In trapezoid $A B C D(B C \| A D)$, it is known that $A D=3 \cdot B C$. A line intersects the lateral sides of the trapezoid at points $M$ and $N, A M: M B=3: 5, C N: N D=2: 7$. Find the ratio of the areas of quadrilaterals $M B C N$ and $A M N D$.

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Continue the lateral sides of the trapezoid until they intersect at point $P$. Find the ratios $\frac{P M}{P B}, \frac{P N}{P C}$ and express the areas of the specified quadrilaterals in terms of the area of triangle $P B C$.

## Solution

Extend the lateral sides of the trapezoid until they intersect at point $P$. Let $B C=a, A D=3 a, B M=5 x, A M=3 x, C N$ $=2 y, N D=7 y$.

From the similarity of triangles $P B C$ and $P A D$, it follows that $B P=\frac{1}{3} A P$ and $C P=\frac{1}{3} D P$. Therefore,

$$
B P=\frac{1}{2} A B=\frac{1}{2}(3 x+5 x)=4 x, P C=\frac{1}{2} C D=\frac{1}{2}(2 y+7 y)=\frac{9 y}{2}.
$$

Then

$$
P M=B P+B M=4 x+5 x=9 x, P N=C P+C N=\frac{9 y}{2}+2 y=\frac{13 y}{2}
$$

Let $S_{\Delta \mathrm{PBC}}=s$. Then

$$
S_{\Delta \mathrm{PMN}}=\frac{P M}{P B} \cdot \frac{P N}{P C} \cdot s=\frac{9 x}{4 x} \cdot \frac{\frac{13 y}{2}}{\frac{9 y}{2}} \cdot s=\frac{9}{4} \cdot \frac{13}{9} \cdot s=\frac{13}{4} s
$$

The ratio of the areas of similar triangles is equal to the square of the similarity coefficient, so $S_{\Delta \mathrm{PAD}}=9 s$.

Therefore,

$$
\frac{S_{M B C N}}{S_{A M N D}}=\frac{S_{\triangle P M N}-S_{\triangle P B C}}{S_{\triangle A P D}-S_{\triangle P M N}}=\frac{\frac{13}{4} s-s}{9 s-\frac{13}{4} s}=\frac{\frac{9}{4}}{\frac{23}{4}}=\frac{9}{23}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-38.jpg?height=641&width=746&top_left_y=1675&top_left_x=653)

## Answer

$\frac{9}{23}$.