Book IV. another; therefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.

b 32. 1.

PROB. XV. PROB.

TO inscribe an equilateral and equiangular hexagon in a given circle.

Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it.

Find the centre G of the circle ABCDEF, and draw the diameter AGD; from D as a centre, at the distance DG, describe the circle EGCH; join EG, CG, and produce them to the points B, F; join AB, BC, CD, DE, EF, FA. The hexagon ABCDEF is equilateral and equiangular.

F

A

B

Because G is the centre of the circle ABCDEF, GE is equal to GD; and because D is the centre of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral; therefore its three a Cor. 5. 1. angles EGD, GDĒ, DEG are equal to one another; therefore the angle EGD is the third part of two right angles. In the same manner it may be demonstrated that the angle DGC is the third part of two right angles. Because the straight line GC makes with EB the adjacent angles EGC, CGB equal to two right angles, the remaining angle CGB is the third part of two right angles; therefore the angles EGD, DGC, CGB are equal to one another. To these the vertical angles BGA, AGF, FGE are also equald; therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. Therefore the six arches

c 13. 1.

d 15. 1.

EX

H

AB, BC, CD, DE, EF, FA are equal to one another; Book IV. therefore the six straight lines are equal to one another, and

the hexagon ABCDEF is equilateral. It is also equiangular; e 26. 3. for, since the arch AF is equal to ED, to each of these add f 29. 3. the arch ABCD; therefore the whole arch FABCD will be equal to the whole EDCBA. Now the angle FED stands upon the arch FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equals to FED. In the same g 27.3. manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; therefore the hexagon is equiangular; it is also equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done.

COR. It appears from the demonstration that the side of the hexagon is equal to the radius of the circle. For the triangle EGD is equilateral.

If through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.

PROP. XVI. PROB.

TO inscribe an equilateral and equiangular quindecagon in a given circle.

Let ABCD be the given circle; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD.

Book IV.

a 2.4.

b 11. 4.

c 30.3.

d 1. 4.

E

Let AC be the side of an equilateral triangle inscribed in the circle, and AB the side of an equilateral and equiangular pentagon inscribed in the same; therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the arch ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC, their difference, contains two of the same parts. Bisect BC in E; therefore BE,

E

D

EC are, each of them, the fifteenth part of the whole circumference ABCD. Therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placedd round in the whole circle, an equilateral and equiangular quindecagon will be inscribed in it. Which was to be done.

In the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon straight lines be drawn touching the circle, an equilateral and equiangular quindecagon may be described about it: likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.

A circle may be inscribed in and described about a given equilateral and equiangular polygon, of any number of sides, as in Prop. XIII and XIV.

Beside the figures described in Book IV, and those arising from them by continual bisections, or taking the differences, no other regular polygon can be described by any known method which is purely geometrical.

The point F*, the common centre of the inscribed and circumscribed circles, may be considered as the centre of the polygon, and therefore the angle CFD, which is formed by

* See the figures in Prop. XIII and XIV.

two radii drawn to the extremities C, D, of the straight line, Book IV. or chord, CD, is called the angle at the centre.

Because all the chords AB, BC, &c. are equal, all the angles at the centre are equal; therefore the value of each may be found by dividing four right angles by the number of sides of the polygon.

A regular polygon of a given number of sides may be inscribed in a given circle by dividing the circumference of the Icircle into as many equal parts as the polygon has sides. For, the arches being equal, the chords AB, BC, &c. are equal; and the triangles ABF, BCF, &c. are equal, because they are mutually equilateral; therefore all the angles ABC, BCD, &c. are equal; therefore ABCDE is a regular polygon.