---
book: 13
number: 12
id: "XIII.12"
kind: "theorem"
uses: ["[[book-4/proposition-15]]", "[[book-3/proposition-31]]", "[[book-1/proposition-47]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.12"
license: "CC-BY-SA-4.0"
---

# XIII.12

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

## Proof

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 of the square on the radius of the circle.

For let the centre *D* of the circle *ABC* be taken, let *AD* be joined and carried through to *E*, and let *BE* be joined.

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 is equal to the radius *DE*. [[book-4/proposition-15|IV. 15, Por.]]

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

But the square on *AE* is equal to the squares on *AB*, *BE*; [[book-3/proposition-31|III. 31]], [[book-1/proposition-47|I. 47]] therefore the squares on *AB*, *BE* are quadruple of the square on *BE*.

Therefore, separando, the square on *AB* is triple of the square on *BE*.

But *BE* is equal to *DE*; therefore the square on *AB* is triple of the square on *DE*.

Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.