Question

The point A,B,C are $z_1,z_2,z_3$ on the circumference of the circle drawn on OA as diameter, O being the origin. If $\angle ABC=\angle BOC$ then prove that $z_2^{2}cos2\theta =z_1{z}_{3}co{s}^2\theta$
Passage : Angle subtended by chord of a circle at the centre is twice the angle subtened at the circumference. If OP is rotated through an angle $\varphi$ an anti - clockwise direction to become OQ.then OQ = OP$e^{i\varphi }$

Answer 1

It is a simple question of rotation but $OA\ne OB\ne OC$ Hence we shall apply the principle of rotation on unit lenghts
$\frac{z_2}{\left|z_2\right|}=\frac{z_1}{\left|z_1\right|}{e}^{i\theta }$ ...(1)
$\frac{z_3}{\left|z_3\right|}=\frac{z_1}{\left|z_1\right|}{e}^{2i\theta }$ ....(2)
Since angles at B and C are ${90}^{\circ }$ as OA is a diameter.
\therefore$\frac{OC}{OA}=\frac{\left|z_3\right|}{\left|z_1\right|}=cos2\theta$....(3)
$\frac{OB}{OA}=\frac{\left|z_2\right|}{\left|z_1\right|}=cos\theta$ .....(4)
From 1st on squaring,
$z_2^2=co{s}^2\theta z_1^2{e}^{2i\theta }$ by (4)
and $z_3=cos2\theta z_1{e}^{2i\theta }$ by (3) and (2)
Dividing,$\frac{z_2^2}{z_3}=\frac{co{s}^2\theta }{cos2\theta }.z_{1}or{z}_2^{2}cos2\theta =z_1{z}_{3}co{s}^2\theta$