AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle.

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


Proof: (i)Given,  AC and BD are chords of a circle which bisect each other
 In ΔABD,  ∠ A = 90° 
      ∴ BD is a diameter                | Since angle in a semi-circle is 90°
In ΔBCD,  ∠ D = 90° 
      ∴ AC is a diameter                | Single angle in a semi-circle is 90°
Thus, AC and BD are diameters.
(ii) Let the chords AC and BD intersect each other at O. Join AB, BC, CD and DA.
       In Δ OAB and Δ OCD,
       OA = OC | Given
      OB = OD | Given
∠ AOB = ∠ COD                         | Vertically opposite angles
∴ Δ OAB ≅ Δ OCD                       | SAS
∴ AB = CD                                | CPCT
⇒ AB ≅ CD              ...(1)
Similarly, we can show that
AD = CB                ...(2)
Adding (1) and (2), we get
AB + AD ≅ CD + CB
⇒ BAD = BCD
⇒ BD divides the circle into two equal parts (each a semi-circle) and the angle of a semi-circle is 90°.
∴ ∠ A = 90° and ∠ C = 90°
Similarly, we can show that
∠ B = 90° and ∠ D = 90°
∴ ∠ A = ∠ B = ∠ C = ∠ D = 90°
⇒ ABCD is a rectangle.