the bisector of angle of B and angle of C of a triangle ABC meet at O, show that angle BOC = 90° + angle of A/2
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In ΔABC, by angle sum property we have 2x + 2y + ∠A = 180° ⇒ x + y + (∠A/2) = 90° ⇒ x + y = 90° – (∠A/2) à (1) In ΔBOC, we have x + y + ∠BOC = 180° 90° – (∠A/2) + ∠BOC = 180° [From (1)] ∠BOC = 180° – 90° + (∠A/2) ∠BOC = 90° + (∠A/2)
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