Again, because DG is greater than AH;

and that GF is not greater than the half of DG, but HK is greater than the half of AH;

therefore the remainder DF is greater than the remainder AK.

But DF is equal to C;

therefore C is greater than AK; that is, AK is less than C. Q.E.D.

And if only the halves be taken away, the same thing may in the same way be demonstrated.

PROPOSITION 1. THEOREM.

Similar polygons inscribed in circles are to one another as the squares on their diameters.

Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHKL; and let BM, GN be the diameters of the circles: the polygon ABCDÉ shall be to the polygon FGHKL as the square on BM is to the square on GM.

Join AM, BE, FN, GL.

Then, because the polygons are similar,

therefore the angle BAE is equal to the angle GFL,

and BA is to AE as GF is to FL.

[VI. Definition 1.

Therefore the triangle BAE is equiangular to the triangle GFL;

[VI. 6.

therefore the angle AEB is equal to the angle FLG. But the angle AEB is equal to the angle AMB, and the angle FLG is equal to the angle FNG;

[III. 21.

therefore the angle AMB is equal to the angle FNG.

And the angle BAM is equal to the angle GFN, for each of them is a right angle.

[III. 31.

B

M

Therefore the remaining angles in the triangles AMB, FNG are equal, and the triangles are equiangular to one another;

therefore BA is to BM as GF is to GN,

[VI. 4.

[V. 16.

and, alternately, BA is to GF as BM is to GN; therefore the duplicate ratio of BA to GF is the same as the duplicate ratio of BM to GN. [V. Definition 10, V. 22.

But the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of BA to GF; [VI. 20. and the square on BM is to the square on GN in the duplicate ratio of BM to GN; therefore the polygon ABCDE is to the polygon FGHKL as the square on BM is to the square on GÑ.

Wherefore, similar polygons &c. Q.E.D.

[VI. 20.

[V. 11.

Circles are to one another as the squares on their diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: the circle ABCD shall be to the circle EFGH as the square on BD is to the square on FH.

For, if not, the square on BD must be to the square on FH as the circle ABCD is to some space either less than the circle EFGH, or greater than it.

First, if possible, let it be as the circle ABCD is to a space S less than the circle EFGH.

In the circle EFGH inscribe the square EFGH.

[IV. 6.

This square shall be greater than half of the circle EFGH.

For the square EFGH is half of the square which can be formed by drawing straight lines to touch the circle at the points E, F, G, H;

and the square thus formed is greater than the circle;

therefore the square EFGH is greater than half of the circle.

Bisect the arcs EF, FG, GH, HE at the points K, L, M, N ;

and join EK, KF, FL, LG, GM, MH, HN, NE. Then each of the triangles EKF, FLG, GMH, HNE shall be greater than half of the segment of the circle in which it stands.

For the triangle EKF is half of the parallelogram which can be formed by drawing a straight line to touch the circle at K, and parallel straight lines through E and F, and the parallelogram thus formed is greater than the segment FEK; therefore the triangle EKF is greater than half of the segment.

And similarly for the other triangles.

Therefore the sum of all these triangles is together greater than half of the sum of the segments of the circle in which they stand.

Again, bisect EK, KF, &c. and form triangles as before; then the sum of these triangles is greater than half of the sum of the segments of the circle in which they stand.

If this process be continued, and the triangies be supposed to be taken away, there will at length remain segments of circles which are together less than the excess of the circle EFGH above the space S, by the preceding Lemma.

Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those which remain, and which are together less than the excess of the circle above S;

therefore the rest of the circle, namely the polygon EKFLGMHN, is greater than the space S.

In the circle ABCD describe the polygon AXBOCP DR similar to the polygon EKFLGMÍN.

Then the polygon AXBOCPDR is to the polygon EKFLGMHN as the square on BD is to the square on

FH,

[XII. 1. that is, as the circle ABCD is to the space S. [Hyp., V. 11. But the polygon AXBOCPDR is less than the circle ABCD in which it is inscribed,

therefore the polygon EKFLGMHN is less than the space S;

but it is also greater, as has been shewn;

which is impossible.

[V. 14.

Therefore the square on BD is not to the square on FH as the circle ABCD is to any space less than the circle EFGH.

In the same way it may be shewn that the square on FH is not to the square on BD as the circle EFGH is to any space less than the circle ABCD.

Nor is the square on BD to the square on FH as the circle ABCD is to any space greater than the circle EFGH.

For, if possible, let it be as the circle ABCD is to a space T greater than the circle EFGH.

Then, inversely, the square on FH is to the square on BD as the space T'is to the circle ABCD.

But as the space T is to the circle ABCD so is the circle EFGH to some space, which must be less than the circle

ABCD, because, by hypothesis, the space T is greater than the circle EFGH

[V. 14.

Therefore the square on FH is to the square on BD as the circle EFGH is to some space less than the circle ABCD;

which has been shewn to be impossible.

Therefore the square on BD is not to the square on FH as the circle ABCD is to any space greater than the circle EFGH.

And it has been shewn that the square on BD is not to the square on FH as the circle ABCD is to any space less than the circle EFGH.

Therefore the square on BD is to the square on FH as the circle ABCD is to the circle EFGH.

Wherefore, circles &c. Q.E.D.