# Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle

## Theorem

In the words of Euclid:

*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.*

(*The Elements*: Book $\text{XIII}$: Proposition $12$)

## Proof

Let $ABC$ be a circle.

Let the equilateral triangle $ABC$ be inscribed within the circle $ABC$.

It is to be demonstrated that the square on one side of $\triangle ABC$ is $3$ times the square on the radius of the circle $ABC$.

Let $D$ be the center of the circle $ABC$.

Let $AD$ be joined and produced to $E$.

Let $BE$ be joined.

We have that $\triangle ABC$ is equilateral.

Therefore the arc $BEC$ is a third part of the circumference of the circle $ABC$.

Therefore the arc $BE$ is a sixth part of the circumference of the circle $ABC$.

Therefore $BE$ is the side of a regular hexagon inscribed within the circle $ABC$.

Therefore from Porism to Proposition $15$ of Book $\text{IV} $: Inscribing Regular Hexagon in Circle:

- $BE = DE$

We have that:

- $AE = 2 \cdot DE$

So:

- $AE^2 = 4 \cdot ED^2 = 4 \cdot BE^2$

But from :

and:

it follows that:

- $AE^2 = AB^2 + BE^2 = 4 \cdot BE^2$

Therefore:

- $AB^2 = 3 \cdot BE^2$

But:

- $BE = DE$

Therefore:

- $AB^2 = 3 \cdot DE^2$

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $12$ of Book $\text{XIII}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions