SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) Because $MN\parallel BC$ so $\angle ANF=\angle FDB=\angle DFB=\angle AFN$, this deduces $AN=AF$. Similarly, $AM=AE=AF=AN$.

Thus, $M$, $N$, $E$, $F$ lie on the circle center $A$. Let $MF$ cut $NE$ at $K$, because $E$, $F$ lie on circle diameter $MN$ so $$\angle KFD=\angle KED=90^{\circ },$$ this means $DK$ is diameter of $(AEF)$. So $K$ lies on $(I)$. Easily seen, $K$ is orthocenter of triangle $DMN$.

2) Let $P$ be the reflection of $K$ through $A$ then $KMPN$ is parallelogram, so $PM\parallel NK\perp MD$ and $PN\parallel KM\perp DN$. Thus, $DP$ is diameter of $(DMN)$. But $DK$ is diameter of $(I)$ so $\angle DLK=90^{\circ }$, we deduce $L$ lies on $AK$. $\square$