Diagonals of a quadrilateral $ABCD$ bisect each other. If $\angle A = 35^{\circ}$,determine $\angle B$. (in $^{\circ}$)

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

$XYZW$ is a rhombus. If the diagonals $XZ$ and $YW$ intersect at $P$,then $\angle XPY = \ldots \ldots \ldots$ (in $^o$)

Points $P$ and $Q$ have been taken on opposite sides $AB$ and $CD$ respectively of a parallelogram $ABCD$ such that $AP = CQ$ (see figure). Show that $AC$ and $PQ$ bisect each other.

The figure obtained by joining the mid-points of the sides of a rhombus,taken in order,is

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