Construction of Equilateral Triangle

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

Construction

 * Euclid-I-1.png

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 Definition I-15 that $AC = AB$.

As $B$ is the center of circle $ACE$, it follows from Definition I-15 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.