<Tri-2024-2147>, hint

Let M be the midpoint of AC, Since OM ㅗ AC OMAE is cyclic  

Since <DFC=<FDC=1/2 <BAC+<ACB and OMEA is cyclic ---><AME=<AOE=<ACB+<BCK

=<ACB+1/2 <BAC.   <DFC=<EMA.   Hence CMEF is cyclic

(Line AF cuts circumcircle of ABC in K)

Since AF , AE are common,  AB=AM, <BAF=<CAF .Triangles BEF  and  MEF are congruent

and <EMF=<ECF. <EBF=<ECF