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# ABCD is a quadrilateral in which $ P, Q, R $ and $ S $ are mid-points of the sides $ \mathrm{AB}, \mathrm{BC}, \mathrm{CD} $ and $ \mathrm{DA} $ (see below figure). AC is a diagonal. Show that :

**(i)** $ \mathrm{SR} \| \mathrm{AC} $ and $ \mathrm{SR}=\frac{1}{2} \mathrm{AC} $

**(ii)** $ \mathrm{PQ}=\mathrm{SR} $

**(iii)** $ \mathrm{PQRS} $ is a parallelogram.

"

Given:

$ABCD$ is a quadrilateral in which \( P, Q, R \) and \( S \) are mid-points of the sides \( \mathrm{AB}, \mathrm{BC}, \mathrm{CD} \) and \( \mathrm{DA} \).

$AC$ is a diagonal.

To do:

We have to show that

(i) \( \mathrm{SR} \| \mathrm{AC} \) and \( \mathrm{SR}=\frac{1}{2} \mathrm{AC} \)

(ii) \( \mathrm{PQ}=\mathrm{SR} \)

(iii) \( \mathrm{PQRS} \) is a parallelogram.

Solution:

$\mathrm{AP}=\mathrm{PB}, \mathrm{BQ}=\mathrm{CQ}, \mathrm{CR}=\mathrm{DR}$ and $\mathrm{AS}=\mathrm{DS}$

(i) In $\triangle A D C$,

$S$ is the mid-point of $AD$ and $R$ is the mid-point of the $DC$.

We know that,

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

This implies,

$\mathrm{SR} \| \mathrm{AC}$.........(i)

$\mathrm{SR}=\frac{1}{2} \mathrm{AC}$.........(ii)

(ii) In $\triangle \mathrm{ABC}$,

$P Q \| A C$.......(iii)

$P Q=\frac{1}{2} A C$........(iv)

From (i) and (iii), we get,

$\mathrm{SR}=\frac{1}{2} \mathrm{AC}$.........(v)

This implies,

$\mathrm{PQ}=\mathrm{SR}$

(iii) From (i) and (iii), we get,

$\mathrm{PQ} \| \mathrm{SR}$

$\mathrm{PQ}=\mathrm{SR}$

If a pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram.

Therefore, $\mathrm{PQRS}$ is a parallelogram.

Hence proved.

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