How do I prove that a triangle with sides a, b, c, has an angle bisector (bisecting angle A) is of length:
$$\frac{2 \sqrt{bcs(s-a)}}{b+c}$$
I have tried using the sine and cosine rule but have largely failed. A few times I have found a way but they are way too messy to work with.
On
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 $2s=a+b+c$
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?