Which of the following is not true for a parallelogram?

$A$ square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

In the given figure,it is given that $BDEF$ and $FDCE$ are parallelograms. Can you say that $BD = CD$? Why or why not?

$P, Q, R$ and $S$ are respectively the mid-points of sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$ in which $AC = BD$ and $AC \perp BD$. Prove that $PQRS$ is a square.

Diagonals of a parallelogram $ABCD$ intersect at $O.$ If $\angle BOC = 90^{\circ}$ and $\angle BDC = 50^{\circ},$ then $\angle OAB$ is (in $^{\circ}$)

In $\triangle ABC$,$AB = 5\, cm$,$BC = 8\, cm$,and $CA = 7\, cm$. If $D$ and $E$ are respectively the mid-points of $AB$ and $BC$,determine the length of $DE$. (in $, cm$)