<Tri-2025-2260>, hint

AF cuts BD at N and BC at P . EF cuts BC at M, then <ANB=<EMB=90 (ABD , EBC are isosceles)

<DEF=90-C, <DBF=<CBF+<DBC=A/2+(90-<BPA), <BPA=<CAP+<ACP=A/2+C

<DBC=A/2+(90-C-A/2)=90-C. Hence <DEF=<DBF=90-C. D,E,B,F are cyclic