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AK bisects BM, <MAC =< PCB, <MPA = <CPK (2018 Kyiv City MO 9.5 10.4.1)

Source:

September 12, 2020
geometrybisects segmentequal angles

Problem Statement

Given a triangle ABCABC, the perpendicular bisector of the side ACAC intersects the angle bisector of the triangle AKAK at the point PP, MM - such a point that ∠MAC=∠PCB\angle MAC = \angle PCB, ∠MPA=∠CPK\angle MPA = \angle CPK, and points MM and KK lie on opposite sides of the line ACAC. Prove that the line AKAK bisects the segment BMBM.
(Anton Trygub)
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