<Tri-2024-2082>, hint

Let R,R' be the radii of triangles ABC, ADF.

Since AE=AF, <AEF=<AFE. 2R=AD/sinAEF, 2R'=AD/sinAFE. Hence R=R'

2R=AE, and <EAD=90. Let the diameter (=2R') be AA' then <ADA'=90

AA'DE is parallelogram and AA'//DE, which means AA' is perpendicular to BC.

AK is also perpendicular to BC.