If $z_{1} + z_{2} + z_{3} + z_{4} = 0$ where $b_{i} \in R$ such that the sum of no two values being zero, and $b_{1} z_{1} + b_{2} z_{2} + b_{3} z_{3} + b_{4} z_{4} = 0$ , where $z_{1}, z_{2}, z_{3}, z_{4}$ are arbitrary complex number such that no three of them are collinear and the four complex numbers are concyclic.
Then prove that $| b_{1} b_{2} | {| z_{1} - z_{2} |}_{1}^{2} = | b_{3} b_{4} | {| z_{3} - z_{4} |}_{3}^{2}$ .

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Solution

$b_{1} + b_{2} = - (b_{3} + b_{4}), b_{1} z_{1} + b_{2} z_{2} = - (b_{3} z_{3} + b_{4} z_{4})$ or $\frac{b_{1} z_{1} + b_{2} z_{2}}{b_{1} + b_{2}} = \frac{b_{3} z_{3} + b_{4} z_{4}}{b_{3} + b_{4}}$
Hence, the line joining the complex numbers $A (z_{1}), B (z_{2})$ and $C (z_{3}), D (z_{4})$ meet.
Let $P (z)$ be the point of intersection.
These points will be concyclic, if $P A \times P B = P C \times P D$ .
Now,
$P A = ∣ ∣ ∣ \frac{b_{2}}{b_{1} + b_{2}} ∣ ∣ ∣ | z_{1} - z_{2} |, P B = ∣ ∣ ∣ \frac{b_{1}}{b_{1} + b_{2}} ∣ ∣ ∣ | z_{1} - z_{2} |$
$P C = ∣ ∣ ∣ \frac{b_{4}}{b_{3} + b_{4}} ∣ ∣ ∣ | z_{-} z_{4} |, P D = ∣ ∣ ∣ \frac{b_{3}}{b_{3} + b_{4}} ∣ ∣ ∣ | z_{3} - z_{4} |$
$⟹ | b_{1} b_{2} | {| z_{1} - z_{2} |}_{1}^{2} = | b_{3} b_{4} | {| z_{3} - z_{4} |}_{3}^{2}$ ( $∴ | b_{1} + b_{2} | = | b_{3} + b_{4} |$ )
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Ifz1+z2+z3+z4=0 where bi∈R such that the sum of no two values being zero, and b1z1+b2z2+b3z3+b4z4=0, where z1,z2,z3,z4 are arbitrary complex number such that no three of them are collinear and the four complex numbers are concyclic.Then prove that |b1b2||z1−z2|2=|b3b4||z3−z4|2.