If $AB,BC,$ and $CD$ are equal chords of a circle with $O$ as center, and $AD$ diameter then $\angle AOB$ equals to

The correct option is B: ${60}^{\circ }$

In a circle chords, $AB=BC=CD$

$O$ is the center of the circle.

We know that equal chord make equal angle at the center.

$\therefore \angle AOB=\angle BOC=\angle COD$

and, $\angle AOB+\angle BOC+\angle COD={180}^{\circ }$ [\because Angles on the same side of straight line add up to ${180}^{\circ }$]

$\Rightarrow 3\angle AOB={180}^{\circ }$

$\therefore \angle AOB=\frac{180}{3}={60}^{\circ }$