Solution:

Let $F$ be the intersection of the two diagonals of the trapezoid and let $M$ be the midpoint of $AB$. Since triangle $AOB$ is isosceles and $OM$ is its median with respect to the base, triangles $AOM$ and $BOM$ are congruent and, in particular, $\widehat{AOM} = \widehat{BOM} = \frac{1}{2} \widehat{AOB}$. Moreover, considering the circle circumscribed about triangle $ABC$, the angles $\widehat{AOB}$ and $\widehat{ACB}$ are respectively the central angle and the inscribed angle subtending the same arc $AB$, so that $\widehat{AOB} = 2 \cdot \widehat{ACB}$ and hence $\widehat{BOM} = \widehat{ACB}$.

It follows that triangles $OBM$ and $CBF$, which are right triangles, have an equal acute angle and therefore the other acute angle is also equal, that is $\widehat{ABE} = \widehat{DBC}$. Furthermore, the angles $\widehat{ABE}$ and $\widehat{BEC}$ are equal because they are alternate interior angles, and so we also have $\widehat{DBC} = \widehat{BEC}$.

Finally, triangles $BCD$ and $ECB$ are similar, because they share the angle at $C$ and have another equal angle. We then have $\overline{BC} : \overline{CD} = \overline{EC} : \overline{CB}$, that is $\overline{BC}^2 = \overline{CD} \cdot \overline{CE}$.