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

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

In $\Delta ABC$,$AD$ is a median and $E$ is the midpoint of $AD$. $BE$ is extended to intersect $AC$ at $F$. Prove that $AF = \frac{1}{3} AC$.

$(1)$ If no three points out of four coplanar points are collinear,then a $\ldots \ldots \ldots$ figure formed by joining these four points in order is called a quadrilateral.
$(2)$ $A$ quadrilateral has $\ldots \ldots \ldots$ pairs of opposite sides.

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

Can the angles $110^{\circ}, 80^{\circ}, 70^{\circ}$ and $95^{\circ}$ be the angles of a quadrilateral? Why or why not?