Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 65 and 31, respectively, and its lateral sides are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AC}$ and $\overrightarrow{BD}$.
Answer: -2015.

Solution. Let $O$ be the point of intersection of the lines containing the lateral sides $AD$ and $BC$. From the similarity of triangles $AOB$ and $DOC$, it follows that $\overrightarrow{OC} = -\frac{31}{65} \overrightarrow{BO}$, and $\overrightarrow{OD} = -\frac{31}{65} \overrightarrow{AO}$.

Let the vector $\overrightarrow{AO}$ be denoted by $\vec{a}$, and the vector $\overrightarrow{BO}$ by $\vec{b}$. Then, from the condition, it follows that $(\vec{a}, \vec{b}) = 0$ and

$$
\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = \vec{a} - \frac{31}{65} \vec{b}, \quad \overrightarrow{BD} = \overrightarrow{BO} + \overrightarrow{OD} = \vec{b} - \frac{31}{65} \vec{a}
$$

From which

$$
(\overrightarrow{AC}, \overrightarrow{BD}) = \left(\vec{a} - \frac{31}{65} \vec{b}, \vec{b} - \frac{31}{65} \vec{a}\right) = -\frac{31}{65}\left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = -\frac{31}{65}|AB|^2 = -2015
$$

where the penultimate equality follows from the fact that triangle $AOB$ is a right triangle.