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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