In a quadrilateral $ABCD$,$\angle A : \angle B : \angle C : \angle D = 4 : 5 : 4 : 5$. Find the measure of each angle of the quadrilateral and state the type of quadrilateral $ABCD$.

(A) In quadrilateral $ABCD$,the sum of the interior angles is $360^{\circ}$.
Given the ratio $\angle A : \angle B : \angle C : \angle D = 4 : 5 : 4 : 5$.
Let the angles be $4x, 5x, 4x$,and $5x$.
Sum of angles: $4x + 5x + 4x + 5x = 360^{\circ}$.
$18x = 360^{\circ}$,which gives $x = 20^{\circ}$.
Therefore,the angles are:
$\angle A = 4 \times 20^{\circ} = 80^{\circ}$
$\angle B = 5 \times 20^{\circ} = 100^{\circ}$
$\angle C = 4 \times 20^{\circ} = 80^{\circ}$
$\angle D = 5 \times 20^{\circ} = 100^{\circ}$
Since $\angle A = \angle C$ and $\angle B = \angle D$,both pairs of opposite angles are equal.
Therefore,$ABCD$ is a parallelogram.