Question 77
Let $P$ be a point inside the triangle $ABC$ satisfying $\angle PAC = \angle PCB$ and $D$ be the midpoint of the line segment $PC$. Let $AP$ intersect $BC$ at $E$ and $ED$ intersect $BP$ at $Q$ such that $P$ is on the line segment $BQ$. Prove that $\angle BAP + \angle BCQ = 180^\circ$.