Question

in the given figure O is center of circle and chords AB and CD intersect at point E inside the circle prove that <AOC+<BOD=2<AEC​

Answer 1

Answer:

Consider arc AC of the circle with centre at O.

Clearly, arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

∴ ∠AOC = 2∠ABC …….(i)

Similarly arc BD subtends ∠BOD at the centre and ∠BCD at the remaining part of the circle.

∴ ∠BOD = 2∠BCD ……….(ii)

Adding (i) and (ii), we get

∠AOC + ∠BOD = 2(∠ABC + ∠BCD)

∠⇒ AOC + ∠BOD = 2∠AEC

[∵ ∠AEC is the exterior angle and ∠ABC and ∠BCD are other interior angles of Δ BEC

∴ ∠ABC + ∠BCD = ∠AEC]

Hence proved.

Answer 2

Step-by-step explanation:

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