In Exercise 4 quad. ABCD is a parallelogram. Find the values of x, y, and z.

According to the provided information, the opposite sides of the provided figure are parallel. That is$\overline{AB}\parallel \overline{CD}$ and $\overline{AD}\parallel \overline{BC}$.

Now, lines$\overline{AD}\parallel \overline{BC}$ and $\overline{AB}$is transversal, so by Theorem 3-2 which states that if two lines are having a separate line that is travelling through them then the angles are congruent which are positioned on the alternate interior sides of the lines.

Write the alternate interior angles.

$\angle z\cong \angle DBC$

That is, $\angle z=\angle DBC=85°$

Also since ABCD is a parallelogram, so by theorem 5-2 states that the relationship between two angles that are related to a parallelogram and positioned on the opposite sides of each other is congruent.

Write the opposite angles of provided figure.

$\angle x\cong \angle DAB$

That is,$\angle x=\angle DAB=56°$

Now consider the triangle $\Delta DAB$, so by theorem 3-11 which states that if all the angles that are positioned in a triangle are added together then they will result in $180°$.

$\angle ADB+\angle DBA+\angle BAD=180°$

$\begin{array}{c}85°+y+56°=180°\\ y+141°=180°\\ y=180°-141°\\ y=39°\end{array}$