Question

in triangle abc bisector of Angle B and angle C meet at O prove that angle BOC = 90 + angle a ÷2

Answer 1

∠a+∠b+∠c=180°
∠b+∠c=180°-∠a
(∠b+∠c)/2 = 90°-∠a/2

∠BOC = 180°-(∠b/2+∠c/2)=180°-(90°-∠a/2) = 90°+∠a/2

hence proved!