Let $M$ be the midpoint of $EF$. By symmetry, $D$ lies on $CM$. Let $BM = x$. Then $FM = 5x, CF = 10x, CM = 5\sqrt{5}x$ and $BC = 2\sqrt{19}x$. It follows that $\frac{AB}{BC} = \frac{1}{19}$. Now $Q$ is the circumcentre of triangle $BCD$. Hence $\angle{BQD} = 2\angle{BCD} = \angle{BCA}$. Since both $QDB$ and $CAB$ are isosceles triangles, they are similar to each other. It follows that $\frac{BD}{QB} = \frac{AB}{BC} = \frac{1}{\sqrt{19}}$, so that $BD=\boxed{\frac{1}{\sqrt{19}}}$.