AN ESSAY For the Discovery of Some NEW Geometrical Problems, (Judged by some Learned Men, Impracticable) Concerning ANGULAR SECTIONS, Beginning with the GEOMETRICAL TRISECTION of any Right Lined Angle, by Plain Geometry of Right Lines and Arches of Circles,

With RVLE and COMPASS only, without all Conick Sections, and Cubick Aequations.

Whether the following Praxis, and apparent demonstration thereof doth not only make it Practicable, but easie to the Understanding of a Tiro, who but understands a little in true Geometrical Learning.

Which layeth a Foundation of a Plain Method how to Sect any Angle into any other Number of Parts required, Even as 4. 6. 8. 10; or Vneven, as 5. 7. 9. 11. &c. As also to divide a Circle into any number Even, or Vneven of equal parts.

All which have great Uses in the Improvement of the Mathematical Sciences, some of which are here specified.

Proposed and Submitted to the Impartial Tryal and Examination of the Right Reason of such Artists, to whose Hands it may come.

By G. K.

London, Printed 1697. And to be Sold by the Author, at his House in Pudding-Lane, at the Sign of he Golden-Ball, near the Monument; And by B. Aylmer at the Three Pigeons over against the Royal-Exchange.

# Some New Geometrical Problems, &c.

ALthough the Trisection of a right lined Angle and also its Se­ction into any parts required, to a true Mathematical exact­ness, is denyed to be practicable by plain Geometry (of right Lines and Arches of Circles, with Rule and Compass only, without all Conick Sections, and Algebra Equations) by some learned Artists; yet others as learned are not so positive, but acknowledge the Practise of it is not as yet discovered, among whom is the learned Is. Barrow, who hath writ thus, in Corol. ad. 9. 1. elem. Eucl.

Methodus vero regula & Circino angulos secandi in aequales quotcun (que) hactenus Geometras latuit. and in Schol. ad 16. 4. Elem. Coeterum divisio circumferentiae in partes datas etiamnum desideratur.

## The Praxis of the Trisection of any given Angle.

Suppose the given Angle to be the Angle of 150 Degrees, made by the addition of 90 d. the Quadrant, and 60 d. the ⅙ of the Circle, whose exact Mathematical quantity is known. And let that Angle be in the subsequent Figure, the Angle BAC. whose Cord is BC, and the Arch answering to that Cord BFC.

[Page 4] 1. Divide the Cord BC into 3 equal Parts, as BE=EE=EC by 10. 6. elem. Euclid. and draw the 2 Perpendiculars EF. EF.

2. With the extent of ⅓ of the Cord as EC measure on the Arch from C to G, and draw the Line or Cord GC=EC.

3. With another Radius less than AC, viz. AK draw a second arch as KL, until it cut the perpendicular EF, and let the Radius AK be so long, that the extent of the Cord betwixt K and L be somewhat longer than EC, and with the same extent EC measure on this second arch KR=EC.

4. From G to R draw a straight Line by post 1. 1. elem. and pro­duce the same by post 2. until it cut the Perpendiculars at I and O.

5. Draw the straight lines AI and AO, and extend them to H on both sides, so that the one line shall be AIH, and the other AOH, which two Lines shall sect the given angle BAC into three parts or angles, viz. BAH, HAH, HAC.

The Question, or Problem to be resolved is, whether these three angles are not equal, and consequently that the angle BAC is tri­sected into equal angles.

The apparent demonstration that is here proposed to prove it, followeth.

In order to the apparent following demonstration, by way of Preparation,

1. Draw the line OI, which shall be parallel to EE by 28. 1. elem. and extend it on both sides, from I to N on the one side, and from O to N on the other, making IN=ON=EC=OI.

2. With Radius IN, and ON, describe the arches on both sides NM until these arches cut the lines AC and AB, so shall the lines or cords drawn betwixt I and M and O and M, viz. IM and OM=EC.

3. Make the angle IMP=angle MIN by 23. 1. elem. and OMP=MON, and extend the line MP, until it cut the line AI at P, and AO at P.

4. Draw the line OP parallel to IM, until it cut MP, whence the perfect Rhombus OMPI shall be formed, having the opposit side; paralel and equal, for PM by construction is parallel with OI, and OP with IM and that the Line OP drawn parallel to IM can cut the Line PM no where but at P on the line AI is proved thus. Suppose it cut the line PM any where else, than on the line AI on either side, it should make PM either longer or shorter than its opposit side (for the arch drawn by the radius MI, and cutting the line AI at P proveth MP=MI by def. 15. 1. elem.) as also OP should be longer or shorter than IM, (as the like arch [Page 5] conceived to be drawn by radius OI, and cutting AI at P, proveth OI=OP by the same, def. 15. 1. elem.) but both these Conse­quences are absurd, making the opposit sides of a Paralellogram to be unequal, contrary to 34. 1. elem.

Because the Figure OPMI is proved to be a perfect Rhombus, having all the sides equal, whereof the line PI is the diagonal, (it be­ing proved that OP and PM terminate on AI) therefore the angle OIP (=OIA)=PIM=(AIM) by 8. and 4. 1. elem.

Lastly, these two Triangles OAI and IAM shall be according to the 4th. prop. 1. elem. and so shall OAI and OAM, for OI=IM as above proved, and AI is common to both, and the Angle OIA=AIM as is above proved, therefore by 4. 1. elem. AM=AO =AI, and consequently by 8. 1. the angles IAM=OAI=OAM, therefore the given angle BAC is Geometrically trisected by the Lines AO and AI extended to H. Q. E. D.

Here Note, let the Radius be ever so much changed betwixt M and C, taking it at any extent from A, and let ever so many Concen­trick Arches be drawn from the Center A betwixt IM and GC, their Cords terminating on the streight Lines MC on the one side, and IG on the other shall all be equal, one to another, and to the Cord GC= [...]BC. as the demonstration above given proveth, for the reason that proveth the Cords of any 3 Concentrick Arches, termi­nating on two streight Lines to be equal within the limits MI and GC, being a Trapezia, proveth any other 3 to be equal within the same limits. But if we draw any Concentrick Arches without the Trapezia IGCM, as with a less Radius, than AM, or with a greater Radius, than AC, the Case is altered, and the equal Cords will not terminate on the streight Line IG, but diviate or depart therefrom, as both true Reason doth prove, and even ocular in­spection doth shew; for though the Eye is not able to judge of a straight Line, yet when a Line is visibly apparent to make an Angle with another Line, these two Lines cannot be a straight Line. I call the Figure IMCG a Trapezia, for it is not any Paralellogram, because the side IG, is not parallel to the side MC, but very unpa­ralel, which makes the side IG to be longer than the side MC, but this doth not hinder that all the Cords of the Concentrick Arches drawn betwixt IG and MC are equal, for let them be ever so many within the limits IM and GC, they are all equal, though their Arches are unequal being from a differing Radius. And though the demonstration above given may seem sufficient to prove it, yet for a further proof, let a Line be drawn, or conceived to be drawn (as in this Figure it is only conceived) from I Paralel to AC, towards [Page 6] FC and beyond it, all the extream Points of the Cords of the Con­centick Arches, above mentioned, terminating on the streight line IG, do gradually still more and more depart from that parallel to­wards the perpendicular IF within the aforesaid limits, IM and GC, on the first Arch, which I call the primitive Arch; but if we draw any Arch, without GC by a longer Radius than AC, or within IM by a shorter Radius than AM, the Cords of those concentrick Ar­ches, if equal, shall deviate from the streight line IG, and come nearer to the Parallel above mentioned, but that within the Tra­pezia above said IMCG, the equal Cords of all these Arches shall terminate on the straight line IGI thus further prove. Draw, (or conceive as drawn in this Figure) a straight Line from C to H pa­rallel to IM, and take a Cord of any of those Concentrick Arches =⅓ of the Cord BC=IM, I say it must terminate on the line IG, and neither go without it, nor within it, otherwise the same extent cutting the paralel Lines, viz. the one parallel to MC, the other to IM, shall make the opposite sides of a Paralellogram unequal contrary to 34. 1. elem. Eucl. as the due Cosideration of the Figure Will show, and the part equal to the whole.

I have in the Example of Trisection used an obtuse Angle, because it is more commodious and easie to be done by Manual Operation in an obtuse Angle than in an Acute as Experience will show, though the demonstration is the same both in the Obtuse and Acute. Nor is it any just exception, that some acute Angles being small, can hardly be trisected by this Method; for when Angles are very acute, they can hardly be bisected. But the proper Remedy for both is, when the Angle is small, double or quadruple it, and then bisect or trisect it, as occasion requires.

The Praxis of the Trisection hath these following great Uses.

By the Trisection of the Angle of 120 degrees we have the true Mathematical Cord of 40 degrees, and seeing it is demonstrated in 11. 4. elem. Euclid. to find the Cord of ⅕ of the Circle,=72 degrees the ½=36 by substracting 36d. from 40d. we have the true Cord of 4 degrees, that bisected gives the Cord of 2 degrees and that a­gain bisected gives the Cord of I degree. Also 40 degrees twice bi­sected gives 10 degrees, and that again bisected gives 5. and thus by the Trisection of 120 or 60 degrees, and certain other bisecti­ons a line of Cords truly Geometrical or Mathematical can be made by plain Geometry, without all conick Sections or Algebra Aequati­ons, and tedious Extractions of Roots, and also without any Table of natural Sines, which never hitherto hath been taught (so far as I ever heard or read) in a plain method, by plain Geometry, so as to [Page 7] be made intelligible to any Tiro or Young Artist; and yet such a method is altogether necessary for the perfection of Geometry and Mathematical Learning, that the way to trisect any Angle be known by plain Geometry, seeing many things in plain Geometry require a true line of Cords, and a true line of Cords cannot be found without the Trisection of certain Angles, or some other Section than Bisecti­on; and to refer a Tiro, or Young Scholar, to the conick Secti­ons, and tedious Algebra Aequations and Resolutions how to trisect an Angle, or understand how to make a true Geometrical Line of Cords is altogether immethodical, even as much or more as to referr him to some of the difficultest Problems in Euclid's Elements to understand the demonstration of one of the first Propositions in those Elements. Seeing therefore Geometry as it is one of the best natural Sciences for certainty and use, so for good Method; and the nature of good Method requireth that in reaching we proceed to the more unknown from and by the more known, therefore to presuppose the knowledge of conick Sections to the knowledge of some necessary Problems in plain Geometry, is greatly incongruous and an immethodical Hysteron Proteron, which this new Praxis doth Remedy.

2. By this Praxis of Trisection a Foundation of Method is Iaid to sect any Angle given into any other equal parts whatsoever, even as 4. 6. 8. 10. or uneven as 5. 7. 9. 11. &c. and an universal Can­non is formed, as by trisection of the Cord of an Angle, that Angle is trisected, so by quinquisection of the Cord of an Angle it is quin­quisected, and the like of all others.

3. By the like method a Circle is divided into any number of e­qual parts even or uneven.

4. And, by the like Method, the quantity of any Right-lined An­gle can be found in degrees, and odd Minutes, without any line of Cords by a line of equal parts only, as the diagonal Scale of Inch, or half Inch or ¼ to a far greater exactness than by a line of Cords.

5. By the same, a way is taught how to protract or project any Angle, whose degrees and odd Minutes are given.

6. Having any one Angle given in any Triangle, and the Ratio of the 2 other Angles, without any side given, to find the other 2 An­gles, and truly to protract them.

7. From a point given on any line given to raise an Isosceles Tri­angle, one of whose sides being produced below that line given shall terminate on any Point, given below the same, the which hath spe­cial use to solve some new Problems in Surveying, Geography, Archi­tecture, Navigation.

[Page 8] 8. As this method of Praxis saveth the pains of finding out Trisection of an Angle by a conick Section, and Algebra Aequations, by teaching the same more easily by plain Geometry, so it is proba­ble it may prove of great use, both in conick Sections, Algebra Que­stions, and other abstruse Theorems to improve the Mathematical Sciences.

9. How to find the natural Cord, Sine, Tangent, Secant of any Angle given in degrees and Minutes, in true Lines, more exactly, than by any common Line of Cords.

## Here followeth another Demonstration of the same Praxis of the Trisection. In a Second Figure.

BY the method of the foregoing Praxis, let the points I and O be found, and the lines AI and AO, which I say shall trisect the Angle NAK=BAC, Geometically into three equal parts.

In order to the demonstration of which, extend the line AI to D on the arch BFC. and AO to X.

Again, with the extent AI on the center I describe the circle ASHT. and with the same Radius describe the arch KLON.

Again, draw IH parallel to AK, extending IH to the Circum­ference of the Circle ASHT.

Again, draw AS parallel and equal to OI, which shall cut the circle at S, because OI=AS cuts the arch KION, having the same Radius.

Again, from the point S on the circumference, draw the line SH parallel to AI, which shall necessarily cut the line IH on the circum­ference at the point H, because by construction IH is parallel to AK, and KH to AI, which makes the Parallogram AKHI, whose opposit sides are equal, but if SH, and IH did meet any where else than on the circumference, as either within, or without it, the op­posit sides of the Parallogram should not be equal contrary to 34. 1. Eucl. elem.

Again, draw the line SI, and extend the line OI from I to K until it cut SH at K, which shall form another Parallogram ASKI, which being divided by the Diagonal line SI, shall give two equal Isosceies Triangles AIS(=OAI) ISK.

## The Demonstration.

In the Isoceles triangle ISH (whose sides IS=IH by 15. def. 1. elem.) the angle IHS=ISH by 5. 1. elem. and IHS=IAK by 8. 1. elem. there­fore IAK=ISH. but ISH=AIS=OAI being alternate Angles be­twixt Parallels by 29. 1. elem. and therefore lasty, OAI=IAK, and by the like method OAI is proved=OAN drawing the like pa­rallel Lines on the other side of the Triangle OAI, and therefore the angle BAC is trisected by 3 equal angles NAO=OAI=IAK= BAX=XAD=DAC. Q. E. D.

## The Praxis of the Section of a Right Lined Angle into Six equal parts.

BEcause of the affinity of the Figure of the Trisection with the Figure of the Sextisection, I shall begin with the Sextisection, and then proceed to the Quinquisection.

In the 3d figure, let the given angle be BAC, as above, whose Cord is BC, and its arch BEFGC.

1. Divide the Cord BC into 6 equal parts, which Divisions are marked with *****. and on the middle division draw the line ADI to the arch BEFGC.

2. With the extent of one of the Divisions, on the given arch measure from C to F, and with the same extent from F to E, and from E to G, marking the points E. F. G. and with the same extent measure from B to E, from E to F, and from F to G marking the points EFG.

3. With a less Radius than AC as AK draw a second Arch as KLHLK. which Radius must be so long that the extent of the 16th of the Cord thrice repeated, may fall short of reaching the line ADI.

4. With the same extent on this 2d. arch, measure from AC, and from AB to K, and from K to L, and from L to H, by a three­fold repitition of the same extent.

5. From the point G on the first arch to the point H on the second arch, draw a straight line, and produce it until it cut the line ADI at the point I.

6. With the Radius AI draw a 3d. concentrick Arch, betwixt the lines AB and AC, I say the extent of the 16th part of the cord BC= MC, being 6 times repeated shall cut the said 3d arch into six Equal parts to every one of which straight lines drawn from the Center A, and produced to the outmost arch, shall divide the given angle BAC into six equal parts orangles, marked with 1. 2. 3. 4. 5. 6.

In order to the demonstration of the Praxis of this Section of an angle into 6 equal parts, on the point 5, making it a center with Radius 5A describe the circle AOPQRS.

2. From the point 5 draw the line 5R parallel to AC.

3. From R draw the line RT parallel to 5A

4. Draw the line AT from A to T

5. Draw the line T5, and draw 5V parallel to AT

## The Demonstration.

In the triangle T5R being an Isosceles, the angle 5TR=5RT. by 5. 1. elem. and 5RT=(5R6) 5A6 therefore 5T6=5A6. [Page 10] Again 5 T6=(5TV) A5T, and A5T=5A4 by 34.1. and therefore lastly 5 A4=5 AV. and as these 2 angles are proved equal by the like reasoning, all the other angles can be proved equal by changing the center of the Circle from 5 to 4, and from 4 to 3, and from 3 to 2, and from 2 to 1. This demonstration having such affinity with the former of Trisection, and which can as easily be demonstrated both ways, as that of Trisection, I shall not en­large upon it, for he who understands and assents to the verity of the former, by the same evidence will assent to the later.

## The Praxis of the Quinquisection.

1. ACcording to the former method, divide the cord of the given angle, BE into 5=parts.

2. On the middle part marked with CD, draw the lines CH and DH cutting the cord at right angles and parallel one to another.

3. On the arch of the given angle BFG take the ⅕ of the cord= CD and set it from E to F, and from F to G, also do the like from B to F, and from F to G.

4. Draw a second arch, with a less radius as AK, so as the radius may be so long that the extent of the ⅕ of the cord twice taken on that second arch may not reach to the perpendicular DH.

5. With the same extent twice taken, as from K to * from * to * measure on the second arch to the second *.

6. From G to * draw a straight line, until it cut the perpendi­cular line DH at H; lastly with the radius AH describe the third arch IHHI, and draw the lines AH AH which shall make the angle HAH=⅕BAE.

The Demonstration of this, or any other Section, being after the same method with the former, as also the Praxis, it were super­fluous to enlarge upon it, or to add any new Problems, showing how to sect any angle into any other parts given, as 7. 8. 9. 10. 11. &c. for he who understands by the foregoing Method and Praxis to sect any angle into 3. 5. 6. as is above shewed, will by the like Method and Praxis be able to sect any angle into 7. 8. 9. 10. &c. equal parts, and also to demonstrate the same.

Wherefore in the next place I shall proceed to shew how a Semi­circle may be sected into any number of equal parts, even as 4. 6. 8. 10, &c. or uneven, as 5. 7. 9. 11.

## The Praxis of a Section of a Semicircle into 9 equal parts.

LEt the Semicircle given be BRSC, whose diameter is BC. and whose center is A.

1. Divide the diameter BC into 9 equal parts.

2. From the center A set off AO and AN, making NO=1/9 of the diameter: and draw the perpendiculars NR. OS.

3. With the extent NO measure from C to I, from 1 to 2, from 2 to three, from 3 to 4, marking the points 1. 2. 3. 4. and do the like from B to 1. 2. 3. 4.

4. With less radius as AP describe the Semicircle PP, and from P to 1 with the same extent on that 2d. arch measure form P, P1= 1. 2=2. 3=3. 4. and do the like from P on the other side. This lesser Radius must be so long, that the extent=NO four times repeat­ed do not reach the Perpendiculars.

5. From 4 to 4 draw a straight line, and produce it until it cut OS at 4. and NR at 9.

6. Draw the 3d arch QQ with radius A 4, and draw the straight lines 4 A 4, which shall make an angle=a 9th part of the semi­circle. The demonstration whereof is apparent from these precedent.

From the above delivered Praxis of the various Sections of angles, into any equal given parts, I shall draw these few plain Corolaries.

1. Corol. Having in any right lined triangles, any one angle given, and the ratio of the other 2 angles to find these 2 angles, in degrees and minuts. e. g. let the given angle of a triangle be 110 deg. and the ratio of the other two angles be as 3 to 2. Q. What are the other 2 angles. Ans. Take the complement of 110 to 180, which is 70, and sect it into 5 equal parts by the method above delivered, ⅕=14, this multiplied by 3 gives 42, and by 2 gives 28, wherefore the grea­ter angle shall be 42, and the less 28. demonstrate 110+42+28 =180 and 3∶ 2∶ 42∶ 28.

2. Cor. From a Point given on a Line given, to erect an Isosceles triangle, one of whose sides produced shall come to a point given. (See the figure) let the point given be A, the line given AF, the limited point given below the line AF let the point be point D. the Praxis.

1. With radius AD describe the semicircle GFDE, so that GA shall cut AF at a right angle, and AE the like.

2. Draw the line AD, and cut the angle EAD into 3=parts.

3. Make the angle GAB=⅓ of EAD, and from B to D draw the line BD, cutting the line AF at C, the triangle ABC is an Isos­celes, whose side BC is produced to D. The demonstration is evi­dent from the trisection, and it hath singular uses in Architecture, and affording new Problems in Geography and Navigation, which I [Page] shall not at present inlarge upon, but leave to the ingenious Stu­dent to devise.

3. Corol. To measure any given angle in degrees and odd minuts, (if it have any) by a line of equal parts without a line of cords. Let the given angle be KAL (see the figure) with radius AL describe the quadrant LKL, and on that arch take the extent of the cord KL, and set it on the other side from L to a point on that arch.

Again, take the cord of that middle arch betwixt the point and K, an all these 3 cords join them together in a straight line, which make the cord of the quadrant (by a longer radius) as BC, which is easily done Geometrically.

Then because in each Quadrant there are 5400 minuts, divide the line BC by the Diagonal Scale into 5400 equal parts, and by the same Diagonal Scale measure on the line BC the length of the cord of KL, and whatever number of parts it giveth, that is the just quan­tity of that angle in minuts, which divided by 60 giveth the de­grees, and if there be any Remainder, they are the odd minuts.

4. Corol. To project any angle without a line of cords, by a Dia­gonal Line of equal parts, whose quantity is given in degrees and minuts. Let the given angle to be projected, contain 36 deg. and 1 min. reduce the degrees into minuts, multiplying 36 by 60, the product is 2160, to which adding the odd minut, the sum is 2161.

## Praxis.

Take off from the cord of the quadrant BC 2161 from C to D. on the point D erect the perpendicular ED, which extend to G, with the extent CD measure on the arch of the quadrant, from C to H.

Again, draw another arch with a less radius as AM (so that the radius be so long that the extent of DC reach not to EDG) and with that extent measure from M to I on that 2d arch.

4. From H to I draw a straight line, and produce it until it cut the perpendicular at K.

5. Draw the line AK, which shall give the angle required KAL=FAC. the de­monstration whereof is apparent from the foregoing Problems.

If any discreet Persons desire further satisfaction or Information about these New Problems, if they please to call at my House, I shall be ready to endeavour to satisfie them; and whereas, to me the Method and Demonstration of these Matters, seem suf­ficiently evident, yet I have called the whole Praxis and Method here delivered, but an Essay, until it be found approved by able Artists, and have proposed it rather Prob­lematically than Thetically, as is usual in such cases.

G. Keith.