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angle chasing, 3 angles 120^o (V Soros Olympiad 1998-99 Round 1 11.8)

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May 25, 2024
geometryangles

Problem Statement

Inside triangle ABCABC, point PP is taken so that angles ∠ARB=∠BPC=∠CPA=120o\angle ARB= \angle BPC = \angle CPA= 120^o. Lines BPBP and CPCP intersect lines ACAC and ABAB at points MM and KK. It is known that the quadrilateral AMPKAMPK has same areq with the triangle BCPBCP. What is the angle ∠BAC\angle BAC?
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