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Midpoint of base in isosceles triangle [<APM + <BPC = 180°]

Source: Poland National Olympiad 2000, Day 1, Problem 2

January 25, 2005
geometrytrigonometrycircumcircleanalytic geometrygeometric transformationreflectioncyclic quadrilateral

Problem Statement

Let a triangle ABCABC satisfy AC=BCAC = BC; in other words, let ABCABC be an isosceles triangle with base ABAB. Let PP be a point inside the triangle ABCABC such that ∠PAB=∠PBC\angle PAB = \angle PBC. Denote by MM the midpoint of the segment ABAB. Show that ∠APM+∠BPC=180∘\angle APM + \angle BPC = 180^{\circ}.
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