Question

In a $△ABC,BC=a,CA=b$ and $\angle BCA={120}^o,CD$ is angle bisector of $\angle BCA$ which meets $AB$ on $D$ find the length of $CD$

Answer 1

Given In $△ABC,BC=a,CA=b$ and $\angle BCA=\angle C={120}^{\circ }$

as CD is angle bisector of $\angle C$

$⟹$length of $CD=\frac{2ab}{a+b}\cos \frac{C}{2}=\frac{2ab}{a+b}\cos {60}^{\circ }=\frac{ab}{a+b}$