Let $A, B, C, D$ be four concyclic points in order in which $A D : A B = C D : C B$ . If $A, B, C$ are represented by complex numbers $a, b, c$ , then vertex $D$ can be represented as

\frac{2 a c + b (a - c)}{a + c + 2 b}

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\frac{2 a c + b (a + c)}{a + c + 2 b}

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\frac{2 a c + b (a + c)}{a + c - 2 b}

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\frac{2 a c - b (a + c)}{a + c - 2 b}

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Solution

The correct option is D $\frac{2 a c - b (a + c)}{a + c - 2 b}$

Let complex number representing point $^{'} D^{'}$ is $d$ and $∠ D A B = θ$ So, $∠ B C D = π - θ$ .
Now, applying rotation formula on $A$ and $C$ we get
$\frac{b - a}{d - a} = \frac{A B}{A D} e^{i θ} and \frac{d - c}{b - c} = \frac{C D}{C B} e^{i (π - θ)}$
Multiplying these two, we get
$(\frac{b - a}{d - a}) (\frac{d - c}{b - c}) = \frac{A B \times C D}{A D \times C B} e^{i π}$
$\Rightarrow \frac{d (b - a) - c (b - a)}{d (b - c) - a (b - c)} = - 1 (∵ \frac{A D}{A B} = \frac{C D}{C B})$
$\Rightarrow d = \frac{2 a c - b (a + c)}{a + c - 2 b}$

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