$\triangle ABC$ is a equilateral triangle. Line $xy$ passes through vertex $A$ (but doesn't intersect any sides of triangle). $H$ is a point on line $xy$ which is angle bisector of $\angle HAB$ and exterior angle of $B$ intersects each other at $M$. $I$ is a point on line $xy$ which is angle bisector of $\angle IAC$ and exterior angle of $C$ intersects each other at $N$. Prove $AN = AM$.
Here is the figure for for clarification:
Things I have done so far: I tried using similar triangles like $ACK$ and $ABG$ and using angle bisector theorem to prove $AN=AM$ but I was not succesful.