<Tri-2025-2284>, hint

AI  and B'C'  cut BC at A', X respectively.

Then By Ceva (AC'/C'B)x (BA'/A'C)x(CB'/B'A)=1

By Menelaus (AC'/C'B)x (BX/XC)x(CB'/B'A)=1.

Hence AX is external  bisector of <A, Since <CAX=60 B' is the incenter of ABX

and B'X bisects <AXC.