Bisection of Arc

Theorem

It is possible to bisect an arc of a circle.

In the words of Euclid:

To bisect a given circumference.

(The Elements: Book $\text{III}$: Proposition $30$)

Construction

Let $AB$ be the given arc.

Join the line $AB$ and bisect it at $C$.

Construct the line $CD$ perpendicular to $AB$ at $C$.

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

Proof

Join $AD$ and $BD$.

We have that $AC = CB$ and $CD$ is common.

We also have that $\angle ACD = \angle BCD$ as they are both right angles.

So from Triangle Side-Angle-Side Congruence $\triangle ACD = \triangle BCD$ and so $AD = BD$.

But from Straight Lines Cut Off Equal Arcs in Equal Circles, the arc $AD$ equals the arc $BD$.

That is, the arc $AB$ has been bisected at $D$.

$\blacksquare$

Historical Note

This proof is Proposition $30$ of Book $\text{III}$ of Euclid's The Elements.

Sources

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