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 $DC$.


 * 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.

$(1)$ 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 Parallel 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 Parallel 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 Triangles have Two Equal Angles, $\angle AEC = \angle ACE$.

But from Parallel Implies Equal Corresponding Angles, $\angle AEC = \angle BAD$.

Also, from Parallel Implies Equal Alternate Interior Angles we have that $\angle ACE = \angle CAD$.

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