Proving $\frac { { { A }_{ 1 }G } }{ G{ { B }_{ 1 } } } +\frac { { A }_{ 2 }G }{ G{ B }_{ 2 } } +...+\frac { { A }_{ n }G }{ G{ B }_{ n } } =n$ 
Let ${ A }_{ 1 }{ { A }_{ 2 }{ A }_{ 3 }...{ A }_{ n } }$ be a $n$-gon with centroid $G$ inscribed in a circle. The lines $\\ \\ { A }_{ 1 }G,{ A }_{ 2 }G,...{ A }_{ n }G$ intersect the circle again at ${B}_{1},{B}_{2},...{B}_{n}$.Prove that $\frac { { { A }_{ 1 }G } }{ G{ { B }_{ 1 } } } +\frac { { A }_{ 2 }G }{ G{ B }_{ 2 } } +...+\frac { { A }_{ n }G }{ G{ B }_{ n } } =n$

I am completely stuck and am not able to find an approach to the problem. Please help, Thanks.
 A: Using power of the point we find $$|A_jG| \cdot |GB_j| = r^2-|OG|^2$$ for all $j=1,\ldots,n$, where $O$ is the center of the circle and $r$ its radius. Therefore
$$\frac{|A_1G|}{|GB_1|} + \ldots + \frac{|A_nG|}{|GB_n|} = \frac{|A_1G|^2}{|A_1G| \cdot |GB_1|} + \ldots + \frac{|A_nG|^2}{|A_nG| \cdot |GB_n|} = \frac{|A_1G|^2 + \ldots + |A_nG|^2}{r^2-|OG|^2}.$$
So we need to prove that $$|A_1G|^2 + \ldots + |A_nG|^2 = n (r^2 - |OG|^2).$$
Now place the $n$-gon in complex plane such that its vertices lie on the unit circle. I will use respective lowercase letters for the complex numbers corresponding to our points. Then $|a_1|=\ldots=|a_n|=r=1$ and $ng = a_1+\ldots+a_n$. For any $j=1,\ldots,n$ we have 
$$|A_jG|^2 = |a_j-g|^2 = |a_j|^2+|g|^2-(a_j\overline{g}+\overline{a_j}g) = 1 + |g|^2 -(a_j\overline{g}+\overline{a_j}g).$$
Summing up we get 
$$\begin{align*}\sum_{j=1}^n|A_jG|^2 & = n + n|g|^2 - \sum_{j=1}^n(\overline{a_j}g + a_j\overline{g}) = \\
&=n + n|g|^2 - \overline{\left(\sum_{j=1}^n a_j\right)}g - \sum_{j=1}^na_j\overline{g} = \\
& = n + n|g|^2 - \overline{ng}g - ng \overline g = \\
& = n - n|g|^2 = \\
& = n(r^2-|OG|^2).
\end{align*}$$
