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Q.

A(z1), B(z2),C(z3) are the vertices of a triangle ABC inscribed in the circle |z|=2.  Internal angle bisector of the angle A, meet the circumcircle again at D(z4) then:

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a

z42=z2z3

b

z4=z2z3z1

c

z4=z1z2z3

d

z4=z1z3z2

answer is A.

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Detailed Solution

∠BAD=∠CAD ⇒arc  BD=arc  CD ⇒    ∠BOD=∠COD=Az4z2=eiA=z3z4⇒z42=z2z3
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