# Prove In Equilateral Triangle $ABC$ $AM$ and $AN$ are equal

$\triangle ABC$ 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.

• Hint: $N\in(MJ)$ is the incenter of $ACJ$. – Lucian Aug 18 '14 at 23:52

## 1 Answer

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 $\angle ANM=\angle AMN (=30^\circ).$

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