Question

Bisectors of angle B and C of a traingle ABC intersect each other at the point O.Prove that angle BOC=90+1/2angleA

Answer 1

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)