Let $D$ be the midpoint of $BC$. Let $G'$ be the reflection of $G$ over $D$. As $BD=DC$ and $GD=DG'$, $GBG'C$ is a parallelogram. Therefore, $GB=G'C$ and $GC=G'B$, whence, $AB+BG'=AC+CG'$. Thus, $B$ and $C$ lie on an ellipse $e$ with foci $A$ and $G'$. Let $e'$ be the reflection of $e$ over $D$. As $BD=DC$, $B$ and $C$ also lie on $e'$, and thus on $e\cap e'$. As $e$ and $e'$ are distinct ellipses symmetric about $AD$, $BC\perp AD$, whence $AB=AC$.
(Note that we didn't use the fact that $G$ is the centroid of $ABC$ in the proof, we just used the fact that $G$ lies on $AD$. So, the same proof proves the following more general statement: If $G$ is a point on $AD$ such that $AB+GC=AC+GB$, then, $\triangle ABC$ is isosceles.)