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IZhO 2015 Problem 2

    Inside the triangle ABCA B C a point MM is given. The line BMB M meets the side ACA C at NN. The point KK is symmetrical to MM with respect to ACA C. The line BKB K meets ACA C at PP. If ∠AMP=∠CMN\angle A M P=\angle C M N, prove that ∠ABP=∠CBN\angle A B P=\angle C B N.

    Dataset text NuminaMath olympiads. Quoted with source.

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