MHB Proving isosceles using centroid and medians

[slwarrior64]

I can definitely do this in the opposite direction, but

 

[mrtwhs]

Letting $AB=c$, $AC=b$ and noting that the usual property of the centroid tells us that $GC = \frac{2}{3}$ of the median from $C$ and $GB = \frac{2}{3}$ of the median from $B$ we can use Stewart's Theorem to write everything in terms of the three sides $a,b,c$. It is an involved equation requiring two squarings to get rid of all radicals. I used Wolfram Alpha to double check my computations. After simplifying everything I end up with

$(b-c)^2 \left[ (b-c)^2-a^2 \right] =0$.

The bracket fails the triangle inequality which only leaves $b=c$ or $AB=AC$. I can't see any synthetic proof.