Question

In the figure, if $AB=CD$ and $\angle AOB={90}^0$ find $\angle COD.$

Answer 1

Since, $AB=CD$

$AO=BO$
therefore $\Delta AOB$ is an isosceles triangle

similarly, $\Delta COD$ is an isosceles triangle
$\angle AOB=\angle COD$ [ Since $\Delta AOB\cong \Delta COD$]