Show that the bisectors of angles of a parallelogram form a rectangle.

In the adjoining figure, $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is a parallelogram in which $\mathrm{\text{'}}\mathrm{P}\mathrm{\text{'}}$ and $\mathrm{\text{'}}\mathrm{Q}\mathrm{\text{'}}$ are midpoints of opposite side $\mathrm{A}\mathrm{B}$ and $\mathrm{C}\mathrm{D}$. If $\mathrm{A}\mathrm{Q}$ intersects $\mathrm{D}\mathrm{P}$ at $\mathrm{S}$ and $\mathrm{B}\mathrm{Q}$ intersects $\mathrm{C}\mathrm{P}$ at $\mathrm{R}$, then show that $\mathrm{P}\mathrm{S}\mathrm{Q}\mathrm{R}$ is parallelogram.

$\mathrm{l}, \mathrm{m}$ and $\mathrm{n}$ are three parallel lines intersected by transversal $\mathrm{\text{'}}{\mathrm{p}}^{\mathrm{\text{'}}}$and $\mathrm{\text{'}}\mathrm{q}\mathrm{\text{'}}$, such that $\mathrm{l}, \mathrm{m}$ and $\mathrm{}\mathrm{n}$ cut off equal intercepts $\mathrm{A}\mathrm{B}$ and $\mathrm{B}\mathrm{C}$ on $\mathrm{p}$ (see figure). Show that $\mathrm{l}, \mathrm{m}$ and $\mathrm{n}$ cut off equal intercept $\mathrm{D}\mathrm{E}$ and $\mathrm{E}\mathrm{F}$ on $\mathrm{q}$ also.