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 Interior 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 Line in Triangle Cuts Sides Proportionally, $BD : DC = BA : AE$.

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

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

Join $AD$.

Using the same construction, we have that $BD : DC = AB : AE$ from Parallel Line in Triangle Cuts Sides Proportionally.

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 Interior Angles:
 * $\angle ACE = \angle CAD$.

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