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.

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Solution

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 getTherefore, $PQ=SR$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

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