We claim that $DFCB$ is cyclic indeed : (with directed angles) $\angle DFB=\angle DFE=\angle DME=\angle BME=\angle BAE=\angle BAD=\angle DCB.$ Hence $\angle DCF= \angle DBF = \angle MBE= \angle MAE= \angle CAD=\angle ACB.$ Best regards. RH HAS

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.

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