7. In trapezoid $ABCD$, the bases $AD=9, BC=2$, angles $A$ and $D$ at the base are $\operatorname{arctg} 4$ and $\operatorname{arctg}(2 / 3)$, respectively. Find the radius of the circle circumscribed around triangle $CBE$, where $E$ is the point of intersection of the diagonals of the trapezoid.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-28.jpg?height=496&width=874&top_left_y=1288&top_left_x=88)

$$
\begin{aligned}
& B F \perp A D ; C G \perp A D ; B F=C G=h, F G=B C . \\
& \text { Let } A F=x \Rightarrow \\
& x \cdot \operatorname{tg} A=((A D-B C)-x) \operatorname{tg} D \\
& 4 x=\frac{2}{3}(A D-B C-x) \\
& 14 x=2(A D-B C), x=\frac{A D-B C}{7} \\
& x=1, A G=x+B C=3, F D=A D-x=8 \\
& h=4 x=4, A C=\sqrt{A G^{2}+h^{2}}=5 \\
& B D=\sqrt{F D^{2}+h^{2}}=4 \sqrt{5}
\end{aligned}
$$

$\triangle B C E \approx \triangle A E D ; \quad \frac{B E}{E D}=\frac{C E}{A E}=\frac{B C}{A D}=\frac{2}{9} \Rightarrow S_{C B E}=\frac{1}{2} B C \cdot \frac{2}{11} h=\frac{8}{11} ; \quad B E=\frac{2}{11} B D=\frac{8 \sqrt{5}}{11} ;$

$E C=\frac{2}{11} A C=\frac{10}{11} ; \quad R_{o n}=\frac{B C \cdot B E \cdot E C}{4 S_{C B E}}=\frac{5 \sqrt{5}}{11}$.

Answer: $\frac{5 \sqrt{5}}{11}$.