2011 AMC 12B Problem 10

Because $AB\parallel CD,$ we have $\angle CDM=\angle AMD.$ Combined with $\angle AMD=\angle CMD,$ this gives $\angle CDM=\angle CMD,$ so $\triangle CMD$ is isosceles with $CM=CD=6.$ Then $\triangle MBC$ is right-angled at $B$ with hypotenuse $CM=6$ and leg $BC=3,$ so it is a $30$-$60$-$90^\circ$ triangle with $\angle BMC=30^\circ.$ Finally, $\angle AMD+\angle CMD$ $+\angle BMC=180^\circ,$ so $2\angle AMD+30^\circ=180^\circ,$ giving $\angle AMD=75^\circ.$ Thus, the correct answer is E.