51st Ukrainian National Mathematical Olympiad, 3rd Round

Lines $LN$ and $AC$ are parallel, hence $\angle NLA=\angle LAC$. We have to show the following implication (fig. 7): $$\angle ANC=\angle ALB\iff \angle ABM=\angle NLA.$$ Let $G$ be a centroid, then we have the following equivalences. $\angle ANC=\angle ALB\iff \angle BNC+\angle ALB=\pi \iff BNLG$ - cyclic $\iff \angle ABM=\angle NLA$, since they share the common segment. The statement is proved