Question

In figure AB = CB and O is the centre of the circle. Prove that BO bisects $\angle ABC$. [1 MARK]

Answer 1

Concept: 0.5 Mark
Application: 0.5 Mark

In $\Delta$'s $AOB$ and $COB$, we have

AB = CB [Given]

OB = OB [Common]

OA = OC [Each equal to radius]

So, by SSS criterion of congruence

$\Delta AOB\cong \Delta COB$

$\angle OBA=\angle OBC$ [C.P.C.T]

OB bisects $\angle ABC$.