Two parallel lines are intersected by a transversal. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

(N/A) Let the two parallel lines be $l$ and $m$,and let $n$ be the transversal intersecting them at points $A$ and $B$ respectively.
Let the interior angles formed be $\angle PAB$,$\angle QAB$,$\angle RBA$,and $\angle SBA$.
The bisectors of these four interior angles form a quadrilateral $EFGH$.
Since $l \parallel m$,the sum of consecutive interior angles is $180^{\circ}$,i.e.,$\angle PAB + \angle RBA = 180^{\circ}$.
Dividing by $2$,we get $\frac{1}{2}\angle PAB + \frac{1}{2}\angle RBA = 90^{\circ}$.
In $\triangle EAB$,$\angle EAB + \angle EBA = 90^{\circ}$,which implies $\angle AEB = 90^{\circ}$.
Similarly,all other angles of the quadrilateral $EFGH$ can be shown to be $90^{\circ}$.
Since all interior angles are $90^{\circ}$,the quadrilateral $EFGH$ is a rectangle.