Let $N$ be the midpoint of $BC$; then $CKN$ and $BAC$ are clearly similar so as $CM / MN = BM / MC$ the angle equality follows as desired.

Through $K$, draw a line which is parallel to $AM$ which intersects $BC$ at $E$. We can prove that $BE=EM=CM$ and $angle BAM=angle BKE$. We can also prove that $BCK$ is an isosceles triangle with $BK=CK$. Hence we can prove that $BKE$ and $CKM$ are congruent triangles. That's also lead to our desired result.

Attachments:

Take $D$ reflection of $A$ across $BC$; $M$ is the centroid of $\triangle ABD$ and $K-M-D$ are collinear. $CKBD$ is a trapezoid, $BD\parallel CK\implies\widehat{CKD}=\widehat{BDK}=\widehat{BAM}$, done. Best regards, sunken rock