Bisection of Straight Line

Theorem

It is possible to bisect a straight line segment.

In the words of Euclid:

To bisect a given finite straight line.

(The Elements: Book $\text{I}$: Proposition $10$)

Construction

Let $AB$ be the given straight line segment.

We construct an equilateral triangle $\triangle ABC$ on $AB$.

We bisect the angle $\angle ACB$ by the straight line segment $CD$.

Then $AB$ has been bisected at the point $D$.

Proof

As $\triangle ABC$ is an equilateral triangle, it follows that $AC = CB$.

The two triangles $\triangle ACD$ and $\triangle BCD$ have side $CD$ in common, and side $AC$ of $\triangle ACD$ equals side $BC$ of $\triangle BCD$.

The angle $\angle ACD$ subtended by lines $AC$ and $CD$ equals the angle $\angle BCD$ subtended by lines $BC$ and $CD$, as $\angle ACB$ was bisected.

So triangles $\triangle ACD$ and $\triangle BCD$ are equal.

Therefore $AD = DB$.

So $AB$ has been bisected at the point $D$.

$\blacksquare$

Historical Note

This proof is Proposition $10$ 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