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Question
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the centre O is equal to twice the angle APC.
Solution
Given: Two chords AB and CD intersect each other at P inside the circle. OA, OB, OC and OD are joined.
To prove: ∠AOC + ∠BOD = 2∠APC
Construction: Join AD.
Proof: Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.
∠AOC = 2∠ADC ...(1)
Similarly,
∠BOD = 2∠BAD ...(2)
Adding (1) and (2),
∠AOC + ∠BOD = 2∠ADC + 2∠BAD
= 2(∠ADC + ∠BAD) ...(3)
But ΔPAD,
Ext. ∠APC = ∠PAD + ∠ADC
= ∠BAD + ∠ADC ...(4)
From (3) and (4),
∠AOC + ∠BOD = 2∠APC
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