Let two chords AB and CD are intersecting each other at point O.
In ΔAOB and ΔCOD,
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
AD = CB (By CPCT)
Since in quadrilateral ACBD, opposite sides are equal in length, ACBD 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°
∠A = 90°
As ACBD 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 as ∠A = 90°, therefore, BD should be the diameter of the circle. Similarly, AC is the diameter of the circle.
