Let F in BD be such that BF=AB (and hence FD=CD). First note that C, F and A are collinear (this is easy to prove). Let G in the ray CD be such that DG=CD. Note that C, F and G belong to the circle centered at D with diameter CG. Hence angle CFG=90º. Note that BE is parallel to MD, for if C' is the orthogonal projection of C into MD then C' is the midpoint of CE, hence MC' is parallel to BE. This also gives BE perpendicular to CE, and this in turn implies that the quadrilateral CFEG is cyclic. Hence angle FEB=angle GCF. But angle GCF=angle CAB, so the quadrilateral FEAB is cyclic. Hence angle AEB=angle BFA=angle ACD, as desired.

Let $AM$ intersect $CD$ at $X$ and $AC$ intersect $BD$ at $Y$. Since $M$ is the midpoint of $BC$ and $AB \| CD$, it follows that $ABXC$ is a parallelogram and hence that $BD=AB+CD=CX+CD=DX$. Hence $DBX$ is isosceles and, since $Y$ is the intersection of $AC$ and $BD$ and $AC \| BX$, it follows that $DYC$ is isosceles. Since $E$ is the reflection of $C$ across $DM$, $DE=DC=DY$ and $E$, $Y$ and $C$ lie on a circle with center $D$. Hence $\angle{AYE}=180^\circ - \angle{EYC}=\angle{MDC}$ since $DM$ bisects $\angle{EDC}$. Let $DM$ intersect $EC$ at $Z$. Then $Z$ is the midpoint of $EC$ which implies that $ZM \| EB$ and hence that $\angle{AYE}=\angle{MDC}=\angle{ABE}$. Hence $ABYE$ is cyclic and $\angle{AEB}=\angle{AYB}=\angle{YAB}=\angle{ACD}$.

Apply http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=419352; the trapezoid $ATMS$ is our $ABCD$, $M$ in our problem is $NP\cap MT$ from that problem and $E$ from our problem is the midpoint of $DE$ in that problem. Best regards, sunken rock

let us note that $DE=DC=DP$,hence D is the circumcenter of triangle EPC,so $\angle APE=\frac{1}{2}\angle EDC=\angle MDC=\angle ABE$, ABPE concyclic what remains is trivial.