Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_6d5fca905c76f828a3d8g-26.jpg?height=460&width=828&top_left_y=732&top_left_x=617)

Fig. 1

Let the point of intersection of the diagonals be $O$ (Fig. 1).

Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.

From the similarity of triangles $A O B$ and $D O C$, we have:

$\frac{A O}{O C}=\frac{41}{24}$ and $\overrightarrow{O C}=\frac{24}{41} \overrightarrow{A O}=\frac{24}{41} \vec{a}$

$\frac{B O}{O D}=\frac{41}{24}$ and $\overrightarrow{O D}=\frac{24}{41} \overrightarrow{B O}=\frac{24}{41} \vec{b}$

Then

$\overrightarrow{A D}=\vec{a}+\frac{24}{41} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{24}{41} \vec{a}$

Let's find the dot product

$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{24}{41} \vec{b}\right) \cdot\left(\vec{b}+\frac{24}{41} \vec{a}\right)=\vec{a} \vec{b}+\frac{24}{41} \vec{a} \cdot \vec{a}+\frac{24}{41} \vec{b} \cdot \vec{b}+\left(\frac{24}{41}\right)^{2} \vec{a} \vec{b}$

Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.

By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$.

Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=41^{2}$.

Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{24}{41}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{24}{41} \cdot 41^{2}=984$.

Answer. 984.