**Given : **ABCD is a cyclic parallelogram.

**To prove :** ABCD is a rectangle.

**Proof :** ∠ABC = ∠ADC ...(i) [Opposite angles of a ||gm are equal]

But, ∠ABC + ∠ADC = 180° ... (ii)

[Sum of opposite angles of a cyclic quadrilateral is 180°]

⇒ ∠ABC = ∠ADC = 90° [From (i) and (ii)]

∴ ABCD is a rectangle [A ||gm one of whose angles is 90° is a rectangle]

Hence, a cyclic parallelogram is a rectangle. **Proved.**