they are as the squares on the homologous sides HL and AF (VI. 20, Cor. 3), that is, because the triangles HLG and AFG are similar, as the square on HG to the square on AG, that is, on GK; but the triangles HGK and ACF have been proved to be similar, and therefore the square on AC is to the square on CF as the polygon M to the polygon N; and by conversion, the square on AC is to its excess above the square on CF; that is, to the square on AF as the polygon M to its excess above the polygon N; but the square on AC, that is, the square described about the circle ABC, is greater than the equilateral polygon of eight sides described about the circle, because it contains that polygon; and, for the same reason, the polygon of eight sides is greater than the polygon of sixteen, and so on; therefore the square on AC is greater than any polygon described about the circle by the continual bisection of the arc AB; it is therefore greater than the polygon M. Now it has been demonstrated that the square on AC is to the square on AF as the polygon M to the difference of the polygons; therefore since the square on AC is greater than M, on AF is greater than the difference of the polygons. difference of the polygons is therefore less than the square on AF; but AF is less than D; still more then is the difference of the polygons less than the square on D; that is, than the

given space.

the square

The

COR. 1.-Because the polygons M and N differ from one another more than either of them differs from the circle, the difference between each of them and the circle is less than the given space-namely, the square of D; and, therefore, however small any given space may be, a polygon may be inscribed in the circle, and another described about it, each of which shall differ from

the circle by a space less
than the given space.

COR. 2.-The space B, which

is greater than any polygon
that can be inscribed in the
circle A, and less than
any polygon that can be
described about it, is equal
to the circle A.

If not,

let them be unequal; and first let B exceed A by the
space C;
then, because the polygons described about the
circle A are all greater than B, by hypothesis, and because
B is greater than A by the space C, therefore no polygon

In the same

can be described about the circle A, but what must exceed it
by a space greater than C, which is absurd.
manner, if B be less than A by the space C,

it is shewn

that no polygon can be inscribed in the circle A, but is less than A by a space greater than C, which is also absurd; therefore A and B are not unequal; that is, they are equal

to one another.

Every circle is equal to the rectangle contained by the semidiameter, and a straight line equal to half the circumference. Let ABC be a circle of which the centre is D, and the diameter AC; if in AC produced there be taken AH equal to half the circumference; it is required to prove that the circle is equal to the rectangle contained by DA and AH.

(Const.) Let AB be the side of any equilateral polygon inscribed in the circle ABC; bisect the circumference AB in G, and through G draw EGF, touching the circle, and meeting DA produced in E, and DB produced in F; EF will be the side of an equilateral polygon described about the circle ABC (5). In AC produced take AK equal to half the perimeter of the polygon whose side is AB, and AL equal to half the perimeter of the polygon whose side is EF;

then AK will be less, and AL greater than the straight line AH (2, Cor. 3).

(Dem.) Now because in

E

F

GAM

B

KH L

N

P

the

the triangle EDF, DG is drawn perpendicular to the base, triangle EDF is equal to the rectangle contained by DG and the. half of EF; and as the same is true of all the other equal triangles having their vertices in D, which make up the polygon described about the circle; therefore the whole polygon is equal to the rectangle contained by DG and AL, half the perimeter of the polygon, or by DA and AL; but AL is greater than AH, therefore the rectangle DAAL is greater than the rectangle DA AH; the rectangle DA AH is therefore less than the rectangle DA AL; that is, than any polygon described about the circle ABC.

Again, the triangle ADB is equal to the rectangle contained by DM the perpendicular and one-half of the base AB, and it is therefore less than the rectangle contained by DG or DA, and the half of AB; and as the same is true of all the other triangles having their vertices in D, which make up the inscribed polygon,

therefore the whole of the inscribed polygon is less than the rectangle contained by DA, and AK half the perimeter of the polygon. Now the rectangle DA AK is less than DAAH; therefore the polygon whose side is AB is still less than

DA AH; and the rectangle DA· AH is therefore greater than any polygon inscribed in the circle ABC; but the same rectangle DA AH has been proved to be less than any polygon described about the circle ABC; therefore the rectangle DAAH is equal to the circle ABC (6, Cor. 2). Now DA is the semidiameter of the circle ABC, and AH the half of its circumference.

COR. 1.-Hence a polygon may be described about a circle, the perimeter of which shall exceed the circumference of the circle by a line that is less than any given line. Let NO

be the given line. Take in NO the part NP less than its half, and less also than AD, and let a polygon be described about the circle ABC, so that its excess above ABC may be less than the square of NP (6, Cor. 1). Let the side of this polygon be EF; and since, as has been proved, the circle is equal to the rectangle DA AH, and the polygon to the rectangle DA · AL, the excess of the polygon above the circle is equal to the rectangle DAHL, therefore the rectangle DA HL is less than the square of NP; - and therefore since DA is greater than NP, HL is less than NP, and twice HL less than twice NP, wherefore twice HI is still less than NO; but HL is the difference between half the perimeter of the polygon whose side is EF, and half the circumference of the circle; therefore twice HL is the difference between the whole perimeter of the polygon and the whole circumference of the circle. The difference, therefore, between the perimeter of the polygon and the circumference of the circle, is less than the given line NO. COR. 2.-Hence, also, a polygon may be inscribed in a circle, such that the excess of the circumference above the perimeter of the polygon may be less than any given line. This is proved like the preceding.

COR. 3.-Hence (VI. 33), the area of a sector is equal to half the rectangle under its arc and the radius.

Circles are to one another in the duplicate ratio, or as the squares on their diameters.

Given two circles ABD and GHL, of which the diameters are AD and GL; to prove that the circle ABD is to the circle GHL as the square on AD to the square on GL.

(Const.) Let ABCDEF and GHKLMN be two equilateral polygons of the same number of sides, inscribed in the circles ABD and GHL; and let Q be such a space that the square

on AD is to the square on GL as the circle ABD to the space Q. (Dem.) Because the polygons ABCDEF and GHKLMN are equilateral, and of the same number of sides, they are similar (4), and are to one another as the squares of the diameters of the circles in which they are inscribed; therefore as the square

on AD to the square on GL, so is the polygon ABCDEF to the polygon GHKLMN; but as the square on AD to the square on GL, so is the circle ABD to the space Q; therefore the polygon ABCDEF is to the polygon GHKLMN as the circle ABD to the space Q; but the polygon ABCDEF is less than the circle ABD, therefore GHKLMN is less than the space Q (V. 14); wherefore the space Q is greater than any polygon inscribed in the circle GHL.

In the same manner it is demonstrated that Q is less than any polygon described about the circle GHL; Qis equal to the circle GHL (4, Cor. 2).

wherefore the space Now, by hypothesis,

the circle ABD is to the space Q as the square on AD to the therefore the circle ABD is to the circle GHL

square on GL;

as the square on AD to the square on GL.

COR. 1.-Hence the circumferences of circles are to one another as their diameters.

Xx

Y

Given the straight line X equal to half the circumference of the circle ABD, and the straight line Y to half the circumference of the circle GHL; and because the rectangles AO X and GPY are equal to the circles ABD and GHL (7); therefore the rectangle AO X is to the rectangle GP Y as the square on AD to the square on GL, or as the square on AO to the square on GP; therefore alternately the rectangle AO X is to the square on AO as the rectangle GP Y to the square on GP; but rectangles that have equal altitudes are as their bases, therefore X is to AO as Y to GP; and again, alternately, X is to Y as AO to GP, and taking the doubles of each, the circumference ABD is to the circumference GHL as the diameter AD to the diameter GL.

N

COR. 2. The circle that is described upon the side of a rightangled triangle opposite to the right angle, is equal to the two circles described on the other two sides; for the circle described upon SR is to the circle described upon RT as the square on SR to the square on RT; and the circle described upon TS is to the circle described upon RT as the square on ST to the square on RT; wherefore the circles described

on SR and ST are to the circle

described on RT as the squares on SR and ST to the square on RT (V. 24); but the squares on RS and ST are equal to the square on RT; therefore the circles described on RS and ST are equal to the circle described on RT.

Equiangular parallelograms are to one another as the products of the numbers proportional to their sides.

Let AC and DF be two equiangular parallelograms,

M, N, P, and Q, be four numbers,

and let

such that AB: BC:: M: N;

AB: DE: M: P, and AB: EF:: M: Q, and therefore,

by equality, BC: EF:: N: Q.

parallelogram DF as MN to PQ.

The parallelogram AC is to the

Let NP be the product of N into P, and the ratio of MN to PQ will be compounded of the ratios of MN to NP, and of NP to PQ; but the ratio of MN to NP

is the same with that of M to
P, because MN and NP are
equimultiples of M and P;
and for the same reason the
ratio of NP to PQ is the same
with that of N to Q; there-

B

D

E

fore the ratio of MN to PQ is compounded of the ratios of M

A

to P, and of N to Q. Now, the ratio of M to P is the same with that of the side AB to the side DE; and the ratio of N to Q the same with that of the side BC to the side EF; therefore the ratio of MN to PQ is compounded of the ratios of AB to DE, and of BC to EF; and the ratio of the parallelogram AC to the parallelogram DF is compounded of the same ratios (VI. 23); therefore the parallelogram AC is to the parallelogram DF as MN, the product of the numbers M and N, to PQ the product of the numbers P and Q.

COR. 1.-Hence, if GH be to KL as the number M to number