Problem 1221

5.22. Let $B\left(x_{1} ; y_{1}\right), C\left(x_{2} ; y_{2}\right), D\left(x_{3} ; y_{3}\right)$ be the unknown vertices of the trapezoid (Fig. 5.12). Since point $E(6 ;-1)$ is the midpoint of base $A B$, we have the system $\frac{3+x_{1}}{2}=6, \frac{0+y_{1}}{2}=-1$, from which $x_{1}=9$,

!

Fig. 5.12

$y_{1}=-2$, i.e., $B(9 ;-2)$. But $B C \| O y$ and, therefore, $x_{2}=9$, and the equation of $B C$ is $x=9$. Next, point $F(7 ; 2)$ is the midpoint of base $D C$, from which $\frac{x_{3}+9}{2}=7$, i.e., $x_{3}=5$. The equation of line $A B$ is $y+1=k(x-6)$; since $A \in(A B)$, we have $0+1=k(3-6) \Rightarrow k=-\frac{1}{3}$. Therefore, $k_{D C}=-\frac{1}{3}$, and we obtain the equation of $D C$:

$$y-2=-\frac{1}{3}(x-7), \text { or } y=-\frac{1}{3} x+\frac{13}{3}$$

Solving the system of equations of lines $B C$ and $D C$, i.e., $x=9$ and $y=-\frac{1}{3} x+\frac{13}{3}$, we find $y=\frac{4}{3}$, i.e., $C\left(9 ; \frac{4}{3}\right)$. Finally, the ordinate of point $D$ is found from the equality $\frac{y_{3}+\frac{4}{3}}{2}=2$, from which $y_{3}=\frac{8}{3}$, i.e., $D\left(5 ; \frac{8}{3}\right)$. Thus, $B C=y_{2}-y_{1}=\frac{4}{3}+2=\frac{10}{3}$, $A D=\sqrt{2^{2}+\left(\frac{8}{3}\right)^{2}}=\frac{10}{3}$, i.e., $B C=A D$ and, therefore, trapezoid $A B C D$ is isosceles.

Let $\angle D A B=\alpha$ and use the formula $\cos \alpha=\frac{\overline{A D} \cdot \overline{A B}}{|\overline{A D}||\overline{A B}|}$. We have $\overline{A D}\left(2 ; \frac{8}{3}\right), \overline{A B}(6 ;-2),|\overline{A B}|=2 \sqrt{10}$, and, therefore,

$$\cos \alpha=\frac{2 \cdot 6+\frac{8}{3}(-2)}{\frac{10}{3} \cdot 2 \sqrt{10}}=\frac{1}{\sqrt{10}}$$

Answer: $\arccos \frac{1}{\sqrt{10}}$.