# Construction of Equilateral Triangle

## Theorem

On a given straight line segment, it is possible to construct an equilateral triangle.

In the words of Euclid:

*On a given finite straight line to construct an equilateral triangle.*

(*The Elements*: Book $\text{I}$: Proposition $1$)

## Construction

Let $AB$ be the given straight line segment.

We construct a circle $BCD$ with center $A$ and radius $AB$.

We construct a circle $ACE$ with center $B$ and radius $AB$.

From $C$, where the circles intersect, we draw a line segment to $A$ and to $B$ to form the straight line segments $AC$ and $BC$.

Then $\triangle ABC$ is the equilateral triangle required.

## Proof

As $A$ is the center of circle $BCD$, it follows from Book $\text{I}$ Definition $15$: Circle that $AC = AB$.

As $B$ is the center of circle $ACE$, it follows from Book $\text{I}$ Definition $15$: Circle that $BC = AB$.

So, as $AC = AB$ and $BC = AB$, it follows from Common Notion $1$ that $AC = BC$.

Therefore $AB = AC = BC$.

Therefore $\triangle ABC$ is equilateral.

$\blacksquare$

## Historical Note

This proof is Proposition $1$ of Book $\text{I}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions