Question

In the given figure, a circle with centre O, chords AB and CD intersect inside a circle at E. Prove that
$\angle AOC+\angle BOD=2\angle AEC$

Answer 1

$\angle AOC=2\angle ABC$ (AC chord property)
$\angle BOD=2\angle BCD$ (BD chord property)
$\angle AOC+\angle BOD=2(\angle ABC+\angle BCD)$ ...(1)
Since, $\angle AEC$ is the exterior angle and $\angle ABC$ and $\angle BCD$ are other two interior angles of triangle $\Delta EBC$
$\therefore \angle AOC+\angle BOD=2\angle AEC$