Question 37
Let $L$, $M$ and $N$ be points on sides $AC$, $AB$ and $BC$ of triangle $ABC$, respectively, such that $BL$ is the bisector of angle $ABC$ and segments $AN$, $BL$ and $CM$ have a common point. Prove that if $\angle ALB = \angle MNB$ then $\angle LNM = 90^{\circ}$.