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