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\angle PBA=\angle PCA, AK \perp BC

Source: Oral Moscow Geometry Olympiad 2024, 8-9.6

September 3, 2024
geometry

Problem Statement

Given an acute-angled triangle ABCABC and a point PP inside it such that PBA=PCA\angle PBA=\angle PCA. The lines PBPB and PCPC intersect the circumcircles of triangles PCAPCA and PABPAB secondly at points MM and NN, respectively. Let the rays MCMC and NBNB intersect at a point SS, KK is the center of the circumscribed circle of the triangle SMNSMN. Prove that the lines AKAK and BCBC are perpendicular.