If in the figure,bisectors $AP$ and $BQ$ of the alternate interior angles are parallel,then show that $l \parallel m$.

(N/A) Given: $AP$ is the bisector of $\angle MAB$ and $BQ$ is the bisector of $\angle SBA$. We are given that $AP \parallel BQ$.
Since $AP \parallel BQ$ and $t$ is a transversal,the alternate interior angles are equal:
$\angle 2 = \angle 3$ (Alternate interior angles)
Multiplying both sides by $2$:
$2 \angle 2 = 2 \angle 3$
Since $AP$ and $BQ$ are bisectors:
$\angle 1 = \angle 2$ and $\angle 3 = \angle 4$
Therefore,$\angle 1 + \angle 2 = \angle 3 + \angle 4$
This implies $\angle MAB = \angle SBA$.
Since the alternate interior angles $\angle MAB$ and $\angle SBA$ are equal,the lines $l$ and $m$ must be parallel,i.e.,$l \parallel m$.