Question

In triangle ABC the bisector of angle B and C meets at P .Prove that : < BPC =90° +1/2<A .

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)