$ \mathrm{ABCD} $ is a rectangle and $ \mathrm{P}, \mathrm{Q}, \mathrm{R} $ and $ \mathrm{S} $ are mid-points of the sides $ \mathrm{AB}, \mathrm{BC}, \mathrm{CD} $ and $ \mathrm{DA} $ respectively. Show that the quadrilateral $ \mathrm{PQRS} $ is a rhombus.

Given:

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

To do:
We have to show that the quadrilateral \( \mathrm{PQRS} \) is a rhombus.
 Solution:

$\angle A=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ}$

$\mathrm{AD}=\mathrm{BC}$ and $\mathrm{AB}=\mathrm{CD}$

$P, Q, R$ and $S$ are mid-points of $A B, B C, C D$ and $D A$ respectively.

This implies,

$P Q \| A C$

$P Q=\frac{1}{2} \mathrm{AC}$

$R S \| A C$

$R S=\frac{1}{2} \mathrm{AC}$

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

In $\triangle \mathrm{ASP}$ and $\triangle \mathrm{BQP}$

$\mathrm{AP}=\mathrm{BP}$

$\mathrm{AS}=\mathrm{BQ}$

$\angle A=\angle B=90^o$

Therefore, by SAS congruency, we get,

$\triangle \mathrm{ASP} \cong \triangle \mathrm{BQP}$

This implies,

$\mathrm{SP}=\mathrm{PQ}$           (CPCT)

In $\triangle \mathrm{RDS}$ and $\triangle \mathrm{RCQ}$,

$\mathrm{SD}=\mathrm{CQ}$

$\mathrm{DR}=\mathrm{CR}$

$\angle C=\angle D=90^o$

Therefore, by SAS congruency, we get,

$\triangle \mathrm{RDS} \cong \triangle \mathrm{RCQ}$

This implies,

$S R=R Q$         (CPCT)

Here,

$PQ=QR=RS=SP$

Therefore, the quadrilateral $PQRS$ is a rhombus.

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Updated on: 10-Oct-2022

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