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AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters,

Q: 7    AC and BD are chords of a circle which bisect each other. Prove that

           (i) AC and BD are diameters

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Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA. 

Proof :

               

In \triangle ABD and \triangleCDO,

          AO = OC     (Given )

         \angleAOB = \angleCOD  (Vertically opposite angles )

          BO = DO        (Given )

So,  \triangle ABD\cong \triangleCDO     (By SAS)

       \angleBAO = \angleDCO    (CPCT)

\angleBAO and \angleDCO are alternate angle and are equal .

So,     AB || DC ..............1

   Also  AD || BC  ...............2

From 1 and 2,

            \angle A+\angle C=180 \degree......................3(sum of opposite angles)

           \angleA = \angleC    ................................4(Opposite angles of the parallelogram )

From 3 and 4,

    \angle A+\angle A=180 \degree

\Rightarrow 2\angle A=180 \degree

 \Rightarrow \angle A=90 \degree

BD is a diameter of the circle.

Similarly, AC is a diameter.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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