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Direction Cosines

Az1, Bz2, Cz3...

Question

$A (z_{1}), B (z_{2}), C (z_{3})$ are the vertices of a triangle $A B C$ inscribed in the circle $| z | = 2$ . Internal angle bisector of the angle $A$ meets the circumcircle again at $D (z_{4})$ .
Complex number representing point $D$ is

A

z_{4} = \frac{1}{z_{2}} + \frac{1}{z_{3}}

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B

\sqrt{\frac{z_{2} + z_{3}}{z_{1}}}

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C

\sqrt{\frac{z_{2} z_{3}}{z_{1}}}

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D

z_{4} = \sqrt{z_{2} z_{3}}

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Solution

The correct option is D $z_{4} = \sqrt{z_{2} z_{3}}$
From the given figure
$∠ B O D = 2 ∠ B A D = ∠ A$
$∠ C O D = 2 ∠ C A D = ∠ A$
Hence
$\frac{z_{4}}{z_{2}} = e^{i A}$
$\frac{z_{3}}{z_{4}} = e^{i A}$

Hence

$z_{4}^{2} = z_{2} . z_{3}$

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