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 $\triangle BCA \sim \triangle MCE$. 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.