Two parallel lines $l$ and $m$ are intersected by a transversal $p$ (see Fig). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

(N/A) It is given that $l \parallel m$ and transversal $p$ intersects them at points $A$ and $C$ respectively.
The bisectors of $\angle PAC$ and $\angle ACQ$ intersect at $B$,and the bisectors of $\angle SAC$ and $\angle ACR$ intersect at $D$.
We need to show that quadrilateral $ABCD$ is a rectangle.
Since $l \parallel m$ and $p$ is a transversal,$\angle PAC = \angle ACR$ (Alternate interior angles).
Therefore,$\frac{1}{2} \angle PAC = \frac{1}{2} \angle ACR$,which implies $\angle BAC = \angle ACD$.
These are alternate interior angles for lines $AB$ and $DC$ with $AC$ as a transversal,and since they are equal,$AB \parallel DC$.
Similarly,by considering $\angle ACB$ and $\angle CAD$,we can show that $BC \parallel AD$.
Since both pairs of opposite sides are parallel,$ABCD$ is a parallelogram.
Now,$\angle PAC + \angle CAS = 180^{\circ}$ (Linear pair).
Dividing by $2$,we get $\frac{1}{2} \angle PAC + \frac{1}{2} \angle CAS = 90^{\circ}$,which means $\angle BAC + \angle CAD = 90^{\circ}$,so $\angle BAD = 90^{\circ}$.
Since $ABCD$ is a parallelogram with one angle equal to $90^{\circ}$,it is a rectangle.