$AP$ and $BQ$ are the bisectors of the two alternate interior angles formed by the intersection of a transversal $t$ with parallel lines $l$ and $m$ (see figure). Show that $AP \parallel BQ$.

(N/A) Given: $l \parallel m$ and $t$ is a transversal intersecting $l$ at $A$ and $m$ at $B$.
$AP$ is the bisector of $\angle MAB$ and $BQ$ is the bisector of $\angle SBA$ (where $S$ is a point on line $m$ such that $\angle MAB$ and $\angle SBA$ are alternate interior angles).
Since $l \parallel m$ and $t$ is a transversal,the alternate interior angles are equal:
$\angle MAB = \angle SBA$
Since $AP$ and $BQ$ are bisectors:
$\angle PAB = \frac{1}{2} \angle MAB$ and $\angle QBA = \frac{1}{2} \angle SBA$
Therefore,$\angle PAB = \angle QBA$.
These are alternate interior angles formed by the transversal $t$ with lines $AP$ and $BQ$.
Since the alternate interior angles are equal,the lines must be parallel.
Hence,$AP \parallel BQ$.