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.

Two chords AB and CD intersect at point 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 of ∠APC.

Answer 1

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