## Task B-2.3.

In trapezoid $A B C D$, the shorter base $\overline{C D}$ has a length of 3. A line parallel to the diagonal $\overline{B D}$ intersects the longer base $\overline{A B}$ at point $E$ and the side $\overline{A D}$ at point $F$. If $|A E|=5$ and $|B E|=2$, what is the ratio of the area of triangle $A E F$ to the area of pentagon $E B C D F$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_7713364b1a15726862ddg-07.jpg?height=469&width=574&top_left_y=548&top_left_x=661)

Let $a$ be the length of the longer base of the trapezoid and $v$ be the height of the trapezoid $ABCD$, and $v'$ be the height of the triangle $\triangle AEF$. Then, $a = 2 + 5 = 7$.

The area $P$ of the trapezoid $ABCD$ is

$$
P(ABCD) = \frac{a + 3}{2} \cdot v = \frac{7 + 3}{2} \cdot v = 5v
$$

and the area of the triangle $\triangle AEF$ is

$$
P(\triangle AEF) = \frac{|AE| \cdot v'}{2} = \frac{5v'}{2}
$$

From the similarity of triangles $\triangle ABD$ and $\triangle AEF$, it follows that

$$
|AB| : v = |AE| : v' \Rightarrow v' = \frac{5}{7} v
$$

Thus,

$$
\frac{P(\triangle AEF)}{P(EBCDF)} = \frac{\frac{5v'}{2}}{5v - \frac{5v'}{2}} = \frac{v'}{2v - v'} = \frac{\frac{5}{7} v}{2v - \frac{5}{7} v} = \frac{\frac{5}{7} v}{\frac{9}{7} v} = \frac{5}{9}
$$

The ratio of the area of the triangle $\triangle AEF$ to the area of the pentagon $EBCDF$ is $5:9$.