## Task B-2.3.

On the side $\overline{A B}$ of triangle $A B C$, a point $E$ is given such that $|A E|=1$ and $|E B|=2$. Let point $D$ be on the side $\overline{A C}$ such that the segment $\overline{D E}$ is parallel to the segment $\overline{B C}$, and point $F$ be on the side $\overline{B C}$ such that the segment $\overline{E F}$ is parallel to the segment $\overline{A C}$. Calculate the ratio of the areas of quadrilateral $C D E F$ and triangle $A B C$.
## Second Solution.

Let $\varangle A C B=\gamma$. Then,

$$
\frac{P_{C D E F}}{P_{A B C}}=\frac{|C D||C F| \sin \gamma}{\frac{|A C||B C| \sin \gamma}{2}}=\frac{2|C D||C F|}{|A C||B C|}
$$

Due to the parallelism of the given segments, triangles $A E D, E B F$, and $A B C$ have corresponding angles that are congruent, making them similar to each other.

Then, from the similarity of triangles $E B F$ and $A B C$, it follows that

$$
\frac{|A C|}{|E F|}=\frac{|A B|}{|E B|}=\frac{3}{2} \text {, or }|A C|=\frac{3}{2}|E F|=\frac{3}{2}|C D| \text {. }
$$

From the similarity of triangles $A E D$ and $A B C$, it follows that

$$
\frac{|B C|}{|D E|}=\frac{|A B|}{|A E|}=\frac{3}{1}, \text { or }|B C|=3|E F|=3|C D|
$$

Then,

$$
\frac{P_{C D E F}}{P_{A B C}}=\frac{2|C D||C F|}{|A C||B C|}=\frac{2|C D||C F|}{\frac{3}{2}|C D| \cdot 3|C F|}=\frac{4}{9}
$$