If `((3-z_1)/(2-z_1))((2-z_2)/(3-z_2))=k(k >0)` , then prove that points `A(z_1),B(z_2),C(3),a n dD(2)` (taken in clockwise sense) are concyclic.

Explanation

`((3-z_(1))/(2-z_(1)))((2-z_(2))/(3-z_(2)))=k`
`:." "arg((3-z_(1))/(2-z_(1)).(2-z_(2))/(3-z_(2)))=arg(k)`
Now `kgt0.` So, `arg(k)=0`

`:." "arg((3-z_(1))/(2-z_(1)))+arg((2-z_(2))/(3-z_(2)))=0`
or `arg((3-z_(1))/(2-z_(1)))=arg((3-z_(2))/(2-z_(2)))`
Hence, chord DC subtends same angle at A and B. Therefore, point A, B, C, D are concyclic.