8. In trapezoid $A B C D$, base $A D$ is four times larger than base $B C$, and angle $\angle B C D$ is twice the angle $\angle B A D$. Find the ratio $C D$ : $P Q$, where $P Q-$ is the midline of the trapezoid.
Answer: 6:5. Solution. Let $B C=a, A D=4 a$, then $P Q=(a+4 a) / 2=$ $2.5 a$. Draw the bisector $C L$ of angle $\angle B C D$ (see figure). It will divide the trapezoid into a parallelogram and an isosceles triangle $\triangle C L D$ with sides $C D=L D=3 a$. Therefore, $C D: P Q=3 a: 2.5 a=6: 5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_581b4976cdd354ac9814g-4.jpg?height=461&width=593&top_left_y=924&top_left_x=381)