# 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."

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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.

Updated on 10-Oct-2022 13:41:06