let the midpoint of $CD$ be $N$,and let $ME$ meet $CD$ at $H$, let the circumcenter of $CDE$ be $O$. since $EBA\sim ECD$, $ECN\sim EBM$ $\therefore \angle NEC=\angle MEB=\angle DEH$ it is a well known fact that $\angle DEH=\angle OEC$ $\therefore NEC=\angle OEC$ $\therefore O$ lies on line $EN$ if $O=N$, $\angle DEC=90^{\circ}$, $AC\perp BD$ if $O\neq N$, $EN\perp CD$ $\therefore \angle ECD=\angle EDC=\angle EAB$, $\therefore AB\parallel CD$