**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.**