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

(N/A) We draw the figure as per the given conditions.
It is given that $PQ = RS$ and $PQ \parallel RS$. Therefore,$PQSR$ is a parallelogram.
So,$PR = QS$ and $PR \parallel QS$ ... $(1)$
Now,$PR \parallel QS$.
Therefore,$\angle RPQ + \angle PQS = 180^{\circ}$ (Interior angles on the same side of the transversal).
i.e.,$\angle RPQ + \angle PQM + \angle MQS = 180^{\circ}$ ... $(2)$
Also,$PN \parallel QM$ (By construction).
Therefore,$\angle NPQ + \angle PQM = 180^{\circ}$ ... $(3)$
i.e.,$\angle NPR + \angle RPQ + \angle PQM = 180^{\circ}$.
Comparing $(2)$ and $(3)$,we get $\angle NPR = \angle MQS$ ... $(4)$
Similarly,$\angle NRP = \angle MSQ$ ... $(5)$
Therefore,$\Delta PNR \cong \Delta QMS$ [By $ASA$ congruence rule,using $(1)$,$(4)$,and $(5)$].
So,$PN = QM$ and $NR = MS$ $(CPCT)$.
As $PN = QM$ and $PN \parallel QM$,we have $PQMN$ is a parallelogram.
So,$MN = PQ$ and $NM \parallel PQ$.