AC and BD are chords of a circle which bisect each other. Prove that:
(i) AC and BD are diameters

(ii) ABCD is a rectangle.

Question

AC and BD are chords of a circle which bisect each other. Prove that: (i) AC and BD are diameters (ii) ABCD is a rectangle.

Philoid · Accepted 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

AC and BD are chords of a circle which bisect each other. Prove that:(i) AC and BD are diameters(ii) ABCD is a rectangle.