Question 92
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 $O$ the circumcenter of triangle $BCD$. Given $\angle DMO=90^\circ$, prove $2\angle BDA=\angle CDE$.
