Question 19
Isosceles $\triangle ABC$ ($AC = BC$) is inscribed in a circle $k$. A point $M$ lies on the side $BC$. A point $N$ from the ray $AM$ ($M$ lies between $A$ and $N$) is such that $AN = AC$. The circumcircle of $\triangle MCN$ intersects $k$ at $C$ and $P$, where $P$ is from the arc $BC$, not containing $A$. The lines $AB$ and $CP$ intersect at $Q$. Prove that $\angle QMB = \angle QMN$.
