$[$ Auxiliary similar triangles  Problem 111049 " $\quad$ Triangle inequality (other).  [ Trapezoids (other).  [ Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part

On the base $B C$ of trapezoid $A B C D$, a point $E$ is taken, lying on the same circle with points $A, C$, and $D$. Another circle, passing through points $A, B$, and $C$, is tangent to the line $C D$. Find $B C$, if $A B=12$ and $B E: E C=4$ : 5. Find all possible values of the ratio of the radius of the first circle to the radius of the second under these conditions.
By the theorem on the angle between a tangent and a chord, $\angle A C D=\angle A B C$, and since $B C \| A D$, then $\angle C A D=\angle A C B$. Therefore, triangles $A C D$ and $C B A$ are similar by two angles. Hence, $\angle B A C=\angle C D A$. By the converse of the theorem on the angle between a tangent and a chord, the line $A B$ is a tangent to this circle. Let $B E=4 t$, $E C=5 t$. By the theorem on the tangent and the secant, $A B^{2}=B C \cdot B E$, or $12^{2}=36 t$. From this, $t=2, B C=9 t=18$.

For the conditions to be met, it is necessary and sufficient for the existence of triangle $A B C$ with sides $A B=12$ and $B C=18$ (point $D$ is restored from the equality of corresponding angles, which implies both the parallelism of lines $A D$ and $B C$, and the tangency of lines $A B$ and $C D$ to the corresponding circles, and the division of the secant $B C$ by point $E$ in the required ratio), except for the case $B C=A C$ (when $A C=A D$, and $A B C D$ is a parallelogram). The existence of such a triangle is equivalent to the condition $B C-A B<A C<B C+A B$, or $18-12<A C<18+12$. Therefore,
the ratio of the radii of the circumcircles of similar triangles $A C D$ and $A B C$ satisfies

the relations $1 \neq \frac{r_{1}}{r_{2}}=\frac{A C}{B C} \in\left(\frac{18-12}{18}, \frac{18+12}{18}\right)=\left(\frac{1}{3}, \frac{5}{3}\right)$.

## Answer