## Task A-1.6.

Let $A B C D$ be a rectangle such that $|A B|:|A D|=2: 3$ and let point $E$ be on side $\overline{A D}$ such that $|A E|=|A B|$. Point $F$ is chosen on ray $A B$ so that triangle $A F E$ and quadrilateral $C D E F$ have equal areas. Determine the ratio $|A B|:|B F|$.
## First Solution.

From $|A E|=|A B|$ and $|A B|:|A D|=2: 3$, it follows that $|A E|:|E D|=2: 1$. Triangles $A E F$ and $D E F$ have the same heights, so $P(A E F)=2 P(D E F)$. Since

$$
2 P(D E F)=P(A E F)=P(C D E F)=P(D E F)+P(C D F)
$$

it follows that $P(D E F)=P(C D F)$.

![](https://cdn.mathpix.com/cropped/2024_05_30_f5c8a7635f49ecc51cacg-06.jpg?height=516&width=946&top_left_y=850&top_left_x=475)

Let $|D E|=x$. Then $|D C|=2 x$ and the length of the height from vertex $F$ in triangle $C D F$ is $3 x$. Therefore,

$$
P(C D F)=\frac{2 x \cdot 3 x}{2}=3 x^{2}
$$

On the other hand, we have

$$
P(D E F)=\frac{x \cdot|A F|}{2}
$$

By equating, we get

$$
\frac{x \cdot|A F|}{2}=P(D E F)=P(C D F)=3 x^{2}
$$

from which we obtain $|A F|=6 x$.

Now we have $|B F|=|A F|-|A B|=4 x$ and $|A B|:|B F|=2 x: 4 x=1: 2$.

Note: After showing that $P(D E F)=P(C D F)=3 x^{2}$ (2 points), we can conclude that $P(A E F)=2 P(C D F)$ (1 point). Now it follows that

$$
6 x^{2}=2 P(C D F)=P(A E F)=\frac{2 x \cdot|A F|}{2} \quad(1 \text{point})
$$

i.e., $|A F|=6 x$ (1 point) and $|A B|:|B F|=1: 2$ (1 point).