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 center O is equal to twice the angle APC.

The below figure shows two chords AB and CD intersecting at P inside the circle:

We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord

∠AOC = 2∠ADC ……… (1)

Similarly,

∠BOD = 2∠BAD ……… (2)

Adding (1) and (2), we get

⇒ ∠AOC + ∠BOD = 2∠ADC + 2∠BAD

⇒ ∠AOC + ∠BOD = 2(∠ADC + ∠BAD) ……… (3)

In ∆PAD,

Exterior angle is equal to the sum of two opposite interior angles.

∴ ∠APC = ∠PAD + ∠ADP

∠APC = ∠BAD + ∠ADC ………. (4)

So, from (3) and (4) we have

∠AOC + ∠BOD = 2∠APC.

Hence, proved that the sum of the angles subtended by the arcs AC and BD at the center O is equal to twice the angle APC.