Geometry : $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$ Let $G$ be centriod of $\triangle ABC$ and GA, GB, GC cut the circumcircle of $\triangle ABC$ again at A', B', C' respectively. Prove that $\frac{GA}{GA'}+\frac{GB}{GB'}+\frac{GC}{GC'}=3$.

My attempt :
Let $D, E, F$ be the midpoints of $BC, CA, AB$ respectively.
$\angle GFE = \angle FCD = \angle C'B'G$ so $C'FEB'$ concyclic.
Similarly, $DEB'A'$ concyclic.
By Power of point, $C'G\cdot FG=GE\cdot GB'=GD\cdot GA'$
 A: Note that $GA*GA'=GB*GB'=GC*GC':=s^2$.
It suffices to prove that $\frac{{GA}^{2}+{GB}^{2}+{GC}^{2}}{s^2}=3$.
Note that $s^2=R^2-OG^2$ and by Leibniz's formula we have ${GA}^{2}+{GB}^{2}+{GC}^{2}+3OG^2=OA^2+OB^2+OC^2=3R^2$. 
Substituting the equation in leads to the wanted equation.
(Sorry, my first post, I 'm not good at posting, please forgive.)
A: Say $AD =2x$, $BG = 2y$, $CG=2z$ and $A'D = x'$, $B'E=y'$, $C'F=z'$. 
Finally $AB = 2c$, $BC= 2a$ and $AC=2b$.
By the PoP we have: $x'\cdot 3x= a^2$, $y'\cdot 3y=b^2$, $z'\cdot 3z=c^2$ and $$x(x+x') = y(y+y')= z(z+z') =:k$$
Then $$3x^2+a^2 = 3y^2+b^2= 3y^2+c^2=3k$$
So $3(x^2+y^2+z^2)+a^2+b^2+c^2 =9k$
We are interested in $$I= {2x\over x+x'}+ {2y\over y+y'}+ {2z\over z+z'}= {2x^2+2y^2+2z^2\over k}$$
By parallelogram identity we have 
\begin{eqnarray*}
  (6x)^2+(2a)^2 &=&  2(2b)^2+2(2c)^2\\
  (6y)^2+(2b)^2 &=&  2(2a)^2+2(2c)^2\\
   (6z)^2+(2c)^2 &=&  2(2b)^2+2(2a)^2\\
\end{eqnarray*}
and so $$3(x^2+y^2+z^2)=a^2+b^2+c^2$$ 
Thus $2(a^2+b^2+c^2) =9k$ and thus $I= 3$
