For fixed positions of B(z2) and C(z3) all the bisectors (internal) of ∠A will pass through a fixed point which is
A
H.M. of z2 and z3
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B
A.M. of z2 and z3
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C
G.M. of z2 and z3
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D
none of these
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Solution
The correct option is D G.M. of z2 and z3 From the given figure ∠BOD=2∠BAD=∠A ∠COD=2∠CAD=∠A Hence z4z2=eiA z3z4=eiA Hence z24=z2.z3 ...(i) Now since all the internal bisectors of A will pass through D. Hence The fixed point is the G.M of z2 and z3 ... from(i)