Q7 of 35 Page 186

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.

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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