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

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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

Euclid-XIII-12.png

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 :

Proposition $31$ of Book $\text{III} $: Relative Sizes of Angles in Segments

and:

Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem

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


Sources