Question

the centre of circumcircle triangle ABC is O.prove that angle OBC +angle BAC=90

Answer 1

∠OBC =  ∠OCB ........opposite angles of equal sides in ΔBOC
∠BOC + ∠OBC + ∠OCB = 180'
2∠BOC + ∠BAC = 180
2∠OBC + 2∠BAC = 180
2(∠OBC + ∠BAC) = 180
∠OBC + ∠BAC = 90'
 
I hope it helps........ 

Answer 2

hope it'll help
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