$\angle{CMD}=2\angle{CAD}=\angle{CBD}$ $\implies BMDC$ is a cyclic quadrilateral $\implies \angle{CDM}=90^{\circ}$

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The answer is $\angle CDM=90^{\circ}$. To prove this, let's prove that the quadrilateral $BMDC$ is cyclic. Let $\angle CAD=\alpha$. Then $\angle DBC=2\alpha$, since $B$ is the circumcenter of triangle $ADC$ (from $AB=BC=BD$). The triangle $AMD$ is isosceles, so $\angle DMC=2\alpha$. So we have $\angle MDC=2\alpha=\angle DBC$, meaning the quadrilateral $BMDC$ is cyclic.

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