<s>[INST] 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. 
 [/INST] 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.
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