Question

In triangle ABC the bisector of angle B and angle C meets at point O . Then prove that angle BOC is equal to 90° + angle BAC .

Answer 1

∠ABO=∠OBC=∠ABC2∠ABO=∠OBC=∠ABC2

∠ACO=∠OCB=∠ACB2∠ACO=∠OCB=∠ACB2

In△ABC,In△ABC,

∠ABC+∠ACB+∠BAC=180∘∠ABC+∠ACB+∠BAC=180∘

⟹∠ABC+∠ACB=180∘−∠BAC⟹∠ABC+∠ACB=180∘−∠BAC

Dividingbothsidesby2,Dividingbothsidesby2,

⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)⟹∠ABC+∠ACB2=90∘−∠BAC2...(1)

In△BOC,In△BOC,

∠ABC2+∠ACB2+∠BOC=180∘∠ABC2+∠ACB2+∠BOC=180∘

⟹∠ABC+∠ACB2

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