2010 IMO Shortlist G5

Let $A B C D E$ be a convex pentagon such that $B C \| A E, A B=B C+A E$, and $\angle A B C=$ $\angle C D E$. Let $M$ be the midpoint of $C E$, and let $O$ be the circumcenter of triangle $B C D$. Given that $\angle D M O=90^{\circ}$, prove that $2 \angle B D A=\angle C D E$. (Ukraine)

令 $A B C D E$ 为凸五边形，使得 $B C \| A E、A B=B C+A E$ 和 $\angle A B C=$ $\angle C D E$。设$M$为$C E$的中点，设$O$为三角形$B C D$的外心。给定$\angle D M O=90^{\circ}$，证明$2 \angle B D A=\angle C D E$。 (乌克兰)