But [by 32. 1.] because the angle C D A is equal to D E C, ECD: Therefore will the angle C D A be twice the angle CED. By the fame reason, because CD is equal to CA, &c. the angle A C B will be thrice the angle D C E, and so fince [by conftruction] F G is parallel to EA, the angle ACB [by 29. 1.] will be thrice the angle G F B. Therefore [by 26. 3.] the arch A B will be thrice the arch F D ; that is, the arch A B will be thrice the arch A G, for the arch AG is equal to F D, fince A D, F G are parallel. Therefore the arch B G will be two equal parts of three of the arch A B, or four equal parts of fix of the arch A B; that is, fince A B [by conftruction] is one fixth part of the circumference of the circle, the arch BG will be four equal parts of thirty fix, or two equal parts of eighteen, or one part of nine, of the whole circumference of the circle.

Therefore, &c. Which was to be demonftrated.

Note, This problem, no more than the laft, can be refolved by a right line and a circle,

To defcribe any regular polygon upon a given right line, admitting the divifion of a given arch of a circle into any number of equal parts.

Let there be a given right line a B. It is required to defcribe a regular polygon upon the given right line A в. First, Let the number of fides of the polygon be odd. Continue out the given right line A B to D, fo that A D be equal to a B. About the centre A with the distance A D, defcribe part of a circle D E C. From A [by II. I.] draw A C perpendicular to A D, meeting the circumference of that circle in the point c. Divide the quarter D C of the circumference into the fame number of equal parts as the polygon to be defcribed has fides. Then if the polygon has five fides, let c E be one of thofe parts. If it has feven fides, let c E be three of those parts; if the polygon has nine fides, let c E be five of thofe parts; if it has eleven, let it be feven of thofe parts; and fo on increafing by two, &c. But if the number of fides of the polygon to be defcribed be even, the arch c E, if the poly

gon

gon has fix fides, muft be two of the fix equal parts the quadrant DC is divided into, or one part of three that it is divided into. If the polygon has eight fides, the arch CE must be four of eight equal parts; that is, one half the quadrant D c. If the polygon has ten fides, the arch c E must be fix of ten equal parts, or three of five that the quadrant D C is divided into. If the polygon has twelve fides, the arch CE must be nine parts of twelve, or three parts of four that the quadrant D C is divided into, and so on. &c.

This done, draw the femidiameter AE of the quadrant D C. Bifet [by 9. 1.] the angle E A B by the right line A F, and having made the angle ABH

[by 23. 1.] equal to the D

G

angle E A B, bifect the angle A B H by the right line BF meeting the right line A F in the point F, and describe a circle A G B about the centre F with either of the distances AF, FB; and applying a right line equal to A B or A E from в to H, and from H to G, and from G to E, &c. round the circumference of this circle. I fay there will be described upon the given right line A B the regular polygon required.

For because all the angles of a regular polygon are equal to one another; and fince the fum of all the angles of any right lined figure whatsoever, is [by prop. 3. of addit. to lib. 1.] equal to twice as many right angles as the figure has fides, abating four; therefore all the angles of a pentagon will be equal to fix right angles. Confequenly one of its angles will be one right, and one fifth part of one right angle. In like manner one of the angles of a regular heptagon will be equal to one right angle, and three parts of feven of one right angle; one of the angles of a nonagon will be one right angle, and five equal parts of nine of one right angle; and one of the angles of a regular polygon having eleven equal fides, will be equal to one right angle and feven equal parts of eleven of one right angle; and fo of others. In like manner it is demonftrated that one angle of a regular hexagon will be equal to one right 04

angle

4

angle, and one third part of one right angle; one angle of a regular octagon equal to one right angle and half a right angle; one angle of a regular decagon equal to one right angle and three fifth parts of one right angle; and one angle of a regular dodecagon equal to one right angle and three fourth parts of one right angle, &c.

Therefore [by conftruction] the angle E A B will be one angle of the required regular polygon, and A B н another angle of the polygon. And fince the femidiameter of a circle drawn to any one angle of a regular polygon infcribed in a circle bifects that angle, therefore the point F will be the centre of a circle, in which the regular polygon required is defcribed, and A F or F B will be a femidiameter, &c.

Therefore, . Which was to be done.

If the base of an ifofieles triangle stands upon the arch of a circle, and the vertical angle be in the circumference of that circle, and one angle at the bafe of that triangle be fome number of times the angle at the vertex, and each of the angles at the bafe be divided into angles each equal to that at the vertex by right lines meeting the circumference of the circle. And if the extremities of thefe right lines be joined, there will be conftituted a regular polygon in the circle confifting of as many fides as there are units in twice the number of divifions of one of the angles at the base of the if fceles triangle, together with a unite added.

Let A B C be an ifefceles triangle, whose base A c stands upon the circumference of the circle A E B H C, and vertical angle B is alfo in that circumference; and let one of the angles B A C at the base be four times the angle ABC at the vertex. Let the angles B AC, BCA be divided into four angles each equal to the vertical angle A B C, by the right lines A G, A H, AI; CF, CE, CD meeting the circumference of the circle in the points G, H, I; F, E, D.

Join

Join the right lines B G, GH, HI, IC; B F, F E, E D, D A. I fay the polygon ADEFBGHICA will have nine equal fides.

For because the four angles BAG, GAH, HAI, IAC [by E fuppofition] are each equal to the angle ABC at the circumference,

F

and alfo the four angles F C B,

D

ECF, EC D, DCA at the periphery each equal to the fame

A

B

G

H

I

C

angle ABC. Therefore [by 26. and 29. 3.] the nine right lines BG, GH, HI, IC, AC, AD, DE, EF, FB will be equal to one another. And fince these are equal to one another, the angles DAC, ACI, D, E, F, FBG, G, H, I will be alfo equal to one another. Wherefore the polygon ADEFBGHICA will be a regular nonagon. The demonftration is the fame whatever be the number of times that one of the angles at the base of the ifofceles triangle A B C contains the angle A B C at the vertex.

Therefore, &c. Which was to be demonftrated.

SCHOLIUM.

From this propofition it is evident, that a regular polygon of any number of fides, might be infcribed in a given circle, if an ifofceles triangle could be made fuch that one angle at the bafe fhould contain the angle at the vertex any given number of times. But this is not to be done generally by a right line and a circle; it must be effected in fome cafes by a conic fection, and in others by geometrical curves of an higher order. The order of the geometrical curve necessary to the bufinefs, being higher according as the number of times the angle at the bafe is greater than that at the vertex. But as

the conftruction by thefe curves is ufelefs, by reafon of the difficulty of the folution, the cycloid or fpiral of Archimedes, ty which the problem can be easily refolved, are much preferable.

PROP. IX. THEOR.

Any regular polygon infcribed in a circle, or circumfcribed about a circle, will approach nearer to the circle, according as the number of the fides of the polygon is greater.

Let

Let ABCDEFGH be a regular octagon inscribed in a circle, whofe centre is L.

Bifect [by 10. 1.] the fide A B in the point K, and draw the femidiameter L I, and join A I, I B.

Then because K A is equal to K B, the angles AK I, BKI [by 3. 3.] are right angles, and the fide I K is common; the right lines A I, IB [by 26. 1.] will be equal to one another; and fo [by 28. 3.] the arches A I, IB will be equal to one another. Now the circle will exceed the

octagon A B C D E F G H by eight equal fegments A I B, B c, CD, DE, E F, GF, HG, A H, and the trilineal space A I K contained under A K, IK, and the arch A I will be one half the fegment A I B of the circle, which is evidently greater than either of the equal fegments A I, IB by the right angled triangles A IK, or I BK. Wherefore fixteen

of the fegments A I, whereby the circle differs from an infcribed regular polygon of fixteen equal fides will be less than eight of the fegments A B, whereby the circle differs from the regular polygon A B C D E F G H of eight fides. Confequently a regular polygon of fixteen fides is nearer to the circle in which it is infcribed, than that of eight fides. In like manner a regular polygon of thirty two fides infcribed in the circle, will differ lefs from the circle than one of fixteen fides; and one of fixty four fides will still be nearer to the circle in which it is infcribed, than one of thirty two fides; and fo on ad infinitum. The demonftration is the fame, whatever be the number of the fides of the affumed polygon A B C D E F G H infcribed in the circle. So likewife when a regular polygon circumfcribes the circle, the demonftration of the theorem is much the fame as when the polygon is infcribed in the circle.

Corollary. Hence when the number of the fides of a regular polygon infcribed in a circle, or circumfcribed

about