China Southeastern Mathematical Olympiad

It is easy to see that points $I$, $D$, $E$ and $B$ are concyclic and
$$\angle AID = 90^\circ - \angle IAD, \\ \angle MED = \angle FDA = 90^\circ - \angle IAD.$$
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 $\triangle GAB$ 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
$$\frac{DM}{DN} = \frac{KM}{KN}. \qquad \textcircled{1}$$
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,
$$\frac{KM}{KN} = \frac{KF}{KE}. \qquad \textcircled{2}$$
By ① and ②, we see that $\frac{DM}{DN} = \frac{KF}{KE}$, that is $DM \cdot KE = DN \cdot KF$. $\square$