Let ABC be an isosceles triangle with BC=CA, and let D be a point inside side AB such that AD<DB. Let P and Q be two points inside sides BC and CA, respectively, such that ∠DPB=∠DQA=90∘. Let the perpendicular bisector of PQ meet line segment CQ at E, and let the circumcircles of triangles ABC and CPQ meet again at point F, different from C.
Suppose that P, E, F are collinear. Prove that ∠ACB=90∘. geometryright triangleIsosceles TriangleIMO ShortlistIMO Shortlist 2020