# 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.

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.)
• @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis remains the same under reflection. Jan 25, 2019 at 15:34
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.