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