<Tri-2025-2258>, hint

Let vectors GA,GB,GC be a,b,c and G is the centroid of ABC. Then a+b+c=o

BA'/A'C=x/1-x, CB'/B'A=y/1-y, AC'/C'B=/1-z. 

GA'=xc+(1-x)a, GB'=ya+(1-y)c, GC'=zb+(1-z)a, GA"=(GB'+GC')/2={(1+y-z)a+zb+(1-y)c

Since A"A'//AB, A"A'={(1+y-z-1+y)a+(z-1+y)b}/2=t(a-b)      #(a+b+c=0)

From above we get 2z+y-1=0. Similarly 2y+x-1=0, 2x+z-1=0

Hence x=y=z=1/3.