Two lines AB and CD intersect at O.

We know that if two lines intersect then the vertically-opposite angles are equal.

Therefore, $\angle A O C=\angle B O D=50^{\circ}$

Let $\angle A O D=\angle B O C=x^{\circ}$

Also, we know that the sum of all angles around a point is 360°">360°360°.
Therefore, 

$\angle A O C+\angle A O D+\angle B O D+\angle B O C=360^{\circ}$

$\Rightarrow 50+x+50+x=360^{\circ}$

$\Rightarrow 2 x=260^{\circ}$

$\Rightarrow x=130^{\circ}$

Hence, $\angle A O D=\angle B O C=130^{\circ}$

Therefore, $\angle A O D=130^{\circ}, \angle B O D=50^{\circ}$ and $\angle B O C=130^{\circ}$.