Question
In a circle with center O, chords AB and CD intersect inside the circumference at E. Prove that ∠ AOC + ∠ BOD = 2∠ AEC.
✓
Answer
Consider arc AC of the circle with the centre of 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 BEF ∴ ∠ ABC + ∠ BCD = ∠ AEC)
Hence proved.
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 BEF ∴ ∠ ABC + ∠ BCD = ∠ AEC)
Hence proved.
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