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.
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.
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