Mongolian National Mathematical Olympiad

Let $P$ and $Q$ be midpoints of the sides $BE$ and $CD$ respectively. Then $PN$ and $MQ$ are midlines of the triangles $BDE$ and $BDC$. Thus $$PN=QM=\frac{BD}{2}$$ and similarly $$NQ=PM=\frac{EC}{2}.$$ From the hypothesis of the problem we have $BD=EC$. So $PNQM$ is a rhombus. Thus $MN$ is a bisector of $\angle PNQ$ and $PN\parallel BA$, $NQ\parallel AC$ implies $AL\parallel MN$.