Proof: By Euclid

Let there be a circle $ABC$, and let the equilateral triangle $ABC$ have been inscribed in it [Prop. 4.2].
I say that the square on one side of triangle $ABC$ is three times the (square) on the radius of circle $ABC$.

fig12e

For let the center, $D$, of circle $ABC$ have been found [Prop. 3.1].
And $AD$ (being) joined, let it have been drawn across to $E$.
And let $BE$ have been joined.
And since triangle $ABC$ is equilateral, circumference $BEC$ is thus the third part of the circumference of circle $ABC$.
Thus, circumference $BE$ is the sixth part of the circumference of the circle.
Thus, straight line $BE$ is (the side) of a hexagon.
Thus, it is equal to the radius $DE$ [Prop. 4.15 corr.] 0.
And since $AE$ is double $DE$, the (square) on $AE$ is four times the (square) on $ED$ - that is to say, of the (square) on $BE$.
And the (square) on $AE$ (is) equal to the (sum of the squares) on $AB$ and $BE$ [Prop. 3.31], [Prop. 1.47].
Thus, the (sum of the squares) on $AB$ and $BE$ is four times the (square) on $BE$.
Thus, via separation, the (square) on $AB$ is three times the (square) on $BE$.
And $BE$ (is) equal to $DE$.
Thus, the (square) on $AB$ is three times the (square) on $DE$.
Thus, the square on the side of the triangle is three times the (square) on the radius [of the circle].
(Which is) the very thing it was required to show.
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