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

$PQ$ and $RS$ are two equal and parallel line segments. Any point $M$ not lying on $PQ$ or $RS$ is joined to $Q$ and $S$. Lines are drawn through $P$ parallel to $QM$ and through $R$ parallel to $SM$,meeting at $N$. Prove that line segments $MN$ and $PQ$ are equal and parallel to each other.

In quadrilateral $ABCD$,$\angle A + \angle D = 180^{\circ}$. What special name can be given to this quadrilateral?

In $\Delta ABC$,$P$,$Q$ and $R$ are the midpoints of $AB$,$BC$ and $CA$ respectively,then $\angle BAC = \angle \ldots \ldots \ldots$

Prove that the diagonals of a square are equal and bisect each other at right angles.

In a parallelogram $ABCD$,$\angle A = 5x - 40^{\circ}$ and $\angle C = 3x + 10^{\circ}$,then $x = \ldots$