<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'