CircleHard

Question

Let z1,z2 & z3 are the vertices of ᐃABC respectively such that is purely imaginary number. A square on side AC is drawn outwardly. If P(z4) is the centre of square, then-

Options

A.|z1 - z2| = |z2 - z3| must necessarily hold
B.Arg + Arg
C.Arg + Arg = 0
D.z1, z2, z3 and z4 lie on a circle.

Solution

       
Let P be the centre of square
⇒ AP = PC and ∠APC =
⇒ ∠ABC =
⇒ ABCP is a cyclic quadrilateral
In circle, chord, AP = AC
⇒ ∠ ABP = ∠PBC
ε BP is angular bisector, so.
Arg + Arg = 0

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