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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.