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<PBM+<CBM=<PCA,<BM=90^o, <ABC+<APC=180^o (2020 Kyiv City MO Round2 10.2)

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September 21, 2020
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Problem Statement

Let MM be the midpoint of the side ACAC of triangle ABCABC. Inside △BMC\vartriangle BMC was found a point PP such that ∠BMP=90o\angle BMP = 90^o, ∠ABC+∠APC=180o\angle ABC+ \angle APC =180^o. Prove that ∠PBM+∠CBM=∠PCA\angle PBM + \angle CBM = \angle PCA.
(Anton Trygub)
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