What part of the area of an isosceles trapezoid $K L M N (K L \| M N)$ is formed by the area of triangle $A B C$, where $A$ is the midpoint of the base $K L$, $B$ is the midpoint of the base $M N$, and $C$ is the midpoint of the side $K N$?
The median $CE$ of trapezoid $KLMN$ intersects segment $AB$ at point $D$, which is the midpoint of both segments $CE$ and $AB$, since $AECB$ is a parallelogram (from the property of medians, it follows that $AC \| LN \| EB$ and $AE \| KM \| CB$, see figure).

![](https://cdn.mathpix.com/cropped/2024_04_17_525835f2a4f0b83ae7a5g-1.jpg?height=460&width=754&top_left_y=2123&top_left_x=662)

Let $|KL| = a$, $|MN| = c$, and $v$ be the height of trapezoid $KLMN$. Then $S_{KLMN} = \frac{1}{2}(a+c)v$, $S_{ABC} = S_{CDB} + S_{CDA}$, and $|CD| = \frac{1}{4}(a+c)$ (half of the median of trapezoid $KLMN$). The heights of both triangles $CDB$ and $CDA$ to side $CD$ are $\frac{1}{2}v$, so

$$
S_{ABC} = 2 \cdot S_{CDB} = 2 \cdot \frac{\frac{a+c}{4} \cdot \frac{v}{2}}{2} = \frac{(a+c)v}{8}
$$

which is $\frac{1}{4}$ of the area of the entire trapezoid $KLMN$.

ANOTHER SOLUTION. Using the notation $a$, $c$, and $v$ as in the previous paragraph. According to the problem, $|KA| = |AL| = \frac{1}{2}a$ and $|NB| = |BM| = \frac{1}{2}c$. Point $C$ lies on the median of trapezoid $KLMN$, so it has the same distance from both base lines $KL$ and $MN$, which is $\frac{1}{2}v$. Therefore, the height of triangle $KAC$ to side $KA$ and the height of triangle $BNC$ to side $BN$ are both $\frac{1}{2}v$. For the respective areas, we have

$$
\begin{aligned}
S_{KLMN} & = \frac{(a+c)v}{2}, \\
S_{ABC} & = S_{KLMN} - S_{KAC} - S_{BNC} - S_{ALMB} = \\
& = \frac{(a+c)v}{2} - \frac{\frac{a}{2} \cdot \frac{v}{2}}{2} - \frac{\frac{c}{2} \cdot \frac{v}{2}}{2} - \frac{\left(\frac{a}{2} + \frac{c}{2}\right)v}{2} = \\
& = \frac{(a+c)v}{2} - \frac{av}{8} - \frac{cv}{8} - \frac{(a+c)v}{4} = \frac{(a+c)v}{8}.
\end{aligned}
$$

The area of triangle $ABC$ is $\frac{1}{4}$ of the area of trapezoid $KLMN$.

[6 points for a complete solution]