Quadrilaterals — Maths STD 9 — Question

Consider $\triangle\text{AOD}\ \&\ \triangle\text{COB},$
$AO = CO$ {Diagonals bisects each other}
$OD = OB$ {Diagonals bisects each other}
$\angle\text{AOD}=\angle\text{COB}$ (Opposite angles)
So by SAS property, $\triangle\text{AOD}\cong\triangle\text{COB},$
$\Rightarrow\angle\text{ADO}=\angle\text{CBO}\dots(1)$
$\angle\text{ABD}=180^\circ-\angle\text{A}-\angle\text{ADO}$ $($in $\triangle\text{ADB})$
$=180^\circ-45^\circ-\angle\text{ADO}$
$\angle\text{ABD}=135^\circ-\angle\text{ADO}\dots(2)$
$\angle\text{B}=\angle\text{ABD}+\angle\text{CBO}$
Putting values From eq $(1)$ and $(2)$
$\angle\text{B}=135^\circ-\angle\text{ADO}+\angle\text{ADO}$
$\angle\text{B}=135^\circ$