2. Triangle $A B C$ is isosceles at $C$, and let $\Gamma$ be its circumcircle. Let $M$ be the midpoint of the arc $B C$ of $\Gamma$ that does not contain $A$, and let $N$ be the point where the line parallel to $A B$ through $M$ intersects $\Gamma$ again. It is known that $A N$ is parallel to $B C$. What are the measures of the angles of $A B C$?
We then have that $\angle A N C=\angle B A N$. But $\angle N A C=\angle A C B$ and $\angle A N M=\angle A B C$ because $A N$ and $B C$ are parallel, and $A B$ and $M N$ are also parallel. We then have on the one hand that $\angle B A N=\angle A+\angle C$. Finally, $\angle C N M=\angle C A M=\frac{1}{2} \angle A$ because $M$ is the midpoint of arc $B C$, making $A M$ the angle bisector of $\angle A$, so $\angle A N C=\angle B+\frac{1}{2} \angle A$. Equating both angles, and using that $\angle A=\angle B$ because $A B C$ is isosceles at $C$, we have that $\angle A=\angle B=2 \angle C$, thus $180^{\circ}=\angle A+\angle B+\angle C=5 \angle C$, for $\angle C=36^{\circ}$, and $\angle A=\angle B=72^{\circ}$.