Bisection of Angle

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

Let $\angle BAC$ be the given angle to be bisected.

Let $D$ be an arbitrary point on $AB$.

From Proposition $3$: Construction of Equal Straight Lines from Unequal, let $AE$ be cut off from $AC$ such that $AE = AD$.

From Euclid's First Postulate, let the line segment $DE$ be constructed.

From Proposition $1$: Construction of Equilateral Triangle, let an equilateral triangle $\triangle DEF$ be constructed on $AB$.

From Euclid's First Postulate, let the line segment $AF$ be constructed.

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 from Proposition $8$: Triangle Side-Side-Side Congruence:

$\triangle ADF \sim \triangle AEF$

Thus:

$\angle DAF = \angle EAF$

Hence $\angle BAC$ has been bisected by $AF$.

$\blacksquare$

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.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions