Q. Triangle ABC, A`(z_1)` ,B`(z_2)` ,C`(z_3)` is inscribed in the circle `|z|=2` .If the internal bisector of the angle at A (CAB) meets the circumcircle `|z|=2` again at D`(z_d)` then
A) `z_d^2 = z_2z_3`
D) ` ` none of these
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Call the angle CAD `theta` . Since AD bisects the angle at A (CAB), then angle DAB is equal to CAD, ie `theta` . Using the circle theorems, CAD is equal to CBD, since they are subtended from the same chord. Similarly, DAB is equal to DCB by the same rule.
Therefore triangle BCD is an isoceles triangle with two angles (at B and C) equal to CAD = DAB = CBD = DCB = `theta` . This implies that the lengths BC and BD are equal in length.
This means that the distance from `z_d` to `z_2` round the circumcircle on which they lie is equal to the distance between from `z_3` to `z_d` on the same circle.
Let `z_2 = 2e^(bipi)` (the radius of the circle `z = |2|` is equal to 2) and
`z_3 = 2e^(cipi)` and `z_d = 2e^(dipi)`
then `z_2 = z_de^((b-d)ipi)` and `z_d = z_3e^((d-c)ipi) = z_3e^((b-d)ipi)`
` `(since the angle `b-d` = the angle `d-c` )
which implies that
`z_d = z_3(z_2/z_d)` so that we have that `z_d^2 = z_2z_3`
Answer is A) `z_d^2 = z_2z_3`
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