For reference: In the hexagon equiangle ABCDEF in which: $DE= 2BC$ and $EF = 2CD$.
Calculate $BF$, and we also know that BD =$\sqrt3$
Resolution by geometry
My progress: Try to make the figure as close as possible in function of the data...I don't know if it's correct...and I drew some auxiliary lines
$\triangle IJK$ is equlateral
Get lost in this exercise...I can't see much useful relationships
Hint:
$\triangle DEF \sim \triangle BCD$ and $DF = 2 DB$
Also, $\angle BDF = 60^0$. Can you show how?
As $DF$ is twice $DB$ and angle between them is $60^\circ$, what does that tell about $\triangle DBF$ and hence what is the value of $BF$?