In a circle with centre O,chords AB and CD intersect inside the circumference at E.Prove that ∠AOC + ∠BOD = 2∠AEC

Given, a circle with centre O. The chords AB and CD of a circle intersect inside the circumference at E.

To prove: ∠AOC + ∠BOD =2 ∠AEC

Proof:

In the figure shown above, the same arc AC subtends ∠AOC on the centre and ∠ABC on the other part.

⇒ ∠AOC = 2∠ABC  [Angle subtended on the centre of the circle is double the angle subtended on other part of the circle by same arc.]

Similarly, arc BD subtends ∠BOD and ∠DCB.

⇒ ∠BOD = 2∠DCB

On adding the above results, we get

∠AOC + ∠BOD = 2∠ABC + 2∠DCB = 2(∠ABC + ∠DCB) ... (1)

In triangle ECB, by exterior angle property, we have

∠AEC = ∠ECB + ∠EBC

⇒ ∠AEC = ∠DCB + ∠ABC ... (2)

On putting the value of (2) in (1), we get

∠AOC + ∠BOD = 2∠AEC

[Hence proved]