AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th

AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th
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1 Answer

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Solution :- 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   ≅ arc CD  ...(ii)  Similarly  BC = DA  =>  arc BC  ≅  arcDA  ..(iii) From (ii) and (iii), we have arcAB + arcBC   ≅  arc CD + 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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