28. The area of trapezoid $ABCD$ is $S$, the ratio of the bases $|AD|:|BC|=3$; on the line intersecting
the extension of base $AD$ beyond point $D$, segment $EF$ is located such that $AE\|DF, BE\|CF$ and $|AE|:|DF|=$ $=|CF|:|BE|=2$. Determine the area of triangle EFD.
28. Let $K$ and $M$ be the points of intersection of line $E F$ with $A D$ and $B C$. Suppose $M$ lies on the extension of $B C$ beyond point $B$. If $|A D|=3 a,|B C|=a$, then from the similarity of the corresponding triangles, it follows that $|D K|=|A D|=3 a,|M B|=|B C|=a$ (Fig. 5, a). Additionally, $|M E|=|E F|=|F K|$. If $h$ is the height of the trapezoid, then the distance from $E$ to $A D$ will be $\frac{2}{3} h, S_{\triangle E D K}=$ $=\frac{1}{2} \cdot \frac{2}{3} h a=\frac{h a}{3}, s_{\triangle E D F}=\frac{1}{2} s_{\triangle E D K}=\frac{h a}{6}=\frac{1}{12} s$.

If, however, line $E F$ intersects the base $B C$ at point $M$, then $B M \left\lvert\,=\frac{1}{3} a\right.$ (Fig. 5, b). In this case, $\frac{|E K|}{|M K|}=2: \frac{5}{3}=\frac{6}{5}$ and the distance from $E$ to $A D$ will be $\frac{6}{5} h$, so $S_{\triangle E F D}=\frac{1}{2} S_{\triangle E D K}=$ $=\frac{1}{4} \cdot 3 a \cdot \frac{6}{5} h=\frac{9}{20} s$.

Answer: $\frac{1}{12} s$ or $\frac{9}{20} s$.

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-070.jpg?height=552&width=690&top_left_y=252&top_left_x=273)

Fig. 5.