<Qua-2025-784>, hint

Incircle of triangle ABC,  incenter I touches BC,CA,AB at D,E,F respectively and M be the midpoint of DF .

Line BM cuts EF at K. Prove <BKC=90.

BM/MK=(AM cotB/2)/AM cotC/2 (<MFE=90-C/2)=(cotB/2)/(cotC/2)

BD/DC=(rcotB/2)/(rcotC/2)=(cotB/2)/(cotC/2), r=inradius

Hence DM and CK are parallel and <BMD=<BKC=90.