Question

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

Answer 1

Step $1:$ Drawing the diagram:

$ABCD$ is a rectangle,

$P,Q,R$ and $S$ are mid-points of the sides $AB,BC,CD$ and $DA$ respectively.

Join diagonals $AC$ and $BD$ which intersect at $O$.

Step $2:$ Proving $PQRS$ is a parallelogram:

In, $∆ADC$,

$S$ is the midpoint of $AD$ and $R$ is the midpoint of $CD$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $SR\parallel AC$ and $SR=\frac{1}{2}AC$…………………..(i)

In, $∆ABC$,

$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$……………..(ii)

From (i) and (ii), we get

Therefore, $PQ=SR$

We have, $SR\parallel AC$ and $PQ\parallel AC$

$\therefore SR\parallel PQ$ (Lines parallel to same line are parallel to each other)

And, also $PQ=SR$.

$\therefore PQRS$ is a parallelogram because a pair of opposite side of quadrilateral $PQRS$ is equal and parallel.

Step $3:$Proving $PQRS$ is a rhombus:

In $∆QBP$ and $∆QCR$,

$\begin{array}{rcl}QB& =& QC\left[Q\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}BC\right]\\ \angle QBP& =& \angle QCR\left[\mathrm{Each}90°\right]\\ BP& =& CR\left[\mathrm{Opposite}\mathrm{sides}\mathrm{are}\mathrm{equal},\mathrm{hence}\mathrm{half}\mathrm{length}\mathrm{is}\mathrm{also}\mathrm{equal}\right]\end{array}$

Therefore, By SAS congruency criteria

$\therefore ∆QBP\cong ∆QCR$

$\therefore QR=QP$ (By C.P.C.T.)

As $PQRS$ is a parallelogram and having adjacent sides are equal

$\therefore PQ=QR=RS=SP$

Hence, $PQRS$ is a rhombus.

Hence, proved