Let $AC \cap \odot(\triangle BED)=K$ . Then $\angle DFC=\angle DBE=\angle DKE$. Thus $DCFK$ is cyclic. Hence $\angle ABD=\angle BDC=\angle CKF$. So $KABF$ is cylcic. We are done.

let L is on AC and circumcircle of triangle ABE . and P is point of meeting AC and BD we must show LBED is cyclic . we want to show LP * PE = BP * DP so LP/BP=DP/PE LBFA is cyclic so LP * AP = PB * PF so LP/BP=PF/AP so we must show PF/AP=DP/PE so AP * PD =PF * PE we know AP * PD =BP * PC so we want to show BP * PC = PF * PE we have BE is parallel to CF so BP/PF=PE/PC so BP * PC = PE * PF and we are done

Mentalist wrote: Let $AC \cap \odot(\triangle BED)=K$ . Then $\angle DFC=\angle DBE=\angle DKE$. Thus $DCFK$ is cyclic. Hence $\angle ABD=\angle BDC=\angle CKF$. So $KABF$ is cylcic. We are done. The consideration of $E$ lying on the extension of segment $AC$ should not be omitted. So this proof is incomplete and not fit to every possible configurations.

Case 1: E lies on AC, Let G be the intersection of (BED) and AC then CFD=EBD=CGD so CGFD is concyclic hence EGD=CFB=190-EBF so EBGD is concyclic. Case 2: E lies outside AC, WLOG outside AB without C so F lies on surface CD without AB. Similarly define G we have EGD=180-EBD=CFD(BE//CF) so CDFG is concylic. Finally GFD=ACD=180-GAB which means ABFG is cyclic

A little different:I assume the configuration where $E$ lies inside segment $AC$ You can use directed angles to have the proof OK for othe cases as well First note that If we prove $DE||AF$ then if we define $X$ to be the intersection of circumcircle of $BDE$ and $AC$ then $\angle{BDE}=\angle{BXE}$ so $\angle{BXE}=\angle{AFD}$ and so $AXBF$ is cyclic as well.For showing $DE||AF$ We apply ratio lemma on triangles $ABE$ and $DCF$ we get that if $T=AC \cap BD$ then $\frac{AX}{XE}=\frac{XF}{XD}$ which implies that $DE||AF$.

-$X=AC \cap (BED)$ -$(DCXF)$ -Converse of reims on $(ABF)$ and $(DCXF)$ gives that $X \in (ABF)$

Very very easy barely took me 15 minutes. We define $R=\overline{AC} \cap (BED)$. Now, we can make the following initial observation. Claim : Quadrilateral $RDCF$ is cyclic. Proof : Simply note that, \[\measuredangle RCF = \measuredangle REB = \measuredangle RDB = \measuredangle RDF\]which implies the claim. Now note that due to this claim it follows that, \[\measuredangle FRA = \measuredangle FDC = \measuredangle BDC = \measuredangle DBA\]from which it follows that $FRAB$ is also cyclic. So, it is clear that the circumcircles of $\triangle ABF,\triangle BED$ and the line $AC$ are concurrent, as we desired.