Iranian Mathematical Olympiad

Note that $KA$ is parallel to the tangent line from $C$ to the circumcircle of $ABC$. Therefore, if we denote by $O$ the circumcenter of $ABC$, then $CO\perp KA$. So, denoting by $F$ the intersection point of $KA$ and $CO$, we have $\angle KFC=90^{\circ }$. On the other hand, $\angle KMC=90^{\circ }$, therefore $KCFM$ is cyclic. We then have $\angle AFO=\angle AMO=90^{\circ }$, hence the quadrilateral $OFAM$ is also cyclic. Thus, $$\angle MCE=\angle MCK=\angle MFK=\angle MFA=\angle MOA=\angle BCA.$$ On the other hand, $180^{\circ }-\angle BCA=\angle BAE=\angle MAE$. These facts together imply that $AECM$ is cyclic. Note that $\angle CEM=\angle CAM=\angle CAB$, hence $△BCA\sim △MCE$. Now denote the midpoint of $ME$ by $D$. As $M$ is the midpoint of $BA$, similarity of triangles $BCA$ and $MCE$ implies that $\angle BCM=\angle MCD$. Analogously, $D$ lies on the reflection of line $BC$ with respect to $CM$, as desired.