# Angle Bisector Theorem

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## Theorem

Let $\triangle ABC$ be a triangle.

Let $D$ lie on the base $BC$ of $\triangle ABC$.

Then the following are equivalent:

$(1): \quad AD$ is the angle bisector of $\angle BAC$
$(2): \quad BD : DC = AB : AC$

where $BD : DC$ denotes the ratio between the lengths $BD$ and $DC$.

In the words of Euclid:

If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle.

## Proof

### $(1)$ implies $(2)$ Let $CE$ be drawn through $C$ parallel to $DA$.

Let $BA$ be produced so as to meet $CE$ at $E$.

From Parallelism implies Equal Alternate Angles we have that:

$\angle ACE = \angle CAD$

But by hypothesis:

$\angle CAD = \angle BAD$

and so:

$\angle DAB = \angle ACE$
$\angle BAD = \angle AEC$

But from above $\angle ACE = \angle BAD$, so:

$\angle ACE = \angle AEC$
$AC = AE$

Since $AD \parallel EC$, from Parallel Line in Triangle Cuts Sides Proportionally:

$BD : DC = BA : AE$

But $AE = AC$, so:

$BD : DC = AB : AC$

$\Box$

### $(2)$ implies $(1)$

Now suppose $BD : DC = AB : AC$.

Join $AD$.

Using the same construction, from Parallel Line in Triangle Cuts Sides Proportionally:

$BD : DC = AB : AE$
$BA : AC = BA : AE$

So $AC = AE$ from Magnitudes with Same Ratios are Equal.

$\angle AEC = \angle ACE$
$\angle AEC = \angle BAD$

Also, from Parallelism implies Equal Alternate Angles:

$\angle ACE = \angle CAD$

Therefore $\angle BAD = \angle CAD$ and so $AD$ has bisected $\angle BAC$.

$\blacksquare$

## Historical Note

This proof is Proposition $3$ of Book $\text{VI}$ of Euclid's The Elements.