# Triangle related question 2

In triangle $\triangle ABC$, if $AD$ is the angle bisector of angle $\angle A$ then prove that $BD=\frac{BC \times AB}{AC + AB}$.

Any help/hints to solve this problem would be greatly appreciated.

-
This follows almost directly from this: en.wikipedia.org/wiki/Angle_bisector_theorem – Wonder Jun 7 '12 at 8:05

$$\triangle ABD: \frac{BD}{\sin A/2}=\frac{AD}{\sin B}$$
For $$\triangle ACD: \frac{DC}{\sin A/2}=\frac{AD}{\sin C}$$
From these: $$\frac{BD}{DC} = \frac{\sin C}{\sin B}$$ But from
$$\triangle ABC: \frac{\sin C}{\sin B} = \frac{AB}{AC}$$ So $$\frac{BD}{DC}=\frac{AB}{AC}$$ Finally, we rearrange: $$\frac{BD}{AB} = \frac{DC}{AC}=\frac{BD+DC=BC}{AB+AC} \;$$