Since $\triangle DBE$ is a right triangle, the projection of $M$ onto $\overline{DE}$ will be the midpoint of $\overline{DE}$, which makes $\triangle DME$ isosceles with $\overline{MD} = \overline{ME}$, and $\overline{MB} = \overline{MD} = \overline{ME} \implies \triangle MEB$ is isosceles. Thus, $\angle MEB = \angle MBE = \angle DBA = \angle CAB \implies EM \ || \ AC$, done.

M is the circumcenter of EBD because E is a right angle, which makes angle BEM equal to angle MBE which is equal to angle DBE. Since ABCD is isosceles, angle DBE equals angle CAB, and we are done.

$M$ is the center of $(BED)$ so $\angle BEM=\angle DBA=\angle BAC$.

<MEB = <MBE (because EM=MD=MB, as AM median in right triangle) = <CDB (because DC// AB, as bases of trapezoid) (1) = <DCA (because triangles DCA, DCB are congruent , due to isosceles trapezoid) = <CAB because of (1) again So EM// AC