6. (30 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 15 and 19, respectively. $AH$ and $BG$ are heights to the line $DC$, and $CF$ is a height to the line $AB$. Points $K, L, M$, and $N$ are the midpoints of segments $AB, CF, CD$, and $AH$ respectively. Find the ratio of the area of trapezoid $ABCD$ to the area of quadrilateral $KLMN$, if $DG=17$.
Answer: 2 or $\frac{2}{3}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_de56aaa25ca847a38066g-4.jpg?height=334&width=1506&top_left_y=1101&top_left_x=250)

Solution. Let the trapezoid $ABCD$ be labeled in a clockwise direction, with the bases assumed to be horizontal. Since the problem does not specify where point $G$ lies on line $DC$, we need to consider all possible cases. There are two scenarios: point $G$ lies to the left of point $D$ (see the left diagram) and point $G$ lies to the right of point $D$ (see the right diagram). In the second case, since

$$
DG = 17 < 19 = DC
$$

point $G$ lies on segment $DC$.

Let $h$ be the height of the trapezoid. By the problem's condition, points $N$ and $L$ are the midpoints of the heights of the trapezoid. Note that the area of quadrilateral $KLMN$ is the sum of the areas of triangles $KNL$ and $MNL$, which share a base and have heights equal to half the height of the trapezoid, i.e.,

$$
S_{KLMN} = S_{KNL} + S_{MNL} = 2 \cdot S_{KNL} = 2 \cdot \frac{1}{2} \cdot NL \cdot \frac{h}{2} = NL \cdot \frac{h}{2}
$$

Thus, the desired ratio of areas is

$$
\frac{S_{ABCD}}{S_{KLMN}} = \frac{\frac{AB + DC}{2} \cdot h}{NL \cdot \frac{h}{2}} = \frac{AB + CD}{NL}
$$

$NL$ is the distance between lines $AH$ and $CF$. In the case where point $G$ lies to the left of point $D$, i.e., does not fall on base $CD$, the value of $NL = AB + BG + DC = 15 + 17 + 19 = 51$. In the second case, $NL = HG + GC = AB + (DC - DG) = 15 + (19 - 17) = 17$. Therefore, we get two answers: $\frac{AB + CD}{NL} = \frac{34}{51} = \frac{2}{3}$ and $\frac{AB + CD}{NL} = \frac{34}{17} = 2$.