Proving the length of angle bisector 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.
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 $\displaystyle 0&lt;A&lt;\pi,0&lt;\frac A2&lt;\frac\pi2\implies \cos\frac A2&gt;0\implies \cos \frac A2=+\sqrt{\frac{1+\cos A}2}$ as $\cos A=2\cos^2\frac A2-1$

Use $\displaystyle\cos A=\frac{b^2+c^2-a^2}{2bc}$  and $2s=a+b+c$
The length of the angle bisector of a triangle is (2bc*cosa/2)/b+c
Just prove this using cosine rule and you will be done because $\cos a= \sqrt{s(s-a)bc}$