Let  $A\left(Z_1\right),B\left(Z_2\right),C\left(Z_3\right)$  are the vertices of triangle ABC respectively in anticlockwise sense. $Arg\left(\frac{Z_2-Z_1}{Z_2-Z_3}\right)=\frac{\pi }{2}$  and  $Z_4$  be the point of intersection of  $Arg\left(\frac{Z_1-Z}{Z_1-Z_2}\right)=\frac{A}{2}$ and  $Arg\left(\frac{Z_3-Z}{Z_3-Z_1}\right)=\frac{C}{2}$  where A and C are angles of triangle at vertices $A\left(Z_1\right)$ and $C\left(Z_3\right)$ and $Z_5$ be the point of intersection of  $Arg\left(\frac{Z_4-Z}{Z_4-Z_1}\right)=\frac{\pi }{2}$  and  $Arg\left(\frac{Z-Z_2}{Z-Z_3}\right)=0$  then which of the following option(s) is (are) CORRECT?

ADBI is a cyclic quadrilateral 
 $\overline{)AID}=\overline{)ABD}=\frac{\pi }{2}$
$\overline{)ADI}=\overline{)ABI}$ (chord AI subtends same angle on the circle)
$\overline{)ABI}=\frac{\pi }{4}$ ( BI is angular bisector of angle B)
 $\overline{)DBI}+\overline{)DAI}=\pi$
 $$\begin{gathered} \overline{)DBI}=\overline{)DBA}+\overline{)ABI}=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4} \\ \Rightarrow \overline{)DAI}=\frac{\pi }{4} \end{gathered}$$
AID is an isosceles triangle which is right angle at I
$$\begin{gathered} AI=ID=\sqrt{2}AD \\ \overline{)ADC}=\overline{)ADI}+\overline{)IDC}=\frac{\pi }{4}+\overline{)IDC} \\ =\overline{)DAI}+\overline{)IAC}=\overline{)DAC}(\overline{)IDC}=\overline{)IDB},\overline{)IDB}=\overline{)IAB}=\frac{A}{2}) \\ \overline{)ADC}=\overline{)DAC}\Rightarrow CD=CA \\ BD=CD-CB=CA-CB=b-a \\ A{D}^2=A{B}^2+D{B}^2=c^2+{(b-a)}^2=c^2+b^2+a^2-2ab=2b^2-2ab \\ = 2b (b – a) \\ AD=\sqrt{2}\sqrt{b(b-a)}\Rightarrow \frac{AD}{\sqrt{2}}=AI=DI \\ CAD is isosceles triangle \Rightarrow \overline{)CED}=\frac{\pi }{2} \\ Arg\left(\frac{z_3-z_1}{z_1-z_5}\right)=\frac{-\pi }{2} \end{gathered}$$