Proof: By Euclid
(related to Proposition: Prop. 12.01: Areas of Similar Polygons Inscribed in Circles are as Squares on Diameters)
- For let BE, AM, GL, and FN have been joined.
- And since polygon ABCDE (is) similar to polygon FGHKL, angle BAE is also equal to (angle) GFL, and as BA is to AE, so GF (is) to FL [Def. 6.1] .
- So, BAE and GFL are two triangles having one angle equal to one angle, (namely), BAE (equal) to GFL, and the sides around the equal angles proportional.
- triangle ABE is thus equiangular with triangle FGL [Prop. 6.6].
- Thus, angle AEB is equal to (angle) FLG.
- But, AEB is equal to AMB, and FLG to FNG, for they stand on the same circumference [Prop. 3.27].
- Thus, AMB is also equal to FNG.
- And the right angle BAM is also equal to the right angle GFN [Prop. 3.31].
- Thus, the remaining (angle) is also equal to the remaining (angle) [Prop. 1.32].
- Thus, triangle ABM is equiangular with triangle FGN.
- Thus, proportionally, as BM is to GN, so BA (is) to GF [Prop. 6.4].
- But, the (ratio) of the square on BM to the square on GN is the square of the ratio of BM to GN, and the (ratio) of polygon ABCDE to polygon FGHKL is the square of the (ratio) of BA to GF [Prop. 6.20].
- And, thus, as the square on BM (is) to the square on GN, so polygon ABCDE (is) to polygon FGHKL.
- Thus, similar polygons (inscribed) in circles are to one another as the squares on the diameters (of the circles).
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"