Length of Angle Bisector/Proof 2

Theorem
Let $\triangle ABC$ be a triangle.

Let $AD$ be the angle bisector of $\angle BAC$ in $\triangle ABC$.


 * LengthOfAngleBisector.png

Let $d$ be the length of $AD$.

Then $d$ is given by:


 * $d^2 = \dfrac {b c} {\left({b + c}\right)^2} \left({\left({b + c}\right)^2 - a^2}\right)$

where $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively.

Proof

 * LengthOfAngleBisector2.png

Let $u$ and $v$ be the segments of $BC$ created by the angle bisector.

From Length of Angle Bisector: Proof 1, we have:


 * $BD = \dfrac {a c} {b + c}$


 * $DC = \dfrac {a b} {b + c}$

Then we have:

Then from Triangles with Two Equal Angles are Similar we have:
 * $\triangle ABD \sim \triangle AFC$

So:

Now we use the Intersecting Chord Theorem, which gives us $BD \cdot DC = d \cdot DF$.