# 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 =...

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`

B) `z_d^2=z_1z_3`

C) `z_d^2=z_2z_1`

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` **