Proving the length of angle bisector

Assuming the usage of Trigonometry is allowed,

Let $AD$ be the bisector of $\angle BAC$

$\triangle ABC=\frac12bc\sin A $

$\triangle ABD=\frac12\cdot c\cdot AD\sin\frac A2$ and $\triangle ADC=\frac12\cdot b\cdot AD\sin\frac A2$

$\triangle ABC=\triangle ABD+\triangle ADC$

$\sin A=2\sin\frac A2\cos\frac A2$

As $\displaystyle 0<A<\pi,0<\frac A2<\frac\pi2\implies \cos\frac A2>0\implies \cos \frac A2=+\sqrt{\frac{1+\cos A}2}$ as $\cos A=2\cos^2\frac A2-1$

Use $\displaystyle\cos A=\frac{b^2+c^2-a^2}{2bc}$ and $2s=a+b+c$

A method where no trigonometry is used.

Consider triangle $ABC$. Let $AD$, the angle bisector, intersect the circumcircle at $L$. Join $LC$. Consider triangle $ABD$ and triangle $ALC$.

Triangle $ABD$ is similar to triangle $ALC$ (by A.A similarity theorem). Therefore, $$\frac{AD}{AC} = \frac{AB}{AL}$$

i.e, $$AD\cdot AL = AC\cdot AB$$ $$= AD(AD+DL) = AC\cdot AB$$ $$=AD\cdot AD + AD\cdot DL = AC \cdot AB \text{ ... (1)}$$

By power of point of point result: $$AD\cdot DL = BD\cdot DC$$ $$BD = BC\cdot \frac{AB}{AB+AC}$$ $$DC = BC\cdot\frac{AC}{AB+AC}$$

In $(1)$ , $$AD\cdot AD = AC\cdot AB-BC^2\cdot AB\cdot \frac{AC}{(AB+AC)^2}$$ $$AD^2 = AC\cdot AB\Bigl(1-\frac{BC^2}{(AB+AC)^2}\Bigr)$$ If $AB = c , BC = a , AC = b$: $$AD^2 = bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ Hence proved.

Refering tp http://en.wikipedia.org/wiki/Stewart%27s_theorem we have $n=\frac{b}{b+c}a$ and $m=\frac{c}{b+c}$, hence $$b^2\frac{ac}{b+c}+c^2\frac{ab}{b+c}=a\left(d^2+\frac{ab}{b+c}\frac{ac}{b+c}\right),$$ where $d$ denotes the length of the bisector. From here we can conclude $$d^2=bc\Bigl( 1-\frac{a^2}{(b+c)^2}\Bigr)$$ using $\frac{b^2c}{b+c}+\frac{bc^2}{b+c}=bc$. Nice formula. Can you get from here to your version?