Quadrilateral $ABCD$ has right angles at $B$ and $C$, $△ABC\sim △BCD$, and $BC>AB$. There is a point $E$ in the interior of $ABCD$ such that $△ABC\sim △CEB$ and the area of $△AED$ is $3$ times the area of $△CEB$. What is $\frac{BC}{AB}$?