Book VIII. arch intercepted between them; and hence, the perimeter of any polygon described about a circle is greater than the circumference of the circle.

It

PRO P. I. THEO R.

F from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half; and fo on: There fhall at length remain a magnitude less than the least of the propofed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and fo on; there fhall at length remain a magnitude less than C.

A

K

H

For C may be multiplied fo as at length
to become greater than AB. Let it be fo
multiplied, and let DE its multiple be
greater than AB, and let DE be divided
into DF, FG, GE, each equal to C. From
AB take BH equal to its half, and from
the remainder, AH take HK equal to its
half, and fo on, until there be as many
di-
vifions in AB as there are in DE: And
let the divifions in AB be AK, KH, HB;
and the divifions in ED be DF, FG, GE. B
And because DE is greater than AB, and

D

E

that EG taken from DE is not greater than its half, but BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is equal to the half of HA; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is lefs than C. Q. E. D.

PROP.

PROP. II. THEOR.

EQUILATERAL polygons of the same number

of fides infcribed in circles, are fimilar, and are to one another as the fquares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the fame number of fides infcribed in the circles ABD, and GHK; ABCDEF and GHIKLM are fimilar, and are to one another as the fquares of the diameters of the circles ABD, GHK.

Find N and O the centres of the circles; join AN and BN, as alfo GO and HO, and produce AN and GO till they meet the circircumferences in D and K.

Book VIII.

M

Because the straight lines AB, BC, CD, DE, EF, FA, are all equal, the arches AB, BC, CD, DE, EF, FA are alfo equal a. For the fame reason, the arches GH, HI, IK, KL, a 28. LM, MG are all equal, and they are equal in number to the others; therefore, whatever part the arch AB is of the whole circumference ABD, the fame is the arch GH of the circumference GHK. But the angle ANB is the fame part of four right angles, that the arch AB is of the circumference ABD b; and the angle GOH is the fame part of four b 47. 3. right angles that the arch GH is of the circumference GHK b, therefore the angles ANB, GOH are each of them the fame

part

€ 6. 6.

Book VIII. part of four right angles, and therefore they are equal to one another. The ifofceles triangles ANB, GOH are therefore equiangular, and the angle ABN equal to the angle GHO; in the fame manner, by joining NC, OE, it may be proved that the angles NBC, OHI are equal

ӨӨ

to one another, and to the angle ABN. Therefore the whole angle ABC is equal to the whole GHK; and the fame may be proved of the angles BCD, HIK, and of the reft. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another; and fince they are equilateral, the fides about the equal angles are proportionals; the polygon d 1. def. 6. ABCD is therefore fimilar to the polygon GHIKLM d. And because fimilar polygons are as the fquares of their homoloe 20. 6. gous fides e, the polygon ABCDEF is to the polygon GHIKLM as the fquare of AB to the fquare of GH; but because the triangles ANB, GOH are equiangular, the fquare of AB is to the fquare of GH as the fquare of AN to the fquare of GO, or as four times the fquare of AN to four times the square g of GO, that is, as the fquare of AD to the fquare of GK h. Therefore alfo, the polygon ABCDEF is to the polygon GHIKLM as the fquare of AD to the fquare of GK; and they have alfo been shewn to be fimilar. Therefore, &c. Q. E. D. Q.E.

f4. 6.

g 15. 5.

1. cor 8.

2,

COR. Every equilateral polygon infcribed in a circle is alfo equiangular. For the ifofceles triangles, which have their com mon vertex in the centre, are all equal and fimilar; therefore, the angles at their bafes are all equal, and the angles of the polygon are therefore also equal.

PROP.

Book VIII.

PRO P. III.

PROB.

THE
HE fide of any equilateral polygon infcribed
in a circle being given, to find the fide of a
polygon of the fame number of fides defcribed a-
bout the circle.

Let ABCDEF be an equilateral polygon infcribed in the circle ABD; it is required to find the fide of an equilateral polygon of the fame number of fides described about the circle.

Find G the centre of the circle; join GA, GB, bifect the arch AB in H; and through H draw. KHL touching the circle in H, and meeting GA and GB produced in K and L; KL is the fide of the polygon required.

Produce GF to N, fo that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join also

HG.

Because the circumference AB is bifected in H, the angle

H

L

b 16.3

AGH is equal to the angle BGH a; and because KL touches a 27. 34 the circle in H, the angles LHG, KHG are right angles b; therefore, there are two angles of the triangle HGK, equal to two angles of the triangle HGL, each to each. But the fide GH is common to these triangles; therefore they are equal c, and GL is equal to GK. Again, in the triangles KGL,KGN, because GN is equal to GL, and GK common, and also the angle LGK

K

G

c 26. 1.

equal to the angle KGN; therefore the base KL is equal to the bafe KNd. But because the triangle KGN is ifof- & 4. 1. celes, the angle GKN is equal to the angle GNK, and the angles GMK, GMN are both right angles by conftruction; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they

Book VIII. arch intercepted between them; and hence, the perimeter of any polygon defcribed about a circle is greater than the circumference of the circle.

IF

PRO P. I. THE O R.

from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half; and fo on: There shall at length remain a magnitude less than the leaft of the propofed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its

half, and fo on; there fhall at length re-
main a magnitude lefs than C.

A

D

A

K

For C may be multiplied fo as at length to become greater than AB. Let it be fo multiplied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From H AB take BH equal to its half, and from the remainder, AH take HK equal to its half, and fo on, until there be as many divifions in AB as there are in DE: And let the divifions in AB be AK, KH, HB; and the divifions in ED be DF, FG, GE. And because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is equal to the half of HA; therefore, the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is lefs than C. Q. E. D.

B C E

PROP.