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.