<Tri-2026-2310>, hint
Suppose line AK cuts BC at P. By Ceva's theorem
(AE/EB )x(BP/PC)x(CF/FA)=1.
since AE=AF , CF/BE=CP/BP, BE=AB sinB sinA/2 cotB, CF=AC sinC snA/2 cotC
CF/BE= (2R sinB cosC)/(2RsinC cosB).
Let P' be the feet of perpendicular from A then BP'=2R snC cosB, CP'=2RsinB cosC
Hence , CF/BE-CP/BP=CP'/BP', P=P'
<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.
