Problem Section #3 c) Let $ABCDE$ be a convex pentagon such that $BC \parallel AE, AB = BC + AE$, and $\angle{ABC} =\angle{CDE}$. Let $M$ be the midpoint of $CE$, and let $O$ be the circumcenter of triangle $BCD$. Given that $\angle{DMO}=90^{o}$, prove that $2\angle{BDA} =\angle{CDE}$.