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Right-angle revelation

Source: Philippine Mathematical Olympiad 2020/4

January 19, 2020
PMOgeometry

Problem Statement

Let △ABC\triangle ABC be an acute triangle with circumcircle Γ\Gamma and DD the foot of the altitude from AA. Suppose that AD=BCAD=BC. Point MM is the midpoint of DCDC, and the bisector of ∠ADC\angle ADC meets ACAC at NN. Point PP lies on Γ\Gamma such that lines BPBP and ACAC are parallel. Lines DNDN and AMAM meet at FF, and line PFPF meets Γ\Gamma again at QQ. Line ACAC meets the circumcircle of △PNQ\triangle PNQ again at EE. Prove that ∠DQE=90∘\angle DQE = 90^{\circ}.
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