If an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle.

Let *ABC* be a circle, and let the equilateral triangle *ABC* be inscribed in it.

I say that the square on one side of the triangle *ABC* is triple the square on the radius of the circle.

Take the center *D* of the circle *ABC,* join *AD* and carry it through to *E,* and join *BE.*

Then, since the triangle *ABC* is equilateral, therefore the circumference *BEC* is a third part of the circumference of the circle *ABC.* Therefore the circumference *BE* is a sixth part of the circumference of the circle. Therefore the straight line *BE* belongs to a hexagon. Therefore it equals the radius *DE.*

And, since *AE* is double *DE,* therefore the square on *AE* is quadruple the square on *ED,* that is, of the square on *BE.*

But the square on *AE* equals the sum of the squares on *AB* and *BE.* Therefore the sum of the squares on *AB* and *BE* is quadruple the square on *BE.*

Therefore, taken separately, the square on *AB* is triple the square on *BE.* But *BE* equals *DE,* therefore the square on *AB* is triple the square on *DE.*

Therefore the square on the side of the triangle is triple the square on the radius.

Therefore, *if an equilateral triangle is inscribed in a circle, then the square on the side of the triangle is triple the square on the radius of the circle.*

Q.E.D.