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