Given: AC and BD are chords of a circle which bisect each other at O. (say).
To Prove: (i) AC and BD are diameters
(ii) ABCD is a rectangle

Construction: Join AB, BC, CD, and DA.
Proof: (i) In ∆OAB and ∆OCD,
OA = OC
| ∵ O is the mid-point of AC
∠AOB = ∠COD
| Vertically opposite angles
OB = OD
| ∵ O is the mid-point of BD
∴ ∆OAB ≅ ∆OCD
| SAS congruence rule
∴ AB = CD    | C.P.C.T

| If two chords of a circle are equal, then their corresponding arcs are congruent
In ∆OAD and ∆OCB,
OA = OC
| ∵ O is the mid-point of AC
∠AOD = ∠COB
| Vertically opposite angles
OD = OB
| ∵ O is the mid-point of BD
∴ ∆OAD = ∆OCB
| SAS congruence rule
∴ AD = CB    | C.P.C.T.

| If two chords of a circle are equal, then their corresponding arcs are congruent
From (1) and (2),



⇒ BD divides the circle into two equal parts.
⇒ BD is a diameter.
Similarly, we can show that AC is a diameter.
(ii) ABCD is a parallelogram
| ∵ AB = DC and AD = BC (A quadrilateral is a parallelogram if both the pairs of opposite sides are equal)
∠ADB = 90°
| Angle in a semi-circle is 90°
∴ ABCD is a rectangle
| A parallelogram with one of its angles 90° is a rectangle