Question 83
Let $ABCDE$ be a convex pentagon in which $BC \parallel AE$, $AB = BC + AE$ and $\angle ABC = \angle CDE$. Let $M$ be the mid-point of $CE$ and let $O$ be the circumcentre of triangle $BCD$. Suppose $\angle DMO = 90^\circ$. Prove that $2\angle BDA = \angle CDE$.
