2011 AMC 10B Problem 18

The angles $\angle AMD$ and $\angle MDC$ are equal since $AB \parallel DC.$ As such, $\angle MDC = \angle DMC ,$ making $MDC$ isosceles and $MC = DC = 6.$ As we can see, $\sin (\angle CMB) = \frac 12,$ making $\angle CMB = 30^\circ .$ Therefore, $\angle AMC = 150^\circ .$ Since $\angle AMD$ is half of that, $\angle AMD = 75^\circ .$ Thus, the correct answer is E .