10.3. In trapezoid $A B C D$, the bases $B C=3$ and $A D=9$, angle $\angle B A D=30$ degrees, and $\angle A D C=60$ degrees. A line through point $D$ divides the trapezoid into two equal areas. Find the length of the segment of this line that is inside the trapezoid.
Answer: $\sqrt{39}$.

Solution: Let $B C=x$, then $A D=3 x$. Drop a perpendicular from point $B$ to the base $A D$, denoted as $B H$. Then the area of triangle $A B D$ is $S_{A B D}=\frac{1}{2} \cdot A D \cdot B H=\frac{3}{2} x h$, and $S_{B C D}=\frac{1}{2} \cdot B C \cdot B H=\frac{1}{2} x h$, i.e., $S_{A B D}=3 \cdot S_{B C D}$. Therefore, the line dividing the trapezoid into two equal areas and starting from point $D$ intersects line $A B$. Let the intersection point be $K$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7d8b551d1395882d9859g-1.jpg?height=457&width=1265&top_left_y=1896&top_left_x=384)

$$
S_{A K D}=\frac{1}{2} \cdot S_{A B C D}=\frac{1}{2} \cdot \frac{A D+B C}{2} \cdot B H=x h
$$

On the other hand, the area $S_{A K D}=\frac{1}{2} \cdot A D \cdot B H_{1}=\frac{3}{2} x h_{1}$, where $h_{1}$ is the height dropped from point $K$ to $A D$. Then $x h=\frac{3}{2} x h_{1}$, from which $\frac{B H_{1}}{B H}=\frac{2}{3}$. From the similarity of triangles $A K H_{1}$ and $A B H$, it follows that $\frac{A K}{A B}=\frac{2}{3}$.

Draw a line from vertex $B$ parallel to $C D$ until it intersects the base $A D$ at point $E$. Then $A E=A D-D E=A D-B C=2 x$. From the right triangle $A B E$, we find $A B=A E \cdot \cos \left(30^{\circ}\right)=x \sqrt{3}$. Then $A K=\frac{2}{\sqrt{3}} x$. Using the cosine rule in triangle $A K D$, we find:
$D K=\sqrt{A K^{2}+A D^{2}-2 \cdot A K \cdot A D \cdot \cos 30^{\circ}}=\sqrt{\frac{4}{3} x^{2}+9 x^{2}-2 \cdot \frac{2}{\sqrt{3}} x \cdot 3 x \cdot \frac{\sqrt{3}}{2}}=x \sqrt{\frac{4}{3}+9-6}=x \sqrt{\frac{13}{3}}$.

Substitute $x=3$, then $D K=\sqrt{39}$.

Criteria: Only the answer - 0 points. 7 points - correct answer with a correct solution.

Partially correct progress in the solution - 3-5 points.