$ABCD$ is a cyclic quadrilateral with diagonals $AC,BD$ meeting at $E$. Then $\angle AOB + \angle COD$ equals—

$\angle AEB$ is the angle between the chords at their intersection and equals half the sum of the two intercepted arcs, i.e. $\tfrac12(\angle AOB+\angle COD)$. Rearranging gives .