Question

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

Answer 1

In given figure $AB=AC$

In figure O is the center of circle and Point A,B,C and D on circumference of circle .

Then OA, OB ,OC and OD is the radius of circle

In $\Delta AOB$ and $COD$

$AB=AC$ (given)

$OA=OC$ (Radius)

$OB=OD$ (Radius)

$\therefore \Delta AOB\cong COD$

$\therefore \angle AOB=\angle COD={90}^0$ Given $(\angle AOB={90}^0)$

Then $(\angle COD={90}^0)$