<Tri-2025-2272>, hint

Let <BAA'=x, <CAA'=y, x+y=<A

Since AM/MA' =2/1, AM=A"M, MBA"C is parallelogram

MB=CA", MC=BA", In triangle BMC" <BMC=x+y=<MC"B, thus BM=BC"

Similarly In triangle CMB", <CMB"=<CB"M=x+y. Triangles CA"B"and BC"A" are congruent

Hence A"B"=A"C".

<Remark> BA' xCA'=(a/2)x(a/2)=A"A' xA'A=(1/4)AA" x (3/4)AA"

Thus a=(V3/2)AA"=2RsinA=d sinA. Suppose AA"=d (diameter), <A=60

Then <A"B"C"=<A"C"B"=x+y=<A=60. A"B"C" is isolateral