Solution:

Let $x$ be $BE$ and $y$ be $AE$. Note that $AEFD$ is a square so $AE = BC = y$. Also, $AB = BE + AE$ so $AB = x + y$. Since $\frac{AB}{BE} = \frac{BE}{BC}$ then $\frac{x + y}{x} = \frac{x}{y}$. Thus, we have $xy + y^2 = x^2$ which yields $x^2 - x y - y^2 = 0$. Solving for $x$ using the quadratic formula gives us $x = \frac{y \pm \sqrt{y^2 - 4(1)(-y^2)}}{2} = \left(\frac{1 \pm \sqrt{5}}{2}\right) y$. However, we will only take $x = \left(\frac{1 + \sqrt{5}}{2}\right) y$ since the other solution will mean that $x < 0$ which is absurd since $x$ is a measure of length. Thus, $\frac{AB}{BC} = \frac{x + y}{y} = \frac{\left(\frac{1 + \sqrt{5}}{2}\right) y + y}{y} = \frac{1 + \sqrt{5} + 2}{2} = \frac{3 + \sqrt{5}}{2}$. Therefore, the answer is $\frac{3 + \sqrt{5}}{2}$.