Angle Bisector Theorem

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 $DC$.

$(1)$ implies $(2)$

 * Euclid-VI-3.png

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:
 * $\angle DAB = \angle ACE$

From Parallelism implies Equal Corresponding Angles:
 * $\angle BAD = \angle AEC$

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

So from Triangle with Two Equal Angles is Isosceles:
 * $AC = AE$

Since $AD \parallel EC$, from Parallel Transversal Theorem:
 * $BD : DC = BA : AE$

But $AE = AC$, so:
 * $BD : DC = AB : AC$

$(2)$ implies $(1)$
Now suppose $BD : DC = AB : AC$.

Join $AD$.

Using the same construction, from Parallel Transversal Theorem:
 * $BD : DC = AB : AE$

From Equality of Ratios is Transitive:
 * $BA : AC = BA : AE$

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

So from Isosceles Triangle has Two Equal Angles:
 * $\angle AEC = \angle ACE$

But from Parallelism implies Equal Corresponding Angles:
 * $\angle AEC = \angle BAD$

Also, from Parallelism implies Equal Alternate Angles:
 * $\angle ACE = \angle CAD$

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