In a circle with centre `O ,` chords `A B\ a n d\ C D` intersect inside the circumference at `E`. Prove that `/_A O C+\ /_B O D=2\ /_A E C`

Explanation

Consider arc `AC` of the circle with centre at O.
Clearly, arc `AC` subtends `angle AOC` at the centre and `angle ABC` at the remaining part of the circle.
`thereforeangle AOC=2angle ABC` ...........(1)
Similarly,arc BD subtends `angleBOD` at the centre and `angle BCD` at the remaining part of the circle.
`angle BOD =2angleBCD` .........(2)
Adding eqs.(1) and (2) , we get
`angle AOC+angleBOD=2(angleABC+angleBCD)`
`rArrangleAOC +angleBOD =2angle AEC`
(`becauseangle AEC` is the exterior angle and `angleABC` and `angleBCD` are other interior angles of `DeltaBECthereforeangleABC+angleBCD=angleAEC`)