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 $[AED]=5[CEB]$. What is $\tfrac{AB}{BC}$?

1. Let $k=\frac{AB}{BC}>1$. The similarities give $\dfrac{[AED]}{[CEB]}=k^2-1+\dfrac1{k^2}$.
2. With $[AED]=5[CEB]$, we get $k^2+\dfrac1{k^2}=6$, so $k+\dfrac1k=2\sqrt2$.
3. The root with $k>1$ is $k=1+\sqrt2$.