Bisection of an Angle

Theorem

It is possible to bisect any given rectilineal angle.

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 is Proposition 9 of Book I of Euclid's The Elements.

There are quicker and easier constructions of a bisection, but this particular one uses only results previously demonstrated.