In any triangle ABC, the base angles at B and C are bisected by BO and CO respectively. Then ∠BOC is

In $$\triangle$$ABC

=> $$\angle$$A + $$\angle$$B + $$\angle$$C = $$\pi$$

=> $$\frac{1}{2} (\angle A + \angle B + \angle C) = \frac{\pi}{2}$$

=> $$\frac{\angle B}{2} + \frac{\angle C}{2} = \frac{\pi}{2} - \frac{\angle A}{2}$$

In $$\triangle$$OBC

=> $$\angle$$OBC + $$\angle$$OCB + $$\angle$$BOC = $$\pi$$

=> $$\frac{\angle B}{2} + \frac{\angle C}{2} + \angle BOC = \pi$$

=> $$\angle BOC = \pi - (\frac{\pi}{2} - \frac{\angle A}{2})$$

= $$\frac{\pi}{2} + \frac{\angle A}{2}$$