Bisection of 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 $$AB$$.

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 $$AF$$.

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.