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Question

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


Solution

Let us draw the figure with given condition as 

In $\triangle ABC$, $P$ and $Q$ are the mid- points of $AB$ and $BC$. 

$PQ \parallel AC$ and by using mid - point theorem

$PQ= \dfrac{1}{2} AC$ ........(1)

Similarly, in $\triangle ADC$, $R$ and $S$ are  the mid- points of $CD$ and $AD$. 

$SR \parallel AC$ and  by using mid point theorem

$SR= \dfrac{1}{2} AC$ .........(2)

From (1) and (2), we get 
$PQ \parallel RS$ and $PQ=SR$

Now, in quadrilateral $PQRS$ its one pair of opposite sides $PQ$ and $SR$ is equal and parallel. 

Therefore, 

$PQRS$ is  a parallelogram 

$AB = BC$ (Sides of a rhombus)

$\Rightarrow \dfrac{1}{2} AB =\dfrac{1}{2}BC$

$PB=BQ$

$\angle 3=\angle 4$

Now, in $\triangle APS$ and $\triangle CQR$, we have 

$AP = CQ$ (Halves of equal sides $AB, BC$)

$AS = CR $ (Halves of equal sides $AD, CD$)

$PS = QR$ (Opp. sides of parallelogram $PQRS$)

Therefore, $\triangle APS \cong \triangle CQR$ $\mid$ using SSS Congruency Theorem

$\angle 1= \angle 2$ (Corresponding parts of congruent triangles are equal)

Now , $\angle 1+ \angle SPQ +\angle 3= 180^o$ (Linear pair axiom)

Therefore, $\angle 1 +\angle SPQ +\angle 3=\angle 2 +\angle PQR +\angle 4$

But, $\angle 1= \angle 2$ and $\angle 3= \angle 4$ 

Therefore, 

$\angle SPQ= \angle PQR$    .....(3)

Since, $SP \parallel RQ$ and $PQ$ intersects them

Therefore, $\angle SPQ + \angle PQR= 180^o$....(4) (Since consecutive interior angles are supplementary)

From  (3) and (4), we get 

$\angle PQR +\angle PQR= 180^o$

$2 \angle PQR =180^o$

$\angle PQR =90^o$ 

$\angle SPQ= \angle PQR=90^o$

Thus, $PQRS$ is a parallelogram whose one angle $\angle SPQ=90^o$.

Hence $PQRS$ is a rectangle.


Mathematics
RS Agarwal
Standard IX

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