imc

It is given that $\angle AMD\sim \angle CMD$. Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB}\parallel \overline{DC}$, $\angle AMD\cong \angle CDM⟶\angle CMD\cong \angle CDM$. Use the Base Angle Theorem to show $\overline{DC}\cong \overline{MC}$. We know that $ABCD$ is a rectangle, so it follows that $\overline{MC}=6$. We notice that $△BMC$ is a $30-60-90$ triangle, and $\angle BMC=30^{\circ }$. If we let $x$ be the measure of $\angle AMD,$ then $$\begin{array}{rl}2x+30& =180\\ 2x& =150\\ x& =\boxed{75}\end{array}$$