Problem

Source: IX International Festival of Young Mathematicians Sozopol, Theme for 10-12 grade

Tags: geometry



On the sides $AC$ and $AB$ of an acute $\triangle ABC$ are chosen points $M$ and $N$ respectively. Point $P$ is an intersection point of the segments $BM$ and $CN$ and point $Q$ is an inner point for the quadrilateral $ANPM$, for which $\angle BQC = 90^\circ$ and $\angle BQP = \angle BMQ$. If the quadrilateral $ANPM$ is inscribed in a circle, prove that $\angle QNC = \angle PQC$.