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 ∠C at the point R (inside the triangle). Prove that ∠ACB+∠PRQ=180o.
Problem
Source: St Petersburg Olympiad 2016, Grade 9, P3
Tags: geometry, perpendicular bisector, angle bisector, equal segments