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BALTIC-WAY 2002 Problem 14

    Let L,ML, M and NN be points on sides AC,ABA C, A B and BCB C of triangle ABCA B C, respectively, such that BLB L is the bisector of angle ABCA B C and segments AN,BLA N, B L and CMC M have a common point. Prove that if ∠ALB=∠MNB\angle A L B=\angle M N B then ∠LNM=90∘\angle L N M=90^{\circ}.

    Dataset text NuminaMath olympiads. Quoted with source.

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