Consider the triangle $ABC$. Let $P, Q$ inside the angle $A$ such that $\angle BAP=\angle CAQ$ and $PBQC$ is a parallelogram. Show that $\angle ABP=\angle ACP.$
Problem
Source: OLCOMA Costa Rica National Olympiad, Final Round, 2013 Shortlist G2 day1
Tags: parallelogram, equal angles, geometry