Question

AC and BD are chords of a circle which bisects each other, prove that AC and BD are diameters.

Answer 1

In triangles AOB and COD, we have

OA = OC

OB = OD

and ∠AOB = ∠COD (Vertically opposite angles)

∴ △AOB ≅ △COD (SAS congruence criterion)

=> AB = CD (CPCT)

=> arc AB ≅ arcCD ...(ii)

Similarly BC = DA

=> arcBC ≅ arcDA ..(iii)

From (ii) and (iii), we have

arcAB + arcBC ≅ arcCD + arcDA =>

=> arcABC = arcCDA

From (ii) and (iii), we have

arcAB + arcBC ≅ arcCD + arcDA =>

=> arcABC = arcCDA

=> AC divides the circle into two equal parts.

=> AC is the diameter of the circle. Similarly, we can prove that BD is also a

diameter of the circle.

Since AC and BD are diameters of the circle.

∴ ∠ABC = 90° = ∠ADC

Also, ∠BAD = 90° = ∠BCD

Also, AB = CD and BC = DA (Proved above)

Hence, ABCD is a rectangle.