Question

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

Answer 1

Given, $ABCD$ is rectangle where $P,Q,RandS$ are mid points of the sides $AB,BC,CD$ and $DA$ respectively.

We have to prove $PQRS$ is a rhombus.

Now join $A$ and $C$

We know rhombus is a parallelogram with all sides equal.

Now, we will prove $PQRS$ is parallelogram, and then prove all

sides are equal.

In $△ABC$

$P$is mid-point of $AB,$

$Q$ is mid-point of $BC$

$\therefore PQ\parallel AC$ and $PQ=\frac{1}{2}AC---\left(1\right)$

Also, In $△ADC$

$R$is mid-point of $CD,$

$S$ is mid-point of $AD$ respectively

$\therefore RS\parallel ACandRS=\frac{1}{2}AC----\left(2\right)$

From (1) & (2)

$PQ\parallel RS$ and $PQ\parallel RS$

In $PQRS$, one pair of opposite side is parallel and equal.

Hence, $PQRS$ is parallelogram.

Now we prove all sides equal

In $△APS\&△BPQ$

$AP=BP$ $P$ is the mid point of $AB$

$\angle PAS=\angle PBQ$

$AS=BQ$

$\therefore △APS\cong △BPQ$ (SAS)congruence rule

$\therefore PS=PQ$ [CPCT]

But $PS=RQ\&PQ=RS$

$\therefore PQ=RS=PS=RQ$

Hence, all sides are equal

Thus, $PQRS$ is a parallelogram with all side equal

so, $PQRS$ is a rhombus

Hence proved.