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
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