AC and bd are two chords of a circle which bisects each other prove that 1 AC and bad are diameter and abcd is a rectangle
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Step-by-step explanation:Let AC and BD bisect each other at point 0. Then,
OA = OC and OB = OD
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
=> 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.
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