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Prove that APM + BPC = 180

Source: 6th Grand Duchy of Lithuania contest

December 31, 2014
geometrycircumcirclegeometry proposed

Problem Statement

An isosceles triangle ABCABC with AC=BCAC = BC is given. Let MM be the midpoint of the side ABAB and let PP be a point inside the triangle such that ∠PAB=∠PBC\angle PAB = \angle PBC. Prove that \angle APM + \angle BPC = 180 \textdegree
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