$ABCD$ is a rhombus such that $\angle ACB = 40^{\circ}$. Then $\angle ADB$ is (in $^{\circ}$)

$(1)$ 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.

$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.

For any convex quadrilateral $ABCD$,prove that $AB + BC + CD + DA > AC + BD$.

$(1)$ $A$ quadrilateral has ......... pairs of consecutive angles.
$(2)$ The sum of measures of all the four angles of a quadrilateral is ...........

The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ intersect each other at the point $O.$ If $\angle DAC = 32^{\circ}$ and $\angle AOB = 70^{\circ},$ then $\angle DBC$ is equal to: (in $^{\circ}$)