Let $B'$ be the reflection of $B$ w.r.t. to $M$ i.e., $ABCB'$ is a parallelogram. We have that $\angle BB'A = \angle B'BC = \angle KBC = \angle BKD \implies KD\parallel B'A.$ This means that $$\frac{BC}{AB} = \frac{B'A}{AB} = \frac{KD}{DB} = \frac{KD}{DC}$$Also, $\angle KDC = \angle DCB = \angle DBC = \angle ABC.$ Therefore, $\triangle ABC \sim \triangle CDK,$ or $\angle KCD = \angle CAB,$ as desired.

Let $N$ be the midpoint of $BC$ and $MN \cap DK=L$. We have $BNM \sim KLM \sim KDB$ so \[\frac{KD}{DC}=\frac{KD}{DB}=\frac{LM}{KL}=\frac{NM}{BN}=\frac{AB}{BC}\]Also $\angle KDC=\angle ABC$ so $KDC \sim ABC$ $\implies \angle KCD=\angle BAC$

It is enough to prove that $ABC$ and $CDK$ triangles are similar. Since $\angle ABC=\angle DCB=\angle KDC$, we will finish by proving $DK/BC=DC/AB$. Let $MB \cap DC=T$ and $ AT\cap BC=L$, since $M$ is midpoint $AC$ we have $DL//AC$, then $DK/BC=BL/BC=BD/AB=DC/AB$ ($DK=BL$ because $BM$ passes through midpoint of $DL$), which completes the problem.