#### 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 ∠AOCat the centre and ∠ADCat 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

Is there an error in this question or solution?

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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. Concept: Arc and Chord Properties - If Two Arcs Subtend Equal Angles at the Center, They Are Equal, and Its Converse.

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