On the side $AB$ of the non-isosceles triangle $ABC$, let the points $P$ and $Q$ be so that $AC = AP$ and $BC = BQ$. The perpendicular bisector of the segment $PQ$ intersects the angle bisector of the $\angle C$ at the point $R$ (inside the triangle). Prove that $\angle ACB + \angle PRQ = 180^o$.