Find the ratio!

Thanks, @Noel Lo for the diagram. Draw a line $DG$ parallel to $AC$ . Then we note that $\triangle EFC$ is similar to $\triangle DFG$ . Therefore, $\dfrac {DG}{CE} = \dfrac {DF}{EF} = \dfrac {DE+EF}{EF} = 3$ , $\implies DG = 3CE = 3a$ , where $CE=a$ .

We also note that $\triangle DBG$ is similar to $\triangle ABC$ . Therefore, $\dfrac {BD}{DG} = \dfrac {AB}{AC} = 2$ , $\implies BD = 2DG = 6a$ .

Now, we have $\dfrac {BD}{CE} = \dfrac {6a}a = \boxed{6}$ .

Use similar triangles

Consider a point $G$ on $BC$ so that $EG$ is parallel to $AB$ . Then $\angle EGC=\angle ABC$ as they are corresponding angles and $\angle ECG=\angle ACB$ as they are common angles. Therefore,

$\frac{GE}{CE}=\frac{AB}{AC}=2$

Moreover, $\angle BDF=\angle GEF$ as they are corresponding angles and $\angle BFD=\angle GFE$ as they are common angles. Therefore,

$\frac{BD}{GE}=\frac{DF}{EF}=\frac{DE+EF}{EF}=\frac{2EF+EF}{EF}=2+1=3$

Finally, $\frac{BD}{CE}=\frac{BD}{GE} \times \frac{GE}{CE} = 3 \times 2 = \boxed{6}$

Alternatively, consider a point $H$ on $BC$ so that $DH$ is parallel to $BC$ . Then $\angle BDH=\angle BAC$ as they are corresponding angles and $\angle DBH=\angle ABC$ as they are common angles. Therefore,

$\frac{BD}{DH}=\frac{AB}{AC}=2$

Moreover, $\angle DHF=\angle ECF$ as they are corresponding angles and $\angle DFH=\angle EFC$ as they are common angles. Therefore,

$\frac{DH}{CE}=\frac{DF}{EF}=\frac{DE+EF}{EF}=\frac{2EF+EF}{EF}=2+1=3$

Finally, $\frac{BD}{CE} = \frac{BD}{DH} \times \frac{DH}{CE} = 2 \times 3 = \boxed{6}$