Given : A circle with chords AB and CD which bisect each other at O.
To Prove : (i) AC and BD are diameters (ii) ABCD is a rectangle.
Proof : In ∆OAB and ∆OCD, we have
OA = OC [Given]
OB = OD [Given]
∠AOB = ∠COD [Vertically opposite angles]
⇒ ∆AOB ≅ ∠COD [SAS congruence]
⇒ ∠ABO = ∠CDO and ∠BAO = ∠BCO [CPCT]
⇒ AB || DC ... (i)
Similarly, we can prove BC || AD ... (ii)
Hence, ABCD is a parallelogram.
But ABCD is a cyclic parallelogram.
∴ ABCD is a rectangle. [Proved in Q. 12 of Ex. 10.5]
⇒ ∠ABC = 90° and ∠BCD = 90°
⇒ AC is a diameter and BD is a diameter
[Angle in a semicircle is 90°] Proved.
