Answer :
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
∠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 BD should be the diameter of the circle
Similarly, AC is the diameter of the circle
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