Construction of Equilateral Triangle

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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

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


Sources