<Tri-2026-2308>, hint

Suppose AM cuts EF at K'

Since [ABM]=[ACM] ;ABxAM sinBAM=ACxAM sin CAM ; 2RsinC  sinBAM=2RsinB sinCAM

sinBAM/sinCAM=sinB/sinC  Hence FK'/EK'=sinB/sinC

2[DKF]=DFxDK sinB/2=2IF cosB/2 xDK sinB/2=IFx DK sinB

Similarly 2[DKE]=IEx DK sinC. IE=IF ( inradius), [DEK]/[DFK]=EK/FK

Hence EK/FK=sinC/sinB=EK'/FK'. Hence K=K' A,K,M are collinear.