Question

PQ and RS are two equal and parallel line segments.
Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N . Prove that line segments MN and PQ are equal and parallel to each other.

Answer 1

Since, $PQ=RS$ and $PQ\parallel RS$, therefore, $PQRS$ is a parallelogram.

Hence, $PR\parallel QS$ and $PR=QS$

Now, $\angle RPQ+\angle PQS={180}^{\circ }$

$\Rightarrow \angle RPQ+\angle PQM+\angle MQS={180}^{\circ }$ ...(i)

Also, $PN\parallel QM$

$\Rightarrow \angle NPQ+\angle PMQ={180}^{\circ }$

$\Rightarrow \angle NPR+\angle RPQ+\angle PQM={180}^{\circ }$...(ii)

From (i) and (ii),

$\Rightarrow \angle NPR=\angle MQS$

Similarly , we can prove that $\angle NRP=\angle MSQ$

Now, in $\Delta NPR$ and $\Delta MQS$,

$\angle NPR=\angle MQS$

$\angle NRP=\angle MSQ$

$PR=QS$

$\therefore \Delta NPR\cong \Delta MQS$ by ASA rule.

$\Rightarrow PN=MQ,NR=MS$ (By CPCT)

Thus, $PQMN$ and $RSMN$ are both the parallelograms with common side $MN$

Hence, $MN\parallel PQ$