Problem 4. The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 155 and 13, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
Answer: 2015.

Solution. Let $O$ be the intersection point of diagonals $AC$ and $BD$. From the similarity of triangles $AOB$ and $COD$, it follows that $\overrightarrow{OC} = \frac{13}{155} \overrightarrow{AO}$, and $\overrightarrow{OD} = \frac{13}{155} \overrightarrow{BO}$.

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

$$
\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD} = \vec{a} + \frac{13}{155} \vec{b}, \quad \overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = \vec{b} + \frac{13}{155} \vec{a}
$$

From which

$$
(\overrightarrow{AD}, \overrightarrow{BC}) = \left(\vec{a} + \frac{13}{155} \vec{b}, \vec{b} + \frac{13}{155} \vec{a}\right) = \frac{13}{155} \left(|\vec{a}|^2 + |\vec{b}|^2\right) + (\ldots) \cdot (\vec{a}, \vec{b}) = \frac{13}{155} |AB|^2 = 2015
$$

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