Without loss of generality, we assume that the vertices of parallelogram ABCD are named conter-clockwise, which implies that vertices A, B, M, E and E, F, M, D appear in counter-clockwise order on the respective circumcircles.
If we prove that
(1) angles FDC = DMA
and
(2) FD : CD = DM : AM,
then triangles CDF and ADM are similar (by S:A:S), which implies that (using angles CAD = ACB in parallelogram ACBD)
angles DCF = MAD = CAD = ACB,
as was to be proved.
Proof of (1): Since quadrilaterals ABME and EFMD are cyclic (and using MBA = DBA = BDC = MDC in parallelogram ABCD),
angles DMA = 180° - AMB = 180° - AEB = 180° - (AEM - FEM) = MBA + FEM = MDC + FDM = FDC.
Proof of (2):
As above, since quadrilaterals EFMD and ABME are cyclic,
angles FDM = FEM = BEM = BAM and angles MFD = MED = 180° - AEM = MBA,
which proves that triangles DFM and ABM are similar (by A:A:A), thus (using AB = CD in parallelogram ABCD)
DM : AM = FD : AB = FD : CD.