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# $ \mathrm{AC} $ and $ \mathrm{BD} $ are chords of a circle which bisect each other. Prove that (i) $ \mathrm{AC} $ and $ \mathrm{BD} $ are diameters, (ii) $ \mathrm{ABCD} $ is a rectangle.

Given:

\( \mathrm{AC} \) and \( \mathrm{BD} \) are chords of a circle which bisect each other.

To do:

We have to prove that

(i) \( \mathrm{AC} \) and \( \mathrm{BD} \) are diameters

(ii) \( \mathrm{ABCD} \) is a rectangle.

Solution:

(i) Let $AC$ and $BD$ be two chords of a circle which bisect each other at $P$.

In $\triangle AOB$ and $\triangle COD$,

$OA = OC$ ($O$ is the mid-point of $AC$)

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

$OB = OD$ ($O$ is the mid-point of $BD$)

Therefore, by SAS congruency,

$\triangle CPD \cong \triangle APB$

This implies,

$\overparen{C D}=\overparen{A B}$.........(i) (CPCT)

Similarly,

In $\triangle A P D$ and $\triangle C P B$, we get,

$\overparen{C B}=\overparen{A D}$.........(ii) (CPCT)

Adding (i) and (ii), we get,

$\overparen{C D}+\overparen{C B}=\overparen{A B}+\overparen{A D}$

$\overparen{B C D}=\overparen{B A D}$

Therefore,

$BD$ divides the circle into two equal parts.

This implies,

$BD$ is a diameter.

Similarly,

$AC$ is a diameter.

(ii) $AC$ and $BD$ bisect each other.

This implies,

$ABCD$ is a parallelogram and

$AC = BD$

Therefore,

$ABCD$ is a rectangle.

Hence proved.

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