Question

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

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)