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

$P$ is the mid-point of the side $CD$ of a parallelogram $ABCD$. $A$ line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$. Prove that $DA = AR$ and $CQ = QR$.

$P, Q, R$ 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.

If bisectors of $\angle A$ and $\angle B$ of a quadrilateral $ABCD$ intersect each other at $P$,of $\angle B$ and $\angle C$ at $Q$,of $\angle C$ and $\angle D$ at $R$,and of $\angle D$ and $\angle A$ at $S$,then $PQRS$ is a

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

$ABCD$ is a rectangle in which diagonal $BD$ bisects $\angle B$. Show that $ABCD$ is a square.