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Kyiv City MO Seniors Round2 2010+ geometry, 2020.10.2

Let $M$ be the midpoint of the side $AC$ of triangle $ABC$. Inside $\vartriangle BMC$ was found a point $P$ such that $\angle BMP = 90^o$, $\angle ABC+ \angle APC =180^o$. Prove that $\angle PBM + \angle CBM = \angle PCA$. (Anton Trygub)

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