**Given :** ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameter of the circle passing through A, B, C and D.

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

**Proof :** In ∆AOD and ∆COB

AO = CO [Radii of a circle]

OD = OB [Radii of a circle]

∠AOD = ∠COB [Vertically opposite angles]

∴ ∆AOD ≅ ∆COB [SAS axiom]

∴ ∠OAD = ∠OCB [CPCT]

But these are alternate interior angles made by the transversal AC, intersecting AD and BC.

∴ AD || BC

Similarly, AB || CD.

Hence, quadrilateral ABCD is a parallelogram.

Also, ∠ABC = ∠ADC ..(i) [Opposite angles of a ||gm are equal]

And, ∠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] **Proved.**