Next, let the base AB be given, A to construct an isosceles triangle of this nature.

A

B

B

Produce AB to C, such that the rectangle AC, CB be equal to the square of AB (II. 19. cor. 2.), and on the base AB, with the distance AC for each of the sides, describe an isosceles triangle.

These isosceles triangles will fulfil the conditions required. For it is evident, from the last Proposition, that isosceles triangles constituted on CB or AB, with each of the angles at the base double the vertical angle, would have AB or AC for their sides, and consequently (I. 2.) must coincide with the triangles now described.

Cor. Hence of such an isosceles triangle the vertical angle is equal to the fifth part of two right angles; for each of the angles at the base being double of the vertical angle, they are both equal to four times it, and consequently this vertical angle is the fifth part of all the angles of the triangle, or of two right angles.

PROP. V. PROB.

On a given finite straight line, to describe a regular pentagon.

Let AB be the straight line, on which it is required to describe a regular pentagon.

On AB erect (IV. 4.) the isosceles triangle ACB, having each of the angles at its base double of its vertical angle, from the centre A with the distance AB describe an

the angles at the base of the triangle ACB, and consequently (IV. 3.) AB is equal to BF and FC, or AG and GC. Again, the triangles BAD and BFC, having the sides AB, BD equal to BF, BC, and the contained angles equal, are themselves equal (I. 3.), and consequently AB is equal to AD, and the angle BAD equal to BFC, or three times ACB. In the same way it is shewn that ́AB is equal to BE, and that the angles round the figure are each equal to thrice the vertical angle of the original isosceles triangle.

PROP. VI. PROB.

On a given finite straight line, to describe a regular hexagon.

Let AB be the given straight line, on which it is required to describe a regular hexagon.

On AB construct (I. 1.) the equilateral triangle AOB, and repeat equal triangles about the vertex O; these triangles will together compose the hexagon required.

Because AOB is an equilateral triangle, each of its an

gles is equal to the third part of two right angles (I. 30. cor. 1.); wherefore the vertical angle AOB is the sixth part of four right angles, or six of such angles may be placed about the

point O. But the bases of the triangles AOB, AOC, COD,

D

E

F

DOE, EOF, and BOF are all equal; and so are the angles at the bases, and which, taken by pairs, form the internal angles of the figure BACDEF. This figure is, therefore, a regular hexagon.

PROP. VII. PROB.

On a given finite straight line, to describe a regular octagon.

Let AB be the given straight line, on which it is required to describe a regular octagon.

Bisect AB (I. 7.) by the perpendicular CD, which make equal to CA or CB, join DA and DB, produce CD until

DO be equal to DA or DB, draw AO and BO, thus forming

(IV. 1.) an angle equal to the half of ADB, and, about the vertex O, repeat the equal triangles AOB, AOE, EOF, FOG, GOH, HOI, IOK, and KOB to compose the octagon.

For the distances AD, BD are

E

G

C

Σ

K

evidently equal; and because CA, CD, and CB are all ́

equal, the angle ADB is contained in a semicircle, and is therefore a right angle (III. 19.). Consequently AOB is equal to the half of a right angle, and eight such angles will adapt themselves about the point O. Whence the figure BAEFGHIK, having eight equal sides and equal angles, is a regular octagon.

PROP. VIII. PROB.

On a given finite straight line, to describe a regular decagon.

Let AB be the straight line, on which it is required to describe a regular decagon.

On AB construct (IV. 4.) an isosceles triangle having each of the angles at its base double of the vertical angle,

to the tenth part of four right angles; whence ten such angles may be formed about the point O. The figure BACDEFGHIK, having therefore ten equal sides and equal angles, is a regular decagon. ́

PROP. IX. PROB.

On a given finite straight line, to describe a regular dodecagon.

Let AB be the straight line, on which it is required to describe a regular twelve-sided figure.

On AB construct (I. 1.) the equilateral triangle ACB, and again (IV. 1.) the isosceles triangle AOB, having its vertical angle equal to the half of ACB, and repeat this triangle AOB about the point O; a regular dodecagon will be thus formed.

For ACB being an equilateral triangle, each of its angles is the third part of two right angles (I. 30. cor. 1.); consequently the angle AOB is the sixth part of two right angles

F

E

H

I

K

M

N

B

or the twelfth part of four right angles, and twelve such angles can, therefore, be placed about the vertex O.

Scholium. Hence a regular twenty-sided figure may be described on a given straight line, by first constructing on it an isosceles triangle having each of the angles at the base double of the vertical angle, and then erecting another isosceles triangle with its vertical angle equal to the half of this. And, by thus changing the elementary triangle, a regular polygon may be always described, with twice the number of sides.