<Tri-2026-2290>, hint

Let areas of triangles  GAD,GDB,GBE,GC,GCF,GFA be z,1,x,1,y,1 respectively.

By Ceva's (BE/EC ) x(CF/FA) x(AD/DB)=x y z=1

[AGB]/AGC]=BE/CE ---> (1+z)/(1+y)=x/1 --->x(1+y)=1+z

Similarly    y(1+z)=1+x, z(1+x)=1+y. From these we get xy+yz+zx=3

Suppose x<=y<=z, then x(1+y)=1+z---> x(1+y)>=1+y  --->(1+y)(x-1)>=0, Thus x>=1

If x>1 then xyz=1 is impossible. Hence x=1 yz=1, Finally x=y=z=1