In the adjoining figure, O is the center of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D. Prove that ∠ABC = 2∠OAD.

Join AD.

Arc AC subtends ∠AOC at the center and ∠ADC at the point D of the circle.

∴ ∠AOC = 2∠ADC (As angle at center = double the angle at the remaining part of the circle)

∠OAD = ∠ADC (∵ alternate angles are equal.)

∴ ∠AOC = 2∠OAD …..(i)

Since, opposite angles are equal in parallelogram,

∴ ∠ABC = ∠AOC

Putting values of ∠AOC in eqn (i) we get,

∠ABC = 2∠OAD.

Hence, proved that ∠ABC = 2∠OAD.