Three angles of a quadrilateral are $75^{\circ}, 90^{\circ}$ and $75^{\circ}$. The fourth angle is (in $^{\circ}$)

Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $O$ such that $OA : OC = 3 : 2$. Is $ABCD$ a parallelogram? Why or why not?

$P$ and $Q$ are the mid-points of the opposite sides $AB$ and $CD$ of a parallelogram $ABCD.$ $AQ$ intersects $DP$ at $S$ and $BQ$ intersects $CP$ at $R.$ Show that $PRQS$ is a parallelogram.

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

$E$ is the mid-point of a median $AD$ of $\triangle ABC$ and $BE$ is produced to meet $AC$ at $F$. Show that $AF = \frac{1}{3} AC$.

In parallelogram $PQRS$,$PQ = 8 \, \text{cm}$ and $QR = 5 \, \text{cm}$,then the perimeter of $PQRS = \ldots \ldots \ldots$ (in $, \text{cm}$)