Question

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

Answer 1

Given, $ABCD$ is rhombus, where $P,Q,R$ and $S$ are the mid points of $AB,BC,CD$ and $DA$ respectively.

We have to prove $PQRS$ is a rectangle

First we will prove PQRS is parallelogram, since parallelogram with one angle ${90}^o$ is rectangle.

In $\Delta$ABC, P is mid point of AB, Q is mid point of BC.

$\therefore PQ||AC$ and $PQ=\frac{1}{2}AC$ ……..$\left(1\right)$

In $\Delta$ADC, R is the mid-point of CD, S is mid point AD.

$\therefore RS||AC$ and $RS=\frac{1}{2}AC$ ………$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

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

Hence, PQRS is a parallelogram.

Now we have to prove PQRS is a rectangle

Since $AB=BC\Rightarrow \frac{1}{2}AB=\frac{1}{2}BC$

So $PB=BQ$

Now, In $\Delta$BPQ, PB$=$BQ

$\therefore \angle 2=\angle 1$ ………$\left(3\right)$

In $\Delta$APS & $\Delta$CQR

$\Rightarrow AP=CQ$

$\Rightarrow AS=CR$

$\Rightarrow PS=QR$

$\therefore \Delta APS\cong \Delta CQR$

$\Rightarrow \angle 3=\angle 4$ by CPCT …………$\left(4\right)$

Now

AB is a line

So, $\angle 3+\angle$SPQ$1={180}^o$ ……….$\left(5\right)$

Similarly for line Bc

$\angle 2+\angle$PQR$+\angle 4={180}^o$

$\angle 1+\angle$PQR$+\angle 3={180}^o$ ……….$\left(6\right)$

From $5$ & $6$

$\angle 1+\angle$SPQ$+\angle 3=\angle 1+\angle$PQR$+\angle 3$

$\therefore \angle$SPQ$=\angle$PQR ……….$\left(7\right)$

Now, PS$=$QR, and PQ is a transversal

So, $\angle$ SPQ$+\angle$PQR$={180}^o$

$\angle$SPQ$+\angle$SPQ$=180$ from $7$

$2\angle$SPQ$={180}^o$

$\angle$SPQ$={90}^o$

So, PQRS is a parallelogram with one angle ${90}^o$

$\therefore$ $PQRS$ is a rectangle.