For $G$ the centroid in $\triangle ABC$, if $AB+GC=AC+GB$, then $\triangle ABC$ is isosceles. (Likewise, for the incenter.) 
Let $G$ be the centroid of $\triangle ABC$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles!

Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians. 
 A: 
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 $AB=AC$.
$\blacksquare$
(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.)
A: In the standard notation we obtain:
$$c+\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=b+\frac{1}{3}\sqrt{2a^2+2c^2-b^2}$$ or
$$3(b-c)=\frac{3(b^2-c^2)}{\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}},$$ which gives $b=c$ or
$$\sqrt{2a^2+2b^2-c^2}+\sqrt{2a^2+2c^2-b^2}=b+c$$ or
$$\sqrt{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}=bc-2a^2,$$ which is impossible for $bc-2a^2\leq0.$
But for $bc>2a^2$ we obtain
$$(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)=(bc-2a^2)^2$$ or
$$(b-c)^2(a+b+c)(b+c-a)=0,$$ which gives $b=c$ again.
The second problem we can solve by the similar way.
