Bisection of Angle

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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

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 $AF$.

$\blacksquare$


Historical Note

This theorem 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