Condition for cyclic quadrilateral given $\sum_{i=1}^{4}{b_i}=0$ and $\sum_{i=1}^{4}{b_iz_i}=0$ 
Question Statement:-
  Given that $\displaystyle\sum_{i=1}^{4}{b_i}=0$ and $\displaystyle\sum_{i=1}^{4}{b_iz_i}=0$, where $b_i$'s are non zero real numbers, no three $z_i$'s form a straight line, prove that the $z_i$'s are concyclic if 
  $$b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|^2$$


Attempt at a solution:-
I was not able to do much to it, so here is my attempt at the solution.
$$\begin{aligned}
&\displaystyle\sum_{i=1}^{4}{b_iz_i}=0
\implies b_1z_1+b_2z_2+b_3z_3+b_4z_4=0 \\ 
&\implies b_1z_1+b_3z_3=-(b_2z_2+b_4z_4) \hspace{5cm} \ldots (1)
\end{aligned}$$
Also, we have the following relation:-
$$\sum_{i=1}^{4}{b_i}=b_1+b_2+b_3+b_4=0 \implies b_1+b_3=-(b_4+b_5) \tag{2}$$
On dividing $(1)$ and $(2)$, we get
$$\dfrac{b_1z_1+b_3z_3}{b_1+b_3}=\dfrac{b_2z_2+b_4z_4}{b_2+b_4}$$
This denotes that the point that divides the line joining $z_1$ and $z_3$ in the ratio $b_3:b_1$ and the point that divides $z_2$ and $z_4$ in the ratio $b_4:b_3$ are same but I am not able to conclude anything more. 
Also other approaches are welcome.
 A: The same argument you used also gives 
\begin{align*}
\dfrac{b_1z_1+b_2z_2}{b_1+b_2}=\dfrac{b_3z_3+b_4z_4}{b_3+b_4}
\end{align*}
Let this point be $P(z)$. This is the point of intersection of line joining $z_1, z_2$ and $z_3, z_4$. We have 
\begin{align*}
z- z_1 = \dfrac{b_1z_1+b_2z_2}{b_1+b_2} - z_1 &= \dfrac{b_2(z_2-z_1)}{b_1+b_2}\\
z- z_2 = \dfrac{b_1z_1+b_2z_2}{b_1+b_2} - z_2 &= \dfrac{b_1(z_1-z_2)}{b_1+b_2}\\
z- z_3 = \dfrac{b_3z_3+b_4z_4}{b_3+b_4} - z_3 &= \dfrac{b_4(z_4-z_3)}{b_3+b_4}\\
z- z_4 = \dfrac{b_3z_3+b_4z_4}{b_3+b_4} - z_4 &= \dfrac{b_3(z_3-z_4)}{b_3+b_4}
\end{align*}
Thus
\begin{align*}
(z-z_1)(z-z_2) &= -\dfrac{b_1b_2(z_1-z_2)^2}{(b_1+b_2)^2} \\
(z-z_3)(z-z_4) &= -\dfrac{b_3b_4(z_3-z_4)^2}{(b_3+b_4)^2} 
\end{align*}
From the condition $b_1b_2|z_1-z_2|^2=b_3b_4|z_3-z_4|^2$, it follows that $b_1b_2$ and $b_3b_4$ have the same sign and hence we have 
\begin{align*}
|b_1b_2||z_1-z_2|^2=|b_3b_4||z_3-z_4|^2
\end{align*}
We thus have 
\begin{align*}
|z-z_1||z-z_2| &= \left|\dfrac{b_1b_2(z_1-z_2)^2}{(b_1+b_2)^2}\right| \\
&= \left|\dfrac{b_3b_4(z_3-z_4)^2}{(b_3+b_4)^2}\right| \\
&= |z-z_3||z-z_4|
\end{align*}
If the given points are $A,B,C,D$, then we have $PA \cdot PB = PC\cdot PD$ and hence $A,B,C,D$ are concyclic.
