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 themTherefore, $\angle SPQ + \angle PQR= 180^o$....(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.MathematicsRS AgarwalStandard IX

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