China Southeastern Mathematical Olympiad

It is easy to see that points $I$, $D$, $E$ and $B$ are concyclic and $$\begin{gathered} \angle AID=90^{\circ }-\angle IAD, \\ \angle MED=\angle FDA=90^{\circ }-\angle IAD. \end{gathered}$$ So $\angle AID=\angle MED$, thus points $I$, $D$, $E$ and $M$ are concyclic. Hence, five points $I$, $D$, $B$, $E$, $M$ are concyclic and $\angle IMB=\angle IEB=90^{\circ }$, that is $AM\perp BM$. Similarly, points $I$, $D$, $A$, $N$ and $F$ are concyclic and $BN\perp AN$. Let lines $AN$ and $BM$ intersect at point $G$. We see point $I$ is the Fig. 2. 1 orthocenter of $△GAB$ and $ID\perp AB$, so points $G$, $I$ and $D$ are collinear. Since points $G$, $N$, $D$ and $B$ are concyclic, we see that $\angle ADN=\angle G$. Similarly, $\angle BDM=\angle G$. So $DK$ bisects $\angle MDN$, thus $$\frac{DM}{DN}=\frac{KM}{KN}.\stackrel{◯}{1}$$ Since points $I$, $D$, $E$ and $M$ are concyclic, and points $I$, $D$, $N$ and $F$ are concyclic, we see that $$KM\cdot KE=KI\cdot KD=KF\cdot KN.$$ Therefore, $$\frac{KM}{KN}=\frac{KF}{KE}.\stackrel{◯}{2}$$ By ① and ②, we see that $\frac{DM}{DN}=\frac{KF}{KE}$, that is $DM\cdot KE=DN\cdot KF$. $\square$