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

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 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.

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