smc

Let $CD=1$, $BC=x$, and $AB=x^2$. Note that $AB/BC=x$. By the Pythagorean Theorem, $BD=\sqrt{x^2+1}$. Since $△BCD\sim △ABC\sim △CEB$, the ratios of side lengths must be equal. Since $BC=x$, $CE=\frac{x^2}{\sqrt{x^2+1}}$ and $BE=\frac{x}{\sqrt{x^2+1}}$. Let F be a point on $\overline{BC}$ such that $\overline{EF}$ is an altitude of triangle $CEB$. Note that $△CEB\sim △CFE\sim △EFB$. Therefore, $BF=\frac{x}{x^2+1}$ and $CF=\frac{x^3}{x^2+1}$. Since $\overline{CF}$ and $\overline{BF}$ form altitudes of triangles $CED$ and $BEA$, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle $BEC$ can be calculated, as it is a right triangle. Solving for each of these yields: $$\begin{array}{rl}[BEC]& =[CED]=[BEA]=\frac{x^3}{2(x^2+1)}\\ [ABCD]& =[AED]+[DEC]+[CEB]+[BEA]\\ \frac{(BC)(AB+CD)}{2}& =17\ast [CEB]+[CEB]+[CEB]+[CEB]\\ \frac{x^3+x}{2}& =\frac{20x^3}{2(x^2+1)}\\ \frac{x}{x^2+1}& =\frac{20x^3}{x^2+1}\\ (x^2+1{)}^2& =20x^2\\ x^4-18x^2+1& =0⟹x^2=9+4\sqrt{5}=4+2(2\sqrt{5})+5\end{array}$$ Therefore, the answer is $\boxed{2+\sqrt{5}}$