In $$\triangle ABC, \angle A = 70^\circ. AB$$ and $$AC$$ are produced to points $$D$$ and $$E$$ respectively. If the bisectors of $$\angle CBD$$ and $$\angle BCE$$ meet at the point $$O$$, then $$\angle BOC$$ is equal to:

A

$$105^\circ$$

B

$$70^\circ$$

C

$$95^\circ$$

D

$$55^\circ$$