Write $\angle BAC = \alpha$. Since $ABF$ is a right triangle and $E$ is the midpoint of the hypotenuse, it is also the circumcentre of the triangle $ABF$ and $|AE| = |BE| = |FE|$. So, $\angle AFE = \angle EAF = \alpha$ and $\angle EFB = \frac{\pi}{2} - \angle AFE = \frac{\pi}{2} - \alpha$, which implies $\angle FBD = \frac{\pi}{2} - \angle EFB = \alpha$. Also, $\angle ACB = \frac{\pi}{2} - \alpha$, so $\angle CBF = \alpha$ and $\angle DBA = \angle BAC = \alpha$. We see that $\frac{\pi}{2} = \angle CBA = \angle DBA + \angle FBD + \angle CBF = 3\alpha$, which implies $\alpha = \frac{\pi}{6}$. The triangle $ABC$ is one half of an equilateral triangle, so $\frac{|AB|}{|BC|} = \sqrt{3}$.