If diagonals of a

Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.

n ∆AOD and ∆COB 

AO = CO [Radii of a circle] 

OD = OB [Radii of a circle] 

∠AOD = ∠COB [Vertically opposite angles] 

∴ ∆AOD ≅ ∆COB [SAS congruency] 

∴ ∠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 are 90° is a rectangle] 
Hence Proved.