If chords $AB,CD$ of a circle with centre $O$ meet at right angles inside the circle, then $\angle AOD + \angle BOC$ equals—

The central angles $\angle AOD$ and $\angle BOC$ subtend arcs $AD$ and $BC$, whose inscribed angles meet at the right-angle intersection of the chords; this forces the two arcs' central angles to sum to $180^{\circ}$, i.e. two right angles.