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.