Parallelogram and Congruence Let point M be outside the parallelogram ABCD such that $\angle MAB = \angle MCB$. Prove that $\angle AMD = \angle CMB$. 
I am trying to prove $\triangle MDE \sim \triangle MBC$ but I am having trouble seeing how to do this. 
 A: Let $AMKB$ be parallelogram, which gives that $DMKC$ is parallelogram. 
Hence, $\measuredangle MAB=\measuredangle MKB$.
Thus, $\measuredangle MCB=\measuredangle MKB$, which says that $MKCB$ is cyclic.
Hence, $\measuredangle CMB=\measuredangle BKC$ and since $\measuredangle AMD=\measuredangle BKC$, we obtain $\measuredangle AMD=\measuredangle CMB$.
Done!

A: Let $k$ be the circle circumscribed around triangle $BCM$. Produce $AB$ until it intersects the circle $k$ the second time at the point $A^*$ and produce $DC$ until it intersects the circle $k$ the second time at the point $D^*$. As $DC$ is parallel to $AB$ (because $ABCD$ is a parallelogram), the segments $CD^*$ and $BA^*$ are parallel, hence $A^*BCD^*$ is a trapezoid, inscribed in the circle $k$. Therefore $BC = A^*D^*$ as well as  $\angle\, D^*A^*B = \angle \, CBA^*$ and $\angle\, CMB = \frac{1}{2} \text{arc}(CB)= \frac{1}{2}\text{arc}(D^*A^*) =  \angle \, D^*MA^* $. 

However, because $ABCD$ is a parallelogram and $BC \, || \, AD$, we have $A^*D^* = BC = AD$ and $\angle \, CBA^* = \angle \, DAB$. Thus  $$\angle \, D^*A^*B = \angle \, CBA^* = \angle \, DAB$$ Since the polygon $BCD^*A^*M$ is inscribed in $k$, combined with the assumptions of the problem,
$$\angle \, MAB = \angle\, MCB = \angle \, MA^*B = \angle \, MA^*A$$ so triangle $MAA^*$ is isosceles with $A^*M = AM$. Finally, $$\angle \, MA^*D^* = \angle \, MA^*B + \angle \, D^*A^*B = \angle \, MAB + \angle \, DAB = \angle \, MAD$$ which yields that triangles $A^*D^*M$ and $ADM$ are congruent and for that reason $$\angle\, AMD = \angle\, A^*MD^* = \angle \, CMB$$
