Estonian Mathematical Olympiad

Since the points $E,K,L$, and $D$ are concyclic (Fig. 28), it follows that $\angle CKL=\angle DKL=\angle DEL=\angle MEB$. From the problem conditions, $\angle CME=90^{\circ }$, and by Thales' theorem, $\angle CKE=\angle DKE=90^{\circ }$. Therefore, the points $C,M,K,E$ are also concyclic. Consequently, $\angle CKM=\angle CEM$. Since point $E$ lies on the perpendicular bisector of side $BC$, the triangles $BEM$ and $CEM$ are congruent, thus $\angle MEB=\angle CEM$. From the previous result, $\angle CKL=\angle CKM$. Since the points $L$ and $M$ lie on the same side of line $CK$, the points $K,L$, and $M$ are collinear.

Fig. 28