<Qua-2024-726>, hint

Let P be a point  on arc EF 

Let <BAD=A, <ABC=B, <ADB=D. In triangle ABE <DBE=<ADB=D, <AEB=<ADB=D

Because A,D,E, B are concyclic. Hence <EAB=180-(B+2D)

Similarly <FAD=180-(2B+D) and <EAF= A-<EAB-<FAD=180-2A, (A+B+D=180)

Since A,E,P,F re concyclic <EPF=180-<EAF=2A. Since <EPF=2 <ECF  the circumcenter of

Triangle ECF lie on the arc EF, where the perpendicular bisector of EF cuts the arc EF at the circumcenter of triangle ECF.