Hint

Let $D$ be the midpoint of $\overline{ME}$, let $O$ be the circumcenter of $\triangle ABC$ and $F=\overline{AO}\cap \overline{KF}$. Note that the given angle condition is equivalent to saying that $K$ lies on the line through $A$ parallel to $C$-tangent to $(ABC)$. Hence $\overline{CO}\perp \overline{KA}$, i.e. $\measuredangle KFC=90^\circ$. As $\measuredangle KMC=90^\circ$, $KCFM$ is cyclic. Also, $OAFM$ is cyclic since $\measuredangle AFO=90^\circ=\measuredangle AMO$. Thus, \begin{align*} \measuredangle MCE=\measuredangle MCK=\measuredangle MFK=\measuredangle MFA=\measuredangle MAO=\measuredangle BCA=\measuredangle BAE=\measuredangle MAE, \end{align*}which means that $AECM$ is cyclic. Thus, also $\measuredangle CEM=\measuredangle CAM=\measuredangle CAB$, hence $\triangle BCA\sim\triangle MCE$. As $M$ is the midpoint of $\overline{BA}$ and $D$ is the midpoint of $\overline{ME}$, we conclude that $\measuredangle BCM=\measuredangle MCD$, we are done. [asy][asy] import geometry;import olympiad; size(8cm);defaultpen(fontsize(10pt)); pair A,B,C,M,O,E,F,K,D; C=dir(120); B=dir(200); A=dir(340); M=midpoint(A--B);O=(0,0);E=(A-C)/(C-B)*(C-M)+C;F=foot(A,C,O);K=extension(C,E,A,F);D=midpoint(M--E); draw(A--B--C--cycle, red+1);draw(circumcircle(A,B,C),royalblue);draw(circumcircle(C,M,A),royalblue+dotted);draw(circumcircle(O,M,F),royalblue); draw(K--C,orange);draw(F--K,orange);draw(C--M,orange);draw(C--F,orange);draw(M--K,orange);draw(M--E,orange); dot("$O$",O,dir(90)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(M)); dot("$K$",K,dir(K)); [/asy][/asy]

Let $O$ be center of $ABC$ and $N$ be midpoint of $ME$. Claim $: MAK$ and $MOC$ are similar. Proof $:$ Note that $\angle KMA = \angle 90 - \angle AMC = \angle OMA - \angle AMC = \angle OMC$ and $\angle MOC = \angle MOA + \angle AOC = \angle 90 - \angle MAO + 2\angle B = \angle 180 - \angle MAE + \angle EAC + \angle 180 - \angle KAC = \angle KAM$. So we have that $MAO$ and $MKC$ are similar so $\angle MCB = \angle ECA$. Claim $: EAC$ and $MCB$ are similar. Proof $:$ Note that $\angle ECA = \angle MCB$ and $\angle EAC = \angle ABC = \angle MBC$. So we have that $CEM$ and $CAB$ are similar so since $N$ and $M$ are midpoints of $EM$ and $AB$ then $\angle NCM = \angle MCB$ we're Done.

Interesting problem. K' is a point such that $\angle K'MC = \angle 90$ and $\angle K'CM = \angle ACB$. Point H lie on line CB such that $\angle AHC = \angle 90$ now we know that AHC and K'MC are similar. We notice that CHM and CAK' are similar too. We have : $\angle CAK' = \angle CHM = \angle 180- ABC$. now we know that K and K' are the same points. $\angle ECM + \angle EAM = 180$ and now we know that EAMC is cyclic and ECM and ACB are similar and this fact solves the problem.