The area of trapezoid $ABCD$ is 30. Point $P$ is the midpoint of side $AB$. Point $R$ on side $CD$ is chosen such that $2CD = 3RD$. Lines $AR$ and $PD$ intersect at point $Q$. Find the area of triangle $APQ$, given that $AD = 2BC$.
Find the ratio $P Q: Q D$.

## Solution

Let $T$ be the point of intersection of lines $A R$ and $B C$. From the similarity of triangles $C R T$ and $D R A$, we find that $C T = D R / R C \cdot A D = 1 / 2 A D$.

Therefore, $B T = A D$, and $A B T D$ is a parallelogram.

From the similarity of triangles $A P Q$ and $T D Q$, we find that $P Q: Q D = A P: T D = A P: A B = 1: 2$.

Thus, $S_{A P Q} = 1 / 3 S_{A P D} = 1 / 6 S_{A B D} = 2 / 3 \cdot 1 / 6 S_{A B C D} = 10 / 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-37.jpg?height=303&width=855&top_left_y=815&top_left_x=607)

## Answer

$10 / 3$.