Length of Angle Bisector

Theorem
The length of an angle bisector is given by:


 * $b_\alpha^2 = \dfrac{cb}{(c+b)^2}[(c+b)^2-a^2]$


 * $b_\gamma^2 = \dfrac{ab}{(a+b)^2}[(a+b)^2-c^2]$


 * $b_\beta^2 = \dfrac{ac}{(a+c)^2}[(a+c)^2-b^2]$

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

and $b_\alpha$, $b_\beta$, and $b_\gamma$ are the angle bisectors from $A$, $B$, and $C$ respectively.

Proof 1
We look at one of the angle bisectors, WLOG $b_\gamma$:

We now plug these results in to Stewart's Theorem, also noting that $CP = b_\gamma$:

A similar argument can be used to show that the statement holds for the others bisectors.

Proof 2
We look at one of the angle bisectors, WLOG $b_\gamma$:



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


 * $u=\dfrac{bc}{a+b}$


 * $v=\dfrac{ac}{a+b}$

we have

Then by AA similarity we have $\triangle CAD \sim \triangle CFB$

Now we use the Intersecting Chord Theorem, which gives us $v \cdot u=b_\gamma \cdot DF$.

A similar argument can be used to show that the statement holds for the other bisectors.