Spring Mathematical Tournament

Let $AN\cap BL=P$.

a) Since $\angle LAM=\angle MAN=\angle LBM=\angle MBN$, the quadrilaterals $ABED$ and $ABFG$ are cyclic. Then $\angle AED=\angle ABL=\angle ANL$ and hence $DE\parallel LN$. Analogously $FG\parallel LN$.

b) Since $\frac{DP}{LP}=\frac{DE}{LN}=\frac{GF}{LN}=\frac{CF}{CN}$, then $\frac{LD}{DP}=\frac{NF}{FC}$. Hence $$\frac{LA}{AP}=\frac{LD}{DP}=\frac{NF}{FC}=\frac{NA}{AC}.$$ It follows that $△APL\sim △ACN$ which gives $\angle APL=\angle ACB$, i.e. $LPNC$ is a cyclic quadrilateral. Then $$\begin{array}{rl}180^{\circ }& =\angle APB+\angle ACB=180^{\circ }-\angle PAB-\angle PBA+180^{\circ }-\angle CAB-\angle CBA\\ & =2\cdot 180^{\circ }-(\angle PAB+\angle CAB)-(\angle PBA+\angle CBA)\\ & =2(180^{\circ }-\angle MAB-\angle MBA)=2\angle AMB,\end{array}$$ i.e. $\angle AMB=90^{\circ }$. So $DF\perp EG$ and therefore the parallelogram $DEFG$ is a rhombus.