[asy][asy] import olympiad; size(7cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair A = dir(20), D = dir(160), E = dir(235), B = dir(305), M = (B+E)/2, C = B+D-A, G = extension(D,M,E,E+dir(B--A)), C1 = 2M-C, F = D+E-C; draw(C--D--C1^^E--G--B, linewidth(1)); draw(C1--B--D--A--B--C^^M--D--E); draw(C--A--E, dotted); draw(G--F, dotted); dot("$A$", A, dir(45)); dot("$B$", B, dir(270)); dot("$C$", C, dir(225)); dot("$D$", D, dir(135)); dot("$M$", M, dir(270)); dot("$E$", E, dir(270)); dot("$C'$", C1, dir(315)); dot("$G$", G, dir(180)); dot("$F$", F, dir(90)); [/asy][/asy] Let $C'$ be the reflection of $C$ over $M$. Note that both $ABED$ and $AC'CD$ are isosceles trapezoids. Also $\overline{EG} \parallel \overline{CD}$ because $CEFD$ is a parallelogram, so $\overline{BG} \parallel \overline{C'D}$ from $$\frac{MG}{MD} = \frac{ME}{MC} = \frac{MB}{MC'}.$$Finally, by symmetry $$\angle EAC = \angle BDC' = \angle GBD$$as desired.

There shall be 2 cases, where $AB > BC$ and $AB < BC$

It's actually sufficient to prove that $OM$, $BG$ and $AE$ are concurrent at some point $X$, where $O$ is the intersection point of the diagonals $AC$ and $BD$ (then $OX$ is parallel to $DE$, so $<XOB=<EDB=<EAB$ ($EBAD$ is isosceles trapezoid), and thus $XBOA$ is cyclic, which gives what we want). Has anyone proof for this concurrency?

Also this problem could be solved by "complex numbers" easily.