Question 24
Quadrilateral $ABCD$ has right angles at $B$ and $C,$ $\triangle ABC \sim \triangle BCD,$ and $AB \gt BC.$ 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 $\dfrac{AB}{BC}?$