8. A circle with center $O$ is inscribed in a rectangular trapezoid with bases $B C$ and $A D$, where $B C: A D=2: 3$. The lines $D O$ and $B C$ intersect at point $E$. In what ratio does the tangent to the circle, drawn from point $E$, divide the side $A D$ of the trapezoid, counting from point A.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_9fc9f23f10a928994848g-03.jpg?height=443&width=677&top_left_y=2097&top_left_x=724)

Let $BC = 2a, AD = 3a$. We need to find the radius of the inscribed circle in the trapezoid. The center of the circle, point $O$, is the intersection of the angle bisectors of the trapezoid. The angle bisectors of the angles at vertices $C$ and $D$ intersect at a right angle, so triangle $COD$ is a right triangle with height $OL$, equal to the radius of the circle $r$. Therefore: 
\[ r^2 = (2a - r)(3a - r), \quad r = \frac{6a}{5} \]

Let the tangent to the circle from point $E$ intersect side $AD$ at point $N$. Then $\angle NEO = \angle OEC$ (since $EO$ is the angle bisector), and $\angle OEC = \angle ADO$, so triangle $END$ is isosceles, and therefore $NO$ is both the angle bisector and the altitude.

Consider the right triangle $NOD$ with height $OH$:

\[
\begin{gathered}
OH^2 = NH \cdot HD \\
\left(\frac{6a}{5}\right)^2 = NH \cdot \left(3a - \frac{6a}{5}\right) \\
NH = \frac{4a}{5} \Rightarrow AN : NH = 1 : 2 \\
AN : ND = \frac{2a}{5} : \left(3a - \frac{2a}{5}\right) = 2 : 13
\end{gathered}
\]

Answer: $2: 13$.