Book 12
BOOK XII. PROPOSITIONS.
PROPOSITION 1.
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Similar polygons inscribed in circles are to one another as the squares on the diameters.
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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.
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For let BE, AM, GL, FN be joined.
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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. [VI. Def. I]
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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]
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Therefore the angle AEB is equal to the angle FLG.
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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.
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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]
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Therefore the triangle ABM is equiangular with the triangle FGN.
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Therefore, proportionally, as BM is to GN, so is BA to GF. [VI. 4]
