In the given figure, a circle with centre O, chords AB and CD intersect inside a circle at E. Prove that
$∠ A O C + ∠ B O D = 2 ∠ A E C$

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Solution

$∠ A O C = ∠ A B C$ (AC chord property)
$∠ B O D = ∠ B C D$ (BD chord property)
$∠ A O C + ∠ B O D = 2 (∠ A B C + ∠ B C D)$ ...(1)
Consider $Δ C E B$ ,
$∠ C E B + ∠ E B C + ∠ E C B = 180^{\circ}$
$∠ C E B + ∠ A B C + ∠ D C B = 180^{\circ}$
$∴ ∠ A B C + ∠ D C B = 180^{\circ} - ∠ C E B$
$= ∠ A E B$ ...(2)
Substitute (2) in (1)
$∠ A O C + ∠ B O D = 2 ∠ A E B$ Proved

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

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