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.