Prove that the diagonals of a parallelogram bisect each other

In parallelogram $ABCD,\hspace{0.17em}E$  is the mid-point of $AD$ and $F$ is the mid-point of $BC$. Prove that, $BFDE$ is a parallelogram.

In the diagram, the bisectors of the interior angles of the parallelogram $PQRS$ enclose a quadrilateral $ABCD$. Show that, $\angle PSB+\angle SPB=90°$.

Thus the bisectors of the angles of a parallelogram enclose a rectangle.

In the diagram, the bisectors of the interior angles of the parallelogram $PQRS$ enclose a quadrilateral $ABCD$. Show that, $\angle PBS=90°$.

Thus the bisectors of the angles of a parallelogram enclose a rectangle.

In the diagram, the bisectors of the interior angles of the parallelogram $PQRS$ enclose a quadrilateral $ABCD$. Show that, $\angle ABC=90°$.

Thus the bisectors of the angles of a parallelogram enclose a rectangle.