AMC 12 2017b Problem 24

Quadrilateral $ABCD$ has right angles at $B$ and $C$, $\triangle ABC \sim \triangle BCD$, and $AB > BC$. There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB$. What is $\frac{AB}{BC}$?

\textbf{(A)}\ 1 + \sqrt{2} \textbf{(B)}\ 2 + \sqrt{2} \textbf{(C)}\ \sqrt{17} \textbf{(D)}\ 2 + \sqrt{5} \textbf{(E)}\ 1 + 2\sqrt{3}