In trapezoid $A B C D$, where $\angle B A D=45^{\circ}, \angle C D A=60^{\circ}$, base $A D$ equals 15, base $B C$ equals 13, the perpendicular to side $A B$, drawn from point $M$, the midpoint of side $A B$, intersects with the perpendicular to side $C D$, drawn from point $N$, the midpoint of side $C D$, at some point $L$. Find the ratio of the area of triangle $M N L$ to the area of trapezoid $A B C D$.
Find the heights of the trapezoid and triangle $M N L$.

## Solution

Let $B P=C Q=h$ be the height of the trapezoid $A B C D$, and $L T=h_{1}$ be the height of the triangle $M N L$. Then

$$
\frac{S_{\triangle M N L}}{S_{A B C D}}=\frac{M N \cdot h_{1}}{(A D+B C) h}=\frac{14 h_{1}}{28 h}=\frac{h_{1}}{2 h}
$$

From the right triangles $A P B$ and $D Q C$, we find that

$$
A P=h, D Q=\frac{h}{\sqrt{3}}
$$

Since $A P+Q D=A D-B C=15-13=2$, then $h+\frac{h}{\sqrt{3}}=2$. Therefore, $h=\frac{2 \sqrt{3}}{1+\sqrt{3}}$.

Since $\angle L M N=45^{\circ}$ and $\angle M N L=30^{\circ}$, then

$$
M T=T L=h_{1}, T N=T L \sqrt{3}=h_{1} \sqrt{3}
$$

Since $M T+T N=M N$, then $h_{1}+h_{1} \sqrt{3}=14$. From this, we find that $h_{1}=\frac{14}{1+\sqrt{3}}$. Therefore,

$$
\frac{S_{\triangle M N L}}{S_{A B C D}}=\frac{h_{1}}{2 h}=\frac{\frac{14}{1+\sqrt{3}}}{\frac{2 \sqrt{3}}{1+\sqrt{3}}}=\frac{7 \sqrt{3}}{6}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-38.jpg?height=380&width=874&top_left_y=856&top_left_x=592)

Answer

$\frac{7 \sqrt{3}}{6}$.