The bisectors of the opposite angles of a cyclic quadrilateral intersect the corresponding circle at the points A and B respectively. Prove that AB is the diameter of the circle.
 

PQRS is a cyclic quadrilateral PA and PB are bisectors of PQRS is a cyclic quadrilateral PA and PB are bisectors of ∠SPQ and ∠QRS respectively.
To Prove: AB is diameter of the circle.
Construction: Join BP and BQ
Proof: ∠2 = ∠3         (Angles in the same segment of the circle are equal)  ............... (i)
As PQRS is cyclic.
∴ ∠SPQ + ∠QRS = 180°
⇒ 2∠1 + 2∠3 = 180° 
⇒ ∠1 + ∠3 = 90°                  ............... (ii)
⇒ ∠1 + ∠2 = 90° (from equations (i) and (ii))
Hence, ∠APB = 90°
∴ AB is a diameter (∵ Angle in a semicircle is 90°)