Bisection of Angle

Theorem
It is possible to bisect any given rectilineal angle.

Construction

 * Euclid-I-9.png

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 $AF$.