Let two chords AB and CD are intersecting each other at point O In
ΔAOBandΔCOD,
OA = OC (Given)
OB = OD (Given)
∠AOB=∠COD (Vertically opposite angles)
∴ΔAOB≅ΔCOD (SAS congruence rule)
AB = CD (By CPCT) Similarly, it can be proved that
ΔAOD≅ΔCOB and AD = CB (By CPCT) Since in quadrilateral ABCD, opposite sides are equal in length, ABCD is a parallelogram. We know that opposite angles of a parallelogram are equal.
∴∠A=∠C However,
∠A+∠C=180∘ (ABCD is a cyclic quadrilateral)
⇒∠A+∠A=180∘ ∴2∠A=180∘ i,e
∠A=90∘ As ABCD is a parallelogram and one of its interior angles is
90∘, therefore, it is a rectangle.
∠A is the angle subtended by chord BD and
∠A=90∘, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.