Bisection of Angle

Theorem

It is possible to bisect any given rectilineal angle.

In the words of Euclid:

To bisect a given rectilineal angle.

(The Elements: Book $\text{I}$: Proposition $9$)

Construction

Let $\angle BAC$ be the given angle to be bisected.

Take any point $D$ on $AB$.

We cut off from $AC$ a length $AE$ equal to $AD$.

We draw the line segment $DE$.

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

We draw the line segment $AF$.

Then the angle $\angle BAC$ has been bisected by the straight line segment $AF$.

Proof

We have:

$AD = AE$

$AF$ is common

$DF = EF$

Thus triangles $\triangle ADF$ and $\triangle AEF$ are equal.

Thus $\angle DAF = \angle EAF$.

Hence $\angle BAC$ has been bisected by $AF$.

$\blacksquare$

Historical Note

This proof is Proposition $9$ of Book $\text{I}$ of Euclid's The Elements.
There are quicker and easier constructions of a bisection, but this particular one uses only results previously demonstrated.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions