Solution

a) From $\angle D A P=\angle P A F$ and $A P \perp D F$, it follows that triangle $A D F$ is isosceles and that $P$ is the midpoint of $D F$. Since $E Q$ and $D F$ are perpendicular to $A E$, we have $D F \| E Q$ and $\angle C E Q=\angle C D F=\angle D F A$. In the parallelogram $A B C D, \angle D A B=\angle D C B$ and thus triangles $A D F$ and $E C Q$ are similar. Therefore, $C E=C Q$ and

$$
\frac{C Q}{A D}=\frac{E Q}{D F}
$$

Since the opposite sides of quadrilateral $D E Q P$ are parallel, it follows that it is a parallelogram. Consequently, $D P=E Q$ and

$$
\frac{C Q}{A D}=\frac{D P}{2 \cdot D P}=\frac{1}{2}
$$

b) Let $x$ be the length of segment $A D$. From $\angle D A E=\angle E A B=\angle D E A$, we can conclude that triangle $A D E$ is isosceles and thus $A D=D E=x$. From the previous item, it follows that $C Q=x / 2$. Thus

$$
\begin{aligned}
C D & =D E+E C \\
20 & =x+x / 2 \\
& =3 x / 2
\end{aligned}
$$

Finally, $x=40 / 3 \mathrm{~cm}$.
In the parallelogram $A B C D$, the angle $B A D$ is acute and the side $A D$ is shorter than the side $A B$. The bisector of the angle $\angle B A D$ intersects the side $C D$ at $E$. Through $D$, a perpendicular to $A E$ is drawn, intersecting $A E$ at $P$ and $A B$ at $F$. Through $E$, a perpendicular to $A E$ is drawn, intersecting the side $B C$ at $Q$. Additionally, the segment $P Q$ is parallel to $A B$ and the length of $A B$ is $20 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-05.jpg?height=386&width=725&top_left_y=595&top_left_x=740)

a) Find the value of $\frac{C Q}{A D}$.

b) Find the measure of the length of the side $A D$.