In the picture, A

Let O be the center of a circle

Join BC, OA, OB, OC and OD

Given: AB⊥ CD

In Δ ABK

∠ ABC + ∠ BCD + ∠ BKC = 180° (Angle sum property)

⇒ ∠ ABC + ∠ BCD + 90° = 180°

⇒ ∠ ABC + ∠ BCD = 90° …..(1)

∠AOC = 2∠ ABC (∵ angle at center twice angle in

segment) ...(2)

∠BOD = 2∠ BCD (∵ angle at center twice angle in

segment) ...(3)

Adding (2) and (3)

∠AOC + ∠BOD = 2∠ ABC + 2∠ BCD

∠AOC + ∠BOD = 2 (∠ ABC + ∠ BCD)

∠AOC + ∠BOD = 2 × 90° (from (1))

∠AOC + ∠BOD = 180°

Hence, (angle at center is half of 360° )