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<APM + <BPC = 180^o , isosceles, < PAB =<PBC

Source: Indian Postal Coaching 2007 set 6 p1

May 25, 2020
anglesgeometryisoscelesequal angles

Problem Statement

Let ABCABC be an isosceles triangle with AC=BCAC = BC, and let MM be the midpoint of ABAB. Let PP be a point inside the triangle such that ∠PAB=∠PBC\angle PAB =\angle PBC. Prove that ∠APM+∠BPC=180o\angle APM + \angle BPC = 180^o.
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