Proof: By Euclid

Let $ABC$ and $FGH$ be circles, and let $ABCDE$ and $FGHKL$ be similar polygons (inscribed) in them (respectively), and let $BM$ and $GN$ be the diameters of the circles (respectively).
I say that as the square on $BM$ is to the square on $GN$, so polygon $ABCDE$ (is) to polygon $FGHKL$.

fig01e

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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