Proof: By Euclid
(related to Proposition: Prop. 13.12: Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"