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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 $\triangle$CDO,

AO = OC     (Given )

$\angle$AOB = $\angle$COD  (Vertically opposite angles )

BO = DO        (Given )

So,  $\triangle$ ABD$\cong$ $\triangle$CDO     (By SAS)

$\angle$BAO = $\angle$DCO    (CPCT)

$\angle$BAO and $\angle$DCO 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)

$\angle$A = $\angle$C    ................................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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