My question is:
In $\Delta ABC$, let $AE$ be the angle bisector of $\angle A$. If $\frac{1}{AE} = \frac{1}{AC} + \frac{1}{AB}$ then prove that $\angle A = 120^\circ$.
What I tried: I extended side $AB$ and took a point $M$ on it such that $AC$ is congruent to $AM$. Then I proved that $AE$ is parallel to $MC$. I was trying to prove that $\triangle AMC$ is equilateral so that I get $\angle MAC=60^\circ$. But I am not able to prove it.
Any help to solve this question would be greatly appreciated!