Hong Kong competition problems, 2012

We have $\frac{DB}{DC} = \frac{7}{2}$.
Firstly, since $\angle EDB = 90^\circ$ and $BF = EF$, we have $DF = BF$. This implies $\angle FDB = \angle FBD = \theta$, and hence $\angle DFE = 2\theta$. It follows that $C$ lies on the segment $BD$, and
$$\angle CDF = 3\theta - 2\theta = \theta.$$
Let $P$ be the intersection point of $AD$ and $FC$, and let $M$ be the midpoint of $BD$. Note that $FM \perp BD$ since $\triangle FBD$ is isosceles. This shows
$$\angle DMF = 90^\circ = \angle DPF.$$
Therefore, $F, M, P, D$ are concyclic. Since $\angle CDF = \angle CFD$, $FMPD$ is an isosceles trapezoid.
Now, applying Menelaus' theorem to $\triangle BCF$, we obtain
$$\frac{BD}{DC} \times \frac{CP}{PF} \times \frac{FA}{AB} = 1.$$

Note that $BD = 2MD = 2PF$ and $\frac{FA}{AB} = \frac{2}{3}$. Therefore, we get $\frac{CP}{DC} = \frac{3}{4}$. This implies $\frac{CM}{DC} = \frac{CP}{DC} = \frac{3}{4}$, and hence
$$\frac{DB}{DC} = \frac{2DM}{DC} = \frac{2(DC + CM)}{DC} = \frac{7}{2}.$$