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 ABCDE 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.
