$\angle FBA=\angle FEA=\angle KEA$ (1) $2\angle KEA=\angle KBA$ (inscribed and central angles of C1) Then $\angle FBK=(1)$ and $BF\perp AC$, $EKDB$ is cyclic (2). $BA=BE$ and $O$ is the circumcenter, so $BOM \perp AE\rightarrow ABEM$ is a kite. $\angle MAE=\angle AKD=\angle MEA$ then $MKDE$ (3) is a cyclic. From (2) and (3) $B,D,K,M,E$ are concyclic (solution ii) If $(1)=\alpha$ then $\angle AMD=\alpha$ and $FDM=(90-\alpha)$, also, $\angle ELB=\angle MLF= (90-\alpha)$ then $LMFD$ is cyclic (solution i).

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Invert with center $B$ and radious $BA$.