Question

$PQRS$ is a parallelogram. $PS$ is produced to meet $M$ so that $SM=SR$ and $MR$ is produced to meet $PQ$ produced at $N$. Prove that $QN=QR.$

Answer 1

In $△SMR,$

$\Rightarrow$ $SM=SR$ [ Given ]

$\therefore$ $\angle SMR=\angle SRM$ [ Angles opposite to equal sides are equal ]

Let $\angle SMR=\angle SRM=x$ ----- ( 1 )

As $\angle PSR$ is an exterior angle of $△SMR$

$\Rightarrow$ So, $\angle PSR=\angle SMR+\angle SRM$

$\Rightarrow$ $\angle PSR=x+x=2x$ ---- ( 2 )

$\Rightarrow$ $\angle PSR=\angle PQR$ [ Opposite angles of parallelogram $PQRS$ ]

$\Rightarrow$ So, $\angle PQR=2x$ ----- ( 3 )

As $PM\parallel QR$

$\Rightarrow$ So, $\angle PSR+\angle QRS={180}^o.$

$\Rightarrow$ $2x+\angle QRS={180}^o.$

$\Rightarrow$ $\angle QRS={180}^o-2x$ ----- ( 4 )

As, $\angle QRS+\angle SRM+\angle QRN={180}^o$

$\Rightarrow$ $({180}^o-2x)+x+\angle QRN={180}^o$ [ From ( 1 ) and ( 4 ) ]

$\Rightarrow$ $({180}^o-x)+\angle QRN={180}^o$

$\Rightarrow$ $\angle QRN=x$ ---- ( 5 )

Also, $\angle PQR$ is an exterior angle of $△QRM$

So, $\angle PQR=\angle QRN+\angle QNR$

$\Rightarrow$ $2x=x+\angle QNR$ [ From ( 5 ) and ( 3 ) ]

$\Rightarrow$ $\angle QNR=x$ ---- ( 6 )

In $△QNR,$

$\Rightarrow$ $\angle QRN=\angle QNR$ [ From ( 5 ) and ( 6 ) ]

$\therefore$ $QR=QN$ [ Angles opposite to sides are equal ]