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