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

Question

Accepted Answer

$2\angle AEB$

Answer

$\angle AEB$

Answer

$\frac{1}{2}\angle AEB$

Answer

$4\angle AEB$

ABCDABCDABCD is a cyclic quadrilateral with diagonals AC,BDAC,BDAC,BD meeting at EEE. Then ∠AOB+∠COD\angle AOB + \angle COD∠AOB+∠COD equals—