to the line HC): and the line BC is sextuple to the line CH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> Wherfore six such lines as BH is, are equal to fiue such lines as BC are: and in the same pro<g ref="char:EOLhyphen"/>portion are their doubles: and 20. such lines as the line DG is, are equal to .10. such lines as the line DM is: for the line DM is double to the line DG. Wherfore as 10. such lines as BC is, are to 10. such lines as DM is, that is, as the line BC is to the line DM, so is the superficies of the Dodecahedron to the superficies of the Icosahedron. But the line BC is the side of the cube, and the line DM the side of the Icosahedron: wherefore (by the 11. of the fifth) as the superficies of the Dodecahedron is to the superficies of the Icosahedron, so is the line BC to the line DM, that is, the side of the cube to the side of the Icosahedron.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">The <hi>7<gap reason="illegible" resp="#PDCC" extent="1 letter">
                                 <desc>•</desc>
                              </gap>
                           </hi> propo<g ref="char:EOLhyphen"/>sition after Flussas.</note>Nowe will we proue that a right line being deuided by an extreme and meane proportio<g ref="char:cmbAbbrStroke">̄</g>, what proportio<g ref="char:cmbAbbrStroke">̄</g> the line co<g ref="char:cmbAbbrStroke">̄</g>taining in power the squares of the whole line and of the greater segment, hath to the line containing in power the squares of the whole line and of the lesse segment, the same proportion hath the side of the cube to the side of the Icosahedron, being both described in one and the selfe same sphere.</head>
                     <p>
                        <note place="margin">Construction.</note>SVppose that AB be a circle conta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ning both the pentagon of a Dodecahedron &amp; the triangle of an Icosahedron described bothe in one and the selfe same sphere. Take the centre of the circle, and let the same be C. And from the point C extend to the circumference a right line at all auentures, and let the same BC. And (by the 30. of the sixth) deuide the line BC by an extreme and meane proportion in the point D, and let the greater segment therof be CD. Wherfore the line CD is the side of a Decago<g ref="char:cmbAbbrStroke">̄</g> de<g ref="char:EOLhyphen"/>scribed in the same circle (by the corollary of the 9. of the thirtenth). Take the side of an Ico<g ref="char:EOLhyphen"/>sahedron, and let the same be the line E, and the side of a Dodecahedron, and let the same be the line F, and the side of a cube &amp; let the same be the line G.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Wherfore the line E is the side of an equilater triangle, and F of an equaliter pentagon described in one and the selfe same circle. And the line G being deuided by an extreme and meane proportion, his greater seg<g ref="char:EOLhyphen"/>ment is the line F, by the corollary of the 17. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th. Now forasmuch as the line E is the side of an equilater triangle, but (by the 12. of the thirtenth) the side of an equilater triangle is in power treble to the line BC, (which is drawne from the center to the circum<g ref="char:EOLhyphen"/>ference) therefore the square of the line E is treble to the square of the line BC: but the squares of the line BC and BD are (by the 4. of the thirtenth) treble to the square of the line CD. Wherfore as the square of the line E is to the square of the line CB, so are the squares of the lines CB and BD to the square of the line CD. Wherefore alternately (by the 16. of
<pb n="420" facs="tcp:7062:449"/>
the fifth) as the square of the line E is to the squares
<figure/>
of the lines CB and BD, so is the square of the line CB to the square of the line CD. <note n="*" place="margin">Here againe is required the Assumpt af<g ref="char:EOLhyphen"/>terward pro<g ref="char:EOLhyphen"/>ued in this <hi>4.</hi> proposition.</note> But as the square of the line BC is to the square of the line CD, so is the square of the line G (the side of the cube) to the square of the line F, the side of Dodecahedron. For the line F is the greater segme<g ref="char:cmbAbbrStroke">̄</g>t of the line G (as was before proued.) Wherfore (by the .11. of the fift) as the square of the line E is to the squares CB and BD, so is the square of the line G, to the square of the line F. Wherefore alternately (by the 16. of the fifth) &amp; also by conuersion (by the corollary of the 4. of the fift) as the square of the line G, is to the square of the line E, so is the square of the line F, to the squares of the lines CB &amp; BD. But vnto the square of the line F are equal the squares of the lines BC &amp; CD, for the side of a pentagon co<g ref="char:cmbAbbrStroke">̄</g>taineth in power both the side of a sixe angled figure, and the side of a ten angled figure (by the 10. of the thirtenth.) Wherfore as the square of the line G, is to the square of the line E, so are the squares of the lines BC and CD to the squares of the lines CB and BD. But as the squares of the lines CB and CD are to the squares of the lines CB &amp; BD,<note n="†" place="margin">As may by the Assumpt afterward in this propositio<g ref="char:cmbAbbrStroke">̄</g> be plainely proued.</note> so (any right line what so euer it be, being diuided by an extreme and meane proportion) is the line containing in power the squares of the whole line, and of the greater segme<g ref="char:cmbAbbrStroke">̄</g>t, to the line containing in power the squares of the whole line, and of the lesse segment: wherfore (by the 11. of the fifth) as the square of the line G (the side of the cube) is to the square of the line E, so (any right line being deuided by an extreme and meane proportion) is the line containing in power the squares made of the whole line, and of the greater segme<g ref="char:cmbAbbrStroke">̄</g>t, to the line containing in power the squares made of the whole line, and of the lesse segment: but the line G is the side of the Cube, and the line E of the Icosahedron (by supposition.) If therfore a right line be deuided by an extreeme and meane proportion, as the line co<g ref="char:cmbAbbrStroke">̄</g>taining in power the squares of the whole line, and of the greater segment, is to the line containing in power the squares of the whole line and of the lesse segment: so is the side of the cube to the side of the Icosahedron, being both described in one and the selfe same sphere.</p>
                  </div>
                  <div type="subsection">
                     <head>Now will we proue that as the side of the Cube is to the side of the Ico<g ref="char:EOLhyphen"/>sahedron,<note place="margin">The <hi>8.</hi> pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>i<g ref="char:EOLhyphen"/>tion a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ter Flussas.</note> so is the solide of the Dodecahedron to the solide of the Icosa<g ref="char:EOLhyphen"/>hedron.</head>
                     <p>FOrasmuche as equal circles comprehend both the pentagon of a Dodecahe<g ref="char:EOLhyphen"/>dron, and the triangle of an Icosahedron, being both described in one and the selfe same sphere, by the 2. of this booke: but in a sphere equal circles are equally distant from the centre (for the perpendicular lines drawn from the centre of the sphere to the plaine superficieces of the circles are equal, and do fall vpon the centres of the circles. <note n="†" place="margin">By the Co<g ref="char:EOLhyphen"/>rollary added by Flussas af<g ref="char:EOLhyphen"/>ter has As<g ref="char:EOLhyphen"/>sumpt put af<g ref="char:EOLhyphen"/>ter the <hi>17.</hi> pro<g ref="char:EOLunhyphen"/>position of the <hi>12.</hi> booke.</note>Wherfore perpendicular lines drawne from the centre of the sphere, to the centre of the circle, comprehending bothe the triangle of an Icosahedron, and the pentagon of a Dodecahedron are equal: wherefore the pyramides, whose bases are the pentagons of the Dodecahedron, are of equal altitude with the piramides whose bases are the triangles of the Icosahedron. But piramids of equal altitude, are in that proportion the one to the other, that their bases are (by the 5. of the twelfth) wherefore as the pentagon is to the triangle, so is the pyramis whose base is the pentagon of the Dodecahe<g ref="char:EOLhyphen"/>dron and toppe the centre of the sphere, to the pyramis whose base is the triangle and top the
<pb facs="tcp:7062:450"/>
centre also of the sphere. Wherfore (by the 15. of the fifth) as 12. pe<g ref="char:cmbAbbrStroke">̄</g>tagons are to 20. triangles, so are 12. pyramids hauing pentagons to theyr bases to 20. pyramids hauing tria<g ref="char:cmbAbbrStroke">̄</g>gles to their bases. But 12. pentagons are the superficies of the D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>decahedron, and 20. triangles are the superficies of the Icosahedron. Wherefore as the superficies of the Dodecahedron is to the su<g ref="char:EOLhyphen"/>perficies of the Icosahedron, so are 12. pyramids hauing pentagons to their bases to 20. pyra<g ref="char:EOLhyphen"/>mids hauing triangles to their bases. But 12. pyramids hauing pentagons to their bases, are the solide of the Dodecahedron, and 20. pyramids hauing triangles to their bases are the so<g ref="char:EOLhyphen"/>lide o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the Icosahedron.<note place="margin">Corollary of the <hi>8.</hi> after Flussas.</note> Wherfore (by the 11. of the fifthe) as the superficies of the Dodecahedron is to the superficies of the Icosahedron <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>o is the solide of the Dodecahedron to the solide of the Icosahedron. But as the superficies of the Dodecahedron, is to the superficies of the Icosahedron, so haue we proued that the side o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the cube is to the side of the Icosahedron. Wherfore, by the 11. of the fifth, as the side of the cube is to the side of the Icosahedron, so is the solide of the Dodecahedron to the solide of the Icosahedron.</p>
                  </div>
                  <div type="subsection">
                     <head>
                        <note place="margin">This Assumpt is the <hi>3.</hi> propositio<g ref="char:cmbAbbrStroke">̄</g> a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ter <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>lussas, and is it which <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> times hath bene taken a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>aunted in this booke, and o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ce also in the last proposition of the <hi>13.</hi> booke: as we haue be<g ref="char:EOLhyphen"/>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ore noted.</note>If two right lines be diuided by an extreame and meane proportion, they shall euery way be in like proportion: which thing is thus demonstrated.</head>
                     <p>LEt the line AB be (by the 30. of the sixth) diuided by an extreame and meane pro<g ref="char:EOLhyphen"/>portion in the poynt C, and let the greater segment thereof be the line CA. And likewise also let the line DE be diuided by an extreame and meane proportion in the poynt F, and let the greater segment thereof be the line DF. Then I say that as the whole line AB is to the greater segment thereof AC, so is the whole line DE to the grea<g ref="char:EOLhyphen"/>ter segment thereof DF. For forasmuch as that which is contayned vnder the lines AB and BC is equall to the square of the line AC (by the diffinition of a line diuided be an ex<g ref="char:EOLhyphen"/>treame and meane proportion):<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> and that which is contayned vnder the lines DE and EF is also equall to the square of the line DF (by the same diffinition): therefore as that which is contayned vnder the lines AB and BC is to the square of the line AC, so is that which is contayned vnder the lines DE and EF to the square of the line DF. For in eche is the pro<g ref="char:EOLhyphen"/>portion of equalitie. Wherfore as that which is contayned vnder the lines AB and BC fower times, is to the square of the line
<figure/>
AC, so is that which is contaynd vnder the lines DE and EF fower times to the square of the line DF (by the 15. of the fifth). Wherfore by composition (by the 18. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>th) as that which is contayned vnder the lines AB and BC fower times, together with the square of the line AC, is to the square of the line AC, so is that which is contayned vnder the lines DE and EF fower times, together with the square of the line DF, to the square of the line DF. Wherefore as the square which is made of the lines AB and BC ad<g ref="char:EOLhyphen"/>ded together and made one line (which square by the 8. of the second is equall to that which is contayned vnder the lines AB and BC fower times together with square of the line AC) is to the square of the line AC, so is the square made of the lines DE &amp; EF added together and made one line (which square is also, by the same, equal to that which is contayned vnder the lines DE and EF fower times together with the square of the line DF) to the square of the line DF. Wherefore also as the lines AB &amp; BC added together are to the line AC, so are the lines DE &amp; EF added together to the line DF (by the 22. of the sixt). Wherefore by co<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>position (by the 18. of the fifth) as both the lines AB &amp; BC added the one to the other, toge<g ref="char:EOLhyphen"/>ther with the line AC, that is, as two such lines as AB is, are to the line AC, so are both the lines DE and EF added the one to the other together with the line DF, that is two such
<pb n="421" facs="tcp:7062:450"/>
lines as DE is to the line DF. And in the same proportion are the halues of the antecedents by the <hi>15.</hi> of the fifth. Wherefore as the line AB is to the line AC, so is the line DE to the line DF. (And therefore by the 19. of the fifth, as the line AB is to the line BC, so is the line DF to the line FE. Wherefore also by diuision by the 17. of the fifth, as the line AC is to the line CB, so is the line DF to the line DE).</p>
                     <p>Now that we haue proued, <note n="*" place="margin">In the <hi>4.</hi> sec<g ref="char:EOLhyphen"/>tion <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>f this proposition.</note> that, any right line whatsoeuer being diuided by an extreame and meane proportion, what proportion the line contayning in power the squares made of the whole line and of the greater segment added together, hath to the line contayning in power the squares made of the whole line and of the lesse segment added together, the same propor<g ref="char:EOLhyphen"/>tion hath the side of the cube to the side of the Icosahedron: Now also that we haue proued, <note n="†" place="margin">In the 1. and <hi>3</hi> section of the same propositio<g ref="char:cmbAbbrStroke">̄</g>.</note> that as the side of the cube is to the side of the Icosahedron, so is the superficies of the Dode<g ref="char:EOLhyphen"/>cahedron to the superficies of the Icosahedron, being both described in one and the selfe same sphere: and moreouer seing that we haue proued, <note n="†" place="margin">In the <hi>5.</hi> sectio<g ref="char:cmbAbbrStroke">̄</g> of the same proposition.</note>that as the superficies of the Dodecahedron is to the superficies of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, so is the Dodecahedro<g ref="char:cmbAbbrStroke">̄</g> to the Icosahedron, for that both the pentagon of the Dodecahedron, and the triangle of the Icosahedron are comprehended in one and the selfe same circle:<note place="margin">A Corollary.</note> All these thinges I say being proued, it is manifest, that if in one and the selfe same sphere be described a Dodecahedron, and an Icosahedron, they shall be in proportion the one to the other, as, a right line whatsoeuer being diuided by an extreame and meane pro<g ref="char:EOLhyphen"/>portion, the line contayning in power the squares of the whole line and of the greater segment added together, is to the line containing in power the squares of the whole line and of the lesse segment added together. For for that as the Dodecahedron is to the Icosahedron, so is the superficies of the Dodecahedron to the superficies of the Icosahedron, that is, the side of the cube to the side of the Icosahedron: but as the side of the cube is to the side of the Icosahedron, so, any right line what so euer being diuided by an extreame and meane proportion, is the line contayning in power the squares of the whole line and of the greater segment added together, to the line contayning in pow<g ref="char:EOLhyphen"/>er the squares of the whole line and of the lesse segment added together. Wherefore as a Dodecahedron is to an Icosahedron described in one and the selfe same sphere, so, any right line what so euer being diuided by an extreame and meane proportion, is the line contayning in power the squares of the whole line &amp; of the greater segment added together, to the line contayning in power the squares of the whole line and of the lesse segment added together.</p>
                  </div>
               </div>
            </div>
            <trailer>The ende of the fourtenth Booke of Euclides Elementes after Hypsicles.</trailer>
         </div>
         <div n="14" type="book">
            <pb facs="tcp:7062:451"/>
            <head>¶The fourtenth booke of Euclides Elementes after Flussas.</head>
            <p>FOr that the fouretenth Booke, as it is set forth by Flus<g ref="char:EOLhyphen"/>sas, containeth in it moe Propositions then are found in Hypsicles, &amp; also some of those Propositions which Hypsicles hath, are by him somewhat otherwise de<g ref="char:EOLhyphen"/>monstrated, I thought my labour well bestowed for the readers sake to turne it also all whole, notwithstan<g ref="char:EOLhyphen"/>ding my trauaile before taken in turning the same booke after Hypsicles. Where note ye, that here in this 14. booke after Flussas, and in the other bookes follo<g ref="char:EOLhyphen"/>wing, namely, the 15. and 16. I haue in alleadging of the Propositions of the same 14. booke, followed the order and number of the Propositions, as Flussas hath placed them.</p>
            <div type="part">
               <head>¶The first Proposition. A perpendicular line drawen from the centre of a circle, to the side of a Pentagon inscribed in the same circle:<note place="margin">
                     <hi>The first pro<g ref="char:EOLhyphen"/>position after Campane.</hi>
                  </note> is the halfe of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle.</head>
               <p>TAke a circle ABC, and inscribe in it the side of a pentagon, which let be BC, and take the centre of the circle, which let be the point D:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and fro<g ref="char:cmbAbbrStroke">̄</g> it draw vnto the side BC a perpendicular line DE: which produce to the point <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And vnto the line E F put the line EG equall. And draw these right lines CG, CD, and CF. Then I say, that the right line DE (which is drawen from the centre to BC the side of the pentagon) is the halfe of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the decagon and hexagon, taken together.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Forasmuch as the line DE is a perpendicular <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nto the line BC: therefore the sections BE and EC shall be equall (by the 3. of the third): and the line EF is common vnto them both; and the angles FEC and FEB, are right an<g ref="char:EOLhyphen"/>gles, by supposition. Wherefore the bases BF and FC are equall (by the 4. of the first). But the line BC is the side of a pentagon, by constructi<g ref="char:EOLhyphen"/>on. Wherefore FC which subtendeth the halfe of the side of the
<figure/>
pentagon, is the side of the decagon inscribed in the circle ABC. But vnto the line FC is, by the 4. of the first, equall the line CG, for they subtend right angles CEG, and CEF, which are con<g ref="char:EOLhyphen"/>tained vnder equall sides. Wherefore also the angles CGE, and CFE, of the triangle CFG, are equall, by the 5. of the first. And forasmuch as the arke FC is subtended of the side of a decagon, the arke CA shall be quadruple to the arke CF: Wherefore also the angle CDA shall be quadruple to the angle CDF (by the last of the six<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>). And forasmuch as the same angle CDA, which is set at the center, is double to the angle CFA, which is set at the circumference, by the 20. of the third: therefore the angle CFA, or CFD, is double to the angle CD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, namely, the halfe of qua<g ref="char:EOLhyphen"/>druple. But vnto the angle CFD or CFG, is proued equall the angle CGF: Wherefore the outward angle CGF, is double to the angle CDF. Wherefore the angles CDG and DCG, shall be equall. For vnto those two angles the angle CGF is equall, by the 32. of the first. Wherefore the sides GC and GD, are equall, by the 6. of the first. Wherefore also
<pb n="422" facs="tcp:7062:451"/>
the line GD is equall to the line FC, which is the side of the decagon. But vnto the right line FE is e<g ref="char:EOLhyphen"/>quall the line EG, by construction. Wherefore the whole line DE is equall to the two lines C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and FE. Wherefore those lines taken together (namely, the lines DF and FC) shall be double to the line DE. Wherefore the line DE (which is drawen from the centre perpendicularly to the side of the pen<g ref="char:EOLhyphen"/>tagon) shal be the halfe of both these lines taken together, namely, of DF the side of the hexagon, and CF the side of the decagon. For the line DF which is drawen from the centre, is equall to the side of the hexagon, by the Corollary of the 15. of the fourth. Wherefore a perpendicular line drawen from the center of a circle, to the side of a pentagon inscribed in the same circle: is the halfe of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>If a right line drawen perpendicularly from the centre of a circle to the side of a pentagon, be diuided by an extreme and meane proportion: the greater seg<g ref="char:EOLhyphen"/>ment shall be the line which is drawen from the same c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tre to the side of an equilater triangle inscribed in the same circle.</hi> For, that li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> (drawen to the side of the triangle) is (by the Corollary of the 12. of the thirtenth) the halfe of the line drawen from the centre to the circumference, that is, of the side of the hexagon: Wherefore the residue shall be the halfe of the side of the decagon. For the whole line is the halfe of the two sides, namely, of the side of the hexagon, and of the side of the decagon. But of the side of a decagon and of an hexagon taken together, the greater segment is the side of the hexagon (by the 9. of the thirtenth). Wherefore the greater segment of their halfes shall be the halfe of the hexagon, by the 15. of the fift: which halfe is the perpendicular line draw<g ref="char:EOLhyphen"/>en from the centre to the side of the triangle, by the Corollary of the 12. of the thirtenth.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The second Proposition. If two right lines be diuided by an extreme and meane proportion:<note place="margin">
                     <hi>The</hi> 2. <hi>propo<g ref="char:EOLhyphen"/>sition after Campane.</hi>
                  </note> they shall be diuided into the selfe same proportions.</head>
               <p>SVppose that these two right lines AB and DE be eche cut by an extreme and meane pro<g ref="char:EOLhyphen"/>portion in the pointes F and Z. Then I say, that these two lines are diuided into the selfe same proportions, that is, that the line AB is in the point F diuided in like sort as the line DE is in the point Z. For if they be not in like sort cut, let one of them, namely, DE, be cut like vnto the line AB in the point C.<note place="margin">
                     <hi>Demonstrati<g ref="char:EOLhyphen"/>on leading to an impossibi<g ref="char:EOLhyphen"/>litie.</hi>
                  </note> So that let the line DE be to DC the greater part, as the greater part DC is to CE the lesse part, by the 3. definitio<g ref="char:cmbAbbrStroke">̄</g> of the sixt. But (by suppositi<g ref="char:EOLhyphen"/>tion) the line DE is to the line DZ, as the
<figure/>
line DZ is to the line ZE. Wherefore the right line DE is diuided by an extreme and meane proportion in two pointes C and Z. But the proportion of DE to DC the lesse line, is greater then the propor<g ref="char:EOLhyphen"/>tion of the same DE to DZ the greater line, by the 2. part of the 8. of the fift. But as DE is to DC, so is DC to CE: Wherefore the proportion of DC to CE, is greater then the proportion of DZ to ZE. And forasmuch as DZ is greater then DC, the proportion of DZ to CE shall be greater then the pro<g ref="char:EOLhyphen"/>portion of DC to CE, by the 8. of the fift. Wherefore the proportion of DZ to CE, is much greater then the proportion of DZ to ZE. Wherefore one and the selfe same magnitude, namely, DZ, hath to CE the greater line, a greater proportion then it hath to ZE the lesse line, contrary to the second part of the 8. of the fift: which is impossible. Wherfore the right lines AB &amp; DE, are not cut vnlike. Wher<g ref="char:EOLhyphen"/>fore they are cut like, and into the selfe same proportions. And the same demonstration also will serue, if the point C fall in any other place. For alwaies some one of them shall be the greater. If therefore two right lines be cut by an extreme and meane proportion: they shall be cut into the selfe same propor<g ref="char:EOLhyphen"/>tions: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The third Proposition. If in a circle be described an equilater Pentagon:<note place="margin">
                     <hi>The</hi> 4. <hi>propo<g ref="char:EOLhyphen"/>sition after Campane.</hi>
                  </note> the squares made of the side of the Pentagon and of the line which subtendeth two sides of the
<pb facs="tcp:7062:452"/>
Pentagon, these two squares (I say) taken together, are quintuple to the square of the line drawen from the centre of the circle to the circu<g ref="char:cmbAbbrStroke">̄</g>ference.</head>
               <p>SVppose that in the circle BCG the side of a
<figure/>
Pentagon be BG: and let the line BC sub<g ref="char:EOLhyphen"/>tend two sides thereof.<note place="margin">
                     <hi>Constru<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on.</hi>
                  </note> And let the line BG be diuided into two equall partes by a right line drawen from the centre D: namely, by the diameter CDE produced to the point Z. And drawe the right line BZ. Then I say, that the right lines BC and BG, are in power quintuple to the right line DZ, which is drawen from the centre to the circumference. For forasmuch as (by the 47. of the first) the squares of the lines CB and BZ,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> are equall to the square of the diameter CZ: therefore they are quadruple to the square of the line DZ, by the 20. of the sixt (for the line CZ is double to the line DZ). Wherefore the right lines CB, BZ, and ZD, are in power quintuple to the line ZD. But the right line BG con<g ref="char:EOLhyphen"/>taineth in power the two lines BZ and ZD, by the 10. of the thirtenth. For DZ is the side of an hexagon, &amp; BZ the side of a decagon. Wherefore the lines BC and BG (whose powers are equall to the powers of the lines CB, BZ, ZD) are in power quintuple to the line DZ. If therefore in a circle be described an equilater Pentagon: the squares made of the side of the Pentagon and of the line which subtendeth two sides of the Pentagon, th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se two squares (I say) taken together, are quintuple to the square of the line drawen from the centre of the circle to the circumference.</p>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <hi>If a Cube and a Doderahedron be contained in one and the selfe same Sphere:</hi>
                     <note place="margin">
                        <hi>This Corolla<g ref="char:EOLhyphen"/>ry Campane also <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>utteth after the</hi> 4. <hi>proposition in his order.</hi>
                     </note> 
                     <hi>the side of the Cube, and the side of the Dodecahedron, are in power quintuple to the line which is drawen from the centre of the circle which contai<g ref="char:EOLhyphen"/>neth the Pentagon of the Dodecahedron.</hi> For it was proued in the 17. of the thirtenth, that the side of the Cube subtendeth two sides of the Pentagon of the Dodecahedron, where the sayd solides are contained in one and the selfe same Sphere. Wherfore the side of the Cube subtending two sides of the Pentagon, and the side of the same Pentagon, are contained in one and the selfe same circle. Wherefore, by this Proposition, they are in power quintuple to the line which is drawen from the cen<g ref="char:EOLhyphen"/>tre of the same circle which containeth the Pentagon of the Dodecahedron.</p>
               </div>
            </div>
            <div type="part">
               <head>The 4. Proposition. One and the selfe same circle containeth both the Pentagon of a Dodeca<g ref="char:EOLhyphen"/>hedron,<note place="margin">
                     <hi>The</hi> 5. <hi>propo<g ref="char:EOLhyphen"/>sition after Campane.</hi>
                  </note> and the triangle of an Icosahedron described in one and the selfe same sphere.</head>
               <p>LEt the diameter of the sphere geuen be AB, and let the bases of the Icosahedron and Do<g ref="char:EOLhyphen"/>decahedron described in it,<note place="margin">
                     <hi>Construction.</hi>
                  </note> be the triangle MNR, and the pentagon FKH, and a<g ref="char:EOLhyphen"/>bout them let there be described circles. by the 5. and 14. of the fourth. And let the lines drawne from the centres of those circles to the circumferences be LN and OK. Then I say that the lines LN and OK are equal, and therfore one and the selfe same circle contai<g ref="char:EOLhyphen"/>neth both those figures. Let the right line AB, be in power quintuple to some one right line, as to the line CG (by the Corollary of the 6. of the tenth.) And making the ce<g ref="char:cmbAbbrStroke">̄</g>tre the poynt C, &amp; the space CG, describe a circle DZG. And let the side of a pentagon inscribed in that circle (by the 11. of the fourth) be the line ZG. And let EG (subtending halfe of the arke ZG) be the side of a Decagon inscribed in that circle. And by the 30. of the sixt, diuide the line CG by an extreme &amp; meane proportion in the poynt I.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Now forasmuche as in the 16. of the thirtenth, it was proued, that this line CG (vnto whome the diameter AB of the sphere is in power quintuple) is the line which is drawne from the centre of the circle, which containeth fiue angles of the Icosahedron, and the side of the pentagon described
<pb n="423" facs="tcp:7062:452"/>
in that circle DZG, namely the line ZG
<figure/>
is side of the Icosahedron described in the Sphere, whose diameter is the line AB: therefore the right line ZG, is equal to the line MN, which was put to be the side of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, or of his tria<g ref="char:cmbAbbrStroke">̄</g>gular base. Moreouer, by the 17. of the thirtenth, it was manifest that the right line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>H (which subtendeth the angle of the pentagon of the Dodecahedron inscribed in the fore<g ref="char:EOLhyphen"/>sayde sphere) is the side of the Cube, in<g ref="char:EOLhyphen"/>scribed in the self same sphere. (For vpon the angles of the cube, were made the an<g ref="char:EOLhyphen"/>gles of the Dodecahedron.) Wherefore the diameter AB is in power triple to F<g ref="char:EOLhyphen"/>H, the side of the Cube (by the 15. of the thirtenth). But the same line AB is (by supposition) in power quintuple to the line CG. Wherefore fiue squares of the line CG, are equal to thre squares of the line FH: (for eche is equal to one and the self same square of the line AB). And forasmuche as EG the side of the Deca<g ref="char:EOLhyphen"/>gon, cutteth the right line CG by an ex<g ref="char:EOLhyphen"/>treme and meane proportion (by the co<g ref="char:EOLhyphen"/>rollary of the 9. of the thirtenth): Like<g ref="char:EOLhyphen"/>wise the line HK, cutteth the line FH, the side of the Cube by an extreeme and meane proportion (by the Co<g ref="char:EOLhyphen"/>rollary of the 17. of the thirtenth): therfore the lines CG and FH, are deuided into the self same pro<g ref="char:EOLhyphen"/>portions, by the second of this booke: and the right lines CI and EG, which are the greater segmentes of one and the selfe same line CG, are equal: And forasmuche as fiue squares of the line <hi>CG</hi> are equal to thre squares of the lines FH: therefore fiue squares of the line GE, are equal to thre squares of the line HK (for the lines GE and HK are the greater segme<g ref="char:cmbAbbrStroke">̄</g>ts of the lines CG and FH). Wherefore fiue squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CG &amp; GE are equal to the squares of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>H &amp; HK, by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ift. But vnto the squares of the lines CG and GE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>qual the squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ine ZG, by the 10. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th: and vnto the line ZG the line MN was equal: wherfore fiue squares of the line MN, are equall to three squares of the lines FH, HK. But the squares of the lines <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> and HK, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> quintuple to the square of the line OK (which is drawne from the centre) by the third of this booke. Wherfore thre squares of the lines FH and HK make 15. squares of the line OK. And forasmuch as the square of the line MN is triple to the square of the line LN (which is drawne from the centre) by the 12. of the thirtenth, ther<g ref="char:EOLhyphen"/>fore fiue squares of the line MN are equal to 15. squares of the line LN. But fiue squares of the line MN are equal vnto thre squares of the lines FH and HK. Wherefore one square of the line LN is equall to one square of the line OK (being eche the fiuetenth part of equal magnitudes) by the 15. of the fif<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. Wherfore the lines LN and OK, which are drawne from the centers, are equal. Wherefore also the circles NRM, and FKH which are described of those lines, are equal. And those circles contayne (by supposition) the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ses of the Dodecahedron and of the Icosahedron described in one and the selfe same sphere. Wherfore one and the selfe same circle, &amp;c. a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> pro<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>sition: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>The 5. Proposition. If in a circle be inscribed the pentagon of a Dodecahedron, and the trian<g ref="char:EOLhyphen"/>gle of an Icosahedron,<note place="margin">
                     <hi>This is the</hi> 6. <hi>and</hi> 7. <hi>propo<g ref="char:EOLhyphen"/>sitions after Campane.</hi>
                  </note> and from the centre to one of theyr sides, be drawne a perpendicular line: That which is contained 30. times vnder the side, &amp; the perpendicular line falling vpon it, is equal to the superficies of that so<g ref="char:EOLhyphen"/>lide, vpon whose side the perpendicular line falleth.</head>
               <p>SVppose that in the circle AGE, be described the pentagon of a Dodecahedron, which let be ABGDE, and the triangle of an Icosahedron described in the same sphere, which let be AFH.<note place="margin">
                     <hi>Construction.</hi>
                  </note> And let the centre be the poynt C. <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>on which draw perpendicularly the line CI to the side of the Pentagon, and the line CL to the side of the triangle. Then I say that the rectangle figure contained vnder the lines CI and GD 30. times, is equal to
<pb facs="tcp:7062:453"/>
the superficies of the Dodecahedron: and that that
<figure/>
which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines CL &amp; AF 30. times is equal to the super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>icies of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> described in the same sphere. Draw these right lines CA, CF, CG and CD.<note place="margin">
                     <hi>Demons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Now forasmuch as that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vn<g ref="char:EOLunhyphen"/>der the base GD &amp; the altitude IC, is double to the triangle GCD, by the 41. of the first: And fiue triangles like and equal to the triangle GCD do make the pen<g ref="char:EOLhyphen"/>tagon ABGDE of the Dodecahedron: wherfore that which is contained vnder the lines GD and IC fiue times is equal to two pentago<g ref="char:cmbAbbrStroke">̄</g>s. Wherfore that which is contained vnder the lines GD and IC <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. times is equal to the 12. pentagons, which containe the superfi<g ref="char:EOLhyphen"/>cies of the Dodecahedron.</p>
               <p>Againe that which is contained vnder the lynes CL and AF, is double to the triangle ACF: where<g ref="char:EOLhyphen"/>fore that which is contained vnder the lines CL and AF three times is equal to two suche triangles as AF<g ref="char:EOLhyphen"/>H is, which is one of the bases of the Icosahedron (for the triangle ACF, is the third part of the triangle AF<g ref="char:EOLhyphen"/>H, as it is easie to proue, by the 8. &amp; 4. of the first.) Wherfore that which is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the lines CL and AF. 30 times times, is equall to 10. such triangles as AFH i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, which containe the superficies of the Icosahedron. And forasmuch as one and the selfe same spher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> containeth the Dodecahedron of this pentagon, and the Icosahedron of this triangle (by the 4. of this booke <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) and the line CL falleth per<g ref="char:EOLhyphen"/>pendicula<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ly vpon the side of the Icosahedron, and the line CI vpon the side of the Dodecahedron: that which is 30. times contained vnder the side, and the perpendicular line falling vpon it, is equal to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uperficies of that solide, vpon whose side the perpendicul<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r falleth. If therefore in a circle &amp;c. as in the proposition: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <note place="margin">This Corolla<g ref="char:EOLhyphen"/>ry Campane also addeth after the 7. proposition i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> his order.</note>The superficieces of a Dodecahedron and of an Icosahedron described in one and the selfe same sphere are the one to the other, as that which is contained vn<g ref="char:EOLunhyphen"/>der the side of the one and the perpendicular line drawne vnto it from the cen<g ref="char:EOLhyphen"/>tre of his base, to that which is contained vnder the side of the other, and the perpendicular line drawne to it from the centre of his base. <hi>For a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> thirty<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> tim<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s is to thirty times, so is once to once by the 15. of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> fifth.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>The 6. Proposition. The superficies of a Dodecahedron, is to the superficies of an Icosahe<g ref="char:EOLhyphen"/>dron described in one and the selfe same sphere,<note place="margin">
                     <hi>The 5. propo<g ref="char:EOLhyphen"/>sition a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ter Campane.</hi>
                  </note> in that proportion, that the side of the Cube is to the side of the Icosahedron contained in the self same sphere.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that there be a circle ABG, &amp; in it (by the 4. of this boke) let there be inscribed the side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of a Dodecahedron and of an Icosahedron contained in on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the selfe same sphere. And let the side o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the Dodecahedron be AG, and the side of the Icosahedron be DG. And let the centre be the poynt E: from which draw vnto those s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>des, perpendicular lines EI and EZ. And produce the line EI to the poynt B, and draw the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BG. And let the side of the cube con<g ref="char:EOLhyphen"/>tained in the self same sphere be GC. Then I say that the superficies of the Dodecahedron i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the su<g ref="char:EOLhyphen"/>perficies of the Icosahedron, as the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> GD. For forasmuche as the line EI bein<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> diuided by an extreme and meane proportion, the greater segment th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rof shall be the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> EZ, by the corollary of the first of this booke:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and the line CG being diuided by an extreme and meane pro<g ref="char:EOLhyphen"/>portion, his greater segment is the line AG, by the corollary of the 17. of the thirtenth: Where<g ref="char:EOLhyphen"/>fore the right lines EI and CG <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cut proportionally by the second of this b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oke. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore as the line CG, is to the line AG, so is the line EI to the line EZ. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore that which it con<g ref="char:EOLhyphen"/>tained vnder the extreames CG and EZ, is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall to that which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> contayn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d vnder the meanes AG and EI. (by the 16. of the sixth.) But as that which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> contained vnder the lin<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> CG and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>Z is to that which is contained vnder the lines DG and EZ, so (by the first of the sixth) i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CG to the
<pb n="424" facs="tcp:7062:453"/>
line DG, for both those parallelogrames haue o<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> and the
<figure/>
selfe same altitude, namely the line EZ. Wherfore as that which is contained vnder the lines EI and AG (which i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> proued equal to that which is contained vnder the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> CG and EZ) is to that which is contained vnder the lines DG and EZ, so is the line CG to the li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> DG. But as that which is contained vnder the lines EI and AG is to that which is contained vnder the lines DG and EZ, so (by the corollary of the former proposition) is the su<g ref="char:EOLhyphen"/>perficies of the Dodecahedron, to the superficies of the I<g ref="char:EOLhyphen"/>cosahedron. Wherfore as the superficies <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the Dodecahe<g ref="char:EOLhyphen"/>dron is to the superficies of the Icosahedron, so is CG the side of the cube, to GD the side of the Icosahedron. The superficies therefore of a Dodecahedron is to the superfi<g ref="char:EOLhyphen"/>cies<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> &amp;c. as in the proposition, which was required to be proued.</p>
               <div type="section">
                  <head>An Assumpt.</head>
                  <p>
                     <q>The Pentagon of a Dodecahedron,<note place="margin">
                           <hi>This Assumpt Campane also hath after the</hi> 8. <hi>proposition, in his order.</hi>
                        </note> is equall to that which is contained vn<g ref="char:EOLhyphen"/>der the perpendicular line which falleth vpon the base of the triangle of the Icosahedron, and fiue sixth partes of the side of the cube, the sayd three so<g ref="char:EOLhyphen"/>lides being described in one and the selfe same sphere.</q>
                  </p>
                  <p>Suppose that in the circle ABEG, the pentagon of a Dodecahedron be A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>CIG, and let two sides thereof AB and AG be subtended of the right line BG. And let the triangle of the Icosahedron inscribed in the selfe same sphere,<note place="margin">
                        <hi>Construction.</hi>
                     </note> by the 4. of this booke, be AFH. And let the centre of the circle be the poynt D, and let the diameter be ADE, cutting FH, the side of the triangle in the poynt Z, and cut<g ref="char:EOLhyphen"/>ting the line BG in the poynt K. And draw the right line BD. And from the right line KG cut of a third part TG, by the 9. of the sixth. Now then the line BG subtending two sides of the Dodecahedron, shalbe the side of the cube inscribed in the same sphere,<note place="margin">
                        <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                     </note> by the 17. of the thirtenth: and the triangle of the Icosahedron of the same sphere shalbe A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>H by the 4. of this booke. And the line AZ which pas<g ref="char:EOLhyphen"/>seth by the centre D shall fall perpendicularly vpon the
<figure/>
side of the triangle. For forasmuch as the angles GAE &amp; BAE are equall (by the 27. of the third<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> for they are see vpon equall circumferences): therefore the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ases BK and KG are (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the first) equall. Wherefore the line BT contayneth 5. sixth partes of the line BG. Then I say that that which is contayned vnder the lines AZ and B<g ref="char:EOLhyphen"/>T, is equall to the pentagon A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>G. For forasmuch as the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>Z is sesq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ialter to the line AD (for the line D<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> is diuided into two equall partes in the poynt Z, by the corollary of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the thirtenth). Likewise by con<g ref="char:EOLhyphen"/>struction, the line KG is sesquialter to the line KT: there<g ref="char:EOLhyphen"/>fore as the line AZ is to the line AD, so is the line KG to the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>T. Wherefore that which is contayned vnde<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> AZ and KT, is equall to that which is con<g ref="char:EOLhyphen"/>tayned vnder the meanes AD and KG, by the 16. of the sixth. But vnto the line KG is the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>K <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>roued equall. Wherefore that which is contayned vnder the lines AZ and KT is equall to that which is contayned vnder the lines AD and BK. But that which is contayned vnder the lines AD and BK is (by the 41. of the first) double to the triangle ABD. Wherefore that which is contayned vnder the lines AZ and KT is double to the same triangle ABD. And forasmuch as the pentagon ABCIG contayneth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o the triangle ABD, and that which is contayned vnder the lines AZ and KT contayneth two such triangles: therefore the pentagon ABCIG is duple sesquialter to the rectangle parallelogramme contayned vnder the lines AZ and KT. And <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> 1. of the sixth, that which is co<g ref="char:cmbAbbrStroke">̄</g>teyned vnder the lines AZ and BT is to that which is contayned vnder the lines AZ and KT, as the base BT is to the base <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> therefore that which is contayned vnder the lines AZ and BT is duple sesquialter to that which is contayned vn<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>r the line AZ &amp; KT. But vnto that which is contayned vnder the lines AZ and KT the pentagon ABCIG is proued duple sesquialter. Wher<g ref="char:EOLunhyphen"/>fore the pentagon ABCIG of the Dodecahedron is equall to that which is contayned vnder the per<g ref="char:EOLhyphen"/>pendicular
<pb facs="tcp:7062:454"/>
line AZ, and vnder the line BT which is fiue sixe partes of the line BG.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 7. Proposition. <note place="margin">
                     <hi>The 9. propo<g ref="char:EOLhyphen"/>sition after Campane.</hi>
                  </note>A right line diuided by an extreame and meane proportion: what propor<g ref="char:EOLhyphen"/>tion the line contayning in power the whole line and the greater segment, hath to the line contayning in power the whole and the lesse segment: the same hath the side of the cube to the side of the Icosahedron contayned in one and the same sphere.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>TAke a circle ABE: and in it (by the 11. of the fourth) inscribe an equilater pentagon BZEC<g ref="char:EOLhyphen"/>H: and (by the second of the same) an equilater triangle ABI. And let the centre thereof be the poynt G. And drawe a line from G to B. And diuide the line GB by an extreame and meane proportion in the poynt D (by the 30. of the sixth). And let the line ML contayne in power both the whole line GB and his lesse segment B<g ref="char:EOLhyphen"/>D
<figure/>
(by the corollary of the 13. of the tenth). And draw the right line B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e<g ref="char:cmbAbbrStroke">̄</g>di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>g the angle of the pentagon, which shall be the side of the cube (by the corollary of the 17. of the thirtenth) <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the line BI shall be the side of the Icosahedron, and the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>Z the side of the Dodecahe<g ref="char:EOLhyphen"/>dron by the 4. of this booke. Then I say that BE the side of the cube is to BI the side of the Icosahedron, as the line contayning in power the lines BG &amp; GD is to the line contayning in power the lines GB and BD.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For for<g ref="char:EOLhyphen"/>asmuch as (by the 12. of the thirtenth) the line BI is in power triple to the line BG, and (by the 4. of the same) the squares of the line GB &amp; BD are triple to the square of the line GD. Wherefore (by the 15. of the fifth) the square of the line BI is to the squares of the lines GB &amp; BD (namely, triple to triple), as the square of the line B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is to the square of the line GD (namely, as one is to one). But as the square of the line BG is to the square of the GD, so is the square of the line BE to the square of the line BZ. (For the lines BG, GD, and BE, BZ are in one and the same proportion, by the second of this booke. For BZ is the greater segment of the line BE, by the corollary of the 17. of the thirtenth). Wherefore the square of the line BE is to the square of the line BZ, as the square of the line BI is to the squares of the lines BG and BD. Wherefore alternately the square of the line BE is to the square of the line BI, as the square of the line BZ is to the squares of the lines GB and BD. But the square of the line BZ is equall to the squares of the lines BG and GD (by the 10. of the thirtenth). For the line BG is equall to the side of the hexagon, and the line GD to the side of the decagon, by the corollary of the 9. of the same. Wherefore the squares of the lines BG and GD, are to the squares of the lines G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and BD, as the square of the line BE is to the square of the line BI. But the line ZB contayneth in power the lines BG and GD, and the line ML contayneth in power the lines GB and BD by construc<g ref="char:EOLhyphen"/>tion. Wherefore as the line ZB (which contayneth in power the whole line BG and the greater seg<g ref="char:EOLhyphen"/>ment GD) is to the line ML (which contayneth in in power the whole line GB and the lesse segment BD) so is BE the side of the cube to BI the side of the Icosahedron, by the 22. of the sixth. Wherefore a right line diuided by an extreame and meane proportion: what proportion the line contayning in power the whole line and the greater segment, hath to the line contayning in power the whole line and the lesse segment: the same hath the side of the cube to the side of the Icosahedron co<g ref="char:cmbAbbrStroke">̄</g>tayned in one and the same sphere which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 8. Proposition. <note place="margin">
                     <hi>This Campane putteth a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> a Co<g ref="char:EOLhyphen"/>rollary in the 9. proposition af<g ref="char:EOLhyphen"/>ter his order.</hi>
                  </note>The solide of a Dodecahedron is to the solide of an Icosahedron: as the side of a Cube is to the side of an Icosahedron, all those solides being described in one and the selfe same Sphere.</head>
               <p>
                  <pb n="425" facs="tcp:7062:454"/>FOrasmuch as in the 4. of this booke, it hath bene proued, that one and the self same cir<g ref="char:EOLhyphen"/>cle containeth both the triangle of an Icosahedron, and the pentagon of a Dodecahe<g ref="char:EOLhyphen"/>dron described in one and the selfe same Sphere: Wherefore the circles, which co<g ref="char:cmbAbbrStroke">̄</g>taine those bases, being equall, the perpendiculars also which are drawen from the centre of the Sphere to those circles, shall be equall (by the Corollary of the Assumpt of the 16 of the twelfth). And therefore the Pyramids set vpon the bases of those solides haue one and the selfe same altitude: For the altitudes of those Pyramids concurr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the centre. Wherefore they are in proportion as their bases are, by the 5. and 6. of the twelfth. And therefore the pyramids which compose the Dodecahedron, ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the pyramids which compose the Icosahedron, as the bases are, which bases are the superficieces of those solides. Wherefore their solides are the one to the other, as their superficieces are. But the superficies of the Dodecahedron is to the superficies of the Icosahedron, as the side of the cube is to the side of the Icosahedron, by the 6. of this booke. Wher<g ref="char:EOLhyphen"/>fore by the 11. of the fifth, as the solide of the Dodecahedron is to the solide of the Icosahedron, so is the side of the cube to the side of the Icosahedron, all the said solides being inscribed in one and the selfe same Sphere. Wherefore the solide of a Dodecah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron is to the solide of an Icosahedron: as the side of a cube is to the side of an Icosahedron, all those solides being described in one and the self same Sphere: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>The solide of a Dodecahedron is to the solide of an Icosahedron,<note place="margin">This Corolla<g ref="char:EOLhyphen"/>ry is the 9. proposition after Cam<g ref="char:EOLhyphen"/>pane.</note> as the su<g ref="char:EOLhyphen"/>perficieces of the one are to the superficieces of the other, being described in one and the selfe same Sphere: <hi>Namely, as the side of the cube is to the side of the Icosahedron, as was before manifest: for they are resolued into pyramids of one and the selfe same altitude.</hi>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 9. Proposition. If the side of an equilater triangle be rationall: the superficies shall be irra<g ref="char:EOLhyphen"/>tionall,<note place="margin">
                     <hi>The</hi> 12. <hi>pro<g ref="char:EOLhyphen"/>position after Campane.</hi>
                  </note> of that kinde which is called Mediall.</head>
               <p>SVppose that ABG be an equilater triangle: and from the point A draw vnto the side BG a perpendicular line AD: and let the line AB be rationall. Then I say, that the superfi<g ref="char:EOLhyphen"/>cies ABG is mediall.<note place="margin">
                     <hi>Construction.</hi>
                  </note> Forasmuch as the line AB is in power sesquitertia to the line AD (by the Corollary of the 12. of the thirtenth): of what partes the line AB containeth in power 12, of the same partes the line AD containeth in power 9:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> wherefore the resi<g ref="char:EOLhyphen"/>due BD containeth in power of the same partes 3. (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or the line AB co<g ref="char:cmbAbbrStroke">̄</g>taineth in power the lines AD and BD, by the 47. of the first). Wherfore the
<figure/>
lines AD and DB are rationall and commensurable to the rationall line set AB, by the 6. of the tenth. But forasmuch as the power of the line AD is to the power of the line DB in that proportion that 9. a square number is to 3. a number not square: therfore they are not in the proportion of square numbers, by the Corollary of the 25. of the eight. And ther<g ref="char:EOLhyphen"/>fore they are not commensurable in length, by the 9. of the tenth. Wherefore that which is contained vnder the lines AD and DB, which are rationall lines commensurable in power onely, is mediall, by the 22. of the tenth. But that which is contained vnder the lines AD and DB, is double to the triangle ABD, by the 41. of the first. Wherefore that which is contained vnder the lines AD and DB, is equall to the whole triangle ABG (which is double to the triangle ABD, by the 1. of the sixt). Wherefore the triangle ABG is mediall. If therfore the side of an equilater triangle be ra<g ref="char:EOLhyphen"/>tionall: the superficies shall be irrationall, of that kinde which is called Mediall: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>If an Octohedron and a Tetrahedron be inscribed in a Sphere whose dia<g ref="char:EOLhyphen"/>meter is rationall:</hi>
                     <note place="margin">
                        <hi>The</hi> 13. <hi>propo<g ref="char:EOLhyphen"/>sition after Campane.</hi>
                     </note> their superficieces shall be mediall. For those superficieces consiste of equilater triangles, whose sides are commensurable to the diameter which is rationall, by the 13. and
<pb facs="tcp:7062:455"/>
14. of the thirtenth, and therefore they are rationall. But they are commensurable in power onely to the perpendicular line, and therefore they containe a mediall triangle, as it was before manifest.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 10. Proposition. <note place="margin">
                     <hi>The</hi> 14. <hi>pro<g ref="char:EOLhyphen"/>position after Campane.</hi>
                  </note>If a Tetrahedron and an Octohedron be inscribed in one and the self same Sphere: the base of the Tetrahedron shall be sesquitertia to the base of the Octohedron, and the supersicieces of the Octohedron shall be sesquialtera to the superficieces of the Tetrahedron.</head>
               <p>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</hi>
                  </note>FOrasmuch as the diameter of the Sphere is in power sesquialtera to the side of the Tetrahe<g ref="char:EOLhyphen"/>dron (by the 13. of the thirtenth) and the same diameter is in power duple to the side of the Octohedron (by the 14. of the same booke): therefore of what partes the diameter contai<g ref="char:EOLhyphen"/>neth in power sixe, of the same the side of the Tetrahedron contayneth in power 4, and of the same the side of the Octohedron containeth in power 3. Wherefore the power of the side of the Tetrahedron, is to the power of the side of the Octohedron in the same proportion that 4. is to 3: which is sesquitertia. And like triangles (which are the bases of the solides) described of those sides, shall haue the one to the other the same proportion that the squares made of those sides haue. For both the triangles are the one to the other, and also the squares are the one to the other, in double proporti<g ref="char:EOLhyphen"/>on of that in which the sides are, by the 20. of the sixth.<note place="margin">
                     <hi>Demonstrati<g ref="char:EOLhyphen"/>on of the se<g ref="char:EOLhyphen"/>cond part.</hi>
                  </note> Wherefore of what partes one base of the Te<g ref="char:EOLhyphen"/>trahedron was 4: of the same are fower bases of the Tetrahedron 16: likewise of what partes of the same one base of the Octohedron was 3: of the same are 8. bases of the Octohedron 24. Wherfore the bases of the Octohedron are to the bases of the Tetrahedron, in that proportion that 24. is to 16: which is sesquialtera. If therefore a Tetrahedron and an Octohedron be inscribed in one and the selfe same Sphere: the base of the Tetrahedron shall be sesquitertia to the base of the Octohedron, and the super<g ref="char:EOLhyphen"/>ficieces of the Octohedron shall be sesquialtera to the superficieces of the Tetrahedron: which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 11. Proposition. A Tetrahedron is to an Octohedron
<figure/>
inscribed in one and the selfe same Sphere,<note place="margin">
                     <hi>The</hi> 17. <hi>pro<g ref="char:EOLhyphen"/>position after Campane.</hi>
                  </note> in proportion, as the rectan<g ref="char:EOLhyphen"/>gle parallelogra<g ref="char:cmbAbbrStroke">̄</g>me contained vnder the line, which containeth in power 27. sixty fower partes of the side of the Tetrahedron, &amp; vnder the line which is subsesquiocta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>a to the same side of the Tetrahedron, is to the square of the diameter of the sphere.</head>
               <p>
                  <note place="margin">
                     <hi>Fir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t part of the construc<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>LEt vs suppose a Sphere, whose diame<g ref="char:EOLhyphen"/>ter let be the line AB, and let the cen<g ref="char:EOLhyphen"/>tre be the point H. And in it let there be inscribed a
<figure/>
Tetrahedron ADC, and an Octohedron AEKBG. And let the line NL containe in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7/64 of AC the side of the Tetrahedron. And let the line ML be in le<g ref="char:cmbAbbrStroke">̄</g>gth subsesquioctaua to the same side. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the Tetrahedron ACD is to the Octohedron AEB, as the rectangle parallelogramme contay<g ref="char:EOLhyphen"/>ned vnder the lines NL and LM,<note place="margin">
                     <hi>First part of the Demon<g ref="char:EOLhyphen"/>stration.</hi>
                  </note> is to the square of the line AB. Foras<g ref="char:EOLhyphen"/>much as the line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the angle A by the centre H perpe<g ref="char:cmbAbbrStroke">̄</g>dicular<g ref="char:EOLhyphen"/>ly vpon the base of the Tetrahedron, falleth vpon the ce<g ref="char:cmbAbbrStroke">̄</g>ter T of the cir<g ref="char:EOLhyphen"/>cle which containeth that base, and maketh the right line HT the sixth part of the diameter AB (by the Corollary of the 13. of the thirtenth): therefore the line HA (which is drawen from the centre to the circu<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ference) is triple to the line HT: and therefore the whole line AT is
<pb n="426" facs="tcp:7062:455"/>
to the line AH, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>. Let the Tetrahedron ADC be cut by a plaine GHK passing by the centr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> H, and being parallel vnto the base DTC, by the Corollary of the 11. of the eleuenth. Now then the triangle ADC of the Tetrahedron, shall be cut by the right line KG, which is parallel to the line DC, by the 16. of the eleuenth. Wherfore as the line AT is to the line AH, so is the line AC to the line AG (by the 2. of the sixth). Wherefore the line AC is to the line AG sesquitertia, that is, as 4. to 3. And forasmuch as the triangles ADC, AKG, and the rest which are cut by the plaine KHG, are like the one to the other, by the 5<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the sixth: the pyramids ADC and AKG, shall be like the one to the o<g ref="char:EOLhyphen"/>ther, by the 7. definition of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>leuenth. Wherefore they are in triple proportion of that in which the sides AC and AG are, by the 8. of the twelfth. But the proportion of the sides AC to AG is, as the proportion of 4. to 3. Now then, if, by the 2. of the eight, ye finde out 4. of the lest numbers in continu<g ref="char:EOLhyphen"/>all proportion, and in that proportion that 4. is to 3: which shall be 64.48.36. and 27: it is manifest, by the 15 definition of the fifth, that the extremes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>4. to 27. are in triple proportion of that in which the proportion geuen 4. to 3. is: Or the quantitie of the proportion of 4. to 3. (which is 1. and 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>) being twise multiplied into it selfe, there shall be produced the proportion of 64. to 27. Wherefore the Pyramis or Tetrahedron ADC is to the pyramis AKG, as 64. is to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7: which is triple to the propor<g ref="char:EOLhyphen"/>tion of 4. to 3. And forasmuch as the line AC is vnto the line AG in length sesquitertia: of what partes the line AC containeth in power 64: of the same partes doth the line AG containe in power 36. For (by the 2. of the sixth) the proportion of the powers or squares, is duple to the porportion of the sides which are as 64. is to 48.</p>
               <p>Now then vpon the line RS which let be equall to the line AG, let there be an equilater triangle QRS described (by the first of the first).<note place="margin">
                     <hi>Second par<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n<g ref="char:EOLhyphen"/>struction.</hi>
                  </note> And from the angle Q, draw to the base RS a perpendicular line QT. And extend the line RS to the poynt X. And as 27. is to 64. (so by the corollary of the 6. of the tenth) let the line RS be to the line RX. And diuide <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he line RX into two equall partes in the poynt V, and draw the line QV. And forasmuch as the line RS is equall to the line AG, of what partes the line AC contayneth in power 64. of the same part the line RS contayneth in power 36. for it is proued that the line AG contayneth in power 36. of those partes:<note place="margin">
                     <hi>Second part of the D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>monstration.</hi>
                  </note> And of what partes the line RS contayneth in power 36 of the same partes the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> QT contayneth in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7. by the corollary of the 12. of the thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tenth. Wherfore of what partes the line AZ contayneth in power 64. of the same parts the line Q<g ref="char:EOLhyphen"/>T contayneth in power 27. Wherefore the right line QT shall be equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll to the right line LN by sup<g ref="char:EOLhyphen"/>position. Agayne forasmuch as the line RS is put equall to the line AG: and of what partes the line R<g ref="char:EOLhyphen"/>S contayneth in length 27. of the same parts is the line RX put to contayne in length 64. and of what partes the line RX contayneth in
<figure/>
length 64. of the same the line A<g ref="char:EOLhyphen"/>C (which is in length sesquitertia to the line AG or RS) contay<g ref="char:EOLhyphen"/>neth 36. Wherefore the line RV (which is the halfe of the line R<g ref="char:EOLhyphen"/>X) containeth in le<g ref="char:cmbAbbrStroke">̄</g>gth of the same partes 32. of which the line AC contayned in length 36. Where<g ref="char:EOLhyphen"/>fore the line RV is to the line A<g ref="char:EOLhyphen"/>C subsesquioctaua: and therefore the line RV is equall to the line LM which is also subsesquioctaua to the same line AC. And forasmuch as the line NL is equall to the line QT, and the line LM to the line RV (as before hath bene proued) the rectangle parallelogramme contayned vnder the lines QT and RV, shall be equall to the rectangle parallelogramme, contayned vnder the line NL which is in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7/64 to the side AC, and vnder the line LM, which is in length subsesquioctaua to the same side A<g ref="char:EOLhyphen"/>C. But that which is contayned vnder the lines QT and RV is double to the triangle QVR by the 41. of the first: and to the same triangle QVR is the triangle QXR duple by the first of the sixth. Where<g ref="char:EOLhyphen"/>fore the whole triangle QXR is equall to that which is contayned vnder the lines QT and RV, and therefore is equall to the parallelogramme MN. And forasmuch as the line RX by supposition contay<g ref="char:EOLhyphen"/>neth in length 64. of those partes of which the line RS contayneth 27: and the triangles QRX, and Q<g ref="char:EOLhyphen"/>RS are, by the first of the sixth, in the proportion of their bases, that is, as 64. is to 27: but as 64. is to 27. so is the pyramis or tetrahedron ADC to the pyramis AKG: wherefore as the parallelogramme NM or the triangle QRX, is to the triangle QRS, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the pyramis ADC to the pyramis AKG. And for<g ref="char:EOLhyphen"/>asmuch as the semidiameter AH is the altitude of the pyramis AKG, and also of the two equall and like pyramid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the octohedron which haue their common base in the square of the octohedron (by the corollary of the 14. of the thirtenth): therefore as the base of the pyramis AKG (which is the tri<g ref="char:EOLhyphen"/>angle QRS) is to two squares of the octoh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron, that is, to the square of the diameter AB, which is e<g ref="char:EOLhyphen"/>quall to those squares (by the 47. of the first), so is the pyramis AKG to the octohedron AEB, by the 6. of the twelfth. And forasmuch as the parallelogramme MN is to the base QRS, as the pyramis ADC is to the pyramis AKG, and the base QRS is to the square of the line BE, as the pyramis AKG is to the octohedron AEB: therefore by p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oportion of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quality ta<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ing away th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> meane<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (by the 22. of the fifth) as the parallelogramme NM is to the square of the line BE, so is the pyramis ADC to the octo<g ref="char:EOLhyphen"/>hedron
<pb facs="tcp:7062:456"/>
AEB inscribed in one and the selfe same sphere. But the parallelogramme NM is contayned vnder the line NL which by supposition is in power <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>7/<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> to AC the side of the tetrahedron ADC, and vnder the line LM which is also by supposition in length subsesquioctaua to the same line AC. Wher<g ref="char:EOLhyphen"/>fore a tetrahedron &amp; an octohedron inscribed in one and the selfe same sphere, are in proportio<g ref="char:cmbAbbrStroke">̄</g>, as the rectangle parallelogramme contayned vnder the line, which contayneth in power 27. sixty fower parts of the side of the Tetrahedron, and vnder the line which is subsesquioctaua to the same side of the Te<g ref="char:EOLhyphen"/>trahedron, is to the square of the diameter of the sphere: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶The 12. Proposition. <note place="margin">
                     <hi>The</hi> 18. <hi>pro<g ref="char:EOLhyphen"/>position after Campane.</hi>
                  </note>If a cube be contayned in a sphere: the square of the diameter doubled, is e<g ref="char:EOLhyphen"/>quall to all the superficieces of the cube taken together. And a perpendicu<g ref="char:EOLhyphen"/>lar line drawne from the centre of the sphere to any base of the cube, is e<g ref="char:EOLhyphen"/>quall to halfe the side of the cube.</head>
               <p>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the first part.</hi>
                  </note>FOr forasmuch as (by the 15. of the thirtenth) the diameter of the sphere is in power triple to the side of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cube: therefore the square of the diameter doubled is sex<g ref="char:EOLhyphen"/>tuple to the base of the same cube. But the sextuple of the power of one of the sides contayneth the whole superficies of the cube <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> or the cube is composed of sixe square superficieces (by the 2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. diffinition of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> eleuenth) whose sides therefore are equall: wherefore the square of the di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>meter <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oubled is equall to the whole super<g ref="char:EOLhyphen"/>ficies of the cube. And forasmuch as the diameter of the cube, and the line which falleth perpendicularly vpo<g ref="char:cmbAbbrStroke">̄</g> the opposite bases of the cube,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion of the second part.</hi>
                  </note> do cut the one the other into two equall partes in the centre of the sphere which containeth the cube (by the 2. corollary of the 15. of the thirtenth) and the whole right line which coupleth the centres of the opposite bases, is equall to the side of the cube by the 33. of the first, for it coupleth the equall and parallel semidiameters of the bases: therefore the halfe thereof shall be equall to the halfe of the side of the cube by the 15. of the fifth. If therefore a cube be contayned in a sphere: the square of the diameter doubled is equall to all the superficieces of the cube taken together. And a perpendicular line drawne from the centre of the sphere to any base of the cube, is equall to halfe the side of the cube: which was required to be prou<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d.</p>
               <div type="section">
                  <head>¶ A Corollary.</head>
                  <p>
                     <hi>If two thirds of the power of the diameter of the sphere be multiplyed into the perpendicular line equall to halfe the side of the cube,</hi>
                     <note place="margin">
                        <hi>The Corolla<g ref="char:EOLhyphen"/>ry of the</hi> 8. <hi>proposition after Cam<g ref="char:EOLhyphen"/>pane.</hi>
                     </note> 
                     <hi>there shall be produced a solide equall to the solide of the cube.</hi> For it is before manifest that two third partes of the power of the diameter of the sphere are equall to two bases of the cube. If therefore vnto eche of those two thirds be applyed halfe the altitude of the cube, they shall make eche of those solides equall to halfe of the cube, by the 31. of the eleuenth: for they haue equall bases. Wherefore two of those solides are equall to the whole cube.</p>
                  <p>You shall vnderstand (gentle reader) that <hi>Campane</hi> in his 14. booke of <hi>Euclides</hi> Ele<g ref="char:EOLhyphen"/>mentes hath 18. propositio<g ref="char:cmbAbbrStroke">̄</g>s with diuers corollaries following of them. Some of which propositions and corollaries I haue before in the twelfth and thirtenth bookes added out of Flussas as corollaries (which thing also I haue noted on the side of those corol<g ref="char:EOLhyphen"/>laries, namely, with what proposition or corollary of Campanes 14. booke they doo agree). The rest of his 18. propositions and corollaries are contained in the twelue for<g ref="char:EOLhyphen"/>mer propositions and corollaries of this 14. booke after Flussas: where ye may see on the side of eche proposition and corollary with what proposition and corollary of Campanes they agree. But the eight propositions following together with their corol<g ref="char:EOLhyphen"/>laries, Flussas hath added of him selfe, as he him selfe affirmeth.</p>
               </div>
            </div>
            <div type="part">
               <pb n="427" facs="tcp:7062:456"/>
               <head>The 13. Proposition. One and the self same circle containeth both the square of a cube, and the triangle of an Octohedron described in one and the selfe same sphere.</head>
               <p>SVppose that there be a cube ABG, and an Octohedron DEF described in one and the selfe same sphere, whose diameter let be AB, or DH. And let the lines drawne from the ce<g ref="char:cmbAbbrStroke">̄</g>tres (that is the semidiameters of the circles which cto<g ref="char:cmbAbbrStroke">̄</g>aine the bases of those solides)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be CA and ID. Then I say that the lines CA and ID are equal. Forasmuch as AB the diameter of the sphere which containeth the cube,<note place="margin">
                     <hi>Dem<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> is in power triple to BG the side of the cube (by the 15. of the thirtenth) vnto which side, AG the diameter of the base of the cube, is in power double (by the 47. of the first): which line AG is also the diameter of the circle, which co<g ref="char:cmbAbbrStroke">̄</g>tai<g ref="char:EOLhyphen"/>neth the base (by the 9. of the fourth:) therfore AB the diameter of the sphere is in power sesquialter to the line AG: namely, of what partes the line AB, containeth in power 12. of the same the line AG, shal containe in pow<g ref="char:EOLhyphen"/>er
<figure/>
8. And therfore the right line AC whiche is drawn from the ce<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tre of the circle to the circumference, contei<g ref="char:EOLhyphen"/>neth in power of the same partes 2. Where<g ref="char:EOLhyphen"/>fore the diameter of the sphere is in power sextuple to the lyne which is drawne from the centre to the cir<g ref="char:EOLhyphen"/>cumference of the cir<g ref="char:EOLhyphen"/>cle whiche containeth the square of the cube But the Diameter of the selfe same Sphere whych containeth the Octohedron, is one and the selfe same with the diameter of the cube, namely, DH, is equall to AB: and the same diameter is also the diameter of the square which is made of the sides of the Octohedron: wherefore the saide diameter is in power double to the side of the same Octohedron, by the 14. of the thirtenth. But the side DF is in power triple to the line drawne from the centre to the circumference of the cir<g ref="char:EOLhyphen"/>cle which containeth the triangle of the octohedron (namely to the line ID) by the 12. of the thirtenth. Wherfore the selfe same diameter AB or DH, which was in power sextuple to the line drawne from the centre to the circumference of the circle which containeth the square of the cube, is also sextuple to the line ID drawne from the centre to the circumference of the circle, which containeth the trian<g ref="char:EOLhyphen"/>gle of the Octohedron: Wherefore the lines drawne from the centres of the circles to the circumfe<g ref="char:EOLhyphen"/>rences which containe the bases of the cube and of the octohedron are equal. And therfore the circles are equal, by the first diffinition of the third. Wherfore one and the selfe same circle containeth &amp;c. as in the proposition: which was required to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>Hereby it is manifest, that perpendiculars coupling together in a sphere, the centres of the circles which containe the opposite bases of the cube and of the Octohedron, are equal.</hi> For the circles are equal, by the second corollary of the assumpt of the 16. of the twelfth: and the lines which passing by the centre of the sphere, couple together the centres of the bases, are also equal, by the first corollary of the same. Wherfore the perpendicular which coupleth together the opposite bases of the Octohedron, is equal to the side of the cube. For either of them is the altitude erected.</p>
               </div>
            </div>
            <div type="part">
               <head>The 14. Proposition. An Octohedron is to the triple of a Tetrahedron contained in one and the
<pb facs="tcp:7062:457"/>
selfe same sphere, in that proportion that their sides are.</head>
               <p>SVppose that there be an octohedron ABCD, and a Tetrahedron EFGH: vpon whose base FGH erect a Prisme,<note place="margin">
                     <hi>Construction.</hi>
                  </note> which is done by erecting from the angles of the base perpendicu<g ref="char:EOLhyphen"/>lar lines equal to the altitude of the Tetrahedron: which prisme shalbe triple to the Tetra<g ref="char:EOLhyphen"/>hedron EFGH, by the first corollary of the 7. of the twelfth. Then I say that the octohedron ABCD is to the prisme which is triple to the Tetrahedron, EFGH, as the side BC is to the side FG. For forasmuch as the sides of the opposite bases of the octohedron,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> are right lines touching the one the other, and are parellels to other right lines touching the one the other, for the sides of the squares which are co<g ref="char:cmbAbbrStroke">̄</g>posed of the sides of the octohedro<g ref="char:cmbAbbrStroke">̄</g>, are opposite: Wherfore the opposite plaine triangles, namely, ABC &amp; KID, shalbe parallels, and so the rest
<figure/>
by the 15. of the eleuenth. Let the diameter of the Oc<g ref="char:EOLhyphen"/>tohedron, be the line AD. Now then the whole Octo<g ref="char:EOLhyphen"/>hedron is cut into foure equal and like pyramids set vp<g ref="char:EOLhyphen"/>on the bases of the octohedron, and hauing the same al<g ref="char:EOLhyphen"/>titude with it, &amp; being about the Diameter AD: name<g ref="char:EOLhyphen"/>ly the pyramis set vpon the base BID, and hauing his toppe the poynt A, and also the pyramis set vppon the base BCD, hauing his top the same poynt A. Likewise the pyramis set vpo<g ref="char:cmbAbbrStroke">̄</g> the base IKD, &amp; hauing his toppe the same poynt A: and moreouer the pyramis set vpon the base CKD, and hauing his toppe the former poynt A: which pyramids shalbe equal by the 8. diffinition of the eleuenth (for they eche consist of two bases of the octohedron, and of two triangles contained vnder the diameter AD and two sides of the octohedro<g ref="char:cmbAbbrStroke">̄</g>). Wher<g ref="char:EOLhyphen"/>fore the prisme which is set vpon the base of the Octo<g ref="char:EOLhyphen"/>hedron, and hauing the same altitude with it, namelye, the altitude of the parallel bases, as it is manifest by the former, is equal to thre of those pyramids of the Octo<g ref="char:EOLhyphen"/>hedron, by the first corollary of the seue<g ref="char:cmbAbbrStroke">̄</g>th of the twelft. Wherefore that prisme shall haue to the other prisme vnder the same altitude, composed of the 4. pyramids of the whole octohedron, the proportion of the triangular bases, by the 3. corollary of the same. And forasmuch as 4. pyramids are vnto 3. pyramids in sesquitercia propor<g ref="char:EOLhyphen"/>tion, therefore the trianguler base of the prisme which containeth 4. pyramids, is in sesquitertia proportion to the base of the prisme which containeth thre pyramids of the same octohedron, and are set vpon the base of the Octohedron and vnder the altitude thereof: that is, in sesquitercia proportion to the base of the Octohe<g ref="char:EOLhyphen"/>dron. But the base of the same octohedron is in sesqui<g ref="char:EOLhyphen"/>tertia proportion to the base of the pyramis, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth of this booke: Wherefore the triangular bases, namely, of the prisme which co<g ref="char:cmbAbbrStroke">̄</g>taineth four pyramids of the octohedron, and is vnder the altitude thereof, are equal to the triangular bases of the prisme, which con<g ref="char:EOLhyphen"/>taineth three pyramids vnder the altitude of the pyra<g ref="char:EOLhyphen"/>mis EFGH. But the prisme of the octohedron is equal to the octohedron: and the prisme of the pyramis EF<g ref="char:EOLhyphen"/>GH is proued triple to the same pyramis EFGH. Now then the prismes set vpo<g ref="char:cmbAbbrStroke">̄</g> equal bases, are the one to the other as their altitudes are (by the corollary of the 25. of the eleuenth) namely, as are the parallelipidedons their doubles, by the corollary of the 31. of the eleuenth. But the altitude of the Octohedron is equal to the side of the cube contained in the same sphere, by the corollary of the 13. of this booke. And the side of the cube is in power to the altitude of the Tetrahedon in that proportion that 12. is to 16, by the 18. of the thirtenth: And the side of the octo<g ref="char:EOLhyphen"/>hedron is to the side of the pyramis in that proportion that 18. is to 24. (by the same 18. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th) which proportion is one &amp; the self same with the proportio<g ref="char:cmbAbbrStroke">̄</g> of 12. to 16. Wherfore that prisme which is equal to the Octohedron, is to the prisme which is triple to the Tetrahedron, in that proportio<g ref="char:cmbAbbrStroke">̄</g> that the altitudes, or that the sides are. Wherfore an octohedro<g ref="char:cmbAbbrStroke">̄</g> is to the triple of a Tetrahedron co<g ref="char:cmbAbbrStroke">̄</g>tained in one and the selfe same sphere, in that proportion that their sides are: which was required to be de<g ref="char:EOLhyphen"/>monstrated.</p>
               <div type="section">
                  <pb n="428" facs="tcp:7062:457"/>
                  <head>A Corollary.</head>
                  <p>
                     <hi>The sides of a Tetrahedron &amp; of an Octohedro<g ref="char:cmbAbbrStroke">̄</g> are proportionall with their altitudes.</hi> For the sides &amp; altitudes were in power sesquitercia. Moreouer the diameter of the sphere is to the side of the Tetrahedron, as the side of the Octohedron is to the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>de of the cube<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> namely, the powers of eche is in sesquialter proportion, by the 18. of the thirtenth.</p>
               </div>
            </div>
            <div type="part">
               <head>The 15. Proposition. If a rational line containing in power two lines, make the whole and the greater segment, and again containing in power two lines, make the whole and the lesse segment: the greater segment shalbe the side of the Icosahe<g ref="char:EOLhyphen"/>dron, and the lesse segment shalbe the side of the Dodecahedron contayned in one and the selfe same sphere.</head>
               <p>SVppose that AG be the diameter of the sphere which containeth the Icosahedron ABGC. And let BG subtend the sides of the pentagon described of the sides of the Icosahedron (by the 16. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th.) Moreouer vpon the same diameter AG, or DF equal vnto it,<note place="margin">
                     <hi>Construction.</hi>
                  </note> let ther be described a dodecahedron DEFH, by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the thirtenth, whose opposites sides ED and FH let be cut into two equal partes in the poynts I and K, and draw a line from I to K. And let the line EF couple two of the opposite angles of the bases which are ioyned together. The<g ref="char:cmbAbbrStroke">̄</g> I say that AB the side of the Icosahedron is the greater segment which the diameter AG containeth in power together with the whole line, and line ED is the lesse segment, which the same diameter AG or DF containeth in power together with the whole.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For forasmuche as the
<figure/>
opposite sides AB and GC of the Icosahedron being cou<g ref="char:EOLhyphen"/>pled by the diameters AG and BC, are equal &amp; parallels, by the 2. corollary of the 16. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th: the right lines BG &amp; AC which couple the<g ref="char:cmbAbbrStroke">̄</g> together are equal &amp; paral<g ref="char:EOLhyphen"/>lels by the 33. of the first. Moreouer the angles BAC &amp; A<g ref="char:EOLhyphen"/>BG being subtended of equal diam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ters, shall by the 8. of the first be equal, &amp; by the 29 of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, they shal be right angles. Wherfore the right line AG <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> in power the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>wo lines AB and BG, by the 47. of <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>. And forasmuch as the line BG subtendeth the angle of the pentagon com<g ref="char:EOLhyphen"/>posed of the sides of the Icosahedron, the greater segment of the right line BG, shalbe the right line AB, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the thirtenth: which line AB, toge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>her with the whole line BG, the line AG containeth in power. And forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line IK coupling the opposite and parallel sides ED and FH of the Dodecahedron, maketh at those poyntes right angles, by the 3. corollary of the 17. of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e thirtenth: the right line EF which coupleth together equal and pa<g ref="char:EOLhyphen"/>rallel lines EI &amp; FK, shalbe equal to the same line IK, by the 33. of the first. Wherfore the angle DEF shalbe <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> right angle by the 29. of the first. Wherefore the diameter DF co<g ref="char:cmbAbbrStroke">̄</g>taineth in power the two lines ED and EF. But the lesse segment of the line IK is ED the side of the Dodecahe<g ref="char:EOLhyphen"/>dron, by the 4. corollary of the 17<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the thirtenth. Wher<g ref="char:EOLhyphen"/>fore the same line ED is also the lesse segment of the line EF (which is equal vnto the line IK): wherfore the diam<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLunhyphen"/>ter DF containing in power the two lines ED and EF (by the 47. of the first) containeth in pow<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> ED the side of the dodecahedron, the lesse segment, together with the whole. If therfore a rational line AG or DF containing in power two lines AB and BG, doo make the whole line and the greater s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>gment, and agai<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e containing in power two lines EF and ED, do make the whole line and the lesse segment: the greater segment AB, shall be the side of the Icosahe<g ref="char:EOLhyphen"/>dron, and the lesse segment ED shall be the side of the Dodecahedron contained in one and the selfe same sphere.</p>
            </div>
            <div type="part">
               <head>The 16. Proposition. If the power of the side of an Octohedron be expressed by two right line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <pb facs="tcp:7062:458"/>
ioyned together by an extreme and me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne proportion: the side of the Ico<g ref="char:EOLhyphen"/>sahedron contained in the same sphere, shalbe duple to the lesse segment.</head>
               <p>LEt AB the side of the Octohedron ABG containe in power the two lines C and H, which let haue that proportion that the whole hath to the greater segment (by the corollarye of the first proposition added by <hi>Flussas</hi> after the last propositio<g ref="char:cmbAbbrStroke">̄</g> of the sixth booke).<note place="margin">
                     <hi>Construction.</hi>
                  </note> And let the Icosahedron contained in the same sphere be DEF, whose side let be DE, and let the right line subtending the angle of the pentagon made of the sides of the Icosahedron be the line EF. Then I say that the side ED is in power double to the line H the lesse of those segmentes. Foras<g ref="char:EOLhyphen"/>much as by that which was demonstrated in the 15. of this booke,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> it was manifest that ED the side of the Icosahedron is the great<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r segment of the line EF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and that the diameter DF containeth in power the two lines ED and EF, namely, the whole and the greater segment: but by suppo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ition the side AB co<g ref="char:cmbAbbrStroke">̄</g>taineth in po<g ref="char:EOLhyphen"/>wer
<figure/>
the two lines C &amp; H ioined to<g ref="char:EOLhyphen"/>gether in the self same proportio<g ref="char:cmbAbbrStroke">̄</g>. Wherefore the line EF is to the line ED, as the line C, is to the line H, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> this boke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> And alt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rna<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>y by the 16. of the fiueth, the line EF is to the line C, as the line ED, is to the line H. And for<g ref="char:EOLhyphen"/>asmuche as the line DF containeth in power the two lines ED and EF, and the line AB containeth in power the two lines C and H: therefore the squares of the lines EF and ED are to the square of the line DF, as the squares of the lines C and H to the square AB. And alternately, the squares of the lines EF and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D, are to the squares of the lines C and H, as the square of the line DF is to the square of the line AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> But DF the diameter is (by the 14. of the thirten<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h) i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> power double to AB the side of the octo<g ref="char:EOLhyphen"/>hedron inscribed (by supposition) in the same sphere. Wherefore the squares of the lines EF and ED, are double to the squares of the lines C and H. And therfore one square of the line ED is double to one square of the line H by the 12. of the fifth. Wherfore ED the side of the Icosahedron is in power duple to the line H, which is the lesse segment. If therfore the powe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the side of an octohedron be expres<g ref="char:EOLhyphen"/>sed by two right lines ioyned together by an extreme and meane proportion: the side of the Icosahe<g ref="char:EOLhyphen"/>dron contained in the same sphere, shalbe duple to the lesse segment.</p>
            </div>
            <div type="part">
               <head>The 17. Proposition. If the side of a dodecahedron, and the right line, of whome the said side is the lesse segment, be so set that they make a right angle: the right line which containeth in power halfe the line subtending the angle, is the side of an Octohedron contained in the selfe same sphere.</head>
               <p>SVppose that AB be the side of a Dodecahedron, and let the
<figure/>
right line of which that side is the lesse segment be AG, name<g ref="char:EOLhyphen"/>ly which coupleth the opposite sides of the Dodecahedron, by the 4. corollary of the 17. of the thirtenth:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and let those lines be so set that they make a right angle at the point A. And draw the right line BG. And let the line D containe in power halfe the line BG (by the first proposition added by <hi>Flussas</hi> after the laste of the sixth). Then I say that the line D is the side of an Octohedron contayned in the same sphere. Forasmuche as the line AG maketh the greater segment GC the side of the cube contained in the same sphere (by the same 4. corol<g ref="char:EOLhyphen"/>lary of the 17. of the thirtenth):<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and the squares of the whole line AG. and of the lesse segment AB are triple to the square of the greater segment GC, by the 4. of the thirtenth: Moreouer the diameter of the sphere, is in power triple to the same line GC the side of the cube (by the 15. of the thirtenth:
<pb n="429" facs="tcp:7062:458"/>
Wherfore the line BG is equal to the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>. For it con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ineth in power the two lines AB and AG (by the 47. of the first,) and therefore it containeth in power the triple of the line GC. But the side of the Octohedron contained in the same sphere, is in power triple to halfe the diameter of the sphere by the 14. of the thirtenth. And by suppo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tion the line D contai<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> in pow<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the halfe of the line BG. Wherefore the line D (containing in power the halfe of the same diameter) is the side of an octohe<g ref="char:EOLhyphen"/>dron. If therfore the side of a Dodecah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron and the right line of whome the said side is the lesse seg<g ref="char:EOLhyphen"/>ment, be so set that they make a right angle: the right line which containeth in power halfe the line subtending the angle, is the side of an Oc<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>edron contained in the selfe same sphere: Which was re<g ref="char:EOLhyphen"/>quired to be proued.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>Vnto what right line the side of the Octo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>edron is in power sesquialter: vnto the same line the side of the Dodecahedron inscribed in the same sphere, is the greater segment.</hi> For the side of the Dodecahedron is the greater segment of the segment CG, vn<g ref="char:EOLhyphen"/>to which D the side of the Octohedron is in power sesqui<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>lter, that is, is halfe of the power of the line BG, which was triple vnto the line CG.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 18. Proposition. If the side of a Tetrahedron containe in power two right lines ioyned to<g ref="char:EOLhyphen"/>gether by an extreme and meane proportion: the side of an Icosahedron described in the selfe same Sphere, is in power sesquialter to the lesse right line.</head>
               <p>SVppose that ABC be a Tetrahedron, and let his side be AB,<note place="margin">
                     <hi>Construction.</hi>
                  </note> whose power let be diuided into the lines AG and GB, ioyned together by an extreme and meane proportion: name<g ref="char:EOLhyphen"/>ly, let it be diuided into AG the whole line, and GB the greater se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ment (by the Corolla<g ref="char:EOLhyphen"/>ry of the first Proposition added by Flussas after the last of the sixth). And let ED be the side of the Icosahedron EDF contained in the selfe same Sphere. And let the line which subtendeth the angle of the Pentagon described of the sides of the Icosahedron be EF. Then I say, that ED the side of
<figure/>
the Icosahedron is in power sesquialter to the lesse line GB.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Forasmuch as (by that which was demonstrated in the 15. of this booke) the side ED is the gre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter segment of the line EF which subtendeth the angle of the Pentagon. But as the whole line EF is to the greater seg<g ref="char:EOLhyphen"/>ment ED, so is the same gr<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>ter segment to the lesse (by the 30. of the sixth): and by suppo<g ref="char:EOLhyphen"/>sition, AG, was the whole line, and G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the greater segment: Wherefore as EF is to ED, so is AG to G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, by the second of the foure<g ref="char:EOLhyphen"/>tenth. And alternately, the line EF is to the line AG, as the line ED is to the line GB. And forasmuch as (by supposition) the line AB containeth in power the two lines AG and GB: therefore (by the 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the first) the angle AGB is a right angle. But the angle DEF is a right angle, by that which was de<g ref="char:EOLhyphen"/>monstrated in the 15. of this booke. Wherefore the triangles AG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and FED, are equiangle, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the sixth. Wherefore their sides are proportionall: namely, as the line ED is to the line GB, so is the line FD to the line AB, by the 4. of the sixth. But by that which hath before bene demonstrated, FD is the diameter of the Sphere which containeth the Icosah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron: which diameter is in power sesqui<g ref="char:EOLhyphen"/>alter to AB the side of the Tetrahedron inscribed in the<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> same Sphere, by the 13. of the thirtenth. Wher<g ref="char:EOLhyphen"/>fore the line ED the side of the Icosahedron, is in power sesquialter to G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the greater segment or lesse line. If therefore the side of a Tetrahedron containe in power two right lines ioyned together an ex<g ref="char:EOLhyphen"/>treme and meane proportion: the side of an Icosahedron described in the selfe same Sphere, is in pow<g ref="char:EOLhyphen"/>er sesquialter to the lesse right line.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:459"/>
               <head>¶ The 19. Proposition. The superficies of a Cube is to the superficies of an Octohedron inscribed in one and the selfe same Sphere, in that proportion that the solides are.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that ABCDE be a Cube, whose fower diameters let be the lines AC, BC, DC, and EC produced on ech side. Let also the Octohedron inscribed in the selfe same Sphere be FGHK: whose three diameters let be FH, GK, and ON. Then I say, that the cube ABD is to the Octohedron FGH, as the superficies of the cube is to the super<g ref="char:EOLhyphen"/>ficies of the Octohedron. Drawe from the centre of the cube to the base ABED, a per<g ref="char:EOLhyphen"/>pendicular line CR. And from the centre of the Octohedron draw to the base GNH, a perpendicular line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>L.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> And forasmuch as the three diameters of the cube do passe by the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> C, therefore, by the 2. Corollary of the 15. of the thirtenth, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>here shall be made of the cube sixe pyramids, as thys pyramis ABDEC, equall to the whole cube. For there are in the cube <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ixe bases, vpon which fall equall perpendiculars from the cen<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> by the Corollary of the Assump<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the 16. of the twelfth, for the bases are contained in equall circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the Sphere. But in the Octohedron the three diameters do make vpon the 8. bases, 8. pyramids, hauing their toppes in the centre, by the 3. Corollary of the 14<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the thirtenth. Now the bases of the cube and of the Octohedron are contained in equall circles of the
<figure/>
Sphere, by the 13. of this booke. Wherefore they shall be equally distant from the centre, and the per<g ref="char:EOLhyphen"/>pendicular lines CR and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, shall be equall, by the Corollary of the Assumpt of the 16. of the twelfth. Wherefore the pyramids of the cube shall be vnder one and the selfe same altitude with the pyramids of the Octohedron, namely, vnder the perpendicular line drawen from the centre to the bases. Wher<g ref="char:EOLhyphen"/>fore sixe pyramids of the cube, are to 8. pyramids of the Octohedron being vnder one and the same al<g ref="char:EOLhyphen"/>titude, in that propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tion that their bases are, by the 6. of the twelfth: that is, one pyramis set vpon sixe bases of the cube, and hauing to his altitude the perpendicular line, which pyramis is equall to the sixe pyramids, by the same 6. of the twelfth, is to one pyramis set vpon the 8. bases of the Octohedron, being equall to the Octohedron, and also vnder on<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the selfe same altitude, in that proportion that sixe bases of the cube, which containe the whole superficies of the cube, are to 8. bases of the Octohedron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which containe the whole superficies of the Octohedron. For the solides of those pyramids are in pro<g ref="char:EOLhyphen"/>portion the one to the other, as their bases are, by the selfe same 6. of the twelfth. Wherefore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he su<g ref="char:EOLhyphen"/>perficies of the cube is to the superficies of the Octohedron inscribed in one and the selfe same Sphere, in that proportion, that the solides are: which was required to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Proposition. If a Cube and an Octohedron be contained in one &amp; the selfe same Sphere: they shall be in proportion the one to the other, as the side of the Cube is to
<pb n="430" facs="tcp:7062:459"/>
the semidiameter of the Sphere.</head>
               <p>SVppose that the Octohedron AECDB be inscribed in the Sphere ABCD: and let the cube inscribed in the same Sphere be FGHIM: whose diameter let be HI,<note place="margin">
                     <hi>Construction.</hi>
                  </note> which is e<g ref="char:EOLhyphen"/>quall to the diameter AC, by the 15. of the thirtenth: let the halfe of the diameter be AE. Then I say, that the cube FGHIM is to the Octohedron AECDB, as the side MG is to the semidiameter AE. Forasmuch as the diameter AC is in power double to BK the side of the Octohedron (by the 14. of the thirtenth) and is in power triple to MG the side of the cube (by the 15. of the same): therefore the square BKDL shall be sesquial<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er to FM the square of the cube. From the line AE cut of a third part AN, and fro<g ref="char:cmbAbbrStroke">̄</g> the line MG cut of like<g ref="char:EOLhyphen"/>wise a third part GO, by the 9. of the sixth.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Now then the line EN shall be two third partes of the line AE, and so also shall the line MO be of the line MG. Wherefore the parallelipipedon set vpon the base BKDL, and hauing his altitude the line EA, is triple to the parallelipipedon set vpon the same base, and hauing his altitude the line AN, by the Corollary of the 31. of the eleuenth: but it is also tri<g ref="char:EOLhyphen"/>ple to the pyramis ABKDL which is set vpon the same base, and is vnder the same altitude (by the se<g ref="char:EOLhyphen"/>cond Corollary of the 7. of the twelfth). Wherefore the pyramis ABKDL is equall to the parallelipi<g ref="char:EOLhyphen"/>pedon, which is set vpon the base BKDL, and
<figure/>
hath to his altitude the line AN. But vnto that pa<g ref="char:EOLhyphen"/>rallelipipedo<g ref="char:cmbAbbrStroke">̄</g>, is double the parallelipipedon which is set vppon the same base BKDL, and hath to his altitude a line double to the line EN, by the Corollary of the 31. of the first and vnto the pyra<g ref="char:EOLhyphen"/>mis is double the Octohedron ABKLDC, by the 2. Corollary of the 14. of the thirtenth. Where<g ref="char:EOLhyphen"/>fore the Octohedron ABKDLC is equall to the parallelipipedon set vpon the base BKLD, &amp; ha<g ref="char:EOLhyphen"/>uing his altitude the line EN (by the 15. of the fifth). But the parallelipipedon set vpon the base BKDL, which is sesquialter to the base FM, and hauing to his altitude the line MO, which is two third partes of the side of the cube MG, is equall to the cube FG: by the 2. part of the 34. of the ele<g ref="char:EOLhyphen"/>uenth. (For it was before proued that the base BKDL is sesquialter to the base FM). Now then these two parallelipipedons, namely, the paralleli<g ref="char:EOLhyphen"/>pipedon which is set vpo<g ref="char:cmbAbbrStroke">̄</g> the base BKDL (which is sesquialter to the base of the cube) and hath to his altitude the line MO (which is two third partes of MG the side of the cube) which paral<g ref="char:EOLhyphen"/>lelipipedon is proued equall to the cube, and the parallelipipedon set vpon the same base BKDL, and hauing his altitude the line EN (which paral<g ref="char:EOLhyphen"/>lelipipedon is proued equall to the Octohedron): these two parallelipipedons (I say) are the one to the other, as the altitude MO, is to the altitude EN (by the Corollary of the 31. of the eleuenth). Wherefore also as the altitude MO, is to the alti<g ref="char:EOLhyphen"/>tude EN, so is the cube FGHIM, to the Octo<g ref="char:EOLhyphen"/>hedron ABKDLC, by the 7. of the fifth. But as the line MO is to the line EN, so is the whole line MG to the whole line EA, by the 18. of the fifth. Wherefore as MG the side of the cube, is to EA the semidiameter, so is the line FGHIM to the Octohedron ABKDLC inscribed in one &amp; the selfe same Sphere. If therefore a cube and an Octohedron be contained in one and the selfe same Sphere. they shall be in proportion the one to the other, as the side of the cube is to the semidiameter of the Sphere: which was required to be demonstrated.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <q>Distinctly to notefie the powers of the sides of the fiue solides by the power of the diameter of the sphere.</q>
                  </p>
                  <p>The sides of the tetrahedron and of the cube doo cut the power of the diameter of the sphere in<g ref="char:EOLhyphen"/>to two squares which are in proportion double the one to the other. The octohedron cutteth the
<pb facs="tcp:7062:460"/>
power of the diameter into two equall squares. The Icosahedron into two squares, whose proportion is duple to the proportion of a line diuided by an extreame and meane proportion, whose lesse segme<g ref="char:cmbAbbrStroke">̄</g>t is the side of the Icosahedron. And the dodecahedron into two squares, whose proportion is quadru<g ref="char:EOLhyphen"/>ple to the proportion of a line diuided
<figure/>
by an extreame and meane proportion, whose lesse segment is the side of the dodecahedron. For AD the diameter of the sphere, contayneth in power AB the side of the tetrahedron, and BD the side of the cube, which BD is in power halfe of the side AB. The diameter also of the sphere contayneth in power AC and C<g ref="char:EOLhyphen"/>D two equall sides of the octohedron. But the diameter contayneth in power the whole line AE and the greater seg<g ref="char:EOLhyphen"/>ment thereof ED, which is the side of the Icosahedron, by the 15. of this booke. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>fore their powers being in duple proportio<g ref="char:cmbAbbrStroke">̄</g> of that in which the sides are, by the first corollary of the 20. of the sixth, haue their proportion duple to the proportion of an extreame &amp; meane proportio<g ref="char:cmbAbbrStroke">̄</g>. Farther the diameter co<g ref="char:cmbAbbrStroke">̄</g>tayneth in power the whole line AF, and his lesse segment FD, which is the side of the dodecahedron, by the same 15. of this booke. Wherefore the whole hauing to the lesse, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> double proportion of that which the extreame hath to the meane, namely, of the whole to the greater segment, by the 10. diffinition of the fifth, it followeth that the proportion of the power is double to the doubled proportion of the sides, by the same first corollary of the 20. of the sixth: that is, is quadru<g ref="char:EOLhyphen"/>ple to the proportion of the extreame and of the meane, by the diffinition of the sixth.</p>
                  <div type="subsection">
                     <head>An aduertisment added by <hi>Flussas.</hi>
                     </head>
                     <p>By this meanes therefore, the diameter of a sphere being geuen, there shall be ge<g ref="char:EOLhyphen"/>uen the side of euery one of the bodies inscribed. And forasmuch as three of those bo<g ref="char:EOLhyphen"/>dies haue their sides commensurable in power onely, and not in length, vnto the dia<g ref="char:EOLhyphen"/>meter geuen (for their powers are in the proportion of a square number to a number not square: wherefore they haue not the proportion of a square number to a square number, by the corollary of the 25. of the eight: wherefore also their sides are incom<g ref="char:EOLhyphen"/>mensurabe in length by the 9. of the tenth): therefore it is sufficient to compare the powers and not the lengths of those sides the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> which powers are con<g ref="char:EOLhyphen"/>tained in the power of the diameter: namely, from the power of the diameter, let there ble taken away the power of the cube, and there shall remayne the power of the Tetra<g ref="char:EOLhyphen"/>hedron: and taking away the power of the Tetrahedron, there remayneth the power of the cube: and taking away from the power of the diameter halfe the power thereof, there shall be left the power of the side of the octohedron. But forasmuch as the sides of the dodecahedron and of the Icosahedron are proued to be irrationall (for the side of the Icosahedron is a lesse line, by the 16. of the thirtenth: and the side of the dedocahe<g ref="char:EOLhyphen"/>dron is a residuall line, by the 17. of the same) therfore those sides are vnto the diame<g ref="char:EOLhyphen"/>ter which is a rationall line set, incommensurable both in length and in power. Where<g ref="char:EOLhyphen"/>fore their comparison can not be diffined or described by any proportion expressed by numbers, by the 8. of the tenth: neither can they be compared the one to the other: for irrational lines of diuers kindes are inco<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable the one to the other: for if they should be commensurable, they should be of one and the selfe same kinde, by the 103. and 105. of the tenth, which is impossible. Wherefore we seking to compare them to the power of the diameter, thought they could not be more aptly expressed, then by such proportions, which cutte that rationall power of the diameter according to their sides: namely, diuiding the power of the diameter by lines which haue that proportio<g ref="char:cmbAbbrStroke">̄</g>, that the greater segment hath to the lesse, to put the lesse segment to be the side of the Icosahedron: &amp; deuiding the sayd power of the diameter by lines hauing the propor<g ref="char:EOLhyphen"/>tion of the whole to the lesse segment, to expresse the side of the dodecahedron by the lesse segment: which thing may well be done betwene magnitudes incommensurable.</p>
                  </div>
               </div>
            </div>
            <trailer>The ende of the fourtenth Booke of Euclides Elementes after Flussas.</trailer>
         </div>
         <div n="15" type="book">
            <pb n="431" facs="tcp:7062:460"/>
            <head>¶The fiftenth booke of Euclides Elementes.</head>
            <p>THis finetenth and last booke of Euclide, or rather the second boke of <hi>Appollonius</hi> or <hi>Hypsicles,</hi>
               <note place="margin">
                  <hi>The argu<g ref="char:EOLhyphen"/>ment of the</hi> 15. <hi>booke.</hi>
               </note> teacheth the in<g ref="char:EOLhyphen"/>scription and circumscriptio<g ref="char:cmbAbbrStroke">̄</g> of the fiue regular bodies one within and about an other: a thing vndoutedly ple<g ref="char:EOLhyphen"/>sant and delectable in minde to contemplate, and also profitable and necessary in act to practise. For without practise in act, it is very hard to se and conceiue the con<g ref="char:EOLhyphen"/>structions and demonstrations of the propositions of this booke, vnles a man haue a very depe, sharpe, &amp; fine imagination. Wherfore I would wish the diligent stude<g ref="char:cmbAbbrStroke">̄</g>t in this booke, (to make the study thereof more pleasant vnto him) to haue presently before his eyes, the bodyes formed &amp; framed of pasted paper (as I taught after the diffinitions of the eleuenth booke.) And then to drawe and describe the lines and diuisions, and superficieces, according to the constructions of the propositions. In which descriptions if he be wary and diligent, he shall finde all things in these solide matters, as clere and as manifest vnto the eye, as were things be<g ref="char:EOLhyphen"/>fore taught only in plaine or superficial figures. And although I haue before in the twelfth boke admonished the reader hereof, yet bicause in this boke chiefly that thing is required, I thought it should not be irkesome vnto him, againe to be put in minde thereof.</p>
            <p>Farther this is to be noted, that in the Greke exemplars are found in this 15. booke only 5. propositions, which 5. are also only touched and set forthe by Hypsicies: vnto which Campane addeth 8. and so maketh vp the number of 13. Campane vndoubted<g ref="char:EOLhyphen"/>ly although he were very well lerned, and that generally in all kinds of learning, yet as<g ref="char:EOLhyphen"/>suredly being brought vp in a time of rudenes, when all good letters were darkned, &amp; barberousnes had ouerthrowne and ouerwhelmed the whole world, he was vtterly rude and ignorant in the Greke tongue, so that certenly he neuer redde <hi>Euclide</hi> in the Greke, nor (of like) translated out of the Greke: but had it translated out of the Ara<g ref="char:EOLhyphen"/>bike tonge. The Arabians were men of great study, and industry, and commonly great Philosophers, notable Phisitions, and in mathematicall Artes most expert, so that all kinds of good learning flourished and raigned amongst them in a manner only. These men turned whatsoeuer good author was in the Greke tonge (of what Art and know<g ref="char:EOLhyphen"/>ledge so euer it were) into the Arabike tonge. And fro<g ref="char:cmbAbbrStroke">̄</g> thence were many of the<g ref="char:cmbAbbrStroke">̄</g> turned into the Latine, and by that meanes many Greeke authors came to the handes of the Latines, and not from the first fountaine the Greke tonge, wherin they were first writ<g ref="char:EOLhyphen"/>ten. As appeareth by many words of the Arabike tonge yet remaining in such bokes: as are <hi>Zenith, nadir, helmuayn, helmuariphe,</hi> and infinite suche other. Which Arabians also in translating such Greke workes, were accustomed to adde, as they thought good, &amp; for the fuller vnderstanding of the author, many things: as is to be sene in diuers authors, as, namely, in <hi>Theodosius de Sphera,</hi> where you see in the olde translation (which was vn<g ref="char:EOLhyphen"/>doubteldy out of the Arabike) many propositions, almost euery third or fourth leafe. Some such copye of <hi>Euclide,</hi> most likely, did <hi>Campanus</hi> follow, wherein he founde those propositio<g ref="char:cmbAbbrStroke">̄</g>s, which he hath more &amp; aboue those which are found in the Greke set out by Hypsicles: and that not only in this 15. boke, but also in the 14. boke, wherin also ye finde many propositions more the<g ref="char:cmbAbbrStroke">̄</g> are founde in the Greeke, set out also by Hypsicles. Likewise in the bookes before, ye shall finde many propositions added, and manye in<g ref="char:EOLhyphen"/>uerted, and set out of order farre otherwise, then they are placed in the Greeke exam<g ref="char:EOLhyphen"/>plars. <hi>Flussas</hi> also a diligent restorer of <hi>Euclide,</hi> a man also which hath well deserued of the whole Art of Geometrie, hath added moreouer in this booke (as also in the former
<pb facs="tcp:7062:461"/>
14. boke he added 8. pro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ositio<g ref="char:cmbAbbrStroke">̄</g>s) 9. propositio<g ref="char:cmbAbbrStroke">̄</g>s of his owne, touching the inscription, and circumscript<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>on <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap> bodies, very si<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>gular <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ndoubtedly and wittye. All which, for that nothing should want to the desirous louer of knowledge, I haue faithfully with no small paines turned. And whereas <hi>Fl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ss<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
               </hi> in the beginning of the eleuenth booke, namely, in the end of the diffinitions there <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>, putteth two diffinitions, of the inscrip<g ref="char:EOLhyphen"/>tion and circumscription of solides or corporall figures, within or about the one the other, which certainely are not to be reiected: yet for that vntill this present 15. boke, there is no mention made of the inscription or circumscription of these bodyes, I thought it not so conuenient th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> to place them, but to referre the<g ref="char:cmbAbbrStroke">̄</g> to the beginning of this 15. booke: where they are in maner of necessitie required to the elucidation of the Proposi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ions and d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>monstration<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> of the same. The diffinitions are these.</p>
            <div type="part">
               <head>Diffinition. 1.</head>
               <p>A solide figure, is then <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>aid to be inscribed in a solide figure, when the an<g ref="char:EOLhyphen"/>gles of the figure inscribed touche together at one time, either the angles of the figure circumscribed, or the superficieces, or the sides.</p>
            </div>
            <div type="part">
               <head>Diffinition. 2.</head>
               <p>
                  <hi>A solide figure is then said to be circumscribed about a solide figure: when together at one time either the angles, or the superficieces, or the sides of the figure circumscribed, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ouch the angles of the figure inscribed.</hi>
               </p>
               <p>IN the four<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> booke in the diffinitions of the inscription or circumscription of playne rectiline figures one with in or about an other, was requ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>red that all the angles of the figu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> inscribed, should at one time touch all the sides of the figure circumscribed: but in the fiue regular solides (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o whome chefely these two diffinitions pertaine) for that the nomber of their angles, superficieces, &amp; sides are not equal, one compared to an o<g ref="char:EOLhyphen"/>ther: it is not of necessitie, that all the angles of the solide inscribed should together at one time touch either all the angles, or all the superficieces, or all the sides of the solide circumscribed: but it is sufficient, that those angles of the inscribed solide which touch, doe at one time together eche touch some one angle of the figure circumscribed, or some one base, or some one side: so that if the an<g ref="char:EOLhyphen"/>gles of the inscribed figure do at one time touche the angles of the figure circumscribed, none of them may at the same time touche either the bases or the sides of the same circumscribed figure: and so if they touch the bases, they may touche neither angles nor sides: and likewise if they touche the sides, they may touch neither angles nor bases. And although sometimes all the angles of the figure inscri<g ref="char:EOLhyphen"/>bed can not touch either the angles, or the bases, or the sides of the figure circumscribed, by reason the nomber of the angles, bases or sides of the said figure circumscri<g ref="char:EOLhyphen"/>bed, wanteth of the nomber of the angles of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>igure inscribed yet shall those angles of the inscribed figure which touch, so touch, that the void places left betwene the inscribed and circumscribed figures shal on euery side be equal and like. As ye may afterwarde in this fiftenth booke most plainely perceiue.</p>
               <div type="section">
                  <pb n="432" facs="tcp:7062:461"/>
                  <head>¶ The 1. Proposition. The 1. Probleme. In a Cube geuen to describe <note n="†" place="margin">In this propo<g ref="char:EOLhyphen"/>sition as also in all the other following, by the name of a py<g ref="char:EOLhyphen"/>ramis vnder<g ref="char:EOLhyphen"/>stand a tetra<g ref="char:EOLhyphen"/>hedron: as I haue before admonished.</note> a trilater equilater Pyramis.</head>
                  <p>SVppose that the
<figure/>
cube geuen be AB<g ref="char:EOLhyphen"/>CDEFGH. In the same cube it is required to in<g ref="char:EOLhyphen"/>scribe a Tetrahe<g ref="char:EOLhyphen"/>dron. Drawe these right lines AC,<note place="margin">Construction.</note> CE, AE, AH, EH, HC.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now it is manifest, that the trian<g ref="char:EOLhyphen"/>gles AEC, AHE, AHC, and CHE, are e<g ref="char:EOLhyphen"/>quilater, for their sides are the diameters of equall squares. Wherfore AECH is a trilater equilater pyramis, or Tetrahedron, &amp; it is in<g ref="char:EOLhyphen"/>scribed in the cube geue<g ref="char:cmbAbbrStroke">̄</g> (by the first definition of this booke): which was required to be done.</p>
               </div>
               <div type="section">
                  <head>¶ The 2. Proposition. The 2. Probleme. In a trilater equilater Pyramis geuen to describe an Octohedron.</head>
                  <p>SVppose that the trilater equi<g ref="char:EOLhyphen"/>later
<figure/>
pyramis geue<g ref="char:cmbAbbrStroke">̄</g> be ABC<g ref="char:EOLhyphen"/>D, whose sides let be diuided into two equall partes in the pointes E, Z, I, K, L, T.<note place="margin">Construction.</note> And draw these 12. right lines EZ, ZI, IE, KL, LT, TK, EK, KZ, ZL, LI, IT, and TE. Which 12. right lines are,<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> by the 4. of the first, e<g ref="char:EOLhyphen"/>quall. For they subtend equall plaine an<g ref="char:EOLhyphen"/>gles of the bases of the pyramis, and those equall angles are contained vnder equall sides, namely, vnder the halfes of the sides of the pyramis. Wherefore the triangles TKL, TLI, TIE, TEK, ZKL, ZLI, ZIB, ZEK, are equilater: and they limitate, and containe the solide TK<g ref="char:EOLhyphen"/>LEZI. Wherefore the solide TKLEZI is an Octohedron: by the 23. definition of the eleuenth. And the angles of the same Octohedron do touch the sides of the pyramis ABCD in the pointes E, Z, I, T, K, L. Wherefore the Octohedron is inscribed in the pyramis (by the 1. definition of this booke). Wherefore in the trilater equilater pyramis geuen, is inscribed an Octohedron: which was required to be done.</p>
                  <div type="subsection">
                     <head>A Corollary added by <hi>Flussas.</hi>
                     </head>
                     <p>
                        <hi>Hereby it is manifest, that a pyramis is cut into two equall partes, by euery
<pb facs="tcp:7062:462"/>
one of the three equall squares, which diuide y<hi rend="sup">e</hi> Octohedron into two equall partes and perpendicularly.</hi> For the three diameters of those squares do in the centre cut the one the other into two equall partes and perpendicularly, by the third Corollary of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. of the thirtenth, which squares, as for example, the square EKLI, do diuide in sunder the pyramids and the prismes, namely, the pyramis KLTD and the prisme KLTEIA from the pyramis EKZB, and the prisme EKZILG, which pyramids are equall the one to the other, and so also are the prismes equall the one to the other: by the 3. of the twelfth. And in like sort do the rest of the squares, namely, KZIT and ZLTE: which squares, by the second Corollary of the 14. of the thirtenth, do diuide the Octohedron into two equall partes.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶ The 3. Proposition. The 3. Probleme. In a cube geuen, to describe an Octohedron.</head>
                  <p>TAke a Cube, namely, ABCDEFGH. And diuide euery one of the sides thereof into two equall partes.<note place="margin">Construction.</note> And drawe right lines coupling together the sections, as for example, these right lines, PQ and RS, which shall be equall vnto the side of the cube (by the 33. of the first) and shall diuide the one the other into two equall parts in the middest of the diameter AG in the point I (by the Corollary of the 34. of the
<figure/>
first). Wherefore the point I is the cen<g ref="char:EOLhyphen"/>tre of the base of the cube. And by the same reason may be found out the cen<g ref="char:EOLhyphen"/>tres of the rest of the bases, which let be the pointes K, L, O, N, M. And drawe these right lines LI, IM, MO, OL, KI, KL, KM, KO, NI, NL, NM, &amp; NO.<note place="margin">Demonstr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>tion.</note> And now forasmuch as the angle IPL is a right angle (by the 10. of the ele<g ref="char:EOLhyphen"/>uenth, for the lines IP and PL are pa<g ref="char:EOLhyphen"/>rallels to the lines RA and AB). And the right line IL subtendeth the right angle IPL, namely, it subtendeth the halfe sides of the cube which containe the right angle IPL, and likewise the right line IM subtendeth the angle IQM which is equall to the same an<g ref="char:EOLhyphen"/>gle IPL, and is contained vnder right lines equall to the right lines which containe the angle IPL. Wherefore the right line IM is equall to the right line IL (by the 4. of the first). And by the same reason may we proue, that euery one of the right lines MO, OL, KI, KL, KM, KO, NI, NL, NM, and NO, which subtend angles equall to the selfe same angle IPL, and are co<g ref="char:cmbAbbrStroke">̄</g>tained vnder sides equall to the sides which containe the angle IPL, are equall to the right line IL. Wherefore the triangles KLI, KLO, KMI, KMO, and NLI, NLO, NMI, NMO, are equilater and equall: and they containe the solide IKLONM. Wherefore IKLONM is an Octo<g ref="char:EOLhyphen"/>hedron, by the 23. definition of the eleuenth. And forasmuch as the angles thereof do alto<g ref="char:EOLhyphen"/>gether in the pointes I, K, L, O, N, M, touch the bases of the cube which containeth it, it followeth that the Octohedron is inscribed in the cube (by the first definition of this booke.) Wherefore in the cube geuen, is described an Octohedron: which was required to be done.</p>
                  <div type="subsection">
                     <pb n="433" facs="tcp:7062:462"/>
                     <head>¶A Co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ollary a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ded by <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>luss<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap>.</hi>
                     </head>
                     <p>
                        <q>Hereby it is manifest, that right lines ioyning together the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>entres of the opposite bases of the cube, do cut the one the other into two equall parts, and perpendicularly, in the centre of the cube, or in the centre of the Sphere which containeth the cube.</q>
                     </p>
                     <p>For forasm<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ch as the right lines LM and IO which kn<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> together the centres of the opposite bases of the cube, do also knit together the opposite angl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s of the Oct<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>hedron inscri<g ref="char:EOLhyphen"/>bed in the cube, it followeth (by the 3. Corollary of the 14. of the thirtenth) that those lines LM and IO, do cut the one the other into two equall partes in a point. But the diameters of the cube do also cut the one the other into two equall partes, by the 39. of the eleuenth. Wher<g ref="char:EOLhyphen"/>fore that point shall be the centre of the sphere which containeth the c<gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>. For making that point the centre, and the space some one of the semidiameters, describe a sphere, and it shall passe by the angles of the cube: and likewise making the same point the centre, and the space halfe of the line LM, describe a sphere, and it shall also passe by the angles of the Octohedron.</p>
                  </div>
               </div>
               <div type="section">
                  <head>¶ The 4. Proposition. The 4. Probleme. In an Octohedron geuen, to describe a Cube.</head>
                  <p>SVppose that the Octohedron geuen be ABGDEZ. And let the two pyramids thereof be ABGDE, and ZBGDE.<note place="margin">Construction.</note> And take the centres of the triangles of the pyramis ABGDE, that is, take the centres of the circles which containe those triangles: and let those centres be the point<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s T, I, K, L. And by these centres let there be drawen parall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>o the sides of the
<figure/>
square BGDE: which parallel <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ig<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> lin<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> let be MTN, NLX, XKO, &amp; OIM.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> And forasmuch as th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se parallel right lines do (by the 2. of the sixth) cut the equall right lines AB, AG, AD, and AE, proportionally, therfore they concurre in the pointes M, N, X, O. Wherefore the right lines MN, NX, XO, and OM, which subtend equall an<g ref="char:EOLhyphen"/>gles set at the point A, &amp; contained vnd<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>r <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>quall right lines, are equall (by the 4. of the first). And moreouer, seing that they are parallels vnto the lines BG, GD, DE, E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, which make a square, therefore MNXO is also a square, by the 10. of the eleuenth. Wherefore also, by the 15. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ame, the square MNXO is parallel to the squar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> BGDE. For all t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e right lines touch the one the other in the pointes of their sections. From the centres T, I, K, L, drawe these right lines TI, IK, KL, LT<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> And drawe the right line AIC. And forasmuch as I is the centre of the equilater triangle ABE, therefore the right line AI being extended, cutteth the right line BE into two equall partes (by the Corollary of the 12. of the thirtenth). And forasmuch as MO is a parallel to BE, therefore the triangle AIO is like to the whole triangle ACE (by the Corollary of the 2. of the sixth). And the right line MO is diuided into two equall partes in the point I (by the 4. of the sixth). And by the same reason may we proue, that the right lines MN, NX, XO, are diuided into two equall
<pb facs="tcp:7062:463"/>
partes in the pointes T, L, K. Wherefore also againe, the bases TI, IK, KL, LT, which sub<g ref="char:EOLhyphen"/>tend the angles set at the pointes M, O, X, N, which angles are right angles, and are contained vnder equall sides, those bases, I say, are equall. And forasmuch as TIM is an Isosceles tri<g ref="char:EOLhyphen"/>angle, therefore the angles set at the base, namely, the angles MTI and MIT, are equal (by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of the first). But the angle M is a right angle: wherefore eche of the angles MIT and MTI, is the halfe of a right angle. And by the same reason the angles OIK &amp; OKI, are equall. Wherefore the angle remayning, namely, TIK,
<figure/>
is a right angle (by the 13. of the first). For the right lines TI and IK are set vpon the line MO. And by the same rea<g ref="char:EOLhyphen"/>son may the rest of the angles, namely, IKL, KLT, LTI, be proued right angles, and they are in one and the self same plaine superficies, namely, MNXO (by the 7. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th). Wherefore the right lines which ioyne together the centres of the plaine superficiall triangles which make the solide angle A, do make the square ITKL. And by the same reason may be proued, that the plaine superficiall triangles of the rest of the fiue solide angles of the Octohedron set at the pointes B, G, Z, D, E, do in the centres of their bases receaue squares, So that there are in number sixe squares, for euery Octohe<g ref="char:EOLhyphen"/>dron hath sixe solide angles: and those squares are equall: for their sides do containe equall angles of inclinations con<g ref="char:EOLhyphen"/>tained vnder equall sides, namely, vnder those sides which are drawen from the centre to the side of the equall triangles (by the 2. Corollary of the 18. of the thirtenth). Wherefore ITKLRPVS is a cube (by the 21. definition of the ele<g ref="char:EOLhyphen"/>uenth) and hath his angles in the centres of the bases of the Octohedron, and therefore is inscribed in it (by the first de<g ref="char:EOLhyphen"/>finition of this booke). Wherefore in an Octohedron geuen is described a cube: which was required to be done.</p>
               </div>
               <div type="section">
                  <head>The 5. Proposition. The 5. Probleme. In an Icosahedron geuen, to describe a Dodecahedron.</head>
                  <p>TAke an Icosahedron, one of whose solide angles let be Z.<note place="margin">Construction.</note> Now forasmuch as (by those thinges which haue bene proued in the 16. of the thirtenth) the ba<g ref="char:EOLhyphen"/>ses of the triangles which contayne the angle of the Icosahedron doo make a pentagon inscribed in a circle, let that pentagon be ABGDE, which is made of the fiue bases of the triangles, whose playne superficiall angles re<g ref="char:EOLhyphen"/>mayning make the solide angle geuen, namely, Z. And take the centres of the circles which contayne the foresaid triangles, which centers let be the poyntes I, T, K, M, L: and draw these right lines IT, TK, KM, ML, LI.<note place="margin">Demonstra<g ref="char:EOLhyphen"/>tion.</note> Now then a perpendicular line drawne from the poynt Z to the playne superficies of the pentagon ABGDE, shall fall vpon the centre of the circle which contayneth the pentagon ABGDE (by those thinges which haue bene proued in the selfe same 16. of the thirtenth). Moreouer perpendicular lines drawne from the centre to the sides of the pentagon ABGDE shall in the poyntes C, N, O where they fall cut the right lines AB, BG, GD into two equall partes (by the 3. of the third). Draw these right lines CN and NO. And forasmuch as the angles CBN and NGO are equall, and are contained vn<g ref="char:EOLhyphen"/>der equall sides, therefore the base CN is equall to the base NO (by the 4. of the first). More<g ref="char:EOLhyphen"/>ouer
<pb n="434" facs="tcp:7062:463"/>
perpendicular lines dr<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>e from the poynt Z to the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>s<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of the pentagon ABGDE, shall likewise cutte the bases into two equall partes (by th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> of the third). For the perpendi<g ref="char:EOLhyphen"/>culars passe by the centre (by the corollary
<figure/>
of the 12. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th): Wherfore th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>se perpendicular lines shall fal vpo<g ref="char:cmbAbbrStroke">̄</g> the points C, N, O. And now forasmuch as the right lines ZI, IG are equall to the right lines ZT, TN, &amp; also to the right lines ZK, KO (by reason of the likenes of the equall tri<g ref="char:EOLhyphen"/>angles): therefore the line IT is a parallell to the line CN, and so also is the line TK to the line NO (by the 2. of the sixt). Wher<g ref="char:EOLhyphen"/>fore the angles ITK, and CNO are equal (by the 11. of the eleuenth). Agayne foras<g ref="char:EOLhyphen"/>much as the triangles CBN, and NGO are Isoscels triangles, therefore the angles BCN and BNC are equall (by the 5. of the first). And by the same reason the angles GNO, and GON are equall. And moreouer the angles BCN and BNC are equall to the angles GNO, and GON, for that the triangles CBN and NGO are equall and like. B<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> the three angles BNC, CNO, ONG, are equall to two right angles (by the 13. of the first): for that vpon the right line B<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> are set the right lines CN &amp; ON. And the three angles of the triangle CBN, namely, the angles BNC, BCN or GNO (for the angle G<g ref="char:EOLhyphen"/>NO is equall to the angle BCN as it hath bene proued) and NBC are also equall to two right angles (by the 32. of the first). Wherefore taking away the angles BNC &amp; GNO, the angle remayning, namely, CNO is equall to the angle remayning, namely, to CBN. Wher<g ref="char:EOLhyphen"/>fore also the angle ITK (which is proued to be equall to the angle CNO) is equall to the an<g ref="char:EOLhyphen"/>gle CBN. Wherefore ITK is the angle of a pentagon. And by the same reason may be pro<g ref="char:EOLhyphen"/>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ed that the rest of the angles, namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the angles TIL, ILM, LMK, MKT, are equall to the rest of the angles, namely to BAE, AED, EDG, DGB. Wherefore ITKML is an e<g ref="char:EOLhyphen"/>quilater and equiangle pentagon (by the 4. of the first) For the equall bases of the pentagon ITKML doo subtend equall angles set at the point Z, and comprehended vnder equall sides. Moreouer it is manifest that the pentagon I<hi>T</hi>KML is in one and the selfe same playne superficies. For foras<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>uch as the angles ONC and NCP are in one and the selfe s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>me playne superficies, namely in the superficies ABGDE: But vnto the same playne super<g ref="char:EOLhyphen"/>ficies the playne superficieces of the angles KTI and TIL are parallels (by the 15. of the ele<g ref="char:EOLhyphen"/>uenth). And the triangles KTI and TIL concurre: wherefore they are in one and the selfe same playne superficies (by the corollary of the 16. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th). And by the same reaso<g ref="char:cmbAbbrStroke">̄</g> so may we proue that the triangles ILM, LMK, MKT are in the selfe same playne super<g ref="char:EOLhyphen"/>ficies wherein are the triangles KTI and TIL. Wherefore the pentagon ITKML is in one and the selfe same playne superficies. Wherefore the solide angle of the Icosahedron, name<g ref="char:EOLunhyphen"/>ly the solide angle at the poynt Z subtendeth an equilater and equiangle pentagon plaine su<g ref="char:EOLhyphen"/>perficies, which pentagon hath his plaine superficiall angles in the centres of the triangles which make the solide angle Z. And in like sort may we proue that the other eleuen solide angles of the Icosahedron, eche of which eleuen solide angles are equall and like to the solide angle Z (by the 16. of the thirtenth) are subtended vnto pentagons equall, and like, and in like sort set to the pentagon ITKML. And forasmuch as in those pe<g ref="char:cmbAbbrStroke">̄</g>tagons the right lines, which ioyne together the centers of the bases, are common sides, it followeth that those 12. pentagons include a solide which solide is therefore a d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>dec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>hedron (by the 24. diffinition of the eleuenth): and is, by the first diffinition of this booke, described in the Icosahedron, fiue
<pb facs="tcp:7062:464"/>
sides wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>reof <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> set vpon the pentagon ABGDE. Wherefore in an Icosahedron geuen i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> inscribed a dodecahedron: which was required to be done.</p>
                  <div type="subsection">
                     <head>An annotation of <hi>Hypsi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>les.</hi>
                     </head>
                     <p>This is to be noted, that if a man should demaund <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> many sides an Icosahedron hath, we may thus answere: It is manifest that an Icosah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>n is contayned vnder 20. triangles, and that euery triangle consisteth of three right lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s. Now then multiply the 20. trian<g ref="char:EOLhyphen"/>gles into the sides of one of the triangles, and so shall there be produced 6<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>he halfe of which is 30. And so many sides hath an Icosahedron. And in like sort in a dodecahedron, foras<g ref="char:EOLhyphen"/>much as 12. pentagons make a dodecahedron, and euery pentagon contayneth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. right lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> multiply <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> into 12. and there shall be produced 60. the halfe of which is 30. And so many are the sides of a dodecahedron. And the reason why we take the halfe, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, for that euery side whe<g ref="char:EOLhyphen"/>ther it be of a triangle or of a pentagon, or of a square as in a cube, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s taken twise. And by the same reason may you finde out how many sides are in a cube, and in a pyramis, and in an octohedron.</p>
                     <p>But now agayne if ye will finde out the number of the angles of euery one of the solide figures, when ye haue done the same multiplication that ye did before, di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>id<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> the same sides, by the number of the plaine superficieces which comprehend one of the angles of the solides As for example, forasmuch as 5. triangles contayne the solide angle of an Icosahedron, diuide 60. by 5. and there will come forth 12. and so many solide angles hath an Icosahed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on. In a dodecahedron, forasmuch as three pentagons comprehend an angle, diuide 60. by 3. and there will come forth 20: and so many are the angles of a dodecahedron. And by the same reason may you finde out how many angles are in eche of the rest of the solide figures.</p>
                     <p>
                        <note place="margin">That which here followeth concerning the inclination of the plaines of the fiue solides, was before tought (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>hough not altogether after the same maner) out of Flussas in the latter <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>nde of the <hi>13</hi> booke.</note>If it be required to be knowne, how one of the plaines of any of the fiue solides being ge<g ref="char:EOLhyphen"/>uen, there may be found out the inclination of the sayd plaines the one to the other, which con<g ref="char:EOLunhyphen"/>tayne eche of the solides. This (as sayth <hi>Isidorus</hi> our greate master) is fo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>d out after this maner. It is manifest that in a cube, the plaines which contayne i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, do<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> 
                        <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> the one the other by a right angle. But in a Tetrahedron, one of the triangles being geuen, let the endes of one of the sides of the sayd triangle be the centers, and let the space be the perpendicular line drawne from the toppe of the triangle to the base, and describe circumfer<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>nces of a circle, which shall cutte the one the other: and from the intersection to the centers draw right lines, which shall containe the inclination of the plaines co<g ref="char:cmbAbbrStroke">̄</g>tayning the Tetrahedron. In an Octo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<g ref="char:EOLhyphen"/>dron, take one of the sides of the triangle ther<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>of, and vpon it describe a square, and draw the diagonall line, and making the centres, the endes of the diagonall line, and the space likewise the perpendicular line drawne from the toppe of the triangle to the base, describe circumfe<g ref="char:EOLhyphen"/>rences: and agayne from the common section to the centres draw right lines, and they shall contayne the inclination sought for. In an Icosahedron, vpon the side of one of the tri<g ref="char:EOLhyphen"/>angles thereof, describe a pentagon, and draw the line which subtendeth one of the angles of the sayd pentagon, and making the centres the endes of that line, and the space the perpendi<g ref="char:EOLhyphen"/>cular line of the triangle, describe circumferences: and draw from the common intersectio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the circumferences, vnto the centres right lines: and they shall contayne likewise the incli<g ref="char:EOLhyphen"/>nation of the plaines of the icosahedron. In a dodecahedron, take one of the pentagons, and draw likewise the line which subtendeth one of the angles of the pentagon, and making the centres the endes of that line, and the space, the perpendicular line drawne from the section into two equall partes of that line to the side of the pentagon, which is parallel vnto it, de<g ref="char:EOLhyphen"/>scribe circumferences: and from the point of the intersection of the circumferences draw vn<g ref="char:EOLhyphen"/>to the centres right lines: and they shall also containe the inclination of the plaines of the do<g ref="char:EOLhyphen"/>decahedron. Thus did this most singular learned man reason, thinking the de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>onstration in euery one of them to be plaine and cleare. But to make the demonstration of them mani<g ref="char:EOLhyphen"/>fest,
<pb n="435" facs="tcp:7062:464"/>
I think it good to declare and make open his wordes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> and first in a T<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>trahedron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </p>
                     <p>Suppose that there be a Pyramis or Te<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rahedro<g ref="char:cmbAbbrStroke">̄</g> ABCD
<figure/>
co<g ref="char:cmbAbbrStroke">̄</g>tained vnder 4. equilater triangles: &amp; let the toppe ther<g ref="char:EOLhyphen"/>of be the point D. And (by the <hi>10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi> of the first) diuide the side AD into two equall parts in the poi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp; draw the lines BE and EG. And forasmuch as ADB and ADC are e<g ref="char:EOLhyphen"/>quilater triangles, and the line AD is diuided into two e<g ref="char:EOLhyphen"/>quall partes, therefore the lines BE and EC fall perpendi<g ref="char:EOLhyphen"/>cularly vpon the line AD, by the 8. of the first. Now I say that the angle BEC is an acute angle. For forasmuch as the line AC is double to the line AE (for by construction the line AD, which is equal to the line AC, is diuided into two equall partes in the point E): therefore the square of the line AC is quadruple to the square of the line AE (by the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rollary of the 20. of the sixt). But the square of the line AC is equall to the squares of the lines AE and EC (by the 47. of the first): and the square of the line AC is to the square of the line CE (sesquitertia) as 4. to 3: (for the square of the line AC is proued quadruple to the square of the line AE): wherefore the square of the line BC (which is equall to the square of the line AC) is lesse then the squares of the two perpendicu<g ref="char:EOLhyphen"/>lars BE &amp; EC (for it is vnto them in subsesquialter proportio<g ref="char:cmbAbbrStroke">̄</g>, namely, as 4. to 6. or 2. to 3.) Wherefore (by the 13. of the second) the angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> EC is an acute angle. Now forasmuch as the line AD is the common intersection of the two plaines ABD, and ADC, and in either of those plaines to one point of the common section are drawne perpendicular lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>E and EC which containe an acute angle BEC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> therefore (by the 5. diffinition of the eleuenth) the an<g ref="char:EOLhyphen"/>gle BEC is the inclination of the plaines, and it is geuen. For the line BC, which is the side of the triangle, being geuen, and any one of the lines BE or EC, which is the perpendicular of the equilater triangle, being also ge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>: make the centres the poynte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> B and C, that is, the endes of one of the sides, and the space the perpendicular of the triangle, and describe circum<g ref="char:EOLhyphen"/>ferences, and they shall cutte the one the other in the poynt <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. And from the poynt B draw to the centres B and C right lines, and they shall containe the inclination of the plain<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>: and this is it which Isidorus before sayd. And now that making the centres the poynts B and C, and the space the perpendicular of the triangle, the circl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>s described shall cutte the one the other, it is manifest, for either of the lines BE and EC i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> greater then half<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the line BC.<note place="margin">The reason of this <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ou shall most plainely see in that which is ad<g ref="char:EOLhyphen"/>ded out of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>roclus a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ter the <hi>22.</hi> propo<g ref="char:EOLhyphen"/>sition of the first booke.</note> Now if the centers were the poynts B and C, and the space the halfe of the line BC, the circles described shall touch the one the other. But if the space <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e lesse then the halfe, they shal neither touch nor cut the one the other: but if it be greater, they shall vndoubtedly cut.</p>
                     <p>Againe suppose that vpon the square ABCD be set a pyramis, hauing his altitude the poynt E, and let the triangles which containe it, be
<figure/>
equilater: wherfore the pyramis ABCDE shalbe the halfe of the Octohedron (by the 2 corollary of the 14. of the thirtenth.) Deuide by the 10<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> of the first) one side of one of the triangles, namely, the line AE, into two equal partes in the poynt F: and draw the lines BF and DF: wherefore the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>es BF and DF are equal and fal perpendicularly vpon the line AE (by the 4. and 8. of the first) Then I say that the angle BFD, is an <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>tuse an<g ref="char:EOLhyphen"/>gle. For draw the line BD. And forasmuch as AC is a square, and the diameter is <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>D: therefo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uare of the line BD is double to the square of the line DA) by the 47. of the first.) But the square of the li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> DA is to the square of the line
<pb facs="tcp:7062:465"/>
DF, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. is to 3.) as was in the former proued.) Wherefore the square of the line DB is to the square of the line FD, as 8. is to 3. (namely, as 2. to 1. and 4. to 3. added together), but the line DF is equal to the line FB. Wherefore the square of the line DB, is greater then the squares of the lines DF and FB (for it is to them, as 8. is to 6.) Wherfore the angle BFD, is an obtuse angle (by the 12. of the second.) And forasmuche as the line AE is the common section of the two plaines ABE and ADE cutting the one the other, and in either of those plaines to a poynt in the common section are drawne perpendicular lines, BF and DF, con<g ref="char:EOLhyphen"/>taining an obtuse angle BFD: wherfore the angle BFD (contained of the right lines BF and DF) is the angle of the inclination of the plaines ABE and ADE. If therefore the angle BFD be geuen, the saide inclination also is g<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>uen. For forasmuch as the triangle of the Octohedron is geuen, and one of the sides of the Octohedron is the line AD, and vpon it is described the square AC, and BD the diameter of that square being geuen, and the lines BF and FD are the perpendiculars of that triangle: wherefore also the angle BFD is ge<g ref="char:EOLhyphen"/>uen. Now then if vpon the side of the triangle be described a square: as the square AC, and the diameter BD be drawne, if also making the centres the poyntes B and D, and the space, the said perpendicular of the triangle, we describe circles, they shall cut the one the other in the poynt F. And the right lines which are drawne from the centres to the poynt F shal con<g ref="char:EOLhyphen"/>taine that inclination, which is comprehended vnder the angle BFD, which is the angle of the inclination of those plaines. And it is manifest that either of the lines BF and FD, is greater then the halfe of the line. For for that, by the demonstration, it was proued that the square of the line BD is to the square of the line FD, as 8. is to 3: therfore the square of half the line BD, is to the square of the line FD, as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. is to thre (for the square of halfe the line BD is the fourth part of the square of the whole line BD, by the 4. of the second). Wherefore either of the lines BF and FD, is greater then the line BD: wherfore the circles which are described by those lines BF and FD, and hauing their centres the poynts B and D shall cut the one the other. And thus much touching the octohedron.</p>
                     <p>As touching the Icosahedron, suppose an equilater
<figure/>
pe<g ref="char:cmbAbbrStroke">̄</g>tagon ABCDE, &amp; vpon it let there be set a pyra<g ref="char:EOLhyphen"/>mis hauing his toppe the poynt F: and let the trian<g ref="char:EOLhyphen"/>gles which co<g ref="char:cmbAbbrStroke">̄</g>taine it, be equilater. Now the<g ref="char:cmbAbbrStroke">̄</g> the pyra<g ref="char:EOLhyphen"/>mis ABCDEF, shal be a part of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>. Let FC one side of one of the triangles be deuided into two equal partes in the poynt G. And draw the lines BG &amp; GD, which shal be equal &amp; fal perpe<g ref="char:cmbAbbrStroke">̄</g>dicular<g ref="char:EOLhyphen"/>ly vpon the line FC (as it is easie to se by the demo<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>stratio<g ref="char:cmbAbbrStroke">̄</g> of the former). The<g ref="char:cmbAbbrStroke">̄</g> I say that y<hi rend="sup">e</hi> angle BGD is an obtuse angle: which thing is manifest. For drawing the line BD, it shall subtend the obtuse an<g ref="char:EOLhyphen"/>gle BCD of the pentagon (which is obtuse, by that which was demonstrated in the ende of the first co<g ref="char:EOLhyphen"/>rollary of the 18. of the 13. booke:) But the angle BGD is greater then the angle BCD, for the lines BG and GD, are lesse then the lines BC and CD: wherefore likewise as in the for<g ref="char:EOLhyphen"/>mer the angle BGD is the inclination of the triangles BFC, and CFD. Wherfore the an<g ref="char:EOLhyphen"/>gle BGD being geuen, the inclination also of the plaines of the Icosahedron shall be geuen. For if vpon the side of the triangle of the Icosahedron be described an equilater pentagon, and then be drawne the line which subtendeth two sides of the pentagon, as in this figure the line BD, if also the perpendiculars BG and GD of the triangles be drawne, the angle BGD shalbe geuen. For if ye make the centres the endes of the line which subtendeth two sides of the pentagon, as the poynts B and D, and the space the perpendicular of the triangle, and so describe circles, they shall cut the one the other in the poynt G, and from the poynt of the in<g ref="char:EOLhyphen"/>ters<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ction
<pb n="436" facs="tcp:7062:465"/>
G, dra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e vnto the centres B and D right <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> they shal containe the angle of the inclination BGD. And it is manifest, by the description of the figure, that either of the lines BG and GD is greater then <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> line BD. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ch <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> may also thus be proued. Sup<g ref="char:EOLhyphen"/>pose an equilater HKL, and vpon K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (one of the sides <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap>) describe an equilater pen<g ref="char:EOLhyphen"/>tagon KMNXL, and draw the line ML. And diuide (<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>y the 10. of the first) the side KL into two equal parts in the poynt O<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> &amp; draw the line H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, which shall be the perpendicular of the triangle HKL (by the 8. of the first.) Then I say that the line HO is greater then half of the line ML, which subtendeth the inclination of the plaines. For from the poynt K draw (by the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. of the first) vnto the line ML a perpendicular line KP: and forasmuche as the angle KLP is greater then the third part of a right angle, that is, then the angle KHO<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (For the angle KLM is two fi<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>th partes of a righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> angle,
<figure/>
by the 4. of the first, and by the assumpt put after the first co<g ref="char:EOLhyphen"/>rollary of the 18. of the thirtenth booke, and the angle K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>O is one third part of one right angle; for the whole angle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>HL, wherof the angle KHO is the half, by the 4. of the first, is one third part of two right angles, by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>. of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>) vpon the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>L, and at the poynt L put vnto the angle KHO an equal angle PL<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> (by the 23. of the first.) Wher<g ref="char:EOLhyphen"/>fore the triangles PLR &amp; OHK, shalbe equiangle, by the corollary of the 32. of the first. Wherefore also the line PL shalbe the perpendicular of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> triangle described vpon the line RL. Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> added by <hi>Flussas</hi> after the 12. proposition of the thirtenth booke) the line RL is in power sesquitercia, that is, as 4. is to 3. to the perpendicular LP. But the <gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap> LR (by the 19. of the first. Wherfore y<hi rend="sup">e</hi> square of the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e K<g ref="char:EOLhyphen"/>L hath to the square of the line LP a greater proportion the<g ref="char:cmbAbbrStroke">̄</g> hath 4. to 3: but it hath to the square of the line HO that proportion that 4. hath to 3. Wher<g ref="char:EOLhyphen"/>fore the line KL hath to the line LP a greater proportion then it hath to the line HO. Wherfore the line HO is greater then the line LP, by the 10. of the fifth.</p>
                     <p>As concerning a Dodecahedron. Take one of the squares of the cube wheron the Dode<g ref="char:EOLhyphen"/>cahedron is described (by the 17. of the thirtenth): and let the same be ABCD: and let the two plaines of the Dodecahedron set vpon it be AEBFG, and GFDHC. Then I say that here also is geuen the inclination of the two
<figure/>
pentagons. Diuide (by the 10. of the first) the side FG into two equal partes in the poynt K. And from the poynt K draw vnto the line FG in either of the plaines AEB<g ref="char:EOLhyphen"/>FG and GFDHC perpendicular lines KL and KM (by the 11. of the first.) And draw the line ML. First I say that the an<g ref="char:EOLhyphen"/>gle MKL is an obtuse angle. For, by the discourse of the demonstration of the 17. proposition of the 13. boke, where is taught the description of the Dodecahedron, it is mani<g ref="char:EOLhyphen"/>fest, that the line drawne perpendicularly from the poynt K to the square ABCD, is equal to halfe the side of the pentagon. Wherefore it is lesse then halfe of the line ML. Wherefore the angle MKL is an obtuse angle. Moreouer by the former discourse of the 17. proposition of the 13. booke, it was manifest that the square of the line KL is equal to the square of half the side of the cube, and to the square of halfe the side of the pentagon. And forasmuche as
<pb facs="tcp:7062:466"/>
the lines KL and KM are equal, and a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>e <gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>he greater then halfe of the line ML<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> wherfore the angle MKL being geuen, there shall also be <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> the inclination of the two plaines of the Dodecahedron. For forasmuch as the side of the square ABCD subtendeth two sides of the pentagon geuen, the pentagon also is geuen, and therefore also is geuen the line ML. But there is also geuen either of the lines MK and KL: for they are drawne perpendicularly fro<g ref="char:cmbAbbrStroke">̄</g> the section into two equal partes of the line AB, which subtendeth two sides of the pentagon vnto the side of the pentagon, which is a parallel vnto the line AB: namely, to the side FG: Wherfore there is geuen the angle LKM, which is the angle of the inclination sought for. And now touching <hi>Isidorus</hi> wordes, he sayeth, that the pentagon being geuen, we must draw the line which subtendeth two sides of the pentagon, which line 15 equal to the side of the cube: and making the centres the endes of that line, and the space the perpendicular line, which is drawne from the section of the same line into two equal parts to the side of the pen<g ref="char:EOLhyphen"/>tagon which is parellel to the said line, as in the former description the line KL, or the line KM, describe circumferences, and from the poynt of the intersection of the circumferences draw vnto the centres right lines which shall containe the angle of the inclination. For by that which was sayd before, namely; touching the Icosahedron, it is manifest that the perpendicular KL, is greater then halfe of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                           <desc>〈◊〉</desc>
                        </gap> ML or CD, which is equal vnto it. And therefore the cir<g ref="char:EOLhyphen"/>cles described by those perpendiculars and hauing to their centres the end of the line CD, shall cut the one the other, as was be<g ref="char:EOLhyphen"/>fore proued.</p>
                  </div>
               </div>
               <trailer>The ende of the fiuetenth Booke of Euclides Elementes after <gap reason="foreign">
                     <desc>〈 in non-Latin alphabet 〉</desc>
                  </gap>
               </trailer>
            </div>
            <div type="part">
               <pb n="437" facs="tcp:7062:466"/>
               <head>¶ The 6. Proposition. The 6. Probleme. In an Octohedron geuen, to inscribe a trilater equilater Pyramis.</head>
               <p>SVppose tha<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the Octohedron where<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the Tetrahedron is required to be ins<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ri<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>, be ABGDEI. Take <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> bases of the Octo<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>dron, that is,<note place="margin">
                     <hi>Construction.</hi>
                  </note> 
                  <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> close in the lowe<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> triangle BGD, namely, AE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, HED, IGD: and let the fourth be AIB, which is opposite to the lowest trian<g ref="char:EOLhyphen"/>gle before put, namely, to EGD. And take the centres of those fower bases, which let be the pointes H, C, N, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And vpon the triangle HCN erecte a pyramis HCNL. Now <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>orasmuch as these two bases of the Octohedron, namely, AGE and ABI are set vpon the right lines EG and BI which are opposite the one to the other<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the square GEBI of the Octohedron, from the poin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> A dra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e by the centres of the bases, namely, by the centres H, L, perpendicular lines AHF, ALK, cutting the lines EG and BI <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> two equall partes in the pointes F, K (by the Corollary of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the thirtenth). Wherfore
<figure/>
a right line drawen fro<g ref="char:cmbAbbrStroke">̄</g> the point F to the point K,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> shall be a parallel and equall to the sides of the Octohedron, namely, to <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> and GI (by the 33. of the first). And the right line HL which cutteth the <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> AF, AK, proportionally (for AH and AL are drawen from the centres of equall circles to the circumferences) is a pa<g ref="char:EOLhyphen"/>rallel to the right line FK (by the 2. of the sixth) and also to the sides of the Octohedron, name<g ref="char:EOLhyphen"/>ly, to E<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and IG (by the 9. of the eleuenth). Wherefore as the line AF is to the line AH, so is the line FK to the line HL (by the 4. of the sixth): For the triangles AFK and AHL are like (by th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Corollary of the 2. of the sixth). But the line AF is in sesquialter proportion to the line AH: (for the side EG maketh HF the halfe of the right line AH, by the Corollary of the 12. of the thirtenth). Wherfore FK or GI the side of the Octohedron, is sesquialter to the righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>line HL. And by the same reason may we proue that the sides of the Octohedron are sesquialter to the rest of the right lines which make the pyramis HNCI, namely, to the right line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, N, NC, CI, LN, and CH: wherefore those right lines are equall, and therefore the triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which are described of them, namely, the triangles HCN, HNL, NCL, and CHL. which make the pyramis HNCL, are equall and equilater. And forasmuch as the angles of the same pyramis, namely, the angles H, C, N, L, do end in the centres of the bases of the Octohedron, therefore it is inscribed <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the same Octohedron, by the first definition of this booke. Wherefore in an Octohedron <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>euen, is inscribed a tril<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter equila<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>amis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which was required to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e don<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>The bases of a Pyramis inscribed in an Octohedron, are parallels to the bases of the Octohedron.</hi> For forasmuch as the sides of the bases of the Pyramis touching the one the other, are parallels to the sides of the Octohedron which also touch the one the other, as for example, HL was proued to be a parallel to GI, and LC to DI, therefore, by the 15. of the eleuenth, the plaine superficies which is drawen by the lines HL and LC, is a parallel to the plaine superficies drawen by the lines GI and DI. And so likewise of the rest.</p>
               </div>
               <div type="section">
                  <head>Second Corollary.</head>
                  <p>
                     <hi>A right line ioyning together the centres of the opposite bases of the Octo<g ref="char:EOLhyphen"/>hedron, is sesquialter to the perpendicular line drawen from the angle of the in<g ref="char:EOLhyphen"/>scribed
<pb facs="tcp:7062:467"/>
pyramis to the base thereof.</hi> For forasmuch as the pyramis and the cube which containeth it do in the selfe same pointes end their angles (by the 1. of this booke): therefore they shall both be inclosed in one and the selfe same Octohedron (by the 4. of this booke). But the diame<g ref="char:EOLhyphen"/>ter of the cube ioyneth together the centres of the opposite bases of the Octohedron, and therefore is the diameter of the Sphere which containeth the cube and the pyramis inscribed in the cube (by the 13. and 14. of the thirtenth): which diameter is sesquialter to the perpendicular which is drawen from the angle of the pyramis to the base thereof: for the line which is drawen from the centre of the sphere to the base of the pyramis, is the sixth part of the diameter (by the 3. Corollary of the 13. of the thir<g ref="char:EOLhyphen"/>tenth). Wherefore of what partes the di<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>meter containeth sixe, of the same partes the perpendicular containeth fower.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 7. Proposition. The 7. Probleme. In a dodecahedron geuen, to inscribe an Icosahedron.</head>
               <p>SVppose that the dodecahedron geuen, be ABCDE. And let the centres of the circles which co<g ref="char:cmbAbbrStroke">̄</g>tayne sixe bases of the same dodecahedron be the polnes L, M, N, P, Q, O.<note place="margin">
                     <hi>Construction.</hi>
                  </note> And draw these right lines OL, OM, ON, OP, OQ, and moreouer these right lines LM, MN, NP, PQ, QL. And now forasmuch as equall and equilater pentagons are contay<g ref="char:EOLhyphen"/>ned in equall circles, therefore perpendicular lines drawne from their centres to the sides shall be equall (by the 14. of the third), and shall diuide the sides of the dodecahedron into two e<g ref="char:EOLhyphen"/>quall partes (by the 3. of the same). Wherefore the foresayde perpendicular lines shall co<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>outre in the point of the section,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> wherein the sides are diuided
<figure/>
into two equall partes, as LF and MF doo. And they also containe equall angles, namely, the in<g ref="char:EOLhyphen"/>clination of the bases of the dodecahedron, (by the 2. corollary of the 18. of the thirtenth). Wher<g ref="char:EOLhyphen"/>fore the right lines LM, MN, NP, PQ, QL, and the rest of the right lines which ioyne together two centres of the bases, and which subtende the equall angles <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ontayned vnder the sayd equall perpendicular lines, are equall the one to the o<g ref="char:EOLhyphen"/>ther (by the 4. of the first). Wherefore the trian<g ref="char:EOLhyphen"/>gles OLM, OMN, ONP, OPQ, OQL, and the rest of the triangles which are set at the cen<g ref="char:EOLhyphen"/>tres of the pentagons, are equilater and equall. Now forasmuch as the 12. pentagons of a dodeca<g ref="char:EOLhyphen"/>hedron containe 60. plaine superficiall angles, of which 60. eue<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>y <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hre make one solide angle of the dodecahedron, it followeth that a dodecahedron hath 20. solide angles: but eche of those solide an<g ref="char:EOLhyphen"/>gles is subte<g ref="char:cmbAbbrStroke">̄</g>ded of ech of the triangles of the Ico<g ref="char:EOLhyphen"/>sahedron, namely, of ech of those triangles which ioyne together the centres of the pentago<g ref="char:cmbAbbrStroke">̄</g>s which make the solide angle, as we haue before proued. Wherefore the 20, equall and equilater triangles which subtende the 20. solide angles of the dodecahedron, and haue their sides which are drawne from the centres of the pentagons common, doo make an Icosahedron (by the 25. diffinition of the e<g ref="char:EOLhyphen"/>leuenth): and it is inscribed in the dodecahedron geuen (by the first diffinition of this booke) for that the angles thereof doo all at one time touch the bases of the dodecahedron. Wherefore in a dodecahe<g ref="char:EOLhyphen"/>dron geuen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> inscribed an Icosahedron: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 8. Proposition. The 8. Probleme. In a dodecahedron geuen, to include a cube.</head>
               <p>DEscribe (by the 17. of the thirtenth) a dodecahedron.<note place="margin">
                     <hi>Construction.</hi>
                  </note> And by the same, take the 12. sides of the cube, eche of which subtend one angle of eche of the 12. bases of the dodecahe<g ref="char:EOLhyphen"/>dron: for the side of the cube subtendeth the angle of the pentagon of the dodecahe<g ref="char:EOLhyphen"/>dron, by the 2. corollary of the 17. of the thirtenth. If therefore in the dodecahedron described (by the selfe same 17. proposition) we draw the 12. right lines sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ended vn<g ref="char:EOLhyphen"/>der the foresayd 12. angles, and ending in 8. angles of the dodecahedron, and concur<g ref="char:EOLhyphen"/>ring
<pb n="438" facs="tcp:7062:467"/>
together in such sort that they be in like sort situate,
<figure/>
as it was plainely proued in that proposition, then shall it be manifest, that the right lines drawne in this dode<g ref="char:EOLhyphen"/>cahedron from the foresayd 8. angles thereof doo make the foresayd cube, which therefore is included in the do<g ref="char:EOLhyphen"/>decahedron, for that the sides of the cube are drawne in the sides of the dodecahedron, and the angles of the same cube are set in the angles of the said dodecahedron. As for example take 4. pentagons of a dodecahedron,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> namely AGIBO, BHCNO, CKEDN and DFAO<g ref="char:EOLhyphen"/>N. And draw these right lines AB, BC, CD, DA. Which fower right lines make a square: for that eche of those right lines doo subtend equall angles of equall penta<g ref="char:EOLhyphen"/>gons, &amp; the angles which those 4. right lines co<g ref="char:cmbAbbrStroke">̄</g>taine are right angles, as we proued in the construction of the do<g ref="char:EOLhyphen"/>decahedron, in the 17. propositio<g ref="char:cmbAbbrStroke">̄</g> before alledged. Wher<g ref="char:EOLunhyphen"/>fore the sixe bases being squares, do make a cube (by the 21. diffinition of the eleuenth) and for that the 8. angles of the sayd cube are set in 8. angles of the dodecaheeron, therefore is the sayd cube inscribed in the dodecahedron (by the first diffinition of this booke). Wherefore in a dodecahedron is inscribed a cube: which was required to be doone.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Proposition. The 9. Probleme. In a Dodecahedron geuen to include an Octohedron.</head>
               <p>SVppose that the dodecahedron ge<g ref="char:EOLunhyphen"/>uen
<figure/>
be ABGD.<note place="margin">
                     <hi>Construction.</hi>
                  </note> Now (by the 3. correllary of the 17. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th take the 6. sides which are opposite the one to the other, those 6. sides, I saye whose sections wherin they are deuided in<g ref="char:EOLhyphen"/>to two equal partes, are coupled by three right lines which in the centre of the sphere, wherin the Dodecahedron is contained, doe cut the one the other perpendicularly. And let the poyntes wherin the forsayde sides are cut into two equal partes be A, B, G, D, C, I. And let the foresaid thre right lines ioyning together the saide sections be AB, GD and CI. And let the centre of the sphere be E.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Now forasmuch as (by the foresaid corrella<g ref="char:EOLhyphen"/>ry) those thre right lines are equal, it folow<g ref="char:EOLhyphen"/>eth (by the 4. of the first) that the right lines subte<g ref="char:cmbAbbrStroke">̄</g>ding the right angles which they make at the centre of the sphere, whiche right an<g ref="char:EOLhyphen"/>gles are contained vnder the halues of the said three right lines, are equal the one to the other: that is, the right lines AG, GB, BD, DA, CA, CG, CB, CD, and IA, IG, IB, ID are equal the one to the o<g ref="char:EOLhyphen"/>ther. Wherfore also the 8. triangles CAG, CGB, CBD, CDA, IAG, IGB, IBD &amp; IDA are equal and equilater. And therefore AGBDCI is an Octohedron by the 23. definition of the eleue<g ref="char:cmbAbbrStroke">̄</g>th.) And the sayd Octahedron is included in the dodecahedron (by the first definition of this booke:) for that all the angles thereof doe at one time touch the sides of the dodecahedron. Wherefore in the dodeca<g ref="char:EOLhyphen"/>hedron geuen, is included an Octohedron: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 10. Proposition. The 10. Probleme: In a Dodecahedron geuen, to inscribe an equilater trilater Pyramis.</head>
               <p>SVppose that the Dodecahedron geuen, be ABCD, of which Dodecahedron take thre bases meting at the poynt S, namely these thre bases ALSIK, DNSLE and SIBRN:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and of those thre bases take the three angles at the poynts A, B, D: and draw these right lines AB, BD and DA: and let the diameter of the sphere containing the dodecahedron, be SO, and
<pb facs="tcp:7062:468"/>
then draw thse right lines AO, BO and DO: Now forasmuch as (by the 17. of the thirtenth) the an<g ref="char:EOLhyphen"/>gles of the dodecahedron are set in the superficies of the sphere described about the Dodecahedron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> therefore if vpon the diameter SO, and by the poynt A, be described a semicircle, it shall make the an<g ref="char:EOLhyphen"/>gle SAO a right angle (by the 31. of the third.) And likewise if the same semicircle be drawne by the poyntes D and B, it shall also make the angles SBO, and SDO right angles. Wherefore the diameter SO containeth in power bothe the lines SA, AO, or the lines SB, BO, or els SD, DO, but the lines SA, SD, SB are equal the one to the other, for they eche subtend one of the angles of equal pentago<g ref="char:cmbAbbrStroke">̄</g>s. Wherfore the other lines remaining, namely, AO, BO, DO are equal the one to the other. And by the same reason may be proued that the diameter HD which subtendeth the two right lines HA, AD, containeth in power both the said two right lines, and also containeth in power bothe the right lines HB and BD, which two right lines it also suhtendeth. And moreouer by the same reason the diameter AC, which subtendeth the right lines CB and BA, containeth in power both the said right lines C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and BA. But the right lines HA, HB and CB
<figure/>
are equal the one to the other, for that eche of them also subtendeth one of the angles of equal pentagons<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the right lines re<g ref="char:EOLhyphen"/>maining, namely, AD, BD, and BA are equal the one to the other. And by the same reason may be proued that eche of those right lines AD, BD and BA is equal to eche of the right lines AO, BO and DO. Wherefore the sixe right lines AB, BD, DA, AO, BO, &amp; DO are equal the one to the other. And therefore the triangles which are made of the<g ref="char:cmbAbbrStroke">̄</g>, namely, the triangles ABD, AOB, AOD and BOD are equal and equilater: which triangles therfore do make a pyramis ABDO, whose base is A<g ref="char:EOLhyphen"/>BD and toppe the poynt O. Eche of the an<g ref="char:EOLhyphen"/>gles of which pyramis, namely, the angles at the pointes A, B, D, O, doe in the selfe same pointes touche the angles of the Dodecahe<g ref="char:EOLhyphen"/>dron. Wherfore the said pyramis is inscribed in the Dodecahedron, (by the first diffinition of this boke.) Wherefore in a Dodecahedron geuen, is inscribed a trilater equilater pyra<g ref="char:EOLhyphen"/>mis: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶ The 11. Proposition. The 11. Probleme. In an Icosahedron geuen, to inscribe a cube.</head>
               <p>IT was manifest by the 7. of this booke, that the angles of a Dodecahedron are set in the cen<g ref="char:EOLhyphen"/>tres of the bases of the Icosahedron. And by the 8. of this boke, it was proued, that the angles of a cube are set in the angles of a Dodecahedron. Wherefore the selfe same angles of the cube, shall of necessitie be set in the centres of the bases of Icosahedron. Wherfore the cube shalbe inscribed in the Icosahedron (by the first diffinition of this boke). Wherfore in an Icosahedron geuen, is included a cube: which was required to be done.</p>
            </div>
            <div type="part">
               <head>¶The 12. Proposition. The 12. Probleme. In an Icosahedron geuen, to inscribe a trilater equilater pyramis.</head>
               <p>BY the former proposition it was manifest, that the angles of a cube are set in the centres of the bases of the Icos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hedron. And (by the first of this booke) it was playne that the foure an<g ref="char:EOLhyphen"/>gles of a pyramis are set in foure angles of a cube. Wherefore it is euident, by the first diffini<g ref="char:EOLhyphen"/>tion of this booke, that a pyramis described of right lines ioyning together these foure cen<g ref="char:EOLhyphen"/>tres of the bases of the Icosahedron, shalbe inscribed in the same Icosahedron. Wherefore in an Icosa<g ref="char:EOLhyphen"/>dron geuen, is inscribed an equilater trilater pyramis: which was required to be done.</p>
            </div>
            <div type="part">
               <pb n="439" facs="tcp:7062:468"/>
               <head>¶The 13. Probleme The 13. Proposition. In a Cube geuen, to inscribe a Dodecahedron.</head>
               <p>TAke a Cube ADFL.<note place="margin">
                     <hi>First part of the construc<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> And diuide euery one of the sides therof into two equall partes in the pointes T, H, K, P: G, L, M, F: and pkQs. And drawe these right lines TK, GF, pQ, Hk, Ps, and LM: which lines againe diuide into two equall partes in the pointes N, V, Y, I, Z, X. And draw these right lines NY, VX, and IZ: Now the three lines NY, VX, and IZ, together with the diameter of the cube, shall cut the one the other into two equall partes in the centre of the cube, by the 3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the eleuenth: let that centre be the point O. And not to
<figure/>
stand long about the demonstration, vn<g ref="char:EOLhyphen"/>derstand all these right lines to be equall and parallels to the sides of the cube and to cut the one the other right angled wise, by the 29. of the first. Let their halfes, namely, FV, GV, HI, and kI, and the rest such like, be deuided by an ex<g ref="char:EOLhyphen"/>treme and meane proportion, by the 30. of the sixth: whose greater segme<g ref="char:cmbAbbrStroke">̄</g>ts let be the lines FS, GB, HC, and kE, &amp;c. and drawe these right lines GI, GE, BC, and BE. Now forasmuch as the line GI is equall to the whole line GV, which is the halfe of the side of the cube:<note place="margin">
                     <hi>First part of the demon<g ref="char:EOLhyphen"/>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ration.</hi>
                  </note> and the line IE is equall to the line BV, that is, to the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t: therfore, the squares of the lines GI and IE, are triple to the square of the line GB, by the 4. of the thirtenth: But vnto the squares of the lines GI and IE, the square of the line GE is equall, by the 47. of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> for the angle GIE is a right angle: Where<g ref="char:EOLhyphen"/>fore the square of the line GE is triple to the square of the line GB. And foras<g ref="char:EOLhyphen"/>much as the line FG is erected perpen<g ref="char:EOLhyphen"/>dicularly to the plaine AGkL, by the 4. of the eleuenth: for it is erected per<g ref="char:EOLhyphen"/>pendicularly to the two lines AG and GI: therefore the angle BGE is a right angle: for the line GE is drawen in the plaine AGkL. Wherefore the line BE, containing in power the two lines BG and GE, by the 47. of the first, is in power quadruple to the line GB (for the line GE was proued to be in power triple to the same line GB): Wherefore the line BE is in length double to the line BG, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. of the sixth. But (by construction) the line CE is double the line IE: Wherefore the halfes GB and IE, are in propor<g ref="char:EOLhyphen"/>tion the one to the other, as their doubles BE and CE: by the 15. of the fifth. Wherefore the line CE is the greater segment of the line BE diuided by an extreme and meane proportion. And forasmuch as the selfe same thing may be proued touching the line BC: therefore the lines BE and BC, are equall, making an Isosceles triangle. Now let vs proue that three angles of the Pentagon of the Dodecahedron are set at the pointes B, C, E: and the other two angles are set betwene the lines BC and BE.</p>
               <p>Forasmuch as the circle which containeth the triangle BCE circumscribeth the Pentagon whose side is the line CE, by the 11. of the fourth:<note place="margin">
                     <hi>Second part of the con<g ref="char:EOLhyphen"/>struction.</hi>
                  </note> Extend the plaine of the triangle BCE, by the parallel lines dB and HE, cutting the line AD, namely, the diameter of AD the base of the cube in the point I: and let it cut the line Ah the diameter of the cube in the point m. And by the point I drawe in the base AD, a parallel line vnto the line Ad: which let be Il. And forasmuch as from the triangle AHN is, by the parallel line lI, taken away the triangle AlI, like vnto the whole triangle AHN,<note place="margin">
                     <hi>Second part of the De<g ref="char:EOLhyphen"/>monstration.</hi>
                  </note> by the Co<g ref="char:EOLhyphen"/>rollary of the 2. of the sixth: the lines Al, and lI, shall be equall. But as the line HA is to the line Ad, so (by the 2. of the sixth) is the line Hl to the line lI, or to the line lA, which is equall to the line lI. And the greater segment of the line HA. (which is halfe the side of the cube) is, as before hath bene proued, the line Ad, that is, the line GB, which is equall to the line Ad (by the 33. of the first).
<pb facs="tcp:7062:469"/>
Wherefore the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> segment of the line Hl is the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>A. And as the whole line Hl is to the grea<g ref="char:EOLhyphen"/>ter segment, so shall the same greater segment Hl be to the lesse segment lA, by the 5. of the thir<g ref="char:EOLhyphen"/>tenth. Wherefore the line HA is di<g ref="char:EOLhyphen"/>uided
<figure/>
by an extreme and meane pro<g ref="char:EOLhyphen"/>portion in the point l. But in the tri<g ref="char:EOLhyphen"/>angle AHN, the line NA, which is drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre of the base AD, is in the point I cut like vnto the line All, by thu parallel line lI (by the same second of the sixth): for the lines HN and Il, are parallels, by construction. Wherefore the line NA is in the point I diuided by an extreme and meane proportion by the superficies dBEH. And forasmuch as the line YON which coupleth the centres of the opposite bases, is a parallel to the line HE: A plaine superficies extended by the line YON, parallel wise to the plaine dB<g ref="char:EOLhyphen"/>EH: the two plaines shall cut the lines AO and AN (the semidiameter of the cube, and the semidiameter of the base AD) into the selfe same proportions in the pointes m and I, by the 17. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th. But the line AN is in the point I diuided by an extreme &amp; meane pro<g ref="char:EOLhyphen"/>portion: Wherefore the semidiameter of the cube is in the point m diuided by an extreme and meane proportion by the plaine of the triangle BCE. And forasmuch as the rest of the triangles de<g ref="char:EOLhyphen"/>scribed in the cube after the like maner, may by the same reasons be proued to be in a plaine which cutteth the semidiameter of the cube by an extreme and means proportion: it is manifest that three plaines of the Dodecahedron shall vnder euery angle of the cube concurre in one &amp; the self same point o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the semidiameter being cut by an extreme and meane proportion. Now resteth to proue that the right lines which couple that point of the semidiameter with the angles of the triangle BEC, are e<g ref="char:EOLhyphen"/>quall: whereby may be proued that the Pentagons are equilater, and equiangle.</p>
               <p>Take the two bases of the cube.
<figure/>
Whereon are set the triangle BCE,<note place="margin">
                     <hi>Third part of the construc<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> namely, the bases AF and Ak, take also the same diameter of the cube that was before, namely, Ah: and let the side set at the poynt n, of the secti<g ref="char:EOLhyphen"/>on of the diameter by an extreame &amp; meane proportion, be the line Cn or Bn: and let the centre of the cube be as before the point O. And extend the line Cn to the line Bd, and let it con<g ref="char:EOLhyphen"/>curre with it in the point a. And foras<g ref="char:EOLhyphen"/>much as the plaine which passeth by the line HCE and the centre O (cut<g ref="char:EOLhyphen"/>ting the cube into two equall partes) is parallel to AF the base of the cube by construction: imagine that by the poynt n, be extended a playne super<g ref="char:EOLhyphen"/>ficies parallel to the former parallel playnes,<note place="margin">
                     <hi>Third part of the demon<g ref="char:EOLhyphen"/>stration.</hi>
                  </note> which shall cutte the semidi<g ref="char:EOLhyphen"/>ameter OA &amp; the line Ca, proporti<g ref="char:EOLhyphen"/>onally in the point n, by the 17. of the eleuenth: For those lines doo touch the extreame parallel plaines exten<g ref="char:EOLhyphen"/>ded by the lines HE and EO, and by the lines Ad and dB. But it is proued that the line OA is diuided by an ex<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>eame and meane proportion in the
<pb n="460" facs="tcp:7062:469"/>
poynt n: wherefore the line Ca, is also <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> by an extr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame and meane proportion in the poynt n. Agayne forasmuch as BCE is an Isoscels triangle, and it is proued that the line BI cutteth the base CE into two equall partes in the poynt I, the angles BIC and BIE<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shall be right angles. Imagine by the line BI and the centre O a plaine to passe (cutting the cube into two equall partes) parallel to the base AD. And vnto those plaines let there be imagined an other parallel plaine passing by the poynt n: which let be ne: which shall cutte the semidiameter AO and the halfe side of the cube, namely, the line lH, like, in the pointes n and c by the 17. of the eleuenth. Wherefore the line IH is in the poynt e diuided by an extreame <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> meane proportio<g ref="char:cmbAbbrStroke">̄</g>. Wherfore the line He is equall to the line CI or IE: namely, ech are lesse segme<g ref="char:cmbAbbrStroke">̄</g>ts. And forasmuch as the line Ie is to the line IC (which is equall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ine EH) as the whole is to the greater segment, take away from the whole line Ie th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> greater segm<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> IC: there shall remayne the lesse segment Ce by the 5. of the thirtenth: Wherefore the line I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is diui<g ref="char:EOLhyphen"/>ded by an extreame &amp; meane proportion in the point C. Againe vnto the same playnes imagine an o<g ref="char:EOLhyphen"/>ther playne to passe by the point a, parallel wise, and let the same be ag. Now then (by the same 17. of the eleuenth) the lines Ca and Cg are in like sort cut in the pointes n and e. But the line Ca was in the point n cutte by an extreame and meane proportion, wherefore the line Cg shall be cutte in the poynt e, by an extreame &amp; meane proportion. But the line IC is to the line Ce, as the greater segment is to the lesse: wherfore the line Ce, is to the line eg, as the greater segment to the lesse: and therefore their proportion is as the whole line IC is to the greater segment Ce, and as the greater segment Ce is to the lesse segment eg: wherefore the whole line Ceg which, maketh the greater segment and the lesse, is equall to the whole line IC or IE. And forasmuch as two parallel plaine superficieces (namely, that which is extended by IOB and that which is extended by the line ag) are cutte by the playne of the triangle BCE, which passeth by the lines ag and IB, their common sections ag and IB shall be parallels (by the 16. of the eleuenth). But the angle BIE or BIC is a right angle, wherefore the angle agC is also a right angle (by the 29. of the first) and those right angles are contayned vnder equall sides, namely, the line gC is equall to the line CI, and the line ag to the line BI, by the 33. of the first: wher<g ref="char:EOLhyphen"/>fore the bases Ca and CB are equall, by the 4. of the first. But of the line CB the line CE was proued to be the greater segment: wherefore the same line CE is also the greater segment of the line Ca: but cn was also the greater segment of the same line Ca. Wherefore vnto the line CE, the line cn which is the side of the dodecahedron, and is set at the diameter, is equall. And by the same reason the rest of the sides, which are set at the diameter may be proued e<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>uall to lines equall to the line CE. Wherfore the pentagon inscribed in the circle where in is contained the triangle BCE is, by the 11. of the fourth equiangle, and equilater. And forasmch as two pentagons, set vpon euery one of the bases of the cube doo make a dodecahedron, and sixe bases of the cube doo receaue twelue angles of the dodecahedron: and the 8. semidiameters doo in the pointes where they are cutte by an extreame and meane proporti<g ref="char:EOLhyphen"/>on receaue the rest: therefore the 12. pentagon bases contayning 20. solide angles doo inscribe the do<g ref="char:EOLhyphen"/>decahedron in the cube: by the 1. diffinition of this booke. Wherefore in a cube geuen is inscribed a dodecahedron: which was required to be done.</p>
               <div type="section">
                  <head>First Corollary.</head>
                  <p>
                     <hi>The diameter of the sphere, which containeth the dodecahedron, containeth in power these two sides, namely, the side of the Dodecahedron, and the side of the cube wherein the Dodecahedron is inscribed.</hi> For in the first figure a line drawne from the centre O, to the poynt B the angle of the Dodecahedron, namely the line OB, containeth in pow<g ref="char:EOLhyphen"/>er these two lines OV the halfe side of the cube, and VB the halfe side of the dodecahedron, by the 47. of the first. Wherefore by the 15. of the fiueth, the double of the line OB, which is the diameter of the sphere containing the Dodecahedron, containeth in power the double of the other lines OV and VB, which are the sides of the cube, and of the dodecahedron.</p>
               </div>
               <div type="section">
                  <head>¶ Second Corollary.</head>
                  <p>
                     <hi>The side of a cube diuided by an extreme and meane proportion, maketh the lesse segment the side of the dodecahedron inscribed in it: and the greater seg<g ref="char:EOLhyphen"/>ment the side of the cube inscribed in the same Dodecahedron:</hi> For it was before pro<g ref="char:EOLhyphen"/>ued, that the side of the dodecahedron is the greater segment of BE the side of the triangle BEC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> but the side BE (which is equall to the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> GB and SF) is the greater segme<g ref="char:cmbAbbrStroke">̄</g>t of GF the side of the cube: which line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>E (subtending th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> angle of the pentagon) was (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of this booke) the side of the cube inscribed in the dodecahedron.</p>
               </div>
               <div type="section">
                  <pb facs="tcp:7062:470"/>
                  <head>Third Corollary.</head>
                  <p>
                     <hi>The side of a cube, is equal to the sides of a Dodecahedron inscribed in it, and circumscribed about it.</hi> For it was manifest by this proposition, that the side of a cube ma<g ref="char:EOLhyphen"/>keth the lesse segment, the side of a Dodecahedron inscribed in it, namely, as in the first figure the line BS the side of the Dodecahedron inscribed, is the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t of the line GF the side of the cube. And it was proued in the 17. of the thirtenth, that the same side of the cube subte<g ref="char:cmbAbbrStroke">̄</g>deth the angle of the penta<g ref="char:EOLhyphen"/>gon of the Dodecahedron circumscribed: and therefore it maketh the greater segment the side of the Dodecahedron or of the pentagon, by the first corollary of the same. Wherefore it is equal to bothe those segments.</p>
               </div>
            </div>
            <div type="part">
               <head>The 14. Probleme. The 14. Proposition. In a cube geuen, to inscribe an Icosahedron.</head>
               <p>SVppose that the cube geuen
<figure/>
be ABC,<note place="margin">
                     <hi>Construction.</hi>
                  </note> the Centres of whose bases let be the points D, E, G, H, I, K: by whiche poyntes draw in the bases vnto the o<g ref="char:EOLhyphen"/>ther sids parallels not touching the one the other. And deuide the lines drawn from the centres, as the line DT. &amp;c. by an extreme and meane pro<g ref="char:EOLhyphen"/>portion in the poyntes A, F: L, M: N, B: P, Q: R, S: C, O: by the 30. of the sixth: and let the greater segmentes be about the ce<g ref="char:cmbAbbrStroke">̄</g>tres. And draw these right lines, AL, AG, AM, and TG. And for<g ref="char:EOLhyphen"/>asmuch as the lines cut are parallels to the sides of the cube:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> they shall make right angles the one with the other by the 29. of the first: and forasmuche as they are equal: their sections shall be e<g ref="char:EOLhyphen"/>qual, for that the sections are like by the 2. of the fourtenth. Wherfore the line TG is equal to the line DT, for they are eche, halfe sides of the cube. Wherfore the square of the whole line TG, and of the lesse segment TA, is triple to the square of the line AD the greater seg<g ref="char:EOLhyphen"/>ment (by the 4. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th). But the line AG containeth in power the lines AT &amp; TG, for the angle ATG is a right angle. Wherefore the square of the line AG is triple to the square of the line AD. And forasmuch as the line MGL is erected perpendicularly to the plain passing by the lines AT, &amp; which is parallel to the bases of the cube (by the corollary of the 14. of the eleue<g ref="char:cmbAbbrStroke">̄</g>th) therfore the angle AGL is a right angle. But the line LG is equal to the line AD, for they are the grea<g ref="char:EOLhyphen"/>ter segments of equal lines: Wherfore the line AG (which is in power triple to the line AD) is in po<g ref="char:EOLhyphen"/>wer triple to the line LG. Wherefore adding vnto the same square of the line AG, the square of the line LG, the square of the line AL, which (by the 47. of the first) containeth in power the two lines A<g ref="char:EOLhyphen"/>G and GL, shalbe quadruple to the line AD or LG. Wherefore the line AL is double to the line AD (by the 20. of the sixth:) and therfore is equal to the line AF, or to the line LM. And by the same rea<g ref="char:EOLhyphen"/>son may we proue that euery one of the other lines which couple the next sections of the lines cut, as the lines AM, PF, PM, MQ, and the rest are equal. Wherfore the triangles ALM, APF, AMP, PMQ and the rest such like, are equal, equiangle, and equilater, by the 4. and eigth of the first. And forasmuch as vpon euery one of the lines cut of the cube are set two triangles, as the triangles ALM, and BLM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> there shalbe made 12. tria<g ref="char:cmbAbbrStroke">̄</g>gles. And forasmuch as vnder euery one of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. angles of the cube, are sub<g ref="char:EOLhyphen"/>tended the other 8. triangles, as the triangle AMP. &amp;c. of 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. and 8. triangles, shall be produced 20. tri<g ref="char:EOLhyphen"/>angles equal and equilater co<g ref="char:cmbAbbrStroke">̄</g>taining the solid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of an Icosahedron, by the 25. diffinition of the eleuenth, which shalbe inscribed in the cube geuen ABC by the first diffinition of this booke. The inuention of the demonstration of this dependeth of the ground of the former. Wherfore in a cube geuen, we haue described an Icosahedron: which was required to be done.</p>
               <div type="section">
                  <pb n="441" facs="tcp:7062:470"/>
                  <head>First Corollary.</head>
                  <p>
                     <hi>The diameter of a sphere which containeth an Icosahedron, containeth two sides, namely the side of the Icosahedron, and the side of the cube which contai<g ref="char:EOLhyphen"/>neth the Icosahedron.</hi> For if we drawe the line AB, it shall make the angles at the poynt A right angles: for that it is a parallel to the sides of the cube: wherfore the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> which coupleth the opposite angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the Icosahedron, at the poynts F and B, co<g ref="char:cmbAbbrStroke">̄</g>taineth in power the line AB (the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> cube) and the line AF (the side of the Icosahedron) by the 47. of the first. Which line FB is equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>l to the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ia<g ref="char:EOLhyphen"/>meter of the sphere, which cont<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ineth the Icosahedron, by the demonstration of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th.</p>
               </div>
               <div type="section">
                  <head>Second Corollary.</head>
                  <p>
                     <hi>The six opposite sides of the Icosahedron deuided into two equal parts: their sections are coupled by three equal right lines, cutting the one the other into two equal partes, and perpendicularly in the centre of the sphere which containeth the Icosahedron.</hi> For those three lines are the three lines which couple the centres of the bases of the cube, which do in suche sort in the centre of the cube, cut the one the other, by the corollary of the third of this booke, and therfore are equal to the sides of the cube. But right lines drawne from the ce<g ref="char:cmbAbbrStroke">̄</g>
                     <g ref="char:EOLhyphen"/>tre of the cube to the angles of the Icosahedron, euery one of them shall subtend the halfe side of the cube, and the halfe side of the Icosahedron (which halfe sides containe a right angle) wherefore those lines are equal. Wherby it is manifest that the foresaid centre is the centre of the sphere which contai<g ref="char:EOLhyphen"/>neth the Icosahedron.</p>
               </div>
               <div type="section">
                  <head>Third Corollary.</head>
                  <p>
                     <hi>The side of a cube deuided by an extreme and meane proportion, maketh the greater segment the side of an Icosahedron described in it.</hi> For the half side of the cube maketh the halfe of the side of the Icosah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>dron the greater segment: wherefore also the whole side of the cube, maketh the whole side of the Icosahedron the greater segment by the 15. of the fifthe, for the sections are like by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>. of the fourtenth.</p>
               </div>
               <div type="section">
                  <head>¶Fourth Corollary.</head>
                  <p>
                     <hi>The sides and bases of the Icosahedron, which are opposite the one to the o<g ref="char:EOLhyphen"/>ther, are parallels.</hi> Forasmuch as euery one of the opposite sides of the Icosahedron, may be in the parallel lines of the cube, namely, in those parallels which are opposite in the cube: and the triangles which are made of parallel lines, are parallels, by the 15. of the eleuenth: therfore the opposite tri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ngle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the Icosahedron, as also the sides, are p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rallels the one to the other.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 15. Probleme. The 15. Proposition. In an Icosahedron geuen, to inscribe an Octohedron.</head>
               <p>SVppose that the Icosahedron geuen be ACDF: and by the former second Corollary,<note place="margin">
                     <hi>Construction.</hi>
                  </note> let there be take<g ref="char:cmbAbbrStroke">̄</g> the three right lines which cut the one the other into two equall partes perpendicularly, and which couple the sections into two equall partes of the sides of the Icosahedron: which let be BE, GH, and KL, cutting the one the other in the point I. And draw<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> these ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ht lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G, GE, EH, and HB. And forasmuch as the an<g ref="char:EOLhyphen"/>gles at the point I are (by construction) right angles,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>
                  <figure/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd are con<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ined under equ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ll lines<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> shall <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> squ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst. Likewyse <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nto tho<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> shall be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall the lines dr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n from <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> pointes K and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to euery one of the poin<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>s <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, G. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, H: And therefore the triangles which <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>ramis DGENK, shall be equall <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>. And by <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>.</p>
            </div>
            <div type="part">
               <pb facs="tcp:7062:471"/>
               <head>¶ The 16. Probleme. The 16. Proposition. In an Octohedron geuen, to inscribe an Icosahedron.</head>
               <p>LEt there be taken an Octohedron, whose 6. angles, let be A, B, C, F, P, L. And draw the lines AC, BF, PL,<note place="margin">
                     <hi>Construction.</hi>
                  </note> cutting the one the other perpendicularly in the point R (by the 2. Corollary of the 14. of the thirtenth). And let euery one of the 12. sides of the Octohe<g ref="char:EOLhyphen"/>dron be diuided by an extreme and meane proportion, in the pointes H, X, M, K, D, S, N, G, V, E, Q, T. And let the greater segmentes be the lines BH, BX, FM, FK, AD, AQ, CS, CT, PN, PG, LV, LE: And drawe these lines HK, XM, GE, NV, DS, QT. Now forasmuch as in the triangle ABF, the sides are cut proportionally, namely, as the line BH is to the line HA, so is the line FK to the line KA (by the 2. of the fouretenth):<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> therefore the line HK shall be a parallel to the line BF (by the 2. of the sixth). And forasmuch as the line AC cutteth the line HK in the point Z, and the line ZK is a parallel vnto the line RF, the line RA shall be cut by an ex<g ref="char:EOLhyphen"/>treme and meane proportion in the point Z: by the 2. of the sixth: namely, shall be cut like vnto the line FA: and the
<figure/>
greater segme<g ref="char:cmbAbbrStroke">̄</g>t ther<g ref="char:EOLhyphen"/>of shall be the line ZR. Vnto the line ZK put the line RO equall, by the 3. of the first: and drawe the line KO: now then, the line KO shall be equall to the line ZR, by the 33. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst. Draw the lines KG, KE, and KI. And forasmuch as the triangles AR<g ref="char:EOLhyphen"/>F, and AZK, are e<g ref="char:EOLhyphen"/>quiangle (by the 6. of the sixth) the sides AZ and ZK, shall be equall the one to the other, by the 4. of the sixth, for the sides AR and RF, are e<g ref="char:EOLhyphen"/>quall. Wherfore the line ZK shall be the lesse segment of the line RA. But if the greater segment RZ be diuided by an ex<g ref="char:EOLhyphen"/>treme &amp; meane pro<g ref="char:EOLhyphen"/>portion, the greater segment therof shall be the line ZK, which was the lesse segment of the whole line RA, by the 5. of the thir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>enth. And for<g ref="char:EOLhyphen"/>asmuch as the two lines FE and FG, are equall to the two lines AH and AK, namely, ech are lesse seg<g ref="char:EOLhyphen"/>mentes of equall sides of the Octohedron, and the angles HAK and EFG are equall, namely, are right angles, by the 14. of the thirtenth: the bases HK and GF shall be equall, by the 4. of the first: And by the same reason vnto them may be proued equall the lines XM, NV, DS, and QT. And forasmuch as the lines AC, BF, and PL, do cut the one the other into two equall parts, and perpendicularly, by con<g ref="char:EOLhyphen"/>struction: the lines HK and GE (which subtend angles of triangles like vnto the triangles whose an<g ref="char:EOLhyphen"/>gles the lines AC, BF, and PL subtend) are cut into two equall partes in the pointes Z and I, by the 4. of the sixth, so also are the other lines NV, XM, DS, QT (which are equall vnto the lines HK &amp; GE) cut in like sort, and they shall cut the lines AC, BF, and PL like. Wherefore the line KO (which is equall to RZ) shall make the greater segment the line RO, which is equall to the line ZK (for the grea<g ref="char:EOLhyphen"/>ter segment of the RZ was the line ZK): and therefore the line OI shall be the lesse segment, when as the whole line RI is equall to the whole line RZ. Wherefore the squares of the whole line KO, and of the lesse segment OI, are triple to the square of the greater segment RO, by the 4. of the thirtenth. Wher<g ref="char:EOLhyphen"/>fore the line KI, which containeth in power the two lines KO and OI, is in power triple to the line RO (by the 47. of the first): for the angle KOI is a right angle. And forasmuch as the lines FE and FG (which are the lesse segmentes of the sides of the Octohedron) are equall: and the line FK is co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>mon
<pb n="442" facs="tcp:7062:471"/>
to them both: and the angles KFG and KFE (of the triangles of the Octohedron) are equall: the bases KG and KE shall (by the 4 of the first) be equall: and therefore the angles KIE and KIG which they subtend, are equall (by the 8. of the first): Wherefore they are right angles, by the 13. of the first. Wherefore the right line KE (which containeth in power the two lines, KI and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E by the 47. of the first) is in power quadruple to the line RO (or IE): for the line RI is proued to be in power triple to the same line RO: But the line GE is double to the line IE: Wherfore the line GE is also in power <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> PF. And by the same reason may be proued, that the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>est of the eleuen solide angles of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, are <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the sections of euery one of the sides of the Octohedron, namely, in the pointes E, N, V, H, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, M, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, D, S, Q, T. Wherefore there are 12. angles of the Icosahedron. Moreouer, forasmuch as euery one of the bases of the Octohedron, do eche containe triangles of the Icosahedron, <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> pyrami<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> ABC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>FP (which is the halfe of the Octohedron) the triangle FCP receaueth in th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> section of his sides the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> triangle GMS: and the triangle CPB containeth the triangle NXS: and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> triangle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>AP contay<g ref="char:EOLhyphen"/>neth the triangle HND: and moreouer the triangle APF containeth the triangle <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DG, and the same may be proued in the opposite pyramis ABCFL: Wherefore there shall be eight triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. And for<g ref="char:EOLhyphen"/>asmuch as besides these triangles, to euery one of the solide angles of the Octohedron <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> subtended two triangles, as the triangles KEG amd MEG, to the angle F: and the triangles HNV and XNV, to the angle B: also the triangles NDS and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DS, to the angle P: likewise the triangle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> DHK and QHK, to the angle A: Moreouer the triangles EQT and VQT, to the angle L: and finally the tri<g ref="char:EOLhyphen"/>angles SXM and TXM, to the angle C: these 12. triangles being added to th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> for <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> triangles, shall produce <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. triangles equall and equil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter coupled together: which shall male an Icosahedron, by the 25. definition of the eleuenth: and it shall be inscribed in the Octohedron geuen ABC<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>L, by the first definition of this booke: for the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. angles thereof are set in 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. like sections of the sides of the Octohedron. Wherefore in an Octohedron geuen, is inscribed an Icosahedron.</p>
               <div type="section">
                  <head>¶ First Corollary.</head>
                  <p>
                     <hi>The side of an equilater triangle being diuided by an extreme and meane proportion: a right line subtending within the triangle, the angle which is con<g ref="char:EOLhyphen"/>tained vnder the greater segment and the lesse: is in power duple to the lesse seg<g ref="char:EOLhyphen"/>ment of the same side.</hi> For the line KE, which subtendeth the angle KFE of the triangle AFL, which angle KFE is contained vnder the two segmentes KF &amp; FE, was proued equall <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the line HK, which containeth in power the two lesse segmentes HA and AK, by the 47. of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>rst fo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> angle HAK is <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>.</p>
               </div>
               <div type="section">
                  <head>Second Corollary.</head>
                  <p>
                     <hi>The bases of the Icosahedron are concentricall (that is, haue one and the selfe same centre) with the bases of the Octohedron which contayneth it.</hi>
                  </p>
                  <p>For suppose that <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> Octohedron <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap>
                     <figure/>
ECD the base of an Icosahedron: and let the centre of the base ABG be the point F. And drawe these right lines FA, FB, FC, and FE. Now then the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> to the two lines FB and BC: for they are lines drawen from the centre, and are also lesse segmentes: and they contayne the <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>.</p>
               </div>
            </div>
            <div type="part">
               <pb facs="tcp:7062:472"/>
               <head>¶ The 17. Probleme. The 17. Proposition. In an Octohedron geuen, to inscribe a Dodecahedron.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that the Octohedron geuen be ABGDEC: whose 12. <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ides let be cut by an ex<g ref="char:EOLhyphen"/>treme and meane proportion, as in the former Proposition. It was manifest that of the right lines which couple th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>se sections, are made 20. triangles, of which 8. are concentri<g ref="char:EOLhyphen"/>call with the bases of the Octohedron, by the second Corollary of the former Proposi<g ref="char:EOLhyphen"/>tion. If therefore in euery one of the centres of the 20. triangles be inscribed (by the 1. of this booke) euery one of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>gles of the Dodecahedron,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> we shall finde, that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. an<g ref="char:EOLhyphen"/>gles of the Dodecahedron
<figure/>
are set in the 8. centres of the bases of the Octohe<g ref="char:EOLhyphen"/>dron: namely, these angles I, u, ct, O, M, a, P, and X: and of the other 12. solide angles there are two in the centres of the two trian<g ref="char:EOLhyphen"/>gles which haue one side common vnder euery one of the solide angles of the Octohedron: namely, vn<g ref="char:EOLhyphen"/>der the solide angle A, the two solide angles, K, Z: vn<g ref="char:EOLhyphen"/>der the solide angle B, the two solide angles H, T: vnder the solide angle G, the two solide angles Y, V: vnder the solide angle D, the two solide angles F, L: vnder the solide angle E, the two solide angles S, N: vnder the solide angle C, the two solide angles Q, R: and forasmuch as in the Octohedron are sixe solide angles, vnder them shall be subtended 12. solide angles of the Dod<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>cahedron: and so are m<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>de 20. solide an<g ref="char:EOLhyphen"/>gles composed of 12. equall and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quil<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ter superficiall pentagons (as it was <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, by the 5. of this booke) which therefore containe a Dodecahedron (by the 24. definition of the eleuenth). And it is inscribed in the Octohedron (by the 1. definition of this booke): for that euery one of the bases of the Octohedron do receaue angles therof. Wherefore in an Octohedron geuen, is inscribed a Dodeca<g ref="char:EOLhyphen"/>hedron.</p>
            </div>
            <div type="part">
               <head>¶ The 18. Probleme. The 18. Proposition. In a trilater and equilater Pyramis, to inscribe a Cube.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that there be a trilater equilater Pyramis, whose base let be ABC, and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oppe the point D. And let it be comprehended in a Sphere<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the 13. of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, And l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> the centre of that Sphere be the point E. And from the solide angles A, B, C, D, draw right lines passing by the centre E, vnto the opposite bases of the pyramis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and they shall <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>all perpendicularly vpon the bases, and shall also fall vpon the centres of the circles which containe the bases, by the Corollary of the 13. of the thirtenth. Let the cen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re of the tri<g ref="char:EOLhyphen"/>angle ABC, be the point G, and let the centre of the triangle ADC be the point H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and of the trian<g ref="char:EOLhyphen"/>gle ADB let the point N be the centre, and finally, let the point F be the centre of the other triangle DBC. And let the right lines falling vpon those ce<g ref="char:cmbAbbrStroke">̄</g>tres be DEG, BEH, CEN, &amp; AEF. And by those centers G, H, N, F, let there be drawen from the angles to the opposite sides these right lines, AGL,
<pb n="443" facs="tcp:7062:472"/>
DHK, BNM, and DFL, which shall fall perpendicularly vpon the sides BC, CA, AD, and CB, by the Corollary of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the thirtenth, and therefore they shall cu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> them into two equall partes in the pointes K, L, M, by the
<figure/>
3. of the third. Agayne let the lines which wer drawen from the solide angles to the opposite bases be diuided into two equal partes, name<g ref="char:EOLhyphen"/>ly, the line DG in the point T, the line CN in the point O, the line AF in the point P, and the line BH in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he point R: and drawe the lines HT, FT, HO, and FO. Now forasmuch as the lines GK,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and GL, which are drawen from the centre of one and the self same trian<g ref="char:EOLhyphen"/>gle ABC to the sides, are equall, and the lines DK and DL are equall, for they are the perpen<g ref="char:EOLhyphen"/>diculars of equal &amp; like triangles:<note place="margin">
                     <hi>Produce in the figure the line TF to the point B.</hi>
                  </note> and the line DG is common to the<g ref="char:cmbAbbrStroke">̄</g>. Wherefore, by the 8. of the first, the angles KD<g ref="char:EOLhyphen"/>G &amp; LDG, are equall. And forasmuch as the lines HD &amp; DF are dra<g ref="char:EOLhyphen"/>wen from the centre of equal circles which co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>taine the equal triangles ADC &amp; DBC, therfore they are equal, &amp; the line DT is co<g ref="char:cmbAbbrStroke">̄</g>mon to the<g ref="char:cmbAbbrStroke">̄</g> both, and they containe equal angles, as before hath bene proued. Wherfore the bases HT and FT are equal by the 4. of the first. And by the same reason if we drawe the lines CF and CH, may we proue that the other lines HO and FO, are equal to the same lines HT and FT, and also the one to the other. Wher<g ref="char:EOLhyphen"/>fore also after the same maner may be proued that the rest of the lines, which couple the centres of the triangles and the sections of the perpendiculars into two equal partes, as the lines NP, GR, GP, RN: NT, PH, GO, and RF, are equal. And forasmuche as from euery one of the centres of the bases are drawne thre right lines to the sections into two equal parts of the perpendiculers, and there are foure centres, it followeth, that these equal right lines so drawne, are twelue in number, of which euery three and three make a solide angle in the foure centres of the bases, and in the foure sections into two equal partes of the perpendiculars: wherfore that solide hath 8. angles, contained vnder 12. equal sides, which make sixe quadrangled figures, namely, HOFT, PGRN, PHOG, GOFR, FRNT, and TNPH. Now let vs proue that those quadrangled figures are rectangle.</p>
               <p>Forasmuch as vpon DC the common base of the triangles ADC and BDC falleth the perpen<g ref="char:EOLhyphen"/>diculars AS and BS, which are drawne by the centres H and F: either of these lines HS and SF shalbe the third part of either of these lines AS and SB: for the line AH is duple to the line HS, and deuideth the base DC into two equal partes by the corollary of the 12. of the thirtenth. Wherefore in the trian<g ref="char:EOLhyphen"/>gle ABS, the sides AS and BS are cut proportionally in the poynts H and F: and therfore the line HF is a parallel to the side AB, by the 2. of the sixth. Wherfore the triangles ASB and HSF are equiangle, by the 6. of the sixth. Wherfore the base HF shalbe the third part of the base AB, by the 4. of the sixth. We may also proue that the line TO is the third part of the line DC, for the lines EC and ED, which are drawne from the centre of the sphere which containeth the pyramis are equal: and the line EN, (which is drawne from the centre to the base) is the third part of the line EC, so also is the line GE the third part of the line ED (by the corollary of the 13. of the thirtenth) for it is the sixth parte of the dia<g ref="char:EOLhyphen"/>meter of the sphere which containeth the pyramis: And the line ON, is the half of the whole line NC wherfore the residue EO is the third part of the line EC<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and so also is the line ET the third part of the line ED. Wherfore the line TO in the triangle DEC is a parallel to the line DC, and is a third parte of the same, by the former 2. and 4. of the sixth, as the line HF was proued the third part of the line AB. But AB and DC being sides of the pyramis are equal. Wherfore the lines HF and TO, being the third partes of equal lines, are equal, by the 15. of the fiueth. Wherfore by the 8. of the first the angles HTF, and TFO are equal: and by the same reason, the angles opposite vnto them, namely, the angles FOH
<pb facs="tcp:7062:473"/>
and OHT are equal the one to the other, and also are equal to the said angles HT<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and TFO: but these foure angles are equal to 4. right angles by the corollary of the 32. of the first: wherefore the an<g ref="char:EOLhyphen"/>gles of the quadrangle HOFT are right angles. And by the same reason may the angles of the other fiue quadrangled figures be proued right angles<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Now resteth to proue that the foresayde quadrangles are ech in one and the selfe same plaine.</p>
               <p>Take the quadra<g ref="char:cmbAbbrStroke">̄</g>gle
<figure/>
HOFT: and foras<g ref="char:EOLhyphen"/>much as in the trian<g ref="char:EOLhyphen"/>gle ASB, the line HF is proued a parallel to the line AB, therefore it cutteth the lines SV, and SB proportional<g ref="char:EOLhyphen"/>ly in the poynts I. and F. by the 2. of the sixth: Now then forasmuch as SF was proued the third parte of the line SB, the line SI, shall also be the thirde part of the line SV. More<g ref="char:EOLhyphen"/>ouer forasmuch as the line VS, whiche cou<g ref="char:EOLhyphen"/>pleth the sections into equal partes of the op<g ref="char:EOLhyphen"/>posite <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ides of the py<g ref="char:EOLhyphen"/>ramis, namely, of the sides AB and DC, is by the centre E deui<g ref="char:EOLhyphen"/>ded into two equal partes, by the corolla<g ref="char:EOLhyphen"/>ry of the seconde of this boke (for it is the diameter of the octo<g ref="char:EOLhyphen"/>hedron inscribed in the pyramis): therfore the line SI is two third partes of the halfe line SE. And by the same reaso<g ref="char:cmbAbbrStroke">̄</g>, forasmuch as in the tria<g ref="char:cmbAbbrStroke">̄</g>gle DEC the line TO is proued to be a parallel to the side DC, it shall in the selfe same triangle cut the lines CE and SE, proportionally in the poynts O and I by the same 2. of the sixth: but the line EO is proued to be a third parte of the line EC. Wherefore the lyne EI is also a third part of the line ES. Wherefore the residue IS shalbe two third partes of the whole line ES. Wherefore the point I cutteth of the lines TO and HF. Wherefore the two lines HIF and TIO cutting the one the other, are in one and the selfe same' plaine, by the 2. of the eleuenth. And therefore the poyntes H, T, F, O are in one &amp; the selfe same plaine. Wherfore<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the rectangle figure HOFT be<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ng quadrilater and equilater, and in one and the selfe same playne, is a square, by the di<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>inition of a square. And by the same reason may the rest of the bases of the solide be proued to be squares equall and plaine or superficial: Now then the solide is comprehended of 6. equal squares (which are contained of 12. equal sides) which squares make 8. solide angles, of which foure are in the ce<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tres of the bases o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the pyramis, and the other 4. are in the midle sections of the foure perdendicu<g ref="char:EOLhyphen"/>lars. Wherfore the solide HOFTPGRN, is a cube by the 21. diffinition of the eleuenth, and is in<g ref="char:EOLhyphen"/>scribed in the pyramis, by the first definition of this boke. Wherfore in a trilater equilater pyramis ge<g ref="char:EOLhyphen"/>uen, is inscribed a cube.</p>
               <div type="section">
                  <head>¶ A Corrollary.</head>
                  <p>
                     <hi>The line which cutteth into two equall partes the opposite sides of the Py<g ref="char:EOLhyphen"/>ramis, is triple to the side of the cube inscribed in the pyramis, and passeth by the centre of the cube.</hi> For the line SEV, whose third part the line SI is, cutteth the opposite sides C<g ref="char:EOLhyphen"/>D and AB into two equll partes: but the line EI (which is drawne from the centre of the cube to the base is proued to be a third part of the line ES: wherefore the side of the cube which is double to the line EI shall be a third part of the whole line VS, which is (as hath bene proued) double to the line ES.</p>
               </div>
            </div>
            <div type="part">
               <head>The 19. Probleme The 19. Proposition. In a trilater equilater Pyramis geuen, to inscribe an Icosahedron.</head>
               <p>
                  <pb n="444" facs="tcp:7062:473"/>SVppose that the pyramis is geuen, <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> euery one of whose s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>des <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> be diuid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d into two equall partes in the poy<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, M, K, L, P, N.<note place="margin">
                     <hi>Construction.</hi>
                  </note> And i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> euery one of the b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ses of that py<g ref="char:EOLhyphen"/>ramis, descride the triangl<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> L<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>, PMN, NKL, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> which triangles shall be e<g ref="char:EOLhyphen"/>quilater by the 4. of the fir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>or the sides sub<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>end equall angles of the pyramis, contayned vnder the halues of the sides of the same pyramis<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the sides of the said triangles are equall. Let those sides be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iuided by an extreame and meane propor<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>on (by the 30. of the sixth) in the poyntes C, E, Q, R, S, T, H, I, O, V, Y, X. Now then those sides are cutte into the selfe same proportions, by the 2. of the fourte<g ref="char:cmbAbbrStroke">̄</g>th: and therfore they make the li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e sectio<g ref="char:cmbAbbrStroke">̄</g>s equall, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. part of the ninth of the fiueth. Now I say, that the foresayd poyn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es do<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> rec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>aue the angles of the Icosahe<g ref="char:EOLhyphen"/>dron inscribed in the pyramis AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D. In the foresayd triangles let there agayne be made other trian<g ref="char:EOLhyphen"/>gles by coupling the sections, and let those triangles be TRS, IOH, CEQ, and VXY, which shall be equilater: for euery one of their sides doo sub<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>d equall angles of equilater triangles, and those sayd equall angles are contayned vnder equall side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> (namely, vnder the greater segmen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and the lesse)<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and therefore the sides which subtend those angles are equall by the 4. of the first. Now let vs proue that at eche of the foresayd poynts, as for example at T, is set the solide angle of an Icosah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Forasmuch as the triangles TRS and TQO are equilater and equall, the 4. right lines TR, TS, TQ and TO shall be equall. And forasmuch as <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>PNK is a square cutting the pyramis AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D into two equall pa<gap reason="illegible" resp="#PDCC" extent="4 letters">
                     <desc>••••</desc>
                  </gap>, by the corollay of the second of this booke<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line TH shall be in power duple to the line TN or N<g ref="char:EOLhyphen"/>H by the 47. of the first.
<figure/>
For the lines TN or NH are equall, for that by con<g ref="char:EOLunhyphen"/>struction they are eche lesse segmentes: and the line RT or TS is in pow<g ref="char:EOLhyphen"/>er duple to the same line TN or NH (by the co<g ref="char:EOLhyphen"/>rollary of the 16. of this booke) for it subtendeth the angle of the triangle contayned vnder the two segmentes. Wherfore the lines TH, TS, TR, TQ, and TO are equall: and so also are the lines HS, SR, RQ, QO, and OH, which subtend the angles at the poynt T, equall. For the line QR contayneth in power the two lines P<g ref="char:EOLhyphen"/>Q and PR the lesse seg<g ref="char:EOLhyphen"/>mentes, which two lines the line TH also contay<g ref="char:EOLhyphen"/>ned in power. And the rest of the lines doo sub<g ref="char:EOLhyphen"/>tend angles (of equilater triangles) contayned vn<g ref="char:EOLhyphen"/>der the greater segment and the lesse. Wherefore the fiue triangles TRS, TSH, THO, TOQ, TQ<g ref="char:EOLhyphen"/>R are equilater and equall making the solide angle of an Icosahedron at the poynt T, by the 16. of the thirtenth, in the side PN of the triangle <hi>P</hi> NM. And by the same reason in the other sides of the 4. tri<g ref="char:EOLhyphen"/>angles PNM, NKL, FMK, &amp; LFP (which are inscribed in the bases of the pyramis) which sides are 12<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in nu<g ref="char:cmbAbbrStroke">̄</g>ber shal be set 12. angles of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> co<g ref="char:cmbAbbrStroke">̄</g>tained vnder 20. equal &amp; equilater triangles of which fowere are set in the 4. bases of the pyramis, namely, these fower triangles, TRS, HOI, CEQ, VXY: 4. triangles are vnder 4. angles of the pyramis: that is, the fower triangles CIX, YSH, ERV, TQO: and vnder euery one of the sixe sides of the pyramis are set two triangles, namely, vnder the side, of the triangles THS and THO<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vnder the side DB the triangles RQE and RQT: vnder the side DA the triangles COQ, and COI: vnder the side AB, the triangles EXC and EXV<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vnder the side BG the triangles SVR and SVY: and vnder the side AG the triangles IYH and IYX. Wherefore the so<g ref="char:EOLhyphen"/>lide being contayned vnder 20. equilater and equall triangles shall be an Icosahedron by the 23. diffiniti<g ref="char:EOLhyphen"/>on of the eleuenth: and shall be inscribed in the pyramis AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>D by the first di<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>nition of this booke, for all his angles doo at one time touch the bases of the pyramis. Wherefore in a trilater equilater py<g ref="char:EOLhyphen"/>ramis geuen, we haue inscribed an Icosahedron.</p>
            </div>
            <div type="part">
               <head>¶ The 20. Proposition. The 20. Probleme. In a trilater equilater Pyramis geuen, to inscribe a dodecahedron.</head>
               <p>
                  <pb facs="tcp:7062:474"/>SVppose that the pyramis geuen be ABGD, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>che of whose sides let be cutte into two e<g ref="char:EOLhyphen"/>quall partes: and draw the lines which couple the sections, which being diuided by an extreame and meane proportion, and right lines being drawne by the sections, shall re<g ref="char:EOLhyphen"/>ceaue 20. triangles making an Icosahedron, as in the former proposition it was manifest. Now then if we take the centres of those triangles, we shall there finde the 20. angles of the dodecahedron inscribed in it by the 5. of this booke. And forasmuch as 4. bases of the foresayd Icosahedro<g ref="char:cmbAbbrStroke">̄</g> are co<g ref="char:cmbAbbrStroke">̄</g>centricall with the bases of the pyramis, as it was proued in the 2. corollary of the 6. of this boke: there shal be placed 4<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> angles of the dodecahedro<g ref="char:cmbAbbrStroke">̄</g>, namely, the 4. angles E, F, H, D, in the 4. centres of the bases: and of the other 16. angles, vnder euery one of the 6. sides of the pyramis are subtended two; namely, vnder the side AD, the angles CK: vnder the side BD the angles LI: vn<g ref="char:EOLhyphen"/>der the side GD the angles M, N: vnder the side AB the angles T, S: vnder the side BG the angles P, O: and vnder the side AG the angles R, Q: so there rest 4. angles, whose true place we will now ap<g ref="char:EOLhyphen"/>poynt. Forasmuch as a cube contayned in one and the selfe same sphere with a dodecahedron, is inscri<g ref="char:EOLhyphen"/>bed in the same dodecahedron, as it was manifest by the 17. of the thirtenth, and 8. of this booke: it fol<g ref="char:EOLhyphen"/>loweth that a cube and a dodecahedron circumscribed about it, are contayned in one and the selfe same bodies, for that their angles concurre in one and the selfe same poyntes. And it was proued in the 18. of this booke, that 4. angles of the cube inscribed in the pyramis are set in the middle sections of the
<figure/>
perpendicular<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which are drawne from the solide angles of the pyramis to the opposite bases: where<g ref="char:EOLhyphen"/>fore the other 4. angles of the dodecahedron are also, as the angles of the cube, set in those middle sec<g ref="char:EOLhyphen"/>tions of the perpendiculars. Namely, the angle V is set in the middest of the perpendicular AH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the an<g ref="char:EOLhyphen"/>gle Y in the middest of the perpendicular BF: the angle X in the middest of the perpendicular GE: and lastly the angle D in the middest of the perpendicular D which is drawne from the toppe of the py<g ref="char:EOLhyphen"/>ramis to the opposite base. Wherefore those 4. angles of the dodecahedron may be sayd to be directly vnder the solide angles of the pyramis, or they may be sayd to be set at the perpendiculars. Wherefore the dodecahedron after this maner set, is inscribed in the pyramis geuen (by the first diffinition of this booke) for that vpo<g ref="char:cmbAbbrStroke">̄</g> euery one of the bases of the pyramis are set an angle of the dodecahedro<g ref="char:cmbAbbrStroke">̄</g> inscri<g ref="char:EOLhyphen"/>bed. Wherefore in a trilater equilater pyramis is inscribed a dodecahedron.</p>
            </div>
            <div type="part">
               <head>The 21. Probleme. The 21. Proposition. In euery one of the regular solides to inscribe a Sphere.</head>
               <p>
                  <pb n="445" facs="tcp:7062:474"/>IN the 13. of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> thirtenth and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> other 4. propositio<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> following,<note place="margin">
                     <hi>This propositi<g ref="char:EOLhyphen"/>on Campane hath, &amp; is the last also in or<g ref="char:EOLhyphen"/>der of the</hi> 15. <hi>booke with him.</hi>
                  </note> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> was declared that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> regular solides<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re so conta<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ed in a sphere, that <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ight lin<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> drawne from the cen<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> of <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> solide in<g ref="char:EOLhyphen"/>scribed, are equall. Which right lines therefore make pyramids, whose <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>oppes are the centre of the sphere, or of the solide, and the bas<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e cu<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> one of the bases of those solides. And <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> solide <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall and like the one to the other, and described in equall circles: those cir<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>les shall cutte the sphere: for the angles which touch the circumference of the circle, touch also the superficies of the sphere. Wherefore perpe<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>diculars drawne from the centre of the sphere to the bases, or to the playne superficieces of the equall circles, are equall, by the corollary of the assumpt of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the twelfth. Wherefore making the cen<g ref="char:EOLhyphen"/>tre the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> of the sphere which <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the solide, and th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> space some one of the equall perpen<g ref="char:EOLhyphen"/>dicular<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>scrib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> a sphere, and it shall touch euery one of the bases of <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> solide: <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>
                  <g ref="char:EOLhyphen"/>perficies of the sphere passe beyond those bases: when as those p<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>pe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>diculars <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> are drawne from the centre to the bases, by the 3. corollary of the sa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>umpt. Wher<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e haue i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> euery one of the regular bodies inscribed a sphere: which regular bo<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> are in number one i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> by the corollary of the 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>The regular figures inscribed in spheres, and also the spheres circumscri<g ref="char:EOLhyphen"/>bed about them, or contayning them, haue one and the selfe same centre.</hi> Namely, their pyramids, the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ngles of whose b<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>ses touch the super<gap reason="illegible" resp="#PDCC" extent="6 letters">
                        <desc>••••••</desc>
                     </gap> of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>here, doo from those angles cause equall right lines to be draw<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap> to one and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>he selfe <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> poyn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> making the top<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap> of the pyra<g ref="char:EOLhyphen"/>mid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> in the same poynt: and therefore they <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> c<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>tres of the spheres in the selfe same toppes when <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the right lines drawne from those angles to the cro<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>ed superficies, wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>rein are <gap reason="illegible" resp="#PDCC" extent="1 word">
                        <desc>〈◊〉</desc>
                     </gap> the an<g ref="char:EOLhyphen"/>gles of the bases of the pyramid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>, are equall<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </p>
               </div>
               <div type="section">
                  <head>An adue<gap reason="illegible" resp="#PDCC" extent="3 letters">
                        <desc>•••</desc>
                     </gap>sment of <hi>Flussas</hi>
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </head>
                  <p>Of these solides, onely the Octohedron receaueth the other solides inscribed one with <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> other. For the Octohedron contayneth the Icosahedron inscribed in it: and the same Icosahedron contayneth the Dodecahedron inscribed in the same Icosahedron: and the same dodecahedron contayneth the cube inscribed in the same Octohedron, and <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> 
                     <gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>r<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>mscribeth the Pyra<g ref="char:EOLhyphen"/>mis inscribed in the sayd Octohedron. But this happ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                     <g ref="char:EOLhyphen"/>neth not in the other solides.</p>
               </div>
            </div>
            <trailer>The ende of the fiuetenth Booke of Euclides Elemen<gap reason="illegible" resp="#PDCC" extent="3 letters">
                  <desc>•••</desc>
               </gap> after <hi>Ca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>pa<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
               </hi> and <hi>
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>.</hi>
            </trailer>
         </div>
         <div n="16" type="book">
            <pb facs="tcp:7062:475"/>
            <head>¶The sixtenth booke of the Elementes of Geometrie added by <hi>Flussas.</hi>
            </head>
            <p>IN the former fiuetenth booke hath bene taught how to inscribe the fiue regular solides one with in an other. Now semeth to rest, to co<g ref="char:cmbAbbrStroke">̄</g>pare those solid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> so inscribed, one to an other, and to set forth their passion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> and pro<g ref="char:EOLhyphen"/>prieties: which thing, <hi>Flussas</hi> considering, in this sixte<g ref="char:cmbAbbrStroke">̄</g>th booke added by him,<note place="margin">
                  <hi>The argu<g ref="char:EOLhyphen"/>ment of the</hi> 16. <hi>booke.</hi>
               </note> hath excellently well and most conningly performed. For which vndoubtedly he hath of all them which haue a loue to the Mathematicals, de<g ref="char:EOLhyphen"/>serued much prayse and commendacion: both for the great tra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>ailes and payn<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s (which it is most likely) he hath ta<gap reason="illegible" resp="#PDCC" extent="2 letters">
                  <desc>••</desc>
               </gap>n in i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>uenting such straunge and wonderfull propositions with their demonstrations, in this booke contayned, as also for participating and communica<g ref="char:EOLhyphen"/>ting abrode the same to others. Which booke also, that the reader should want nothing conducing to the perfection of Euclides Elements: I haue with some trauaile translated, &amp; for the worthines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>hereof haue added it, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> a sixtenth booke to the 15. bookes of Euclide. Vouchsafe therefore gentle reader dili<g ref="char:EOLhyphen"/>gently to read and peyse it, for in it shall you finde no<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap> onely matter strange and delec<g ref="char:EOLhyphen"/>table, but also occasion of inuention of greater things pertayning to the natures of the fiue regular solid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                  <desc>•</desc>
               </gap>
            </p>
            <div type="part">
               <head>¶ The 1. Proposition. A Dodecahedron, and a cube inscribed in it, and a Pyramis inscribed in the same cube, are contained in one and the selfe same sphere.</head>
               <p>FOr the angles of the pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are se<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the ang<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>es of the cube wherein it is inscri<g ref="char:EOLhyphen"/>bed (by the first of the fiuetenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and all the angles of the cube are set in the angles of the dodecahed<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> circumscribed <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> (<gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the 8. of the fiuetenth): And all the angles of the Dodecahedron, are set in the superficies of the sphere, by the 17. of the thirtenth. Wherefore those three solides inscribed one within an other, are contained in one and the selfe same sphere, by the first diffinition of the fiuetenth. A dodecahedron therfore and a cube inscribed in it, and a pyramis inscribed in the same cube, are contained <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>lfe same sphere.</p>
               <div type="section">
                  <head>
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>
                  </head>
                  <p>
                     <hi>These three solides li<gap reason="illegible" resp="#PDCC" extent="1 span">
                           <desc>〈…〉</desc>
                        </gap>elfe same Icosahedron, or Octohedron, or Pyramis.</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>me Icosahedron, by the, 5.11. &amp; 12. of the fiuetenth: and they ar<gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap>ctohedron, by the 4. <hi>6.</hi> and 16. of the same: lastly they are inscribed in <gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the first, 18. and 19. of the same. For the angles of all these solide<gap reason="illegible" resp="#PDCC" extent="1 span">
                        <desc>〈…〉</desc>
                     </gap> the circumscribed Icosahedron, or octohedron, or pyramis.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> The proportion of a Dodecahedron circumscribed about a cube, to a Dodeca<g ref="char:EOLhyphen"/>hedro<g ref="char:cmbAbbrStroke">̄</g>
                  <pb n="446" facs="tcp:7062:475"/>
inscribed in the same cube, is triple to an extreme &amp; meane propartio<g ref="char:cmbAbbrStroke">̄</g>.</head>
               <p>FOrasmuch as in the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> corollary of the 13. of the fiu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>enth, it was proued, that the side of a Dodecahedron inscribed in a cube, is the lesse segment of the side of that cube de<g ref="char:EOLhyphen"/>uided by an extreme and meane proportion: and the side of the dodecahedron <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ircum<g ref="char:EOLhyphen"/>scribed about the same cube, is the greater segment of the side of the same cube (which thing also was taught in the 13. of the fiuetenth) the side of the Dodecahedron circum<g ref="char:EOLhyphen"/>scribed, shalbe to the side of the Dodecahedron inscribed, as the greater segment of a right line deuided by an extreme and meane proportion, is to the lesse segment of the same, which proportion is called an extreme and meane proport<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> by the diffinition, and by the 30. of sixth. But the proportion of like solide prolihedrons, is t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iple to the proportion of the side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of like pro<g ref="char:EOLhyphen"/>portion, by the corollary of the 17. of the twelueth. Wherefore the proportion of the Dodecahedron circumscribed about the cube, is to the dodecahedron inscribed in the same cube, in triple proportion of the sides ioyned together by an extreme and meane proportion: The proportion therefore of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Do<g ref="char:EOLhyphen"/>decahedron circumscribed about a cube to a dodecahedron inscribed in the same cube, is triple to an extreme and meane proportion.</p>
            </div>
            <div type="part">
               <head>The 3. Proposition. In euery equiangle, and equilater Pentagon, a perpendicular drawne from one of the angles to the base, is deuided by an extreme and meane proporti<g ref="char:EOLhyphen"/>on by a right line subtending the same angle.</head>
               <p>SVppose that ABCDF be a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> equiangle and <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quilater
<figure/>
pentagon:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and from one of the angles namely, from A, let there be drawne to the base CD a perpendicu<g ref="char:EOLhyphen"/>lar AG: and let the line BF subtend the angle BAP,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> which line BF let the line AD cut in the poynt I. Then I say that the line BF, cutteth the line AG by an extreme and meane proportion. For forasmuche as the angles GAF and GAB are equal by the 27. of the third, and the angles A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>F and AFB, are equal by the 5. of the first<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>gles remaining at the poynt E, of the triangles AEB and AEF are equal: for that they are the residues of two right angles by the corollary of the 32. of the first. But the angle EGC, is by construction a right angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherfore the lines BF &amp; CD are parallels by the 28. of the first. Wherefore as the line DI is to the line IA, so i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line GE to the line EA, by the 2. of the sixth. But the line DA, is in the poin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> I deuided by an extreme and meane proportion, by the 8. of the thirtenth. Wherfore the line GA is in the poynt E, deuided by an extreme and meane proportion (by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the fourtenth). Wherfore in e<g ref="char:EOLhyphen"/>uery equiangle and equilater pentagon, a perpendicular drawne from one of the angles to the base, is deuided by an ex<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>reme and meane proportion by a right line subte<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ding the same angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
               </p>
               <div type="section">
                  <head>¶ A Corollary.</head>
                  <p>
                     <hi>The line which subtendeth the angle of a pentagon, is a parallel to the side opposite vnto the angle.</hi> As it was manifest in the lines <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>F and CD.</p>
               </div>
            </div>
            <div type="part">
               <head>The 4. Proposition. If fro<g ref="char:cmbAbbrStroke">̄</g> the angles of the base of a <note n="*" place="margin">
                     <hi>By a Pyra<g ref="char:EOLhyphen"/>mis vnder<g ref="char:EOLhyphen"/>stand a Te<g ref="char:EOLhyphen"/>trahedron throughout all this booke.</hi>
                  </note> Pyramis, be drawne to the opposite sides, right lines cutting the sayd sides by an extreme and meane proportion: they shall containe the bise of the Icosahedron inscribed in the Pyramis, which base shalbe inscribed in an equilater triangle, whose angles cut the sides of the base of the Pyramis by an extreme and meane proportion.</head>
               <p>
                  <pb facs="tcp:7062:476"/>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that ABG be the base of a pyramis, in which let be inscribed an equilater triangle FKH, which is done by deuiding the sides into two equal partes. And in <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>his triangle let there be inscribed the base of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in the pyramis: which is described by deuiding the sides FK, KH, HF, by an extreme &amp; meane proportio<g ref="char:cmbAbbrStroke">̄</g> in the poynts C, D, E, by the 19. of the fiuete<g ref="char:cmbAbbrStroke">̄</g>th. Againe let the sides of the pyramis, namely, AB, BG, and GA be deuided by an extreme and me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ne proportion in the poynts I, M, L, by the 30. of the sixth. And drawe these right lines AM, BL, GI.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> Then I say that those lines describe the triangle CDE of the Icosahedron. For forasmuch as the lines BG and FH are parallels, by the 2. of the sixth: by the point D let the line ODN be drawne parallel to either of the lines BG &amp; FH. Wherfore the triangle HDN shalbe like to the triangle HKG, by the corollary of the 2. of the sixth. Wherfore either of these lines DN and NH shall be equal to the line DH, the greater segment of the line KH or FH. And forasmuch as the line FO is a parallel to the line HK, and the line OD to the line FH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the line OD shall be equal to the whole line FH in the pa<g ref="char:EOLhyphen"/>rallelogramme FODH, by the 34. of the
<figure/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst. Wherefore as the whole line FH is to the grea<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>er segment FE, so shall the lines equal to them be, namely, the line OD to the line DN, by the 7. of the fifth. Wherfore the line ON is deuided by an extreme and meane proportion in the poynt D, by the 2. of the fourtenth. But the triangles AOD, AFE, and ABM, are like the one to the other, and so also are the triangles ADN, AEH, and AMG, by the corollary of the seco<g ref="char:cmbAbbrStroke">̄</g>d of the sixth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore as FE is to EH, so is OD to DN, and BM to MG. Whe<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>fore the line AM cutting the lines FH and ON, lyke vnto the line BG in the pointes E, D, M, describeth ED the side of the triangle of the Icosahedron ECD, which is descri<g ref="char:EOLhyphen"/>bed in the sections E, C, D, by suppositio<g ref="char:cmbAbbrStroke">̄</g>. And by the same r<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ason the lines BL and GI shall describe the other sides EC and CD of the same triangle. By the point E, let there be drawne to GI a parallel line PEQ. Now forasmuch as the lines BM and FE are parallels, the line AM is in the poynt E, cut like to the line AB in the poynt F, by the 2. of the sixth. Wherefore the line AE is equal to the line EM: and vnto the line EM also are equal either of the lines GD and DI: which <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re cut l<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ke vnto the forsaid lines. Againe forasmuche as in the triangle ADI the lines DI and EP are parallels, as the line DI is to the line EP, so is the line AD to the line AE: but as the line AD is to the line AE, so is the line DG to the line EQ by the 2. of the sixth: wherefore as the line DI is to the line EP, so is the line DG to the line EQ: and alternately as the line DI is to the line DG, so is the line EP to the line EQ: but the lines DI and IG are equal: wherfore also the lines EP and EQ are equal. And forasmuch as the line AH is equal to the line FH, whose greater segme<g ref="char:cmbAbbrStroke">̄</g>t is the line HN<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therfore the whole line AN, is deuided by an extreme and meane proportion in the poynt H, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the thirtenth. But as the line AN is to the line AH, so is the line AD to the line AE, by the 2. of sixth (for the line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> FH and ON are parallel<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>:) and againe as the line AD is to the line AE, so (by the same) is the line AG to the line AQ, and the line AI to the line AP: for the lines PQ, and GI are parallels: Wherefore the lines AG and AI are deuided by an extreme and meane proportion in the points Q &amp; P: &amp; the line AQ shalbe the greater segme<g ref="char:cmbAbbrStroke">̄</g>t of the line AG or AB. And forasmuch as the whol<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line AG is to the greater segment AQ, as the greater segment AI is to the residue AP: the line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shalbe the lesse segment of the whole line A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> or AG. Wherfore the li<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> PEQ (which by the poynt E passeth parallelwise to the line GI) cutteth the lines AG and BA by an extreme and meane proportion in the poynts Q and P. And by the same reason the line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>R (which by the poynt C, passeth parallelwise to the line AM) shall fall vpon the sections P and R: so also shal the line RQ (which by the poynt D passeth parallelwise to the line BL) fall vpo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the sections RQ. Wherefore either of the lines PE and EQ shalbe equal to the line CD, in the parallelogrammes PD, and QC, by the 34. of the first. And forasmuch as the lines PE and EQ are equal, the lines PC, CR, RD and DQ shalbe likewise equal. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>rfore the triangle PRQ i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quilater, and cutteth the sides of the base of the pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> in the poyntes P, Q, R, by an extreme and meane proportion. And in it is inscribed the base ECD of the Ico<g ref="char:EOLhyphen"/>sahedron contained in the for<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>sayd pyramis. If therefore from the angles of the base of a pyramis, be drawne to the opposite sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s, right lines cutting the sayde sides by an extreme and meane proportion: they shall containe the base of the Icosahedron inscribed in the pyramis, which base shall be inscribed in an equilater triangle, whose angles cut the sides of the base of the pyramis by an extreme &amp; meane propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tion.</p>
               <div type="section">
                  <pb n="447" facs="tcp:7062:476"/>
                  <head>¶ A Corollary.</head>
                  <p>
                     <hi>The side of an Icosahedron inscribed in an Octohedron, is the greater segment of the line, which being drawen from the angle of the base of the Octohedron cutteth the opposite side by an extreame and meane proportion.</hi> For, by the 16. of the fiuetenth, FKH is the base of the Octohedron, which containeth the base of the I<g ref="char:EOLhyphen"/>cosahedron CDE: vnto which triangle FKH, the triangle HKG is equall, as hath bene proued. By the point H draw vnto the line ME a parallel line HT, cutting the line DN in the point S. Wherefore ES, DT, and ET, are parallelogrammes: and therefore the lines EH and MT are equall: and the lines EM and HT are like cut in the pointes D and S, by the 34. of the first. Wherefore the greater segment of the line HT is the line HS, which is equall to ED the side of the Icosahedron. But (by the 2. of the sixth) the line TK is cut like to the line HK by the parallel DM. And therefore (by the 2. of the fourtenth) it is di<g ref="char:EOLhyphen"/>uided by an extreme and meane proportion. But the line TM is equall to the line EH. Wherefore also the line TK is equall to the line EF or DH. Wherefore the residues EH and TG are equall. For the whole lines FH and KG are equall. Wherefore KG the side of the triangle HKG is in the point T diui<g ref="char:EOLhyphen"/>ded by an extreme and meane proportion in the point T, by the right line HT, and the greater segment thereof is the line ED the side of the Icosahedron inscribed in the Octohedron, whose base is the trian<g ref="char:EOLhyphen"/>gle HKG (or the triangle FKH which is equall to the triangle HKG) by the <hi>16.</hi> of the fiuetenth.</p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 5. Proposition. The side of a Pyramis diuided by an extreme and meane proportion, ma<g ref="char:EOLhyphen"/>keth the lesse segment in power double to the side of the Icosahedron in<g ref="char:EOLhyphen"/>scribed in it.</head>
               <p>SVppose that ABG be the base of a pyramis:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and let the base of the Icosahedron inscri<g ref="char:EOLhyphen"/>bed in it, be CDE, described of three right lines, which being drawen from the angles of the base ABG cut the opposite sides by an extreme and meane proportion, by the for<g ref="char:EOLhyphen"/>mer Proposition: namely, of these three lines AM, BI, and GI. Then I say, that AI the lesse segment of the side A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is in power duple to CE the side of the Icosahedron. For, forasmuch as by the former Proposition, it was proued that the triangle CDE is inscri<g ref="char:EOLhyphen"/>bed in an equilater triangle,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> whose angles cut the sides of
<figure/>
ABG the base of the pyramis by an extreme and meane proportion, let that triangle be FHK, cutting the line AB in the point F. Wherefore the lesse segment FA is equall to the segment AI, by the 2. of the fouretenth: (for the lines AB and AG are cut like). Moreouer the side FH of the triangle FHK is in the point D cut into two equall partes, as in the former Proposition it was proued, and FC<g ref="char:EOLhyphen"/>ED also by the same is a parallelogramme: Wherefore the lines CE and FD are equall, by the 33, of the first. And for<g ref="char:EOLhyphen"/>asmuch as the line FH subtendeth the angle BAG of an e<g ref="char:EOLhyphen"/>quilater triangle, which angle is contained vnder the grea<g ref="char:EOLhyphen"/>ter segment AH and the lesse segment AF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> therefore the line FH is in power double to the line AF or to the line AI the lesse segment, by the Corollary of the <hi>16.</hi> of the fiue<g ref="char:EOLhyphen"/>tenth. But the same line FH is in power quadruple to the line CE, by the 4. of the second: (for the line FH is double to the line CE). Wherefore the line AI being the halfe of the square of the line FH is in power duple to the line CE, to which the line FH was in power quadruple. Wherefore the side AG of the pyramis being diuided by an extreme and meane proportion, maketh th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> lesse segment AI in power duple to the side CE of the Icosahedron inscribed in it.</p>
               <div type="section">
                  <head>¶ A Corollary.</head>
                  <p>
                     <hi>The side of an Icosahedron inscribed in a pyramis, is a residuall line.</hi> For the diameter of the Sphere which containeth the fiue regular bodies, being rationall, is in power ses<g ref="char:EOLhyphen"/>quialtera to the side of the pyramis, by the 13. of the thirtenth: and therefore the side of the pyramis is rationall, by the definition: which side being diuided by an extreme and meane proportion, maketh
<pb facs="tcp:7062:477"/>
the lesse segment a residuall line, by the 6. of the thirtenth. Wherefore the side of the Icosahedron be<g ref="char:EOLhyphen"/>ing commensurable to the same lesse segment (for the square of the side of the Icosahedron is the halfe of the square of the said lesse segment) is a residuall line, by that which was added after the 103. of the tenth booke.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 6. Proposition. The side of a Cube containeth in power halfe the side of an equilater trian<g ref="char:EOLhyphen"/>gular Pyramis inscribed in the said Cube.</head>
               <p>FOr forasmuch as the side of the pyramis inscribed in the cube subte<g ref="char:cmbAbbrStroke">̄</g>deth two sides of the cube which containe a right angle, by the 1. of the fiuetenth: it is manifest, by the 47. of the first, that the side of the pyramis subte<g ref="char:cmbAbbrStroke">̄</g>ding the said sides, is in power duple to the side of the cube: Wherefore also the square of the side of the cube is the halfe of the square of the side of the pyramis. The side therefore of a cube containeth in power halfe the side of an equilater triangular pyra<g ref="char:EOLhyphen"/>mis inscribed in the said cube.</p>
            </div>
            <div type="part">
               <head>¶ The 7. Proposition. The side of a Pyramis is duple to the side of an Octohedron inscri<g ref="char:EOLhyphen"/>bed in it.</head>
               <p>FOrasmuch as by the 2. of the fiuetenth it was proued, that the side of the Octohedron in<g ref="char:EOLhyphen"/>scribed in a pyramis coupleth the midle sections of the sides of the pyramis. Wherefore the sides of the pyramis and of the Octohedron are parallels, by the Corollary of the 39. of the first: and therefore, by the Corollary of the 2. of the sixth, they subtend like trian<g ref="char:EOLhyphen"/>gles. Wherfore (by the 4. of the sixth) the side of the pyramis is double to the side of the Octohedron, namely, in the proportion of the sides. The side therefore of a pyramis is duple to the side of an Octohedron inscribed in it.</p>
            </div>
            <div type="part">
               <head>¶ The 8. Proposition. The side of a Cube is in power duple to the side of an Octohedron inscri<g ref="char:EOLhyphen"/>bed in it.</head>
               <p>IT was proued in the 3. of the fiuetenth, that the diameter of the Octohedron inscribed in the cube, coupleth the centres of the opposite bases of the cube. Wherefore the said diameter is equall to the side of the cube. But the same is also the diameter of the square made of the sides of the Octohedron, namely, is the diameter of the Sphere which containeth it, by the 14. of the thirtenth. Wherefore that diameter being equall to the side of the cube, is in power double to the side of that square, or to the side of the Octohedron inscribed in it, by the 47. of the first. The side therefore of a Cube, is in power duple to the side of an Octohedron inscribed in it: which was requi<g ref="char:EOLhyphen"/>red to be proued.</p>
            </div>
            <div type="part">
               <head>¶ The 9. Proposition. The side of a Dodecahedron, is the greater segment of the line which containeth in power halfe the side of the Pyramis inscribed in the sayd Dodecahedron.</head>
               <p>SVppose that of the Dodecahedron ABGD the side be AB: and let the base of the cube inscribed in the Dodecahedron be ECFH, by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> of the fiuetenth. And let the side of the pyramis inscribed in the cube be CH, by the 1. of the fiuetenth.<note place="margin">
                     <hi>Construction.</hi>
                  </note> Wherefore the same pyramis is inscribed in the Dodecahedron, by the 10. of the fiuetenth. Then I say, that AB the side of the Dodecahedron is the greater segment of the line which contai<g ref="char:EOLhyphen"/>neth in power halfe the line CH, which is the side of the pyramis inscribed in the
<pb n="448" facs="tcp:7062:477"/>
Dodecahedron.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For forasmuch as EC the side of the cube be<g ref="char:EOLhyphen"/>ing
<figure/>
diuided by an extreme and meane proportion maketh the greater segment the line AB, the side of the Dodecahedron, by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>rst Corollary of the 17. of the thirtenth: (For they are cont<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ined in one and the selfe same Sphere (by the first of this booke): and the line EC the side of the cube contayneth in power the halfe of the side CH, by the 6. of this booke. Wherefore AB the side of the Dodecahedron, is the greater segment of the line EC, which containeth in power the halfe of the line CH, which is the side of the Dodecahedron inscribed in the pyramis. The side therefore of a Dodecahedron, is the greater segment of the line which con<g ref="char:EOLhyphen"/>taineth in power halfe the side of the Pyramis inscribed in the said Dodecahedron.</p>
            </div>
            <div type="part">
               <head>¶The 10. Proposition. The side of an Icosahedron, is the meane proportionall betwene the side of the Cube circumscribed about the Icosahedron, and the side of the Dode<g ref="char:EOLhyphen"/>cahedron inscribed in the same Cube.</head>
               <p>SVppose that there be a cube ABFD, in which let there be inscribed an icosahedron CL<g ref="char:EOLhyphen"/>IGOR, by the 14. of the fiuetenth.<note place="margin">
                     <hi>Construction.</hi>
                  </note> Let also the Dodecahedron inscribed in the same be EDMNPS, by the 13. of the same. Now forasmuch as CL the side of the Icosahedron is the greater segme<g ref="char:cmbAbbrStroke">̄</g>t of AB the side of the cube circumscribed about it, by the 3. Corolla<g ref="char:EOLhyphen"/>ry of the 14. of the fiuetenth:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and the side ED of the
<figure/>
Dodecahedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in thesame cube is the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t of the same side AB of the cube, by the 2. Corollary of the 13. of the fiuetenth: it followeth that AB the side of the cube be<g ref="char:EOLhyphen"/>ing diuided by an extreme and meane proportion, maketh the greater segment CL the side of the Icosahedron inscribed in it, and the lesse segment ED the side of the Dodecahedron likewise inscrib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d in it. Wherefore as the whole line AB the side of the cube, is to the greater segment CL the side of the Icosahedron, so is the greater segment CL the side of the Icosahedron, to the lesse segment ED<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the side of the Dodecahedron, by the third definition of the sixth. Wherefore the side of an Icosahedron, is the meane proportionall betwene the side of the cube circum<g ref="char:EOLhyphen"/>scribed about the Icosahedron, and the side of the Dodecahe<g ref="char:EOLhyphen"/>dron inscribed in the same cube.</p>
            </div>
            <div type="part">
               <head>¶The 11. Proposition. The side of a Pyramis, is in power <note n="†" place="margin">
                     <hi>That is, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </hi> 18. <hi>to</hi> 1.</note> Octodecuple to the side of the cube in<g ref="char:EOLhyphen"/>scribed in it.</head>
               <p>FOr, by that which was demonstrated in the 18. of the fiuetenth, the side of the pyra<g ref="char:EOLhyphen"/>mis is triple to the diameter of the base of the cube inscribed in it:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and therefore it is in power nonecuple to the same diameter (by the 20. of the sixth). But the dia<g ref="char:EOLhyphen"/>mer is in power double to the side of the cube, by the 47. of the first. And the dou<g ref="char:EOLhyphen"/>ble of nonecuple maketh Octodecuple. Wherefore the side of the pyramis is in power Octodecuple to the side of the cube inscribed in it.</p>
            </div>
            <div type="part">
               <head>¶The 12. Proposition. The side of a Pyramis, is in power Octodecuple to that right line, whose
<pb facs="tcp:7062:478"/>
greater segment is the side of the Dodecahedron inscribed in the Pyramis.</head>
               <p>FOrasmuch as the Dodecahedron and the cube inscribed in it, are set in one and the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>lf<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> same pyramis, by the Corollary of the first of this booke: and the side of the pyramis cir<g ref="char:EOLhyphen"/>cumscribed about the cube is in power octodecuple to the side of the cube inscribed, by the former Proposition: but the greater segment of the selfe same side of the cube, is the side of the Dodecahedron which containeth the cube, by the Corollary of the 17. of the thirtenth. Wherfore the side of the pyramis is in power octodecuple to that right line, namely, to the side of the cube, whose greater segment is the side of the Dodecahedron inscribed in the pyramis.</p>
            </div>
            <div type="part">
               <head>¶ The 13. Proposition. The side of an Icosahedron inscribed in an Octohedron, is in power duple to the lesse segment of the side of the same Octohedron.</head>
               <p>FOrasmuch as in the 17. of the fiuetenth, it was proued, that the side of an Icosahedron inscri<g ref="char:EOLhyphen"/>bed in a pyramis, coupleth together the two sections (which are produced by an extreme and meane proportion) of the side of the Octohedron which make a right angle: and that right angle is contained vnder the lesse segmentes of the sides of the Octohedron, and is subtended of the side of the Icosahedron inscribed: it followeth therefore, that the side of the Icosahedron which subtendeth the right angle, being in power equall to the two lines which containe the said angle, by the 47. of the first, is in power duple to euery one of the lesse segme<g ref="char:cmbAbbrStroke">̄</g>tes of the side of the Octohedron which containe a right angle. Wherefore the side of an Icosahedron inscribed in an Octohedron, is in power duple to the lesse segment of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ide of the same Octohedron.</p>
            </div>
            <div type="part">
               <head>¶The 14. Proposition. The sides of the Octohedron, and of the Cube inscribed in it, are in power the one to the other <note n="†" place="margin">
                     <hi>That i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, as</hi> 9. <hi>to</hi> 2.</note> in quadrupla sesquialter proportion.</head>
               <p>SVppose that ABGDE be an Octohedron, and let the cube inscribed in it be FCHI. Then I say, that AB the side of the Octohedron, is in power quadruple sesquialter to FI the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ide of the cube. Let there be drawen to BE the base of the triangle ABE a perpendicular AN: and againe let there be drawen to the same base in the triangle G<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>E the perpendicular GN: which AN &amp; GN shall
<figure/>
passe by the centres F and I: and the line AF is duple to the line FN, by the Corollary of the 12. of the thirtenth. Wherfore the line AO is duple to the line OE, by the 2. of the sixth. For the lines FO and NE are parallels. And therefore the diameter AG is triple to the line FI. Wherfore the pow<g ref="char:EOLhyphen"/>er of AG is <note n="*" place="margin">
                     <hi>That is, as</hi> 18. <hi>to</hi> 2. <hi>or</hi> 9. <hi>to</hi> 1.</note> noncuple to the power of FI. But the line AG is in power duple to the side AB, by the 14. of the thirtenth. Wherefore the square of the line AB, being ing the halfe of the square of the line AG, which is noncuple to the square of the line FI, i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> quadruple sesqui<g ref="char:EOLhyphen"/>alter
<pb n="449" facs="tcp:7062:478"/>
to the square of the line FI. The sides therefore of the Octohed<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nd of the cube inscribed in it<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> are in power the one to the other, in quadruple sesquialter proportion.</p>
            </div>
            <div type="part">
               <head>¶The 1<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. Proposition. The side of the Octohedron, is in power quadruple sesquialter to that right line, whose greater segment is the side of the Dodecahedron inscribed in the same Octohedron.</head>
               <p>FOrasmuch as in the 14. of this booke, it was proued, that the side of the Octohedron is in power quadruple sesquialter to the side of the cube inscribed in it: but the side of the cube being cut by an extreme and meane proportion, maketh the greater segment the side of the Dodecahedron circumscribed about it, by the 3. Corollary of the 13. of the fiuetenth: there<g ref="char:EOLhyphen"/>fore the side of the Octohedron is in power quadruple sesquialter to that right line (namely, to the side of the cube) whose greater segment is the side of the Dodecahedron inscribed in the cube. But the Do<g ref="char:EOLhyphen"/>decahedron and the cube inscribed one within an other, ar<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> inscribed in one and the selfe same Octo<g ref="char:EOLhyphen"/>hedron, by the Corollary of the first of this booke. The side therefore of the Octohedron, is in power quadruple sesquialter to that right line, whose greater segment is the side of the Dodecahedron inscri<g ref="char:EOLhyphen"/>bed in the same Octohedron.</p>
            </div>
            <div type="part">
               <head>¶ The 16. Proposition. The side of an Icosahedron, is the greater segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Icosahedron.</head>
               <p>SVppose that there be an Icosahedron ABGDFHEC: whose side let be BG or <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let the Octohedron ins<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ib<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>d in it be AKD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>: and let the side therof be AL. Then I say, that the side <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>C is the greater segment of that right line which is in power duple to the side AL. For forasmuch as figures inscribed and circumscribed haue o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e &amp; the selfsame centre, by the Corollary of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>1. of the fiuetenth, let the same be the point I. Now right line<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> drawen by th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> to the midle sections of the opposite sides, namely, the lines AID and KIL, do in the point I <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ut <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap> the other in<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>
                  <figure/>
two <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>quall <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap>, and perpendicularly, by the Corolla<g ref="char:EOLhyphen"/>ry of the 14. of the fiuetenth: and forasmuch as they couple the midle sections of the opposite lines BG and HF, ther<g ref="char:EOLhyphen"/>fore they cut them perpendiularly:<note place="margin">
                     <hi>Draw in the figure a line from B to H.</hi>
                  </note> wherefore also the lines BG <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>, are parallels, by the 4. Corollary of the 14. of the <gap reason="illegible" resp="#PDCC" extent="1 span">
                     <desc>〈…〉</desc>
                  </gap>. Now then draw a line from B to H: and the sayd <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ne BH shall be equall and parallel to the line KL, by the 33. of the first. But the line BH subtendeth <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> sides of the pentagon which is composed of the sides of the Icosahedron, namely, the sides BA and AH: Wherfore the line BH being cut by an extreme and meane proporti<g ref="char:EOLhyphen"/>on maketh the greater segment the side of the pentagon, by the 8. of the thirtenth: which side is also the side of the Ico<g ref="char:EOLhyphen"/>sahedron, namely, EC. And vnto the line BH the line KL<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is equall: and the line KL is in power duple to AL the side of the Octohedron, by the 47. of the first: for in the square AKDL the angle KAL is a right angle. Wherefore EC the side of the Icosahedron, is the greater seg<g ref="char:EOLhyphen"/>ment of the line BH or KL, which is in power duple to AL <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>he side of the Octohedron inscribed in the Icosahedron. Wherefore the side of an Icosahedron, is the greater segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Icosahedron.</p>
            </div>
            <div type="part">
               <head>¶The 17. Proposition. The side of a Cube is to the side of a Dodecahedron inscribed in it, in duple proportion of an extreame and meane proportion.</head>
               <p>
                  <pb facs="tcp:7062:479"/>FOr it was manifes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. corollary of the 13. of the fiuetenth, that the side of a cube diui<g ref="char:EOLhyphen"/>ded by an extreame and meane pr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>portion, maketh the lesse segment, the side of the dode<g ref="char:EOLhyphen"/>cahedron inscribed in it: but the whole is to the lesse segment in duple proportion of that in which it is to the greater, by the 10. diffinitio<g ref="char:cmbAbbrStroke">̄</g> of the fifth. For the whole, the greater segme<g ref="char:cmbAbbrStroke">̄</g>t, and the lesse, are lines in continuall proportion, by the 3. diffinition of the sixth. Wherefore the whole namely the side of the cube, is to the side of the dodecahedron inscribed in it, namely, to his lesse seg<g ref="char:EOLhyphen"/>ment, in duple propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tion of an extreame and meane proportion', namely, <note n="*" place="margin">
                     <hi>What the duple of an extreme and meane pro<g ref="char:EOLhyphen"/>portion is.</hi>
                  </note> of that which the whole hath <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the greater segmen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, by the 2. of the fourtenth.</p>
            </div>
            <div type="part">
               <head>¶ The 18. Proposition. The side of a Dodecahedron is, to the side of a Cube inscribed in it, in con<g ref="char:EOLhyphen"/>uerse proportion of an extreame and meane proportion.</head>
               <p>IT was proued in the 3. corollary of the 13. of the fiuetenth, that the side of a Dodecahe<g ref="char:EOLhyphen"/>d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>on circumscribed about a Cube, is the greater segment of the side of the same Cube. Wherefore the whole side of the Cube inscribed is to the greater segment, namely, to the side of the dodecahedron circumscribed, in an extreame and meane proportion: wherefore by conuersion, the greater segment, that is, the side of the dodecahedron, is to the whole, namely, to the side of the Cube inscribed, in the conuerse proportion of an extreame and meane proportion, by the 13. diffinition of the fiueth.</p>
            </div>
            <div type="part">
               <head>¶ The 19. Proposition. The side of an Octohedron, is sesquialter to the side of a Pyramis inscri<g ref="char:EOLhyphen"/>bed in it.</head>
               <p>FOr (by the corollary of the 14. of the thirtenth) the Octohedron is cu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>te into two quadrilater py<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>amids, one of which let be ABGDF:<note place="margin">
                     <hi>Construction.</hi>
                  </note> and let the centres of the cir<g ref="char:EOLhyphen"/>cles which contayne the 4. bases of the Octohedron be K, E, I, C. And dr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>w these right lines KE, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>I, IC, CK, and EC. Wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>IC is a square, and one of the bases of the cube inscribed in the Octohedron, by the 4. of the fiuetenth. And foras<g ref="char:EOLhyphen"/>much as the angles of a cube and of the pyramis in<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>cribed in it, are for in the centres of the bases of the Octohedron circumscribed about the cube, by the 6<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the fiue<g ref="char:EOLhyphen"/>tenth: and the side of the pyramis coupleth the opposite angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the base of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cube, by the 1. of the fiuetenth, it is manifest that
<figure/>
the line EC is the side of the pyramis inscribed in the Octohedron ABGDF. Then I say that GD the side of the Octohedron, is sesquialter to EC the side of the pyramis inscribed in it. From the poynt A draw to the bases BG and FD perpendi<g ref="char:EOLhyphen"/>culars AN and AM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which (by the corollary of the 12. of the thirtenth) shall passe by the centres E and C.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> And draw the line NM. Now forasmuch, as B<g ref="char:EOLhyphen"/>GDF is a square, by the 14. of the thirtenth, the lines NG and MD shall be parallels and equall. For the lines BG and FD are by the perpendicu<g ref="char:EOLhyphen"/>lars cutte into two equall partes in the poyntes N and M (by the 3. of the third). Wherefore the lines NM and GD shall be parallels and equall, by the 33. of the first. And forasmuch as the lines AN and AM which are the perpendiculars of equall and like<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> triangles are c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>t a like in the poyntes <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and C, the lines EC and NM<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shall be parallels, by the 2. of the sixth: and therefore by the corollary of the same, the triangles AEC, and ANM shall be like. Wherefore as the line AN is to the line AE, so is the line NM to the line EC by the 4. of the sixth. But the line AN is sesquialter to the line AE, for the line AE is duple to the line EN, by the corollary of the 12<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the thirtenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> wherefore the line NM, or the line GD which is equall vnto it, is sesquialter to the line EC. Wherefore GD the side of the Octohedron, is sesquialter to EC the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the pyra<g ref="char:EOLhyphen"/>mis inscribed in it.</p>
            </div>
            <div type="part">
               <pb n="450" facs="tcp:7062:479"/>
               <head>¶ The <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>0. Proposition. If from the power of the diameter of an Icosahedron, be taken away the power tripled of the side of the cube inscribed in the Icosahedron: the power remayning shall be sesquitertia to the power of the side of the I<g ref="char:EOLhyphen"/>cosahedron.</head>
               <p>LEt there be taken an Icosahedron ABGD: and l<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> two bases of the cube inscribed in it, ioyned together be EHKL and LKFC: and let the diameter of the cube be FH and the side be EH, and let the diameter of the Icos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>dron be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>G, and the side be AB. Then I say, that if from the power of the diamet<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> GB, be taken away the power tripled of EH the side of the cube:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> the power remayning, shall be sesquetertia to the power of AB the side of the Icosahedron. For forasmuch as the centres of inscribed and circum<g ref="char:EOLhyphen"/>scribed figures, are in one &amp; the selfe same poynt, by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>rollary of the 21. of the <gap reason="illegible" resp="#PDCC" extent="1 word">
                     <desc>〈◊〉</desc>
                  </gap> the diame<g ref="char:EOLhyphen"/>ters BG and FH shall in one and the selfe same poynt
<figure/>
cutte the one the other into two equall partes: for we haue before by the same corollary taught, that the toppes of equall and like pyramids doo in that poynt concurre, let that poynt be the centre I. Now the an<g ref="char:EOLhyphen"/>gles of the cube, which are at the poyntes F and H are set at the centres of the bases of the Icosahedron by the 11. of the fiuetenth<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore the line FH shall be perpendicular to both the bases of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, by the corollary of the assu<g ref="char:cmbAbbrStroke">̄</g>pt of the 16. of the twelfth. Wherefore the line IB contayneth in power the two lines IH and HB, by the 47. of the first. But the line H<g ref="char:EOLhyphen"/>B, is drawne from the centre of the circle which con<g ref="char:EOLhyphen"/>tayneth the base of the Icosahedron namely, the angle B is placed in the circumference, and the poynt H is the centre. Wh<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>refore the whole line BG contayneth in power the whole lines FH and the diameter of the circle (namely, the double of the line BH) by the 15<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the fiueth. But the diameter which is double to the line HB is in power sesquiterti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the side of the e<g ref="char:EOLhyphen"/>quilater triangle inscribed in the same circle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the corollary of the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. of the thirtenth. For it is in pro<g ref="char:EOLhyphen"/>portion to the side<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> as the side is to the perpendicular, by the corollary of the 8. of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ixth. And FH the diameter of the cube, is in power triple to EH the side of the same cube, by the 15<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the thirtenth. If therefore from the power of the diameter BG, be taken away the power tripled of EH the side of the cube inscribed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> that is<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the power of the line FH: the residue (namely, the power of the diameter of the circle which is duple to the line HB shall be sesquiterti<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the side of the triangl<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> inscribed in that circle: which selfe side is AB the side of the Icosahedron. If therfor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> from the power of the diameter of an Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, be take<g ref="char:cmbAbbrStroke">̄</g> away the power tripled of the side of the cube inscribed in the Icosahedron, the power remayning shall be s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>squitertia <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the power of the side of the Icosahedron.</p>
               <div type="section">
                  <head>A Corollary.</head>
                  <p>
                     <hi>The diameter of the Icosahedron, contayneth in power two lines, namely, the diameter of the cube inscribed, which coupleth the centres of the opposite ba<g ref="char:EOLhyphen"/>ses, and the diameter of the circle which contayneth the base of the Icosahedron.</hi> For it was manifest, that BG the diameter contayneth to power the line FH which doupleth the centres, and the double of the line BH, that is, the diameter of the circle contayning the bas<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> wherein i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> the centre H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>
                  </p>
               </div>
            </div>
            <div type="part">
               <head>¶ The 21. Proposition. The side of a Dodeca<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>edron is the lesse segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Do<g ref="char:EOLhyphen"/>decahedron.</head>
               <p>
                  <pb facs="tcp:7062:480"/>LEt ther be taken a Dodecahedron AB<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>DCT, one of whose sides let be AB. And let the Octohedron inscribed in the Dodecahedron be EFLKI: one of whose sides let be EF. Then I say that AB the side of the Dodecahedron, is the lesse seg<g ref="char:EOLhyphen"/>ment of a certayne right <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ine <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> cut
<figure/>
by an extreame and meane pro<g ref="char:EOLhyphen"/>portion) which is in power duple to EF the side of the Octohedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in the Dodecahedron. Draw the diameters EL and FK of the Octohedron. Now they couple the midle sections of the opposite sides of the dode<g ref="char:EOLhyphen"/>cahedron AB and GD, (by the 9. of the fiuetenth, &amp; 3. corollary of the 17. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th) &amp; euery one of those diameters being diuided by an extreame and meane proportion, doo make the lesse segment, the side of the dodecahedron, by the 4. corollary of the same. Wherefore the side AB is the lesse segment of the line FK. But the line FK contayneth in power the two equall lines EF &amp; EK, by the 47. of the first: for the angle FEK is a right angle of the square FEKL of the Octohedron. Wherfore the line FK is in pow<g ref="char:EOLhyphen"/>er duple to the line EF. Wherefore the line AB (the side of the dodecahedron) is the lesse segment of the line FK, which is in power duple to EF the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the Octohedron. The side therefore of a Dodecahedron i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the lesse segment of that right line, which is in power duple to the side of the Octohedron inscribed in the same Dod<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>cahedron.</p>
            </div>
            <div type="part">
               <head>¶ The 22. Proposition. The diameter of an Icosahedron is in power sesquitertia to the side of the same Icosahedron, and also is in power sesquialter to the side of the Pyra<g ref="char:EOLhyphen"/>mis inscribed in the Icosahedron.</head>
               <p>FOr forasmuch as it hath bene proued (by the 10. of this booke) that if fro<g ref="char:cmbAbbrStroke">̄</g> the power of the diameter of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g> be taken away the triple of the power of the side of the cube inscribed in it, there shalbe left a square sesquitertia to the square of the side of the Icosahedron: But the power of the side of the cube tripled, is the diameter of the same cube, by the 15. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th: And the cube, &amp; the pyramis inscribed in it are contai<g ref="char:EOLhyphen"/>ned in one &amp; the self same sphere, by the first of this booke, and in one &amp; the self same Icosahedron by the corollary of the same. Wherfore one and the selfe same diameter of the cube, or of the sphere which co<g ref="char:cmbAbbrStroke">̄</g>taineth the cube and the pyramis, is in power sesquialter to the side of the pyramis by the 13. of the thirtenth. Wherfore it followeth, that if from the diameter of the Icosahedron, be ta<g ref="char:EOLhyphen"/>ken away the triple power of the side of the cube, or the sesquialter power of the side of the pyramis, which are the powers of one and the selfe same diameter, there shall be left the sesquitertia power of the side of the Icosahedron. The diameter therefore of an Icosahedron is in power sesquitertia to the side of the same Icosahedron, and also is in power sesquialter to the side of the Pyramis inscribed in the Icosahedron.</p>
            </div>
            <div type="part">
               <head>The 23. Proposition. The side of a Dodecahedron is to the side of an Icosahedron inscribed in it, as the lesse segment of the perpendicular of the Pentago<g ref="char:cmbAbbrStroke">̄</g>, is to that line which is drawne from the centre to the side of the same pentagon.</head>
               <p>
                  <note place="margin">
                     <hi>Constrution.</hi>
                  </note>LEt there be taken a Dodecahedron ABGDFSO. Whose side let be AS or SO: and let the Icosahedron inscribed in it be KLNMNE, whose side let be KL. From the two angles of the pentagon<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BAS and FAS of the Dodecahedron, namely, from the angle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> 
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and F, let there be drawne to the common base AS perpendicular lines BC &amp; FC: which shal passe by the centres K &amp; L of the sayd pentagons, by the corollary of the 10. of the thirte<g ref="char:cmbAbbrStroke">̄</g>th. Draw
<pb n="451" facs="tcp:7062:480"/>
the lines BF and RO. Now forasmuche as the
<figure/>
line RO subtendeth the angle OFR of th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> pentagon of the dodecahedron, it shall cut the line FC by an extreme and meane proportion, by the 3. of this booke, let it cut it in the poynt I. And forasmuche as the line KL is the side of the Icosahedron inscribed in the Dodecahe<g ref="char:EOLhyphen"/>dron, it coupleth the ce<g ref="char:cmbAbbrStroke">̄</g>tres of the bases of the dodecahedron: for the angles of the Icosahe<g ref="char:EOLhyphen"/>dron are set in the centres of the bases of the dodecahedron, by the 7. of the fiuetenth. Now I say that SO, the side of the dodecahedron is to KL the side of the Icosahedron, as the lesse segment IF of the perpendicular line CF, is to the line LC which is drawne from the centre L to AS the side of the pentagon. For foras<g ref="char:EOLhyphen"/>much as in the triangle BCF the two sides CB and CF are in the centres L and K cut like proportionally,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> the lines BF and KL shalbe pa<g ref="char:EOLhyphen"/>rellels, by the 2. of the sixth. Wherefore the tri<g ref="char:EOLhyphen"/>angles BCF, and KCL shall be equiangle, by the corollary of the same. Wherfore as the line CL is to the line KL<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> so is the line CF to the line BF, by the 4. of the sixth. But CF maketh the lesse segment the line IF, by the 3. of this booke, and the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BF maketh the lesse segment the line SO, namely, the side of the Dodecahedron, by the 2. corollary of the 13. of the fiuetenth. For the line BF which coupleth the angles B and F of the bases of the dodeca<g ref="char:EOLhyphen"/>hedron, is equall to the side of the cube, which contayneth the dodecahedron, (by the .13. of the fiuetenth). Wherefore as the whole line C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, is to the whole line BF, so is the lesse segment IF to the lesse segment SO (by the 2. of the 14). But as the line CF is to the line BF, so is the line CL proued to be to the line KL. Wherefore as the line IF is to the line SO, so is the line CL to the line KL. Wherefore alternately by the 16. of the fiueth, as the line IF the lesse segment of the per<g ref="char:EOLhyphen"/>pendicular of the pentagon FAS, is to the line LC which is drawne from the centre of the pentagon, to the base, so is the line SO the side of the Dodecahedron to th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> line KL the side of the Icosahe<g ref="char:EOLhyphen"/>dron inscribed in it. The side therfore of a Dodecahedron is to the side of an Icosahedron inscribed in it, as the lesse segment of the perpendicular of the pentagon, is to that line which is drawne from the cen<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re to the side of the same pentagon.</p>
            </div>
            <div type="part">
               <head>¶ The 24. Proposition. If halfe of the side of an Icosahedron be deuided by an extreme &amp; meane proportion: and if the lesse segment thereof be taken away from the whole side, and againe from the residue be taken away the third part: that which remaineth shall be equal to the side of the Dodecahedron inscribed in the same Icosahedron.</head>
               <p>SVppose that ABGDF be a
<figure/>
pentagon,<note place="margin">
                     <hi>Construction.</hi>
                  </note> containing fiue sides of the Icosahedron by the 16. of the thirtenth, and let it be inscribed in a circle, whose centre let be the point E. And vpon the sides of the pentagon, let there be rea<g ref="char:EOLhyphen"/>red vp triangles, making a solide angle of the Icosahedron at the poynt I, by the 16. of the thirtenth. And in the circle ABD, inscribe an equilater triangle AHK. From the centre E drawe to HK the side of the triangle, and GD the side of the penta<g ref="char:EOLhyphen"/>gon, a perpendicular line, which let be E<g ref="char:EOLhyphen"/>CNM. And draw these right lines EG, ED, IG and ID. And deuide the line BG into two equal parts in the poynt T. And
<pb facs="tcp:7062:481"/>
drawe these lines IN, IT, TN, ET. And
<figure/>
forasmuche as in the perpendiculars IT &amp; IN are the centres of the circles which containe the equilater triangles IBG, &amp; IGD, by the corollarye of the first of the thirde. Let those centres be the points S and O. And draw the line SO. Deuide the line TB the half of BG the side of the Icosahedron by an extreme and meane proportion in the poynt R, by the 30. of the sixth, and let the lesse segment therof be RB. And forasmuch as the line SO cou<g ref="char:EOLunhyphen"/>pleth the centres of the triangles IBG, &amp; IGD, it is by the 5. of the fiuetenth, the side of the Dodecahedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in the Icosahedron, whose side is the line BG. From the side BG take away <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>R the lesse segment of the halfe side. And from the residue GR take away the third part GV (by the 9. of the sixth.) Then I say that the residue RV is equal to SO the side of the Dodecahedron inscribed.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For forasmuch as the perpendicular EN is in the poynt C deuided by an extreme and meane proportion, by the corollary of the first of the fourtenth, and the greater segment therof is the line EC, and vnto the line EC the line CM is equal, by the corollary of the 12. of the thirtenth: wherefore the line EC is to the line CN, as the line CM is to the same line CN, by the 7. of the fiueth. But as the line EC is to the line CN, so is the whole line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>N, to the greater segment E<g ref="char:EOLhyphen"/>C, by the 3. diffinition of the sixth. Wherefore (by the 11. of the fiueth), as the whole line EN is to the greater segment EC, so is the line CM to the line CN. Wherefore the line CM, is deuided by an extreme and meane proportion in the poynt N, namely, is deuided like vnto the line EN, by the 2. of the fourtenth. Wherfore the line EM excedeth the line EN by the lesse segment of his halfe, namely, by MN. And forasmuche as EGD is the triangle of an equilater and equiangle pentagon ABGDF, and ETN is likewise the triangle of the like pentagon inscribed in the pentagon ABGDF: Therefore by the 20. of the sixth, the triangle ETN is like to the triangle EGD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Wherefore as the line EG is to the line EN, so by the 4. of the sixth, is the line GD to the line NT. Wherefore the line GD (or BG which is equal vnto it) excedeth the line NT by the lesse segment of the halfe of BG. For the line EG did in like sort excede the line EN. But that lesse segment is the line BR. Wherefore the residue RG is equal to the line TN. And forasmuch as IBG is an equilater triangle: the perpendicular ST shalbe the halfe of the line SI which is drawne from the centre, by the corollary of the 12. of the thirtenth: wher<g ref="char:EOLhyphen"/>fore the line IT excedeth the line IS by his third part. And forasmuche as the line SO which coupleth the sections, is a parallel to the line TN, by the 2. of the sixth. For the equal perpendiculars IT, and IN are cut like in the poynts S &amp; O: therfore the triangles ITN &amp; I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>O, are like by the corollary of the second of the sixth. Wherfore as the line IT is to the line IS, so by the 4. of the sixth is the line TN to the line SO. But the line IT excedeth the line IS by a third part: wherfore the line TN, excedeth the line SO by a third part: but the line TN is proued equal to the line RG. Wherfore the line RG exce<g ref="char:EOLhyphen"/>deth the line SO by a third part of himself, which is GV. Wherfore the residue RV, is equal to the line SO, which is the side of the dodecahedron inscribed in the Icosahedron, whose side is the line BG. If therfore halfe of the side of an Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, be deuided by an extreme &amp; meane proportion: and if the lesse segment therof be taken away from the whole side, and againe from the residue be take<g ref="char:cmbAbbrStroke">̄</g> away the third part: that which remaineth shall be equal to the side of the dodecahedron inscribed in the same Icosahedron.</p>
            </div>
            <div type="part">
               <head>The 25. Proposition. To proue that a cube geuen, is to a trilater equilater pyramis inscribed in it, triple.</head>
               <p>SVppose that the cube geuen, be ABCH: and let the pyramis inscribed in it be AGDF. Then I say that the cube ABCH is triple to the pyramis AGDF. For forasmuche as the base AFD is common to the pyramis AFDB and AFDG, the pyramis AFDB shalbe set without the pyramis AFDG. Likewise the rest of the bases of the inscribed pyramis are common to the rest of the pyramids sorte without: which are these: the pyramis AGDC vppon
<pb n="452" facs="tcp:7062:481"/>
the base AGD: the pyramis AGF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> vpon the base AGF<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>
                  <figure/>
and the pyramis GDFH vpon the base GDF. Which py<g ref="char:EOLhyphen"/>ramids taken without, are foure in number, equal and like the one to the other, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. diffinition of the eleu<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>th. For euery one of them is contained vnder thr<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> halfe squares of the cube, and one of the bases of the pyramis inscribed. Wherfore euery one of the<g ref="char:cmbAbbrStroke">̄</g> is co<g ref="char:cmbAbbrStroke">̄</g>tained vnder the halfe base of the cube, &amp; the altitude of the cube. As the pyramis AL<g ref="char:EOLhyphen"/>GF, hath to his base halfe of the square EH, namely, the tri<g ref="char:EOLhyphen"/>angle EGF, &amp; hath to his altitude, the altitude of the cube, namely, the line AE. Wherfore the sayd pyramis is the sixth part of the cube. For if the cube be deuided into two pris<g ref="char:EOLhyphen"/>mes, by the plaine CBFG, the prisme ACBGEF, shalbe tri<g ref="char:EOLhyphen"/>ple to the pyramis AEGF, hauing one &amp; the selfe same base with it EGF, and one and the selfe same altitude EA, by the first corollary of the 7. of the twelueth. Wherefore the sayd outward pyramis AEGF is the sixth part of the whole cube. Wherfore also the same pyramis together wyth the other thre outwarde pyramids AFDB, AGDC, and GDFH, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hal containe two third partes of the cube. Wherfore the resi<g ref="char:EOLhyphen"/>due, namely, the pyramis inscribed AGDF, shal contain one third part of the cube. And therefore conuersedly the cube shall be triple to it: wherefore we haue proued <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hat a cube geue<g ref="char:cmbAbbrStroke">̄</g> triple to a trilater &amp; equilater pyramis inscribed in it.</p>
            </div>
            <div type="part">
               <head>¶The 26. Proposition. To proue that a trilater equilater Pyramis is duple to an Octohedron in<g ref="char:EOLhyphen"/>scribed in it.</head>
               <p>LEt there be taken a trilater Pyramis ABCD: whose sixe sides let be cut into two e<g ref="char:EOLhyphen"/>quall partes, in the pointes E, K, F, L, G, and H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> inscribing thereby an Octohedron in the pyramis, by the 2. of the fiuetenth. Wherefore the pyramids AEGH, BEFK, CFGL, &amp; DKHL, fall without the Octohedron inscribed, by the same second of the fiuetenth. But the outward Pyramids (namely, AEGH, and the three other) are like vnto the whole pyramis, by the 7. definition of the eleuenth. For the bases of the whole pyramis are by parallel lines drawen in them cut into like triangles, by the Corollary of the 2. of the sixth, of which the foresayd pyra<g ref="char:EOLhyphen"/>mids
<figure/>
are made. Wherefore the whole pyramis is to euery one of them in treble proportion of that in which the sides of like proportion are, by the 8. of the twelfth. But by construction, the proportion of the side A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> to the side A<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is duple. Wherefore the whole pyramis ABCD is octuple to the pyramis AEGH, and so is it to euery one of the pyramids which are equall to AEGH. For duple proportion multiplyed into it selfe twise maketh octuple. Wherefore it followeth that the 4. pyramids AEGH, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>EFK, CFGL, and DKHL, taken together, make the halfe of the whole pyramis ABCD. Where<g ref="char:EOLhyphen"/>fore the residue, namely, the Octohedron EG<g ref="char:EOLhyphen"/>LKHF, is the other half of the pyramis. Wher<g ref="char:EOLhyphen"/>fore the pyramis is duple to the Octohedron. Wherefore we haue proued that a trilater e<g ref="char:EOLhyphen"/>quilater pyramis is duple to an Octohedron in<g ref="char:EOLhyphen"/>scribed in it.</p>
            </div>
            <div type="part">
               <head>¶ The 27. Proposition. To proue that a Cube is sextuple to an Octohedron inscribed in it.</head>
               <p>
                  <pb facs="tcp:7062:482"/>LEt there be taken a cube ABCD, EFGH: whose 4. standing lines AE, BF, CH, &amp; DG, let be cut into two equall partes in the pointes I, K, M, L: and<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by those pointes let there be extended a plaine KLMI: which shall be a square, and parallel to the squares BC &amp; FH, by the 15. of the eleuenth. Wherefore in it shall be the base which is common to the two pyramids of the Octohedron inscribed in the cube, by that which was demonstrated in the third of the fiuetenth. Let that base be NPRQ, coupling
<figure/>
the centres of the bases of the cube: and vpon that base let be set the two pyramids of the Octohedron, which let be NPQRS, and NPQRT. And forasmuch as those two pyramids taken together, haue their altitude equall with the altitude of the whole cube,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> ech of them a part hath to his alt<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ude halfe the altitude of the cube, namely, halfe of the side of the cube, as the line KB. And forasmuch as the square KLMI is double to the square NRQP, by the 47. of the first: the other squares of the cube shall also be dou<g ref="char:EOLhyphen"/>ble to the square NRQP. And forasmuch as the cube, as it was manifest by the last of the fiuetenth, is resolued into sixe pyramids, whose bases are the bases of the cube, &amp; the altitudes the lines drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centre to the bases, which are equall to halfe the side of the cube<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> it followeth that e<g ref="char:EOLhyphen"/>uery one of the sixe pyramids of the cube, hauing his base double to the base of eche of the pyramids of the Octohe<g ref="char:EOLhyphen"/>dron, and the selfe same altitude that the said pyramids of the Octohedro<g ref="char:cmbAbbrStroke">̄</g> haue, is double to either of the pyramids of the octohedro<g ref="char:cmbAbbrStroke">̄</g>, by the 6. of the twelfth. And forasmuch as euery one of the pyramids of the cube is equall to the two pyramids of the Octohedron, the sixe pyramids of the cube shall be sextuple to the whole Octohedron. Wherefore it is manifest, that a cube is sextuple to an Octohedron inscri<g ref="char:EOLhyphen"/>bed in it.</p>
            </div>
            <div type="part">
               <head>¶The 28. Proposition. To proue that an Octohedron is quadruple sesquialter to a Cube inscri<g ref="char:EOLhyphen"/>bed in it.</head>
               <p>SVppose that the Octohedron geuen be ABCDEF: and let the cube inscribed in it be GHIK, VQRS. Then I say, that the Octohedron is quadruple sesquialter to the cube inscribed in it. Forasmuch as the lines drawen from the centre of the Octohedron, or of the Sphere which containeth it, vnto the centres of the bases of the Octohedron, are pro<g ref="char:EOLhyphen"/>ued equall, by the 21. of the fiuetenth: and the angles of the cube are set in the centres of those bases, by the 4. of the fiuetenth: it followeth, that the selfe same right lines are drawen from one and the selfe same centre of the cube and of the Octohedron: for they haue eche one and the selfe same centre, by the Corollary of the 21. of the fiuetenth. Let that centre be the point T. Wherefore the base BDFC, which cutteth the Octohedron into two equall and quadrilater pyra<g ref="char:EOLhyphen"/>mids, by the Corollary of the 14. of the thirtenth, shall also cut the cube into two equall partes, by the Corollary of the 39. of the eleuenth. For it passeth by the centre T, by that which was demonstra<g ref="char:EOLhyphen"/>ted in the 14. of the thirtenth. And forasmuch as the base of the cube is in the 4. centres G, H, I, K, of the bases of the pyramis ABDFC, a plaine LNOM, extended by those pointes, shall be parallel to the plaine BDFC, by that which was demonstrated in the 4. of the fiuetenth, and shall cut the pyra<g ref="char:EOLhyphen"/>mis in the pointes L, N, O, M: and the lines LN, BD, and NO, DF, shall be parallels, so also shall the lines OM, FC, and LM, BC: and the square GHIK of the cube shall be inscribed in the square LNOM, by the same. Wherefore the square LNOM is duple to the square GHIK, by the 47. of the first. From the solide angle A, let there be drawen to the plaine superficies BDFC, a perpen<g ref="char:EOLhyphen"/>dicular, which let fall vppon it in the point T, and let the same perpendicular be AT, cutting the plaine LNOM in the point P. And it shall also be a perpendicular to the plaine LNOM, by the Corollary of the 14. of the eleuenth. Againe from the angle BAD of the triangle ADB, let there be drawen by the centre H of the triangle, to the base a line AHX. Wherefore the line AX is ses<g ref="char:EOLhyphen"/>quialter to the line AH, by the Corollary of the 12. of the thirtenth. Wherefore the line AH is duple to the line HX. But the other lines AB, AD, AF, AC, and the perpendicular APT, are cut like vnto the line AHX, by the <hi>17.</hi> of the eleuenth: Wherefore the line AP is double to the line PT. Wherefore the line AP is the altitude of the cube, for the line PT is the halfe thereof.
<pb n="453" facs="tcp:7062:482"/>
And forasmuch as vpon the base
<figure/>
GHIK of the cube, and vnder the altitude AP of the same cube, is set the pyramis AGHIK: the said pyramis is the third part of the cube, by the Corollary of the 7. of the twelfth. But vnto the pyramis AGHIK the pyramis ALNOM is duple, by the 6. of the twelfth, for the base of the one is double to the base of the other. Wherefore the pyramis ALNOM is two third partes of the cube. And forasmuch as the pyramids ALNOM, and ABDFC, are like, by the 7. defi<g ref="char:EOLhyphen"/>nition of the eleuenth: therefore they are in triple proportion of that in which the sides of like pro<g ref="char:EOLhyphen"/>portion AH to AX, or AL to AB, are, by the Corollary of the 8. of the twelfth. But the side AB is proued to be sesquialter to the side AL. Wherefore the pyramis A<g ref="char:EOLhyphen"/>BCDF is to pyramis ALN<g ref="char:EOLhyphen"/>OM, as 27. is to 8. (that is, in ses<g ref="char:EOLhyphen"/>quialter proportion tripled: for the quantitie or denomination of sesquialter proportion, namely, 1 ½ multiplied into it selfe once maketh 2¼, which againe multiplyed by 1½ maketh 3 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, that is, 27. to 8.). But of what partes the pyramis ALNOM containeth 8, of the same the cube con<g ref="char:EOLhyphen"/>taineth 12: namely, is sesquialter to the pyramis. Wherefore of what partes the cube containeth 12, of the same the whole Octohedron (which is double to the pyramis ABDFC) containeth 54. Which 54. hath to 12. quadruple sesquialter proportion. Wherefore the whole Octohedron is to the cube in<g ref="char:EOLhyphen"/>scribed in it, in quadruple sesquialter proportion. Wherefore we haue proued that an Octohedron ge<g ref="char:EOLhyphen"/>uen is quadruple sesquialter to a cube inscribed in it.</p>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <hi>An Octohedron is to a cube inscribed in it, in that proportion that the squares of their sides are.</hi> For by the 14. of this booke, the side of the Octohedron is in power quadruple sesquialter to the side of the cube inscribed in it.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 29. Proposition. To proue that an octohedro<g ref="char:cmbAbbrStroke">̄</g> geue<g ref="char:cmbAbbrStroke">̄</g>, is <note n="*" place="margin">
                     <hi>That is at</hi> 13. 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> 
                     <hi>is to</hi> 
                     <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>.</note> tre<g ref="char:EOLhyphen"/>decuple sesquialter to a trilater equila<g ref="char:EOLhyphen"/>ter pyramis inscribed in it.</head>
               <p>LEt the octohedron ge<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>en, be AB: in which let
<figure/>
there be inscribed a cube FCED, by the 4. of the fiuetenth, and in the cube let there be inscribed a pyramis FEGD, by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>. of the fiuetenth. And forasmuche as the angles of the pyramis are (by the same first of the fiuetenth) set in the angles of the cube: and the angles of the cube are set in the centres of the bases of the Octohedron, namely, in the poyntes F, E, C, D, G by the 4. of the fiuetenth. Wherfore the angles of the pyramis, are set in the centres F, C, E, D of the octohedron. Where<g ref="char:EOLhyphen"/>fore the pyramis FEDG is inscribed in the octohedron (by the 6. of the fiuetenth.) And forasmuche as the octohe<g ref="char:EOLhyphen"/>dron
<pb facs="tcp:7062:483"/>
AB is to the cube FCED, inscribed in it quadruple
<figure/>
sesquialter (by the former propositio<g ref="char:cmbAbbrStroke">̄</g>): and the cube CDEF is to the pyramis FEDG inscribed in it triple, by the 25. of booke: wherefore three magnitudes being geuen, namely, the octohedron, the cube and the pyramis, the proportion of the extremes (namely, of the octohedron to the piramis) is made of the proportions of the meanes, (namely, of the octohedron to the cube, and of the cube to the pyramis,) as it is easie to see by the declaration vpon the 10. diffinition of the fiueth. Now then multiplying the quantities or denomi<g ref="char:EOLunhyphen"/>nations of the proportions (namely, of the octohedron to the cube which is 4 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, and of the cube to the pyramis, which is 3) as was taught in the diffinition of the sixth, there shalbe produced 13 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, namely, the proportion of the octo<g ref="char:EOLunhyphen"/>hedron to the pyramis inscribed in it. For 4 ½, multiplyed by 3. produce 13 ½. Wherefore the Octohedron is to the pyramis inscribed in it in tredecuple sesquialter proportion. Wherefore we haue proued that an Octohedron is to a tri<g ref="char:EOLhyphen"/>later equilater pyramis inscribed in it, in tredecuple sesqui<g ref="char:EOLhyphen"/>alter proportion.</p>
            </div>
            <div type="part">
               <head>¶ The 30. Proposition. To proue that a trilater equilater Pyramis, is noncuple to a cube inscribed in it.</head>
               <p>SVppose that the pyramis geuen, be ABCD, whose two bases let be ABC, and DBC, and let their centres be the poynts G and I. And from the angle A, draw vnto the base B<g ref="char:EOLhyphen"/>C a perpendicular AE: likewise from the angle D draw vnto the same base BC, a per<g ref="char:EOLhyphen"/>pendicular DE: and they shal concurre in the section E by the 3. of the third and in them shalbe the ce<g ref="char:cmbAbbrStroke">̄</g>tres G and I, by the corollary of the first of the third. And forasmuch as the line AD is the side of the pyramis, the same AD shall be the diameter of the base of the cube which co<g ref="char:cmbAbbrStroke">̄</g>taineth the pyramis, by the 1 of the fiuete<g ref="char:cmbAbbrStroke">̄</g>th.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note>
                  <figure/>
Draw the line GI. And forasmuch as the line GI coupleth the centre<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the bases of the pyramis: the saide line GI shalbe the diameter of the base of the cube inscribed in the pyramis by the 18. of the fiuetenth. And forasmuche as the line AG is double to the line GE, by the corollarye of the twelueth of the thirtenth: the whole line AE shal be triple to the line GE: and so is also the line DE to the line IE. Wherefore the lines AD and GI are parallels, by the 2. of the sixth. And therefore the triangles AED, and GEI are like<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the corollary of the same. And forasmuch as the tri<g ref="char:EOLhyphen"/>angles AED, and GEI are like, the line AD<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> shalbe triple to the line GI, by the 4. of the sixth. But the line AD is the dia<g ref="char:EOLhyphen"/>meter of the base of the cube circumscribed about the py<g ref="char:EOLhyphen"/>ramis ABCD, and the line GI is the diameter of the base of the cube inscribed in the pyramis ABCD: but the diameters of the bases are equemultiplices to the sides (namely, are in power duple). Wherfore the side of the cube circumscribed about the pyramis ABCD, is triple to the side of the cube, inscribed in the same piramis, by the 15. of the fiueth: but like cubes are in triple proportion the one to the other of that in which their sides are, by the 33. of the eleuenth: and the sides are in triple proportion the one to the other: Wherfore triple taken thre times bringeth forth twenty seuencuple, which is 27. to 1: for the 4. termes 27.9.3.1, being set in triple proportion: the proportion of the first to the fourth, namely, of 27. to 1. shalbe triple to the proportion of the first to the second, namely, of 27. to 9, by the 10. diffinition of the fiueth: which proportion of 27. to 1. is the proportio<g ref="char:cmbAbbrStroke">̄</g> of the sides tripled, which proportio<g ref="char:cmbAbbrStroke">̄</g> also is found in like solides. Wherefore of what partes the cube circumscribed containeth 27. of the same, the cube inscribed containeth one: but of what partes the cube circumscribed, containeth 27. of the same, the pyramis inscribed in it, containeth 9. by the 25. of this booke: wherfore of what partes the pyramis AB CD containeth 9. of the same, the cube inscribed in the pyramis, containeth one. Wherefore we haue proued that a trilater and equilater pyramis, is non<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>cuple to a cube inscribed in it.</p>
            </div>
            <div type="part">
               <pb n="454" facs="tcp:7062:483"/>
               <head>¶ The 31. Proposition. An Octohedron hath to an Icosohedron inscribed in it, that proportion, which two bases of the Octohedron haue to fiue bases of the Icosahedron.</head>
               <p>SVppose that the octohedron geuen be ABCD, and let the Icosahedron inscribed in it, be F<g ref="char:EOLhyphen"/>GHMKLIO. Then I say that the octohedron is to the Icosahedron, as two bases of the octohedron, are to fiue bases of the Icosahedron. For forasmuche as the solide of the octohe<g ref="char:EOLhyphen"/>dron consisteth of eight pyramids, set vpon the bases of the octohedron,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> and hauing to theyr altitude a perpendicular line drawne from the centre to the base: let that perpendicular be ER, or ES, being drawne from the centre E (which centre is common to either of the solides, by the corollary of the 21. of the fiuetenth) to the centres of the bases, namely, to the poyntes R and S. Wherefore for that thre pyramids are equal and like, they shalbe equal to a prisme set vpon the selfe same base, and vnder the selfe same altitude, by the corollary of the seuenth of the twelueth. But vnto this prisme is double that prisme which is set vpon the self same base, and hath his altitude duple, namely, the whole line RS by the corollary of the 25. of the eleuenth: for it is equal to the two equal and like prismes whereof it is composed. Wherfore the prisme set vpo<g ref="char:cmbAbbrStroke">̄</g> the base of the octohedron, and hauing to his altitude the line RS is equal to six pyramids, set vpon six bases of the Octohedron, and hauing to their altitude the line ER. So there remaine two pyramids (for in the octohedron are 8. bases) which shall be equal to the prisme which is set vpon the third part of the base of the octohedron, and vnder the altitude RS. For prismes vnder one and the selfe same altitude, are in proportion the one to the other, as are their bases, by the corollary of the 7 of the
<figure/>
twelueth. Wherefore the two prismes which are set vppon the base of the octohedron, and vp<g ref="char:EOLhyphen"/>on a third part therof, and vnder the altitude RS, are equal to the 8. pyramids of the Octohedron, or to the whole solide of the oc<g ref="char:EOLhyphen"/>tohedron. And forasmuch as the Icosahedron inscribed in the oc<g ref="char:EOLhyphen"/>tohedron, hathe his bases set in the bases of the Octohedron, by the 17. of the fiuetenth: it follow<g ref="char:EOLhyphen"/>eth that the pyramids set vppon the bases of the Icosahedron, &amp; hauing to their toppes one and the selfe same centre E, are con<g ref="char:EOLhyphen"/>tained vnder the selfe same alti<g ref="char:EOLhyphen"/>tude, that the pyramids of the octohedron are co<g ref="char:cmbAbbrStroke">̄</g>tained vnder. namely, vnder the line ER, or ES. And therefore a prisme, set vpon the base of the Icosahedron, and hauing his altitude double to the altitude of the pyramis, namely, the whole line RS, is equal to sixe pyramids set vpon the base of the Icosahedron, and vnder the altitude ER or ES, as we haue pro<g ref="char:EOLhyphen"/>ued in the octohedron. Wherfore the 20. pyramids, set vpon the 20. bases of the Icosahedron, are equal to thre prismes set vpon the base of the Icosahedron, and vnder the altitude RS, and moreouer to an o<g ref="char:EOLhyphen"/>ther prisme set vppon a thirde part of the base of the Icosahedron and vnder the same altitude RS, which prisme is a thirde part of the former prisme, by the corollarye of the 7. of the twelueth: for their proportion is as the proportion of the bases. Wherfore two prismes set vpon the base of the octo<g ref="char:EOLhyphen"/>hedron, and a third part therof, and vnder the altitude RS, is to 4. prismes set vpon three bases of the I<g ref="char:EOLhyphen"/>cosahedron, and a third part thereof, and vnder the same altitude RS, in the same proportion that the bases are, that is, as 4. third partes of the base of the Octodron (which are equal to one base, and 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>;) to ten third partes of the base of the Icosahedron (which are equal to thre bases &amp; 1/<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>;) or as two third partes of the base of the Octohedron, are to fiue thirde partes of the base of the Icosahedron. But two thirde partes of the base of the Octohedron, are to fiue thirde partes of the base of the Icosa<g ref="char:EOLhyphen"/>hedron, as two bases are to fiue bases (by the 15. of the fifth, for they are partes of equemultiplices:) And two prismes of the Octohedron are to 4. prismes of the Icosahedron, as the solide of the Octohedron is to the solide of the Icosahedron, when as eche are equal to eche of the solides: Wherefore (by the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>. of the fiueth) the solide of the Octohedron, is to the solide of the Icos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hedron inscribed in it, as two bases of the Octohedron, are to fiue bases of the Icosahedron. An Oc<g ref="char:EOLhyphen"/>tohedron
<pb facs="tcp:7062:484"/>
therfore is to an Ico<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ahedron inscribed in it, in that proportion, that two bases of the Octo<g ref="char:EOLhyphen"/>hedron, are to fiue bases of the Icosahedron.</p>
            </div>
            <div type="part">
               <head>¶ The 32. Proposition. The proportio<g ref="char:cmbAbbrStroke">̄</g> of the solide of an Icosahedron to the solide of a Dodecahe<g ref="char:EOLhyphen"/>dron inscribed in it, consisteth of the proportion of the side of the Icosahe<g ref="char:EOLhyphen"/>dron to the side of the Cube contayned in the same sphere, and of the pro<g ref="char:EOLhyphen"/>portion tripled of the diameter to the line which conpleth the centers of the opposite bases of the Icosahedron.</head>
               <p>
                  <note place="margin">
                     <hi>Construction.</hi>
                  </note>SVppose that there be <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> Dodecahedron, whose diameter let be HI, and let the Icosahe<g ref="char:EOLhyphen"/>dron contained in the same sphere be ABGC, whose dimetient let be AC. And let the right line which coupleth the centres of the opposite base<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> be BG. And let the dodeca<g ref="char:EOLhyphen"/>hedron inscribed in the Icos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hedron be that which is set vpon the diameter BG, by the 5. of the fiuetenth. And let the side of the cube be DE, and let the side of the Icosahe<g ref="char:EOLhyphen"/>dron be D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, both the sayd solides being described in one and the selfe same sphere. The<g ref="char:cmbAbbrStroke">̄</g> I say that the proportion of the solide of the Icosahedron ABCG to the solide of the dodecahedron set vpon the diameter BG, inscribed in it, consisteth of the proportion of the line DF to the line DE, and of the proportion tripled of the line AC to the line BG.<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> For forasmuch as the solide of the Icosa<g ref="char:EOLhyphen"/>hedron ABGC is to the solide of the dodecahedron HI, being contayned in one and the selfe same sphere, as DF is to D<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>,
<figure/>
by the 8. of the fourtenth But the dodecahedron whose diameter is HI, is to the dodecahedron whose diamer is BG, in treble propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tio<g ref="char:cmbAbbrStroke">̄</g> of that in which the diameter HI is to the diameter B<g ref="char:EOLhyphen"/>G, by the corollary of the 17. of the twelfth: &amp; the lines HI and AC are equall by supposition (namely, the diameters of one and the selfe same sphere). Wherefore as HI is to BG, so is AC to BG. Wherefore the pro<g ref="char:EOLhyphen"/>portion of the extremes, namely, of the Icosahe<g ref="char:EOLhyphen"/>dron ABGC to the Do<g ref="char:EOLhyphen"/>decahedron set vppon the diameter BG which coupleth the c<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ntres, co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>sisteth (by the 5. diffiniti<g ref="char:EOLhyphen"/>on of the sixt) of the pro<g ref="char:EOLhyphen"/>portions of the me<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>nes, namely, of the proportio<g ref="char:cmbAbbrStroke">̄</g> of the Icos<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hedron ABCG to the dodecahedron HI (which is one and the same with the proportio<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of DF to DE) and of the proportion of the same HI to the other dodecahedron set vpon the diamete<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> BG, inscribed in the same Icosahedron ABGC, by the same 5. of the fiuetenth: which proportion is triple to the propo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>tio<g ref="char:cmbAbbrStroke">̄</g> of the line HI (or the line AC) to GB which coupleth the centres of the oppo<g ref="char:EOLhyphen"/>site bases of the Icosahedron. The proportion therefore of the solide of the Icosahedron to the solide of a D<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ecahedron inscribed in it, consisteth of the proportion of the side of the Icosahedron to the side of the Cube contayned in the same sphere, and of the proportion tripled of the diameter to the lin<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> which coupleth the centres of the opposite bases of the Icosahedron.</p>
            </div>
            <div type="part">
               <head>¶ The 33. Proposition. The solide of a Dodecahedron excedeth the solide of a Cube inscribed in
<pb n="455" facs="tcp:7062:484"/>
it, by a parallelipipedon, whose base wanteth of the base of the Cube by a third part of the lesse segment, and whose altitude wanteth of the altitude of the Cube, by the lesse segment of the lesse segment, of halfe the side of the Cube.</head>
               <p>FOrasmuch as by the 17. of the thirtenth, and 8. of the fiuetenth, it w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>s manifest, that the base of a cube inscribed in a dodecahedron, doth with his sides subtend t<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap> angles of 4. pentagons co<g ref="char:cmbAbbrStroke">̄</g>curring at one and the selfe same side of the dodecahedron:<note place="margin">
                     <hi>Construction.</hi>
                  </note> let that base of the cube be A<g ref="char:EOLhyphen"/>BCD: and let the side wherat 4. bases of the dodecahedron circumscribed concurre, be EG: which shall contayne a solide AEBDGC set vpon the base ABCD. Diuide the sides AB and DC in<g ref="char:EOLhyphen"/>to two equall partes in the poyntes L and N. And draw the line LN, which is a parallel to the side EG, as it was manifest by the 17. of the thirtenth. The perpendiculars also ER and GO which couple those parallels, are eche equall, to halfe of the side EG, and eche is the greater segment of halfe the side of the cube, and therefore the whole line EG is the greater segment of the whole line LN the side of the cube (by the foresayd 17. of the thirtenth). By the poyntes R and O, draw vnto the sides AB and CD parallel lines FH and IK. And draw these right lines EF, EH, GI and GK. Now forasmuch as the two lines FH &amp; ER touching the one the other are parallels to the two lines IK and GO touching also the one the other,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> &amp; not being in the selfe same playne with the two first lines: therfore the playne super<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>i<g ref="char:EOLhyphen"/>cieces EFH and GIK passing by those lines are parallels, by the 15. of the eleuenth: which playnes do cutte the solide AEBDGC. Whererefore there are made fower quadrangled pyramids set vpon the rectangle parallelogrames LH, LF, NK, and NI, and hauing their toppes the poyntes E and G. And forasmuch as the triangles GOK and ERH are equall and like, by the 4. of the first, namely,<note place="margin">
                     <hi>Extend in the figure a line <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>rom the point E to the point B.</hi>
                  </note> they con<g ref="char:EOLhyphen"/>tayne
<figure/>
equall angles comprehe<g ref="char:cmbAbbrStroke">̄</g>ded vnder equall sides, and they are parallels by construction, being set in the playnes GIK and EFH: the figures GKHE, OKHR, and GORE shall be parallelogrammes, by the di<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>nition of a parallelogramme, and therefore the solide GOKERH is a prisme, by the 11. diffini<g ref="char:EOLhyphen"/>tion of the eleuenth. And by the same reason may the solide GOIERF be proued to be a prisme. And forasmuch as vpon equall bases NOKC, and RLBH, and vnder equall altitudes OG and RE are set pyramids: those pyramids shall be equall to <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>hat pyramis which is set vpon the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ase CKID (which is double to either of the bases NOKC, and RLBH) and vnder the same altitude OG, by the 6. of the twelfth. And forasmuch as the side GE is the greater segment of the line CB, the line KH, which by the 33. of the first, is equall to the line GE, shall be the greater segment of the same line CB, by the 2. of the fourtenth. Wherefore the residues CK and HB shall make the lesse segment of the whole line C<g ref="char:EOLhyphen"/>B. But as the greater segment KH is to the two lines CK and HB the lesse segment, so is the rectangle parallelogramme OH to the two rectangle parallelogrammes OC and HL, by the 1. of the sixt. Wher<g ref="char:EOLhyphen"/>fore the lesse segment of the parallelogramme NB, shall be the two parallelogrammes OC and HL. Put the line KM double to the line KC and draw the line MS parallel to the line CN. Wherefore the parallelogramme OKMS is equall to the parallelogrammes OC and HL, by the 1. of the sixth. Wher<g ref="char:EOLhyphen"/>fore the pyramis set vpon the base OKMS contayneth two third partes of the prisme set vpon the selfe same base, by the 4. corollary of the 7. of the twelfth. Wherfore the prisme which is set vpon two third
<pb facs="tcp:7062:485"/>
partes of the base OKMS is equall to the two pyramids NOKCG and RLBHE. For the sections of a prisme are one to the other, as the sections of the base are, by the first corollary of the 25. of the ele<g ref="char:EOLhyphen"/>uenth. But the sections of the base are as the sections of the line CB or KM, by the 1. of the sixt. Wher<g ref="char:EOLhyphen"/>fore the two pyramids NOKCG and RLBHE, adde vnto the prisme GOKERH two third partes of the prisme set vpon the base OKMS. And forasmuch as the line KM is the lesse segme<g ref="char:cmbAbbrStroke">̄</g>t of the whole line BC (for it is equall to to the two lines CK and H<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>), and the prisme set vpon the base OKHR is cutte like vnto the line KM, namely, in eche are taken two thirdes, as hath before bene proued: the prisme equall to the two pyramids, shall adde vnto the prisme GOKERH, which is set vpon the grea<g ref="char:EOLhyphen"/>ter segment KH, two th<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>ds of the lesse segment. Wherefore in the line BC there shall remayne one third part of the lesse segment: and therefore in the rectangle parallelogramme NB which is halfe the base of the cube, there shall remayne the same third part of the lesse segment. And by the same reason may we proue that in the other pyramids ONDIG, and RLAFE, and in the prisme GOIERF is left the selfe same excesse of the base LAND, namely, the third part of the lesse segment. Wherefore the whole prisme contayned betwene the triangles IGK and FEH, and vnder the length of the greater
<note place="margin">
                     <hi>Extend in the figure a line from the point E to the point B.</hi>
                  </note>
                  <figure/>
segment and two third partes of the lesse segment of BC the side of the cube, is equall to the solide composed of 4. bases of the dodecahedron and set vpon the base of the cube. Wherfore the base of that prisme wanteth of the whole base of the cube onely a third part of the lesse segment: and the altitude of that prisme was the line GO, which is the greater segment of halfe the side of the cube. And foras<g ref="char:EOLhyphen"/>much as vnto the triangle IGK, is double the rctangle parallelogramme set vpon the same base IK, (the side of the cube) and vnder the altitude GO, by the 41. of the first: it followeth that three rectan<g ref="char:EOLhyphen"/>gle parallelogrammes set vpon the same base IK, the side of the cube, and vnder the altitude OG the greater segment of halfe the side of the cube, are sextuple to the triangle IGK. Wherefore those three rectangle parallelogrammes doo make one rectangle parallelogramme set vpon the base IK and vnder the altitude of the line GO tripled. But by the 7. diffinition of the eleuenth, there are sixe prismes equal and like vnto the foresayd prisme, being set vpon euery one of the sixe bases of the cube: which prismes are in proportion the one to the other as their bases are by the 3. corollary of the 7. of twelfth. Where<g ref="char:EOLhyphen"/>fore the solide composed of these sixe prismes, shall want of the base ABCD the third part of the lesse segment, and taking his altitude of the foresayd rectangle parallelogramme, the sayd altitude shall be e<g ref="char:EOLhyphen"/>quall to three greater segmentes (one of which is GO) of halfe the side of the cube.</p>
               <p>Now resteth to proue that these three segmentes want of the side of the cube by the lesse segment of the lesse segment of halfe the sid<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> of the cube. Suppose that AB the side of the cube be diuided into the greater segment AC, and into the lesse segment CB (by the 30. of the sixt). And diuide into two e<g ref="char:EOLhyphen"/>quall partes the line AC in the poynt G, and the line CB in the poynt E. And vnto the line CG put the line CL equall. Now forasmuch as the lines AG and GC are the greater seg<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>tes of halfe the line AB, for <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>che of
<figure/>
them is the halfe of the greater segment of the whole line A<g ref="char:EOLhyphen"/>B: the lines EB and EC shall be the lesse segmentes of halfe the line AB. Wherefore the whole line C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> is the greater segment, and the line C<g ref="char:EOLhyphen"/>E
<pb n="456" facs="tcp:7062:485"/>
is the lesse segment. But as the line CL is to the line CE, so is the line CE to the residue EL. Wher<g ref="char:EOLhyphen"/>fore the line EL is the greater segment of the line CE, or of the line EB which is equall vnto it. Wher<g ref="char:EOLhyphen"/>fore the residue LB is the lesse segment of the same EB (which is the les<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e segment of half<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> the side of the cube). But the lines AG, GC, and CL are three greater segmentes of the halfe of the whole line AB: which thre greater segmentes make the altitude of the foresayd solide: wherefore the altitude of the sayd solide wanteth of AB the side of the cube by the line LB, which is the lesse s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>gment of the line BE. Which line BE agayne is the lesse segment of halfe the side AB of the cube. Wherefore the fore<g ref="char:EOLhyphen"/>sayd solide consisting of the sixe solides, whereby the dodecahedron exceedeth the cube inscribed in it, is set vpon a base which wanteth of the base of the cube by a third part of the lesse segment, and is vnder an altitude wanting of the side of the cube by the lesse segment of the lesse segment of halfe the side of the cube. The solide therefore of a dodecahedron exceedeth the solide of a cube inscribed in it, by a parallelipipedon, whose base wanteth of the base of the cube by a third part of the lesse segment, and whose altitude wanteth of the altitude of the cube, by the lesse segment of the lesse segment of halfe the side of the cube.</p>
               <div type="section">
                  <head>¶A Corollary.</head>
                  <p>
                     <hi>A Dodecahedron is double to a Cube inscribed in it, taking away the third part of the lesse segment of the cube, and moreouer the lesse segment of the lesse segment of halfe of that excesse.</hi> For if there be geuen a cube, from which is cut of a solide set vpon a third part of the lesse segment of the base and vnder one and the same altitude with the cube: that solide taken away hath to the whole solide the proportion of the section of the base to the base, by the 32. of the eleuenth. Wherefo<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e from the cube is taken away a third <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>art of the lesse segment. Far<g ref="char:EOLhyphen"/>ther, forasmuch as the residue wanteth of the altitude of the cube, by the lesse segment of the lesse seg<g ref="char:EOLhyphen"/>ment of halfe the altitude or side, and that residue is a parallelipipedon, if it be cut by a plaine superficies parallel to the opposite plaine superficieces, cutting the altitude of the cube by a point, it shall take away from that parallelipipedon a solide, hauing to the whole the proportion of the section to the altitude, by the 3. Corollary of the 25. of the eleuenth. Wherefore the excesse wanteth of the same cube by the thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>d part of the lesse segment, and moreouer by the lesse segment of the lesse segment of halfe of that excesse.</p>
               </div>
            </div>
            <div type="part">
               <head>¶The 34. Proposition. The proportion of the solide of a Dodecahedron to the solide of an Icosa<g ref="char:EOLhyphen"/>hedron inscribed in it, consisteth of the proportion tripled of the diameter to that line which coupleth the opposite bases of the Dodecahedron, and of the proportion of the side of the Cube to the side of the Icosahedron inscri<g ref="char:EOLhyphen"/>bed in one and the selfe same Sphere.</head>
               <p>SVppose that
<figure/>
AHBCK be a Dode<g ref="char:EOLhyphen"/>cahedron<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> whose diamet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>r let be AB: and let the line which coupleth the ce<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>tres of the opposite ba<g ref="char:EOLhyphen"/>ses be KH<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> and let the Icosahedron inscribed in the Dodecahedron ABC, be DEE: whose diameter let be DE. Now forasmuch a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>e and the selfe same circle co<g ref="char:cmbAbbrStroke">̄</g>taineth the pentagon of a Dodecahedron, &amp; the triangle of an Ico<g ref="char:EOLhyphen"/>sahedro<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> described in one and the selfe same Sphere, by the 14. of the fourtenth: Let that cir<g ref="char:EOLhyphen"/>cle
<pb facs="tcp:7062:486"/>
be IGO. Wherfore
<figure/>
IO is the side of the cube, and IG the side of the Icosahedron, by the same. The<g ref="char:cmbAbbrStroke">̄</g> I say, that the proportion of the Dodecahedron AHB<g ref="char:EOLhyphen"/>CK to the Icosahedron DEF inscribed in it, co<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>sisteth of the proportio<g ref="char:cmbAbbrStroke">̄</g> tripled of the line AB to the line KH, and of the proportion of the line IO to the line IG. For <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>asmuch as the I<g ref="char:EOLhyphen"/>cosahedron DEF is in<g ref="char:EOLhyphen"/>scribed in the Dodeca<g ref="char:EOLhyphen"/>hedro<g ref="char:cmbAbbrStroke">̄</g> ABC,<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> by suppo<g ref="char:EOLhyphen"/>sitio<g ref="char:cmbAbbrStroke">̄</g>, the diameter DE shalbe equal to the line KH, by the 7. of the fiue<g ref="char:EOLhyphen"/>tenth. Wherefore the Dodecahedron set vpo<g ref="char:cmbAbbrStroke">̄</g> the diameter KH shall be inscribed in the same Sphere, wherein the I<g ref="char:EOLhyphen"/>cosahedron DEF is in<g ref="char:EOLhyphen"/>scribed: but the Dodecahedron AHBCK is to the Dodecahedron vpon the diameter KH in tri<g ref="char:EOLhyphen"/>ple proportion of that in which the diameter AB is to the diameter KH, by the Corollary of the 17. of the twelfth: and the same Dodecahedron which is set vpon the diameter KH, hath to the Icosahedron DEF (which is set vpon the same diameter, or vpon a diameter equall vnto it, namely, DE) that pro<g ref="char:EOLhyphen"/>portion which IO the side of the cube hath to<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> IG the side of the Icosahedron, inscribed in one &amp; the selfe same Sphere, by the 8 of the fouretenth. Wherefore the proportion of the Dodecahedron AH<g ref="char:EOLhyphen"/>BCK to the Icosahedron DEF inscribed in it, consisteth of the proportion tripled of the diameter AB to the line KH, which coupleth the centres of the opposite bases of the Dodecahedron (which propor<g ref="char:EOLhyphen"/>tion is that which the Dodecahedron AHBCK hath to the Dodecahedron set vpon the diameter KH) and of the proportion of IO the side of the cube to IG the side of the Icosahedron (which is the pro<g ref="char:EOLhyphen"/>portion of the Dodecahedron set vpon the diameter KH to the Icosahedron DEF described in one and the selfe same Sphere) by the 5. definition of the sixth. The proportion therefore of the solide of a Do<g ref="char:EOLhyphen"/>decahedron to the solide of an Icosahedron inscribed in it, con<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>isteth of the proportion tripled of the diameter to that line which coupleth the opposite bases of the Dodecahedron, and of the prop<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>tion of the side of the cube to the side of the Icosahedron inscribed in one and the selfe same Sphere.</p>
            </div>
            <div type="part">
               <head>The 35. Proposition. The solide of a Dodecahedron containeth of a Pyramis circumscribed a<g ref="char:EOLhyphen"/>bout it two ninth partes, taking away a third part of one ninth part of the lesse segment (of a line diuided by an extreme and meane proportion) and moreouer the lesse segment of the lesse segment of halfe the residue.</head>
               <p>IT hath bene proued that the Dodecahedron, together with the cube inscribed in it is contai<g ref="char:EOLhyphen"/>ned in one and the selfe same pyramis, by the Corollary of the first of this booke. And by the Corollary of the 33. of this booke, it is manifest, that the Dodecahedron is double to the same cube, taking away the third part of the lesse segment, and moreouer the lesse segment of the lesse segment of halfe the residue, or of this excesse. But a pyramis is to the same cube inscribed in it nonecuple, by the 30. of this booke. Wherefore the Dodecahedron inscribed in the pyramis, and con<g ref="char:EOLhyphen"/>taining the same cube twise, taking away the selfe same third of the lesse segment, and moreouer the lesse segment of the lesse segment of halfe the residue, shall containe two ninth partes of the solide of the pyramis (of which ninth partes eche is equall vnto the cube) taking away this selfe same excesse. The solide therefore of a Dodecahedron containeth of a Pyramis circumscribed about it two ninth partes, taking away a third part of one ninth part of the lesse segment (of a line diuided by an extmere and meane proportion) and moreouer the lesse segment of the lesse segment of halfe the residue.</p>
            </div>
            <div type="part">
               <pb n="457" facs="tcp:7062:486"/>
               <head>¶The 36. Proposition. An Octohedron exceedeth an Icosahedron inscribed in it, by a parallelipi<g ref="char:EOLhyphen"/>pedon set vpon the square of the side of the Icosahedron, and hauing to his altitude the line which is the greater segment of halfe the semidiame<g ref="char:EOLhyphen"/>ter of the Octohedron.</head>
               <p>SVppose that there be an Octohedron ABCFPL:<note place="margin">
                     <hi>Construction.</hi>
                  </note> in which let there be inscribed an Icosahedron HKEGMXNVDSQT<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>6. of the fiuetenth. And draw the dia<g ref="char:EOLhyphen"/>meters AZRCBROIF, and the perpendicular KO <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>arallel to the line AZR. Then I say, that the Octohedron ABCFPL is greater th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>n the Icosahedron inscribed in it, by a parallelipipedon set vpon the square of the side HK or GE, and hauing to his altitude the line KO or RZ: which is the greater segment of the semidiameter AR. Forasmuch as in the same 16. it hath bene proued, that the triangles KDG and KEQ are described in the bases APF and ALF of the Octohedron:<note place="margin">
                     <hi>Demonstra<g ref="char:EOLhyphen"/>tion.</hi>
                  </note> therefore about the solide angle there remaine vppon the base FEG three triangles KEG, KFE, and KFG, which containe a pyramis KEFG. Vnto which pyramis shall be equall and like the opposite pyramis MEFG set vpon the same base FEG, by the 8. definition of the eleuenth. And by the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ame reason shall there at euery solide angle of the Octohedron remayne two pyramids equall and like: namely, two vpon the base AHK, two vpon the base BNV, two vpon the base DPS, and
<figure/>
moreouer two vp<g ref="char:EOLhyphen"/>on the base QLT. Now the<g ref="char:cmbAbbrStroke">̄</g> there shal be made twelue pyramids, set vpon a base contained of the side of the Ico<g ref="char:EOLhyphen"/>sahedron, and vn<g ref="char:EOLhyphen"/>der two le<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>e seg<g ref="char:EOLhyphen"/>mentes of the side of the Octohedron containing a right angle, as for example the base GEF, And forasmuch as the side GE subte<g ref="char:cmbAbbrStroke">̄</g>
                  <g ref="char:EOLhyphen"/>ding a right angle, is, by the 47. of the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>irst, in power du<g ref="char:EOLhyphen"/>ple to either of the lines EF and FG, and so the <gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>de<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> KH is in power duple to either of the sides AH and AK: and either of the lines AH, AK, or EF, FG, is in pow<g ref="char:EOLhyphen"/>er duple to eyther of the lines AZ or ZK which co<g ref="char:cmbAbbrStroke">̄</g>tayne a right angle, made in the triangle or base AHK by the perpendicular AZ. Wherfore it followeth that the side GE or HK, is in power quadruple to the triangle EFG or AHK. But the pyramis KEFG, ha<g ref="char:EOLhyphen"/>uing his base EFG in the plaine FLBP of the Octohedron, shall haue to his altitude the perpendicu<g ref="char:EOLhyphen"/>lar KO (by the 4. definition of the sixth) which is the greater segment of the semidiameter of the Octohedron, by the 16. of the fiuetenth. Wherfore three pyramids set vnder the same altitude and vpon equall bases, shall be equall to one prisme set vpon the same base, and vnder the same altitude, by the 1. Corollary of the 7. of the twelfth. Wherefore 4. prismes set vpon the base GEF quadrupled (which is equall to the square of the side GE) and vnder the altitude KO (or RZ the greater segment which is equall to KO) shall containe a solide equall to the twelue pyramids, which twelue pyramids make the excesse of the Octohedron aboue the Icosahedron inscribed in it. An Octohedron therefore excedeth an Icosahedron inscribed in it, by a parallelipipedon set vpon the square of the side of the Icosahedron, and hauing to his altitude the line which is the greater segment of halfe the semidiameter of the Octo<g ref="char:EOLhyphen"/>hedron.</p>
               <div type="section">
                  <pb facs="tcp:7062:487"/>
                  <head>¶ A Corollary.</head>
                  <p>
                     <hi>A Pyramis exceedeth the double of an Icosahedron inscribed in it, by a so<g ref="char:EOLhyphen"/>lide, set vpon the square of the side of the Icosahedron inscribed in it, and hauing to his altitude that whole line of which the side of the Icosahedron is the greater segme<g ref="char:cmbAbbrStroke">̄</g>t.</hi> For it is manifest by the 19. of the fiuete<g ref="char:cmbAbbrStroke">̄</g>th, that an octohedro<g ref="char:cmbAbbrStroke">̄</g> &amp; an Icosahedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in it are inscribed in one &amp; the self same pyramis. It hath moreouer bene proued in the 26. of this boke, that a pyramis is double to an octohedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in it. Wherfore the two excesses of the two octohedrons (vnto which the pyramis is equal) aboue the two Icosahedrons (inscribed in the said two octohedrons) being brought into an solide, the said solide shalbe set vpon the selfe same square of the side of the Ico<g ref="char:EOLhyphen"/>sahedron, and shall haue to his altitude the perpendicular KO doubled: whose double coupling the opposite sides HK and XM maketh the greater segment the same side of the Icosahedron, by the first and second corollary of the 14. of the fiu<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>en<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h.</p>
               </div>
            </div>
            <div type="part">
               <head>The 37. Proposition. If in a triangle hauing to his base a rational line set, the sides be commen<g ref="char:EOLhyphen"/>surable in power to the base, and from the toppe be drawn to the base a per<g ref="char:EOLhyphen"/>pendicular line cutting the base: The sections of the base shall be commen<g ref="char:EOLhyphen"/>surable in length to the whole base, and the perpendicular shall be commen<g ref="char:EOLhyphen"/>surable in power to the said whole base.</head>
               <p>SVppose that there be a triangle ABG, whose base BG let be a rational line set of purpose. And let the sides AB and AG be vnto the same BG commensurable at the least in power.<note place="margin">
                     <hi>Construction.</hi>
                  </note> And from the toppe A, draw vnto the base BG, a perpendicular, cutting the base in the point P. Then I say that the sections of the base, are commensurable in lengthe to the whole line BG, and that the perpendicular AP, is vnto the same base BG co<g ref="char:cmbAbbrStroke">̄</g>mensurable at the least in power. Pro<g ref="char:EOLhyphen"/>duce on either side the line BG to the poyntes C and E. And vnto the line AG put the line GE equal, and vnto the line AB put the line BC equal. And vpon the lines CB, BG and GE describe squares B<g ref="char:EOLhyphen"/>K, BD, and GL. And from the greater of the squares of the lines AB or AG, which let be GL cut of a parallelogramme EM equal to the lesse square BK (by the 45. of the first:) And (by the same) vnto the residue GM, let there be applied vpon the line GD an equall rectangle parallelogramme OD.<note place="margin">
                     <hi>First part of the demon<g ref="char:EOLhyphen"/>stration.</hi>
                  </note> Now for<g ref="char:EOLhyphen"/>asmuch as the angles APB and APG are right angles, therfore (by the 47. of the first) the line AG con<g ref="char:EOLhyphen"/>taineth in power the two lines AP and PG, and the line AB the two lines AP and PB. Wherfore how much the line AG containeth in power more then the line AB, so much also doth the line PG contain in power more then the line BP: namely, taking away the common square of AP, there is left the ex<g ref="char:EOLhyphen"/>cesse of the square of PG aboue the square of BP. But the square of AG (which is GL) exceedeth the square of AB (namely, the square BK) by the rectangle parallelogramme GM or OD, by construction. Wherfore the square of PG exceedeth the square of BP, by the rectangle parallelogramme OD. And forasmuch as vnto the squares of AB and AG, are equal the squares of AP and PB, and of AP and P<g ref="char:EOLhyphen"/>G: and their excesse is taken away, namely, the rectangle parallelogramme OD: there shallbe left the squares of AP and PO equal to the squares of AP and PB. And taking away the square of AP which is co<g ref="char:cmbAbbrStroke">̄</g>mon, the residues
<figure/>
namely, the squares of BP and PO shalbe e<g ref="char:EOLhyphen"/>qual: and therefore their sides (namely, the lines BP and PO) are equal. And forasmuch as the squares GL and PK are (by suppositio<g ref="char:cmbAbbrStroke">̄</g>) rational, and therefore co<g ref="char:cmbAbbrStroke">̄</g>me<g ref="char:cmbAbbrStroke">̄</g>surable their ex<g ref="char:EOLhyphen"/>cesse OD, shalbe com<g ref="char:EOLhyphen"/>me<g ref="char:cmbAbbrStroke">̄</g>surable vnto the<g ref="char:cmbAbbrStroke">̄</g> by the 15. of the te<g ref="char:cmbAbbrStroke">̄</g>th. And therfore it is rationall by the 9. diffinition of the te<g ref="char:cmbAbbrStroke">̄</g>th. Wherfore the rational parallelogram OD, being applied vp<g ref="char:EOLhyphen"/>on the rational line G<g ref="char:EOLhyphen"/>D (or BG) maketh the
<pb n="458" facs="tcp:7062:487"/>
bredth OG rational and co<g ref="char:cmbAbbrStroke">̄</g>mensurable in le<g ref="char:cmbAbbrStroke">̄</g>gth to the whole line BG<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> by the 20. of the tenth. But if the whole line BG be commensurable to one of the partes OG, the lines BO, OG, and BG sh<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap> be com<g ref="char:EOLhyphen"/>mensurable, by the same 15. of the tenth. Wherfore also the line OG shalbe commensurable to the half of the line BO, namely, to the line PO, or PB, by the 12. of the tenth. And forasmuch as the two lines PO and OG are commensurable, the whole line PG shalbe commensurable to the line PO, or to the line PB, by the same 15. of the tenth. Wherfore either of the lines PG and PB shall be co<g ref="char:cmbAbbrStroke">̄</g>mensurable vnto the whole line PB, by the same. Wherefore the lines BG, PB and PG haue the one to the other that proportion which numbers haue, by the 5. of thirtenth. Wherfore the sections PB and PG of the base BG are commensurable in length <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the same base, by the 6: of the tenth.</p>
               <p>And now that the perpendicular AP is commensurable in power to the base BG,<note place="margin">
                     <hi>Second part of the De<g ref="char:EOLhyphen"/>monstration.</hi>
                  </note> i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> thus pro<g ref="char:EOLhyphen"/>ued. Forasmuch as the square of AB is by supposition, commensurable to the square of BG: and vnto the rational square of AB is commensurable the rational square of BP (by the 12. of the eleuenth) Wherfore the residue, namely, the square of PA is commensurable to the same square of BP, by the 2. part of the 15. of the eleuenth. Wherefore by the 12. of the tenth, the square of PA is commensurable to the whole square of BG. Wherefore the perpendicular AP is commensurable in power to the base BG, by the 3. diffinition of the tenth: which was required to be proued.</p>
               <p>In demonstrating of this, we made no mention at all of the length of the sides AB and AG, but only of the length of the base BG: for that the line BG is the rational line first set: and the other lines AB and AG are supposed to be commensurable in power only to the line BG. Wherefore if that be plainely demonstrated, when the sides are commensurable in power only to the base, much more easily wil it follow, if the same sides be supposed to be commensurable both in length and in power to the base: that is, if their lengthes be expressed by the rootes of square nombers.</p>
               <div type="section">
                  <head>¶ A Corollary. 1.</head>
                  <p>
                     <hi>By the former things demonstrated, it is manifest that if from the powers of the base, and of one of the sides, be taken away the power of the other side, and if the halfe of the power remaining, be applied vpon the whole base, it shall make the bredth that section of the base which is coupled to the first side.</hi> For from the po<g ref="char:EOLhyphen"/>wers of the base BG, and of one of the sides AG, that is, from the squares BD and GL, the power of the other side AB, namely the square BK (or the parallelogramme EM) is taken away. And of the residue, (namely, of the square BD, and of the parallelogramme OD, or DR, which by supposition is equal vn<g ref="char:EOLhyphen"/>to OD) the halfe (namely<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> of the whole FR, which is PD, for the lines GR and PB are equal to the lines GO and PO) is applied to the whole line BG or GD: and maketh the bredthe the line PG the section of the base BG, which section is coupled to the first side AG. And by the same reason in the o<g ref="char:EOLhyphen"/>ther side, if from the squares BD and BK be taken away the square GL, there shall remaine the rectan<g ref="char:EOLhyphen"/>gle parallelogramme FO: For the parallalelogramme EM is equal to the square BK, and the parallelo<g ref="char:EOLhyphen"/>gramme GM to the parallelogramme OD. Wherefore FP the halfe of the residue FO, maketh the bredth BP, which is coupled to the first side taken AB.</p>
               </div>
               <div type="section">
                  <head>A Corollary. 2.</head>
                  <p>
                     <hi>If a perpendicular drawne from an angle of a triangle do cut the base: the sections are to the other sides in power proportionall by an A<g ref="char:EOLhyphen"/>rithmetical proportion.</hi> For it was proued that the excesse of the powers of the lines AG and AB is one and the same with the excesse of the powers of the lines PG and PB. If therfore the powers do equally excede the one the other, they shall by an Arithmet<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>cal proportion, be proportionall.</p>
               </div>
            </div>
            <trailer>The ende of the sixtenth Booke of the Elementes of Geometrie added <gap reason="illegible" resp="#PDCC" extent="1 span">
                  <desc>〈…〉</desc>
               </gap>.</trailer>
         </div>
         <div type="treatise">
            <pb facs="tcp:7062:488"/>
            <head>A briefe treatise, added by Flussas, of mixt and composed regular solides.</head>
            <p>REgular solides are sayd to be composed and mixt, when ech of them is transformed into other solides, keeping still the forme, number, and inclination of the bases, which they before had one to the other: some of which yet are trans<g ref="char:EOLhyphen"/>formed into mixt solides, and other some into simple. Into mixt, as a Dodecahedron and an Icosahedron: which are transformed or altered, if ye diuide their sides into two e<g ref="char:EOLhyphen"/>quall partes, and take away the solide angles subtended of plaine superficiall figures made by the lines coupling those middle sections: for the solide remayning after the taking away of those solide an<g ref="char:EOLhyphen"/>gles, is called an Icosidodecahedron.<note place="margin">
                  <hi>Icosidodecahe<g ref="char:EOLhyphen"/>dron.</hi>
               </note> If ye diuide the sides of a cube and of an Octohedron into two equall partes, and couple the sections, the solide angles sub<g ref="char:EOLhyphen"/>tended of the plaine superficieces made by the coupling lines, being taken away, there shall be left a solide, which is called an Exoctohedron.<note place="margin">
                  <hi>Exoctohedro<g ref="char:cmbAbbrStroke">̄</g>.</hi>
               </note> So that both of a Do<g ref="char:EOLhyphen"/>decahedron and also of an Icosahedron, the solide which is made, shall be called an Icosidodecahedron: and likewise the solide made of a Cube &amp; also of an Octo<g ref="char:EOLhyphen"/>hedron, shall be called an Exoctohedron. But the other solide, namely, a Pyramis (or Tetrahedron) is transformed into a simple solide: for if ye diuide into two equall partes euery one of the sides of the pyramis, triangles described of the lines which couple the sections, and subtending, and taking away solide angles of the pyramis, are equall and like vnto the equilater triangles left in euery one of the bases: of all which triangles is produced an Octohedron, namely, a simple and not a composed solide. For the Octohedron hath fower bases, like in number, fonne, and mutuall inclination with the bases of the pyramis: and hath the other fower bases with like situation opposite and parallel to the former. Wherefore the application of the pyramis taken twise, maketh a simple Octohedron, as the other solides make a mixt compound solide.</p>
            <div type="part">
               <head>¶ First Definition.</head>
               <p>An Exoctohedron is a solide figure contained of sixe equall squares, and eight equilater and equall triangles.</p>
            </div>
            <div type="part">
               <head>¶ Second Definition.</head>
               <p>
                  <hi>An Icosidodecahedron is a solide figure, contained vnder twelue equi<g ref="char:EOLhyphen"/>later equall and equiangle Pentagons, and twentie equall and equilater triangles.</hi>
               </p>
               <p>For the better vnderstanding of the two former definitions, and also of the two Propositions following, I haue here set two figures, whose formes, if ye first
<pb n="459" facs="tcp:7062:488"/>
describe vpon pasted paper or such like matter, and then cut them and folde them accordingly, they will represent vnto you the perfect formes of an Exoctohedron and of an Icosidodecahedron.</p>
               <figure/>
            </div>
            <div type="part">
               <head>¶The first Probleme. To describe an equilater and equiangle exoctohedron, and to contayne it in a sphere geuen: and to proue that the diameter of the sphere is double to the side of the sayd exoctohedron.</head>
               <p>SVppose that there be a sphere geuen, whose diameter let be AB. And about the diameter AB let there be described a square by the sixth of the fourth:<note place="margin">
                     <hi>Construction of the exocto<g ref="char:EOLhyphen"/>bedron.</hi>
                  </note> and vpon the square let there be described a cube by the 15. of the thirtenth: which let be CDEFQTVR: and let the diameter thereof be QR, and the centre S. And diuide the sides of the cube into two equall partes, in the poyntes G, H, I, K, L, M, N, O, P. &amp;c. And couple the middle sections by the right lines IN, NO, OP, PI and such like, which subtend the angles of the squares or bases of the cube: and they are equall by the 4. first, and contalne right angles, as the angle NIP. For the angle NID which is at the base of the Isosceles triangle NDI, is the halfe of a right angle, and so likewise is the opposite angle R<g ref="char:EOLhyphen"/>IP. Wherefore the residue NIP is a right angle, and so the rest. Wherefore NIP<g ref="char:EOLhyphen"/>O is a square. And by the same reason shall the rest NMLK, KGHI &amp;c. inscribed
<pb facs="tcp:7062:489"/>
in the bases of the cube, be squares: and they shall be sixe in number, according <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>o the nu<g ref="char:cmbAbbrStroke">̄</g>ber of the bases of the cube.
<figure/>
Agayne forasmuch as the triangle KIN subtendeth the solide angle D of the cube, and likewise the tri<g ref="char:EOLhyphen"/>angle KGL the solide angle C, &amp; so the rest, which subtend the eight solide angles of the cube: and these triangles are equall and equilater, namely, being made of equall sides &amp; they are the limmits or borders of the squares, and the squares the limmits or borders of the<g ref="char:cmbAbbrStroke">̄</g>, as hath before bene proued: wherefore L<g ref="char:EOLhyphen"/>MNOPHGK is an exoctohedro<g ref="char:cmbAbbrStroke">̄</g>, by the diffinition, and is equilater, for it is contayned of equall sub<g ref="char:EOLhyphen"/>tendent lines: it is also equiangle, for euery solide angle thereof, is contayned vnder two superficiall angles of two squares, and two superficiall an<g ref="char:EOLhyphen"/>gles of two equilater triangles.</p>
               <p>And now forasmuch as the opposite sides and diameters of the bases of the cube are parallels,<note place="margin">
                     <hi>That the <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>xoc<g ref="char:EOLhyphen"/>tohedron is contayned in a sphere.</hi>
                  </note> the playne extended by the right lines QT, VR, shall be a pa<g ref="char:EOLhyphen"/>rallelogramme. And for that also in that playne lyeth QR the diameter of the cube, and in the same playne also is the line MH, which diuideth the sayd playne into two equall parts, and also coupleth the opposite angles of the exoctohedron this line MH therefore diuideth the diameter into two equall partes, by the co<g ref="char:EOLhyphen"/>rollary of the 34. of the first, and also diuideth it selfe in the same poynt, which let be S, into two equall partes, by the 4 of the first. And by the same reason may we proue that the rest of the lines, which couple the opposite angles of the exoctohe<g ref="char:EOLhyphen"/>dron, doo in S the centre of the cube diuide th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap> one the other into two equall parts. For euery one of the angles of the exoct<gap reason="illegible" resp="#PDCC" extent="2 letters">
                     <desc>••</desc>
                  </gap>edron are set in euery one of the bases of the cube. Whe<gap reason="illegible" resp="#PDCC" extent="3 letters">
                     <desc>•••</desc>
                  </gap>ore making the centre the poynt S, and the space SH or SM, describe a sphere, and it shall touch euery one of the angles eq<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>edistant from the poynt S.</p>
               <p>And forasmuch as AB the diameter of the sphere geuen, is put equall to the diameter of the base of the cube,<note place="margin">
                     <hi>That the exoc<g ref="char:EOLhyphen"/>tohed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on is contayned in the sphere geuen.</hi>
                  </note> namely, to the line RT, and the same line RT is equall to the line MH, by the 33. of the first: which line MH coupling the oppo<g ref="char:EOLhyphen"/>site angles of the exoctohedron, is drawne by the centre: wherefore it is the dia<g ref="char:EOLhyphen"/>meter of the sphere geuen which contayneth the exoctohedron.</p>
               <p>Finally forasmuch as in the triangle RFT, the line PO doth cutte the sides into two equall partes,<note place="margin">
                     <hi>That the dia<g ref="char:EOLhyphen"/>
                        <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>ter of the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>here is do<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                        <g ref="char:EOLhyphen"/>ble to the side <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>f the exocto<g ref="char:EOLhyphen"/>hedron.</hi>
                  </note> it shall cutte them proportionally with the bases namely, as FR is to FP, so shall RT be to PO, by the 2. of the sixth. But FR is double to F<g ref="char:EOLhyphen"/>P, by supposition: wherefore RT, or the diameter HM, is also double to the line PO the side of the exoctohedron. Wherefore we haue described an equilater &amp; e<g ref="char:EOLhyphen"/>quiangle exoctohedron, and comprehended it in a sphere geuen, and haue proued that the diameter of the sphere is double to the side of the exoctohedron.</p>
            </div>
            <div type="part">
               <pb n="460" facs="tcp:7062:489"/>
               <head>¶The 2. Probleme. To describe an equilater &amp; equiangle Icosidodecahedron, &amp; to co<g ref="char:cmbAbbrStroke">̄</g>prehend it in a sphere geuen: and to proue that the diameter being diuided by an ex<g ref="char:EOLhyphen"/>treame and meane proportion, maketh the greater segment double to the side of the Icosidodecahedron.</head>
               <p>SVppose that the diameter of the sphere geuen be NL, and (by the 30. of the sixth) diuide the line NL by an extreame and meane proporti<g ref="char:EOLhyphen"/>on in the poynt I: and the greater segment thereof let be NI.<note place="margin">
                     <hi>Construction of the Icosi<g ref="char:EOLhyphen"/>dodecah<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>dron.</hi>
                  </note> And vpo<g ref="char:cmbAbbrStroke">̄</g> the line NI describe a cube by the 15. of the thirtenth: and about this cube let there be circumscri<g ref="char:EOLhyphen"/>bed
<figure/>
a dode<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>ahedron, by the 17. of the thirtenth: &amp; let the same be ABCDEFHKMO. And diuide euery one of the sides into two equall parts in the poynts Q, R, S, T, V, X, Y, Z, P, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>, G. &amp;c? and couple the sections with right lines, which shall subtend the an<g ref="char:EOLhyphen"/>gles of the pentagons, as the lines PG, GV, VQ, QY, Y<g ref="char:EOLhyphen"/>R, RQ, VT, TX, XV, and so the rest. Now forasmuch as these lines subtend equall an<g ref="char:EOLhyphen"/>gles of the pentagons, and those equall angles are con<g ref="char:EOLhyphen"/>tayned of equall sides (name<g ref="char:EOLhyphen"/>ly of the halues of the sides of the pentagons: therefore those subtending lines are equall, by the 4. of the first. Wherefore the triangles G<g ref="char:EOLhyphen"/>QV, YQR, VXT, and the rest which take away solide angles of the dodecahe<g ref="char:EOLhyphen"/>dron, are equilater. Agayne forasmuch as in euery pentagon are described fiue e<g ref="char:EOLhyphen"/>quall right lines, coupling the middle sections of the sides, as are the lines QV, V<g ref="char:EOLhyphen"/>T, TS, SR, RQ: they describe a pentagon in the playne of the pentagon of the dodecahedron: and the sayd pentagon is contayned in a circle, namely, whose cen<g ref="char:EOLhyphen"/>
                  <gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>re is the centre of a pentagon of the dodecahedron. For the lines drawne from that centre to the angles of this pentagon are equall, for that they are perpendicu<g ref="char:EOLhyphen"/>lars vpon the bases cutte, by the 12. of the fourth. Wherefore the pentagon QRS<g ref="char:EOLhyphen"/>TV is equiangle, by the 11. of the same. And by the same reason may the rest of the pentagons described in the bases of the dodecahedron be proued equall and like. Wherefore those pentagons are 12. in number: And forasmuch as the equall and like t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                     <desc>•</desc>
                  </gap>iangles, doo subtend and take away 20. solide angles of the dodecahedron, therefore the sayd triangles shall be 20. in nu<g ref="char:cmbAbbrStroke">̄</g>ber. Wherfore we haue described an Icosidodecahedro<g ref="char:cmbAbbrStroke">̄</g> by the diffinitio<g ref="char:cmbAbbrStroke">̄</g>, which Icosidodecahedro<g ref="char:cmbAbbrStroke">̄</g> is equilater, for that all the sides of the triangles are equal &amp; co<g ref="char:cmbAbbrStroke">̄</g>mon with the pe<g ref="char:cmbAbbrStroke">̄</g>tagons: and it is also e<g ref="char:EOLhyphen"/>quiangle. For euery one of the solide angles is made of two superficiall angles of an equilater pentagon, and of two superficiall angles of an equilater triangle.</p>
               <p>
                  <pb facs="tcp:7062:490"/>Now let vs proue that it is contained in the Sphere geuen, whose diameter is NL.<note place="margin">
                     <hi>That the Icosi<g ref="char:EOLhyphen"/>dodecahedron is contayned in the sphere geuen.</hi>
                  </note> Forasmuch as perpendiculars drawen fro<g ref="char:cmbAbbrStroke">̄</g> the centres of the Dodecahedron, to the midle sections of his sides, are the halfes of the lines, which couple the op<g ref="char:EOLhyphen"/>posite midle sections of the sides of the Dodecahedron, by the 3. Corollary of the 17. of the thirtenth: which lines also, by the same Corollary, do in the centre diuide the one the other into two equall partes: therefore right lines drawen from that point to the angles of the Icosidodecahedron (which are set in those midle secti<g ref="char:EOLhyphen"/>ons) are equall: which lines are 30. in number according to the number of the sides of the Dodecahedron: for euery one of the angles of the Icosidodecahedron are set in the midle sections of euery one of the sides of the Dodecahedron. Wher<g ref="char:EOLhyphen"/>fore making the centre the centre of the Dodecahedron, and the space any one of the lines drawen from the centre to the midle sections, describe a Sphere, and it shall passe by all the angles of the Icosidodecahedron, and shall containe it.</p>
               <p>And forasmuch as the diameter of this solide, is that right line, whose grea<g ref="char:EOLhyphen"/>ter segment is the side of the cube inscribed in the Dodecahedron, by the 4. Co<g ref="char:EOLhyphen"/>rollary of the 17. of the thirtenth, which side is NI, by supposition. Wherefore that solide is contayned in the Sphere geuen whose diameter is put to be the line NL.</p>
               <p>
                  <note place="margin">
                     <hi>That the dia<g ref="char:EOLhyphen"/>meter being deuided by an extreme and meane propor<g ref="char:EOLhyphen"/>tion. &amp;c.</hi>
                  </note>Now let vs proue that the
<figure/>
greater segment of the diame<g ref="char:EOLhyphen"/>ter is duple to QV the side of this solide. Forasmuch as the sides of the triangle AEB are in the pointes Q and V diui<g ref="char:EOLhyphen"/>ded into two equall partes, the lines QV and BE are pa<g ref="char:EOLhyphen"/>rallels, by the Corollary of the 39. of the first. Wherefore as AE is to AV, so is EB to VQ, by the 2. of the sixth. But the line AE is double to the line AV. Wherefore the line BE is double to the line QV: by the 4. of the sixth. Now the line BE is equall to NI, or to the side of the cube, by the 2. Corollary of the 17. of the thirtenth, which line NI is the greater segment of the diameter NL. Wherefore the greater segment of the diameter geuen, is double to the side of the Icosidodecahedron inscribed in the Sphere geuen. Wherefore we haue described an equilater and equiangle Ico<g ref="char:EOLhyphen"/>sidodecahedron, and contained it in a Sphere geuen, and haue proued that the di<g ref="char:EOLhyphen"/>ameter thereof being diuided by an extreme and meane proportion, maketh hys greater segment double to the side of the Icosidodecahedron.</p>
               <div type="section">
                  <head>¶ An aduertisment of Flussas.</head>
                  <p>To the vnderstanding of the nature of this Icosidodecahedron, ye must well conceaue the passions and proprieties of both those solides, of whose bases i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> con<g ref="char:EOLhyphen"/>sisteth, namely, of the Icosahedron and of the Dodecahedron. And although in it
<pb n="461" facs="tcp:7062:490"/>
the bases are placed oppositely, yet h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>u<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> they one to the other one &amp; the s<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>me in<g ref="char:EOLhyphen"/>clination. By reason wherof there he hidden in it the actions and p<gap reason="illegible" resp="#PDCC" extent="4 letters">
                        <desc>••••</desc>
                     </gap>ons of the o<g ref="char:EOLhyphen"/>ther regular solides. And I would haue thought i<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> not impertinent to the purpose to haue set forth the inscriptions and circumscriptions of this solide, if w<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>nt of t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>h<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> had not hindred. But to the end the reader may the better a<gap reason="illegible" resp="#PDCC" extent="2 letters">
                        <desc>••</desc>
                     </gap>aine to the vnder<g ref="char:EOLhyphen"/>standing therof, I haue here following briefly set forth, how it may in o<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap> about e<g ref="char:EOLhyphen"/>uery one of the fiue regular solides be inscribed or circumscribed: by the helpe whereof <gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e may, with small trauaile or rather none at all, so that he haue well p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>y<g ref="char:EOLhyphen"/>sed and considered the d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>monstrations pertayning to the foresayd fi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                        <desc>•</desc>
                     </gap>e regular so<g ref="char:EOLhyphen"/>lides, demonstrate both the inscription of the sayd solides in it, and the inscription of it in the sayd solides.</p>
                  <div type="subsection">
                     <head>¶Of the inscriptions and circumscriptions of an Icosidodec<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>hedron.</head>
                     <p>An Icosidodecahedron may containe the other fiue regular bodyes. For it will receaue the angles of a Dodecahedron, in the centres of the triangles which subtend the solide angles of the Dodecahedron: which solide angles are 20. in nu<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>ber, and are placed in the same order in which the solide angles of the Dodecahe<g ref="char:EOLhyphen"/>dron taken away or subtended by them, are. And by that reason it shall receaue a Cube and a Pyramis contayned in the Dodecahedron: when as the angles of the one are set in the angles of the other.</p>
                     <p>An Icosidodecahedron receaueth an Octohedron, in the angles cutting the sixe opposite sections of the Dodecahedron, euen as if it were a simple Dode<g ref="char:EOLhyphen"/>cahedron.</p>
                     <p>And it containeth an Icosahedron, placing the 12. angles of the Icosahedron in the selfe same centres of the 12. Pentagons.</p>
                     <p>It may also by the same reason be inscribed in euery one of the fiue regular bo<g ref="char:EOLhyphen"/>dies: namely, in a Pyramis, if ye place 4. triangular bases concentricall with 4. bases of the Pyramis, after the same maner, that ye inscribed an Icosahedro<g ref="char:cmbAbbrStroke">̄</g> in a Pyrami<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>. So likewise may it be inscribed in an Octohedron, if ye make 8. bases thereof con<g ref="char:EOLhyphen"/>centricall with the 8. bases of the Octohedron. It shall also be inscribed in a Cube, if ye place the angles which receaue the Octohedron inscribed in it, in the centres of the bases of the Cube. Moreouer, ye shall inscribe it in an Icosahedron, when the triangles compased in of the Pentagon bases, are concentricall with the trian<g ref="char:EOLhyphen"/>gles, which make a solide angle of the Icosahedron. Finally, it shall be inscribed in a Dodecahedron, if ye place euery one of the angles thereof in the midle secti<g ref="char:EOLhyphen"/>ons of the sides of the Dodecahedron, according to the order of the constructi<g ref="char:EOLhyphen"/>on thereof.</p>
                     <p>The opposite plaine superficieces also of this solide are parallels. For the op<g ref="char:EOLhyphen"/>posite solide angles are subtended of parallel plaine superficieces, as well in the an<g ref="char:EOLhyphen"/>gles of the Dodecahedron subtended by <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ri<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ngle<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> in the angles of the Icosahe<g ref="char:EOLhyphen"/>dron subtended of Pentagons, which thing may easily be d<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>monstra<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ed. More<g ref="char:EOLhyphen"/>ouer in thi<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> solide are infinit<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> properti<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> &amp; p<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>ssion<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>, springing of the solide<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> wher<g ref="char:EOLhyphen"/>of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>t is composed.</p>
                     <p>Wherefor<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap> it is manifest that a Dodecahedron &amp; an Icosahedron, mixed, are
<pb facs="tcp:7062:491"/>
transformed into one &amp; the selfe same solide of an Icosidodecahedron. A cube al<g ref="char:EOLhyphen"/>so and an octohedro<g ref="char:cmbAbbrStroke">̄</g> are mixed and altered into an other solide, namely, into one and the same Exoctohedron. But a pyramis is transformed into a simple and per<g ref="char:EOLhyphen"/>fect solide, namely into an Octohedron.</p>
                     <p>If we will frame these two solides ioyned together into one solide, this onely must we obserue.</p>
                     <p>In the pentagon of a dodecahedron inscribe a like pentagon, so that let the an<g ref="char:EOLhyphen"/>gles of the pentagon inscribed be set in the midle sections of the sides of the pen<g ref="char:EOLhyphen"/>tagon circumscribed, and then vpon the said pentagon inscribed, let there be set a solide angle of an Icosahedron, and so obserue the selfe same order in euery one of the bases of the Dodecahedron: and the solide angles of the Icosahedron set vpon these pentagons shall produce a solide consisting of the whole Dodecahe<g ref="char:EOLhyphen"/>dron, and of the whole Icosahedron. In like sort, if in euery base of the Icosahedro<g ref="char:cmbAbbrStroke">̄</g>, the sides being diuided into two equall partes be inscribed an equilat<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap> triangle, and vpon euery one of those equilater triangles be set a solide angle of a Dodeca<g ref="char:EOLhyphen"/>hedron: there shall be produced the selfe same solide consisting of the whole Ico<g ref="char:EOLhyphen"/>sahedron, &amp; of the whole Dodecahedron.</p>
                     <p>And after the same order, if in the bases of a cube, be inscribed squares subten<g ref="char:EOLhyphen"/>ding the solide angles of an Octohedron, or in the bases of an Octohedron, be in<g ref="char:EOLunhyphen"/>scribed equilater triangles subte<g ref="char:cmbAbbrStroke">̄</g>ding the solide angles of a cube, there shall be pro<g ref="char:EOLhyphen"/>duced a solide consisting of either of the whole solides, namely, of the whole cube and of the whole Octohedron.</p>
                     <p>But equilater triangles inscribed in the bases of a pyramis, hauing their angles set in the midle sections of the sides of the pyramis, and the solide angles of a pyra<g ref="char:EOLhyphen"/>mis set vpon the sayd equilater triangles, there shall be produced a solide, consi<g ref="char:EOLhyphen"/>sting of two equal and like pyramids.</p>
                     <p>And now if in these solides thus composed, ye take away the solide angles, there shalbe restored againe the first composed solides: namely, the solide angles taken away from a Dodecahedron and an Icosahedron composed into one, there shalbe left an Icosidodecahed<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>on: the solide angles take<g ref="char:cmbAbbrStroke">̄</g> away from a cube, and an octohedro<g ref="char:cmbAbbrStroke">̄</g> co<g ref="char:cmbAbbrStroke">̄</g>posed into one solide, there shalbe left an exocthedro<g ref="char:cmbAbbrStroke">̄</g>. Moreouer the solide angles taken away from two pyramids composed into one solide, there shal be left an Octohedron.</p>
                     <p>
                        <hi>Flussas</hi> after this setteth forth certaine passions and properties of the fiue sim<g ref="char:EOLhyphen"/>ple regular bodies: which although he demonstrateth not, yet are they not hard to be demonstrated, if we wel pease and conceiue that, which in the former bookes hath bene taught touching those solides.</p>
                  </div>
                  <div type="subsection">
                     <head>Of the nature of a trilater and equilater Pyramis.</head>
                     <p>
                        <milestone type="tcpmilestone" unit="unspecified" n="1"/> A trilater equilater Pyramis, is deuided into two equal partes, by three equal squares, which in the centre of the pyramis cutte the one the other into two equal partes, and perpendicularly, and whose angles are set in the midle sections of the <milestone type="tcpmilestone" unit="unspecified" n="2"/> sides of the pyramis. From a pyramis are taken away 4. pyramids like vnto the whole, which vtterly take away the sides of the pyramis, and that which is left
<pb n="462" facs="tcp:7062:491"/>
is an octohedro<g ref="char:cmbAbbrStroke">̄</g> inscribed in the pyramys in which all the solides inscribed in the pyramis are contained. A perpendicular drawne from the angle of the pyramis <milestone type="tcpmilestone" unit="unspecified" n="3"/> to the base, is double to the diameter of the cube inscribed in it. And a right line <milestone type="tcpmilestone" unit="unspecified" n="4"/> coupling the midle sections of the opposite sides of the pyramis, is triple to the side of the selfe same cube. The side also of the pyramis is triple to the diameter of the <milestone type="tcpmilestone" unit="unspecified" n="5"/> base of the cube. Wherefore the same side of the pyramis is in power duple to <milestone type="tcpmilestone" unit="unspecified" n="6"/> the right line which coupleth the midle sections of the opposite sides. And it is <milestone type="tcpmilestone" unit="unspecified" n="7"/> in power sesquialter to the perpendicular which is drawne from the angle to the base. Wherefore the perpendicular is in power sesquitertia to the line which cou<g ref="char:EOLhyphen"/>pleth <milestone type="tcpmilestone" unit="unspecified" n="8"/> the midle sections of the opposite sides. A pyramis, and an Octohedron in<g ref="char:EOLhyphen"/>scribed in it, also an Icosahedron inscribed in the same Octohedron, doo containe <milestone type="tcpmilestone" unit="unspecified" n="9"/> one and the selfe same sphere.</p>
                  </div>
                  <div type="subsection">
                     <head>Of the nature of an Octohedron.</head>
                     <p>Foure perpendiculars of an Octohedron, drawne in 4. bases therof from two <milestone type="tcpmilestone" unit="unspecified" n="1"/> opposite angles of the said Octohedron, and coupled together by those 4. bases, describe a Rhombus, or diamond figure: one of whose diameters is in power du<g ref="char:EOLhyphen"/>ple to the other diameter. For it hath the same proportio<g ref="char:cmbAbbrStroke">̄</g> that the diameter of the Octohedron, hath to the side of the Octohedron. An Octohedron &amp; an Icosahe<g ref="char:EOLhyphen"/>dron inscribed in it, do containe one and the selfe same sphere. The diameter of <milestone type="tcpmilestone" unit="unspecified" n="2"/> the solide of the Octohedron, is in power sesquialter to the diameter of the circle <milestone type="tcpmilestone" unit="unspecified" n="3"/> which containeth the base: and is in power triple to the right line which coupleth <milestone type="tcpmilestone" unit="unspecified" n="4"/> the ce<g ref="char:cmbAbbrStroke">̄</g>tres of the opposite bases: and is in power <note n="*" place="margin">
                           <hi>That is as</hi> 8. <hi>103</hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </note> duple superbipartiens tercias to the perpe<g ref="char:cmbAbbrStroke">̄</g>dicular or side of the foresaid Rhombus: and moreouer is in le<g ref="char:cmbAbbrStroke">̄</g>gth triple to the line which coupleth the centres of the next bases. The angle of the incli<g ref="char:EOLhyphen"/>nation <milestone type="tcpmilestone" unit="unspecified" n="5"/> of the bases of the Octohedron, doth with the angle of the inclination of <milestone type="tcpmilestone" unit="unspecified" n="6"/> the bases of the pyramis, make angles equal to two right angles.</p>
                  </div>
                  <div type="subsection">
                     <head>Of the nature of a Cube.</head>
                     <p>The diameter of a cube, is in power sesquialter to the diameter of his base:<milestone type="tcpmilestone" unit="unspecified" n="1"/> and is in power triple to his side: and vnto the line which coupleth the centres of <milestone type="tcpmilestone" unit="unspecified" n="2"/> the next bases, it is in power sextuple. Moreouer the side of the cube is to the side <milestone type="tcpmilestone" unit="unspecified" n="3"/> of the Icosahedron inscribed in it, as the whole is to the greater segment: vnto <milestone type="tcpmilestone" unit="unspecified" n="4"/> the side of the Dodecahedron, it is as the whole is to the lesse segment: vnto the <milestone type="tcpmilestone" unit="unspecified" n="5"/> side of the Octohedron, it is in power duple: and vnto the side of the pyramis, it is <milestone type="tcpmilestone" unit="unspecified" n="6"/> in power subduple. Moreouer the cube is triple to the pyramis: but to the cube <milestone type="tcpmilestone" unit="unspecified" n="7"/> the Dodecahedron is in a maner duple. Wherfore the same Dodecahedron is in a <milestone type="tcpmilestone" unit="unspecified" n="8"/> maner sextuple to the sayd pyramis.</p>
                  </div>
                  <div type="subsection">
                     <head>Of the nature of an Icosahedron.</head>
                     <p>Fiue triangles of an Icosahedron, do make a solide angle, the bases of which <milestone type="tcpmilestone" unit="unspecified" n="1"/> triangles make a pentagon. If therfore from the opposite bases of the Icosahedron
<pb facs="tcp:7062:492"/>
be taken the other pentagon by them described, these pentagons shall in such sort cut the diameter of the Icosahedron which coupleth the forsaid opposite angles, that that part which is contained betweene the plaines of those two pentagons, shalbe the greater segment: and the residue which is drawne from the plaine to <milestone type="tcpmilestone" unit="unspecified" n="2"/> the angle shall be the lesse segment. If the opposite angles of two bases ioyned to<g ref="char:EOLhyphen"/>gether, be coupled by a right line, the greater segment of that right line is the side <milestone type="tcpmilestone" unit="unspecified" n="3"/> of the Icosahedron. A line drawne from the centre of the Icosahedron to the an<g ref="char:EOLhyphen"/>gles, is in power quintuple, to halfe that line which is take<g ref="char:cmbAbbrStroke">̄</g> betwene the pentago<g ref="char:cmbAbbrStroke">̄</g>s, or of the halfe of that line which is drawne from the centre of the circle which co<g ref="char:cmbAbbrStroke">̄</g>
                        <g ref="char:EOLhyphen"/>tayneth <milestone type="tcpmilestone" unit="unspecified" n="4"/> the foresaid pentagon: which two lines are therefore equall. The side of the Icosahedron contayneth in power either of them, and also the lesse segment, <milestone type="tcpmilestone" unit="unspecified" n="5"/> namely, the line which falleth from the solide angle to the pentagon. The diame<g ref="char:EOLhyphen"/>ter of the Icosahedron contayneth in power the whole line, which coupleth the opposite angles of the bases ioyned together, and the greater segment thereof, <milestone type="tcpmilestone" unit="unspecified" n="6"/> namely, the side of the Icosahedron. The diameter also is in power quintuple to the line which was taken betwene the pentagons, or to the line which is drawne from the centre to the circumference of the circle which containeth the pentagon <milestone type="tcpmilestone" unit="unspecified" n="7"/> co<g ref="char:cmbAbbrStroke">̄</g>posed of the sides of the Icosahedron. The dimetient contayneth in power the right line which coupleth the centres of the opposite bases of the Icosahedron, <milestone type="tcpmilestone" unit="unspecified" n="8"/> and the diameter of the circle which contayneth the base. Moreouer the sayd di<g ref="char:EOLhyphen"/>metient contayneth in power the diameter of the circle, which contayneth the pentagon, and also the line which is drawne from the centre of the same circle to the circumference: That is, it is quintuple to the line drawne from the centre to <milestone type="tcpmilestone" unit="unspecified" n="9"/> the circumference. The line which coupleth the centres of the opposite bases, con<g ref="char:EOLunhyphen"/>tayneth in power the line which coupleth the centres of the next bases, and also the rest of that line of which the side of the cube inscribed in the Icosahedron is <milestone type="tcpmilestone" unit="unspecified" n="10"/> the greater segment. The line which coupleth the middle sections of the opposite sides, is triple to the side of the dodecahedron inscribed in it. Wherefore if the <milestone type="tcpmilestone" unit="unspecified" n="11"/> side of the Icosahedron, and the greater segment thereof be made one line, the third part of the whole, is the side of the dodecahedron inscribed in the Icosahe<g ref="char:EOLhyphen"/>dron.</p>
                  </div>
                  <div type="subsection">
                     <head>Of the nature of a Dodecahedron.</head>
                     <p>
                        <milestone type="tcpmilestone" unit="unspecified" n="1"/> The diameter of a dodecahedron contayneth in power the side of the dode<g ref="char:EOLhyphen"/>cahedron, and also that right line, vnto which the side of the dodecahedron is the lesse segment, and the side of the cube inscribed in it is the greater segme<g ref="char:cmbAbbrStroke">̄</g>t: which line is that which subtendeth the angle of the inclination of the bases, contayned <milestone type="tcpmilestone" unit="unspecified" n="2"/> vnder two perpendiculars of the bases of the dodecahedron. If there be taken two bases of the dodecahedron distant the one from the other by the length of one of the sides, a right line coupling their centres, being diuided by an extreame and meane proportion, maketh the greater segment the right line which coupleth the <milestone type="tcpmilestone" unit="unspecified" n="3"/> centres of the next bases. If by the centres of fiue bases set vppon one base, be drawne a playne superficies, and by the centres of the bases which are set vpon the opposite base be drawne also a playne superficies, and then be drawne a right line coupling the centres of the opposite bases, that right line is so cut, that eche of his partes set without the playne superficies, is the greater segment of that part which <milestone type="tcpmilestone" unit="unspecified" n="4"/> is contayned betwene the playnes. The side of the dodecahedron is the greater
<pb n="463" facs="tcp:7062:492"/>
segment of the line which subtendeth the angle of the pentagon. A perpendicular <milestone type="tcpmilestone" unit="unspecified" n="5"/> line drawne from the centre of the dodecahedron to one of the bases, is in power quintuple to half the line which is betwene the playnes: And therfore the whole <milestone type="tcpmilestone" unit="unspecified" n="6"/> line which coupleth the centres of the opposite bases, is in power quintuple to the whole line which is betwene the sayd playnes. The line which subt<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>deth the <milestone type="tcpmilestone" unit="unspecified" n="7"/> angle of the base of the dodecahedro<g ref="char:cmbAbbrStroke">̄</g>, together with the side of the base, are in power quintuple to the line which is drawne from the ce<g ref="char:cmbAbbrStroke">̄</g>tre of the circle, which contayneth the base, to the circumference. A secti<g ref="char:EOLhyphen"/>on <milestone type="tcpmilestone" unit="unspecified" n="8"/> of a sphere contayning three bases of the dodecahedron taketh a third part of the diameter of the sayd sphere. The side of the dodecahedron, and the line which <milestone type="tcpmilestone" unit="unspecified" n="9"/> subtendeth the angle of the pentagon, are e<g ref="char:EOLhyphen"/>quall to the right line which coupleth the middle sections of the opposite sides of the dodecahedron.</p>
                  </div>
               </div>
            </div>
            <trailer>¶The ende of the Elementes of Geometrie, of the most auncient Philosopher <gap reason="illegible" resp="#PDCC" extent="1 word">
                  <desc>〈◊〉</desc>
               </gap> of Megara.</trailer>
         </div>
      </body>
      <back>
         <div type="errata">
            <pb facs="tcp:7062:493"/>
            <head>Faultes escaped.</head>
            <p>
               <gap reason="illegible" resp="#PDCC" extent="1 paragraph">
                  <desc>〈1 paragraph〉</desc>
               </gap>
            </p>
            <p>
               <table>
                  <row>
                     <cell role="label">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>cl.</hi>
                     </cell>
                     <cell role="label">
                        <hi>
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ag.</hi>
                     </cell>
                     <cell role="label">
                        <hi>Line.</hi>
                     </cell>
                     <cell role="label">
                        <hi>Faultes<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </cell>
                     <cell role="label">
                        <hi>Co<gap reason="illegible" resp="#PDCC" extent="7 letters">
                              <desc>〈7 letters〉</desc>
                           </gap>.</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell role="label">Errata Lib. 1.</cell>
                     <cell> </cell>
                  </row>
                  <row>
                     <cell>1</cell>
                     <cell>2</cell>
                     <cell>41</cell>
                     <cell>
                        <hi>point B. at Campane</hi>
                     </cell>
                     <cell>
                        <hi>point C, a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> Campane</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>3</cell>
                     <cell>1</cell>
                     <cell>22</cell>
                     <cell>
                        <hi>a<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>l lines drawne</hi>
                     </cell>
                     <cell>
                        <hi>all righ<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 span">
                              <desc>〈…〉</desc>
                           </gap>
                        </hi>
                     </cell>
                  </row>
                  <row>
                     <cell>3</cell>
                     <cell>1</cell>
                     <cell>28</cell>
                     <cell>
                        <hi>lines drawen to the superficies</hi>
                     </cell>
                     <cell>
                        <hi>right lines drawe<g ref="char:cmbAbbrStroke">̄</g> to the circumference</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>9</cell>
                     <cell>1</cell>
                     <cell>42</cell>
                     <cell>
                        <hi>li<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>es AB and AC,</hi>
                     </cell>
                     <cell>
                        <hi>lines AB and BC</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>15</cell>
                     <cell>1</cell>
                     <cell>35</cell>
                     <cell>
                        <hi>are equall</hi>
                     </cell>
                     <cell>
                        <hi>are proued equall</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>20</cell>
                     <cell>2</cell>
                     <cell>28</cell>
                     <cell>
                        <hi>(by the first</hi>
                     </cell>
                     <cell>
                        <hi>(by the fourth</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>21</cell>
                     <cell>1</cell>
                     <cell>39</cell>
                     <cell>
                        <hi>t<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>e centre C.</hi>
                     </cell>
                     <cell>
                        <hi>the centre E</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>2<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </cell>
                     <cell>2</cell>
                     <cell>
                        <gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </cell>
                     <cell>
                        <hi>I<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap> 
                           <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>ower right</hi>
                     </cell>
                     <cell>
                        <hi>If two right</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>25</cell>
                     <cell>2</cell>
                     <cell>3</cell>
                     <cell>
                        <hi>f<gap reason="illegible" resp="#PDCC" extent="3 letters">
                              <desc>•••</desc>
                           </gap> petition</hi>
                     </cell>
                     <cell>
                        <hi>fiueth petition</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>49</cell>
                     <cell>2</cell>
                     <cell>7</cell>
                     <cell>14.<gap reason="illegible" resp="#PDCC" extent="2 letters">
                           <desc>••</desc>
                        </gap>.32.64 <hi>&amp;c.</hi>
                     </cell>
                     <cell>4.8.16.32.64. <hi>&amp;</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>53</cell>
                     <cell>1</cell>
                     <cell>39</cell>
                     <cell>the triangle NG</cell>
                     <cell>the triangle K<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </cell>
                  </row>
                  <row>
                     <cell>54</cell>
                     <cell>2</cell>
                     <cell>25</cell>
                     <cell>
                        <hi>(by the</hi> 44</cell>
                     <cell>
                        <hi>(by the</hi> 42</cell>
                  </row>
                  <row>
                     <cell>57</cell>
                     <cell>2</cell>
                     <cell>23</cell>
                     <cell>
                        <hi>and C<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>G in the</hi>
                     </cell>
                     <cell>
                        <hi>and CGB is th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>
                        </hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell cols="2">
                        <hi>In stede of <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>lussates through out <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap> whole booke read <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>lus<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>as.</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell role="label">Errata Lib. 2.</cell>
                     <cell> </cell>
                  </row>
                  <row>
                     <cell>60</cell>
                     <cell>2</cell>
                     <cell>29</cell>
                     <cell>
                        <hi>Gnomon FGEH</hi>
                     </cell>
                     <cell>
                        <hi>Gnomon AHKD</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell>30</cell>
                     <cell>
                        <hi>Gnomon EHFG</hi>
                     </cell>
                     <cell>
                        <hi>Gnomon <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>CKD</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>69</cell>
                     <cell>1</cell>
                     <cell>18</cell>
                     <cell>
                        <hi>the whole line</hi>
                     </cell>
                     <cell>
                        <hi>the whole <gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>igure</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>76</cell>
                     <cell>2</cell>
                     <cell>9</cell>
                     <cell>
                        <hi>the diameter CD</hi>
                     </cell>
                     <cell>
                        <hi>the diameter AHF</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell role="label">Errata Lib. 3.</cell>
                     <cell> </cell>
                  </row>
                  <row>
                     <cell>82</cell>
                     <cell>2</cell>
                     <cell>36</cell>
                     <cell>
                        <hi>angle</hi>
                     </cell>
                     <cell>
                        <hi>equall to the angle</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>92</cell>
                     <cell>1</cell>
                     <cell>
                        <hi>last</hi>
                     </cell>
                     <cell>
                        <hi>the line AC is</hi>
                     </cell>
                     <cell>
                        <hi>the line AF <gap reason="illegible" resp="#PDCC" extent="1 word">
                              <desc>〈◊〉</desc>
                           </gap>
                        </hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell role="label">Errata Lib. 4.</cell>
                     <cell> </cell>
                  </row>
                  <row>
                     <cell>110</cell>
                     <cell>2</cell>
                     <cell>10</cell>
                     <cell>
                        <hi>CD toucheth the</hi>
                     </cell>
                     <cell>
                        <hi>ED toucheth the</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell>12</cell>
                     <cell>
                        <hi>side of the other</hi>
                     </cell>
                     <cell>
                        <hi>angle of the other</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>115</cell>
                     <cell>1</cell>
                     <cell>21</cell>
                     <cell>
                        <hi>and HB</hi>
                     </cell>
                     <cell>
                        <hi>and HE</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>117</cell>
                     <cell>2</cell>
                     <cell>44</cell>
                     <cell>
                        <hi>the angle ACD</hi>
                     </cell>
                     <cell>
                        <hi>the angle ACB</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>118</cell>
                     <cell>1</cell>
                     <cell>2</cell>
                     <cell>
                        <hi>into ten equall</hi>
                     </cell>
                     <cell>
                        <hi>into two equall</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>121</cell>
                     <cell>1</cell>
                     <cell>3<gap reason="illegible" resp="#PDCC" extent="1 letter">
                           <desc>•</desc>
                        </gap>
                     </cell>
                     <cell>
                        <hi>CD, and EA,</hi>
                     </cell>
                     <cell>
                        <hi>CD, DE, &amp; EA,</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>
                        <gap reason="illegible" resp="#PDCC" extent="3 letters">
                           <desc>•••</desc>
                        </gap>
                     </cell>
                     <cell>1</cell>
                     <cell>29</cell>
                     <cell>
                        <hi>the first</hi>
                     </cell>
                     <cell>
                        <hi>the third</hi>
                     </cell>
                  </row>
                  <row>
                     <cell> </cell>
                     <cell> </cell>
                     <cell> </cell>
                     <cell role="label">Errata Lib. 5.</cell>
                     <cell> </cell>
                  </row>
                  <row>
                     <cell>126</cell>
                     <cell>1</cell>
                     <cell>43</cell>
                     <cell>
                        <hi>it maketh</hi> 12. <hi>more then</hi> 17. <hi>by</hi> 5.</cell>
                     <cell>
                        <hi>it maketh</hi> 24. <hi>more then</hi> 17. <hi>by</hi> 7.</cell>
                  </row>
                  <row>
                     <cell>129</cell>
                     <cell>1</cell>
                     <cell> </cell>
                     <cell cols="2">
                        <hi>In stede of the figure of the</hi> 6. <hi>definition draw in the mag<gap reason="illegible" resp="#PDCC" extent="2 letters">
                              <desc>••</desc>
                           </gap> a figure like vnto th<gap reason="illegible" resp="#PDCC" extent="1 letter">
                              <desc>•</desc>
                           </gap>s.</hi>
                     </cell>
                  </row>
                  <row>
                     <cell cols="5">
                        <figure/>
                     </cell>
                  </row>
                  <row>
                     <cell>134</cell>
                     <cell>2</cell>
                     <cell>4</cell>
                     <cell>
                        <hi>As AB is to A, so is CD to C</hi>
                     </cell>
                     <cell>
                        <hi>As AB is to B, so is CD to D</hi>
                     </cell>
                  </row>
                  <row>
                     <cell>141</cell>
                     <cell>2</cell>
                     <cell>
                        <hi>last</hi>
                     </cell>
                     <cell>
                        <hi>But if K excede M,</hi>
                     </cell>
                     <cell>
                        <hi>But if H excede M,</hi>
                     </cell>
                  </row>
                  <pb n="464" facs="tcp:7062:493"/>
                  <row>
                     <cell>
                        <gap reason="illegible" resp="#PDCC" extent="1 page">
                           <desc>〈1 page〉</desc>
                        </gap>
                     </cell>
                  </row>
               </table>
            </p>
         </div>
         <div type="colophon">
            <pb facs="tcp:7062:494"/>
            <figure>
               <head>LIEFE IS DEATHE AND DEATH IS LIEFE: AETATIS SVAE: XXXX</head>
            </figure>
            <p>
               <hi>AT LONDON</hi> Printed by Iohn Daye, dwelling ouer Aldersgate beneath Saint Martins.</p>
            <p>¶ These Bookes are to be solde at his shop vnder the gate. 1570.</p>
            <pb facs="tcp:7062:494"/>
         </div>
      </back>
   </text>
</TEI>
