Question 133
Let $P$ and $Q$ be two points in the interior of the triangle $ABC$ such that $\angle PAC = \angle BAQ$ and $\angle CBP = \angle QBA$. Denote by $P_A, P_B$, and $P_C$ the orthogonal projections of the point $P$ onto the sides $BC, CA$, and $AB$, and by $Q_A, Q_B$, and $Q_C$ the orthogonal projections of the point $Q$ onto the sides $BC, CA$, and $AB$. Prove that the points $P_A, P_B, P_C, Q_A, Q_B$, and $Q_C$ lie on the same circle.