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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.

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This follows almost directly from this: – Wonder Jun 7 '12 at 8:05

We are going to use "Law of sines":

$$\triangle ABD: \frac{BD}{\sin A/2}=\frac{AD}{\sin B}$$

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: $$\frac{BD}{AB} = \frac{DC}{AC}=\frac{BD+DC=BC}{AB+AC} \;$$

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