AC and BD are chords of a circle which bisect each other at O. Prove that

(i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Given AC and BD are chords of a circle which bisect each other at O.

(i)AC and BD are diameters.

(ii)ABCD is a rectangle

Proof

(i) In ∆AOS and ∆COD, we have

OA=OC [∵ O is the mid-point of AC]

∠AOB=∠COD

OB = OD [∵ O is the mid-point of BD]

⇒ CD = AB [by CPCT]

∴ CD = AB …(i)

Similarly, in ∆AOD and ∆COS, we have

CB = AD …(ii)

On adding Eqs. (i) and (ii), we get

CD + CB = AB+ AD

⇒ BCD = BAD

Thus, BD divides the circle into two equal parts.

Hence, BD is a diameter.

Similarly, AC is a diameter.

(ii) Since, AC and BD bisect each other.

∴ABCD is & parallelogram.

Also, ∠A =∠C = 90°

[∵ BD is a diameter and angle in a semi-circle is 90°]

and ∠B = ∠D = 90°

[∵AC is a diameter and angle in a semi-circle is 90°]

Hence, ABCD is a rectangle.