In Exercise 5 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 $\overline{AB}\parallel \overline{CD}$and $\overline{BD}$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.

$\angle z\cong \angle ABD$

That is,$\angle z=\angle ABD=70°$

$\angle AOD$and $\angle BOC$are vertical angles, so by theorem 2-3 which states that the relationship between two angles is congruent when those angles are vertical angles,

$\angle x\cong \angle AOD$

That is, $\angle x=\angle AOD=110°$

In the triangle $\Delta AOB$, $\angle BOC$is the exterior angle corresponding to the two remote interior angles $\angle BAO$and $\angle ABO$, so by theorem 3-12 which states that the measurement related to an angle that is positioned on the exterior side of a triangle is the same as the measurement related to two angles that are remote interior angles when the remoter interior angles are added together,

$\angle BAO+\angle ABO=\angle BOC$

$\begin{array}{c}y+70°=110°\\ y=110°-70°\\ y=40°\end{array}$