MathDB

2018.9.5

Part of Kyiv City MO Juniors 2003+ geometry

Problems(1)

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

Source:

9/12/2020
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)
geometrybisects segmentequal angles