CAN YOU PLEASE ANSWER QUICKLY? ITS URGENT
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Proof:
in ΔAOB and ΔCOB,
AB = BC (given)
AO = OC (radii of circle are equal)
OB = BO ( common )
by SSS,
ΔAOB ≅ΔBOC
⇒∠ABO = ∠OBC
⇒ BO bisects ∠ABC
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