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Question

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

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Solution

Concept: 0.5 Mark
Application: 0.5 Mark

In Δ's AOB and COB, we have

AB = CB [Given]

OB = OB [Common]

OA = OC [Each equal to radius]

So, by SSS criterion of congruence

ΔAOBΔCOB

OBA=OBC [C.P.C.T]

OB bisects ABC.


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