Bisection of Angle

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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 $9$)



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 $AF$.


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 $AF$.


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.