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Question

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=2BAD=A
COD=2CAD=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)
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