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$\triangle ABC$ 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:enter image description here

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.

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  • $\begingroup$ Hint: $N\in(MJ)$ is the incenter of $ACJ$. $\endgroup$ – Lucian Aug 18 '14 at 23:52
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It's easy to see that $N$ and $M$ lies on the angle bisector of $\angle BJA$. It follows that $$\angle ANM=\angle NAJ+\angle JAN=30^\circ.$$ Similarly, $$\angle AMN=\angle MAH-\angle AJM=30^\circ.$$ Thus $\angle ANM=\angle AMN (=30^\circ).$

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  • $\begingroup$ How do you get $\angle ANM = \angle NAJ + \angle JAN = 30^\circ$? From exterior angles, I see that $\angle ANM = \angle NAJ + \angle BAM$. $\endgroup$ – Hao Ye Aug 18 '14 at 18:39
  • $\begingroup$ shouldn't that be $\angle AMN=\angle MAH-\angle AJM=30^\circ$ instead? $\endgroup$ – Mick Aug 19 '14 at 1:44
  • $\begingroup$ @Mick: Yes, fixed. $\endgroup$ – Quang Hoang Aug 19 '14 at 1:55

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