4. In an isosceles trapezoid $A B C D (B C \| A D)$, angles $A B D$ and $D B C$ are $135^{\circ}$ and $15^{\circ}$ respectively, and $B D=\sqrt{6}$. Find the perimeter of the trapezoid.
Answer: $9-\sqrt{3}$.

Note that $\angle D B C=\angle A C B=\angle B D A=\angle C A D=15^{\circ}, \angle B A C=30^{\circ}-\angle D B C=15^{\circ}$, so $A C$ is the angle bisector of $\angle A$. Since $\angle B A C=\angle C A B=\angle C D B=15^{\circ}$, triangles $A B C$ and $B C D$ are isosceles and $A B=B C=C D$. By the Law of Sines,

![](https://cdn.mathpix.com/cropped/2024_05_06_a6b3d717349ee776cd18g-2.jpg?height=320&width=808&top_left_y=1136&top_left_x=701)

Fig. 1:

for triangle $A B D: \frac{B D}{\sin 30^{\circ}}=\frac{A D}{\sin 135^{\circ}}$, from which

$$
A D=2 \sqrt{3}
$$

By the Law of Cosines for the same triangle:

$$
\begin{gathered}
A B^{2}+(2 \sqrt{3})^{2}-2 \cdot A B \cdot 2 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=(\sqrt{6})^{2} \Leftrightarrow \\
\Leftrightarrow A B^{2}-6 A B+6=0
\end{gathered}
$$

From the last equation, we find $A B=3-\sqrt{3}$ (the second root does not fit the context, as the side opposite the larger angle should be larger). Finally:

$$
A B+B C+C D+A D=3-\sqrt{3}+3-\sqrt{3}+3-\sqrt{3}+2 \sqrt{3}=9-\sqrt{3}
$$

Evaluation criteria. Correct solution - 20 points; the solution process is correct, but the wrong answer is obtained due to an arithmetic error - 15 points; the perimeter is not found, but it is proven that $A B=B C=C D-6$ points.