Similar polygons inscribed in circles are to one another as the squares on the diameters.
圆内接相似多边形面积之比等于其直径的平方之比。
两圆各内接相似五边形：左 ABCDE 内接于圆 ABC（圆心 P，直径 BM），右 FGHKL 内接于圆 FGH（圆心 Q，直径 GN）。证明面积之比等于直径 BM、GN 平方之比。
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Let ABC, FGH be circles, let ABCDE, FGHKL be similar polygons inscribed in them, and let BM, GN be diameters of the circles; I say that, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL. For let BE, AM, GL, FN be joined. Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and, as BA is to AE, so is GF to FL.
设圆ABC和FGH内接相似多边形ABCDE和FGHKL，直径BM和GN。连接BE、AM、GL、FN。
[VI. Def. I] Thus BAE, GFL are two triangles which have one angle equal to one angle, namely the angle BAE to the angle GFL, and the sides about the equal angles proportional; therefore the triangle ABE is equiangular with the triangle FGL. [VI. 6] Therefore the angle AEB is equal to the angle FLG.
由多边形相似，得角BAE等于角GFL，且BA比AE等于GF比FL，故三角形ABE与FGL等角。
But the angle AEB is equal to the angle AMB, for they stand on the same circumference; [III. 27] and the angle FLG to the angle FNG; therefore the angle AMB is also equal to the angle FNG. But the right angle BAM is also equal to the right angle GFN; [III. 31] therefore the remaining angle is equal to the remaining angle. [I. 32] Therefore the triangle ABM is equiangular with the triangle FGN.
角AEB等于角AMB（同弧所对圆周角），角FLG等于角FNG，故角AMB等于角FNG；又直角BAM等于直角GFN，故三角形ABM与FGN等角。
Therefore, proportionally, as BM is to GN, so is BA to GF. [VI. 4] But the ratio of the square on BM to the square on GN is duplicate of the ratio of BM to GN, and the ratio of the polygon ABCDE to the polygon FGHKL is duplicate of the ratio of BA to GF; [VI. 20] therefore also, as the square on BM is to the square on GN, so is the polygon ABCDE to the polygon FGHKL.
由相似得BM比GN等于BA比GF，而BM平方比GN平方是BM比GN的二倍比，多边形面积比是BA比GF的二倍比，故结论成立。