Find area of rhombus Given the following rhombus, where points E and F divide the sides CD and BC respectively, AF = 13 and EF = 10

I think the length of the diagonal BD is two times EF = 20, but i got stuck from there.

"Straightforward" algebraical solution, in which one don't have to think:
Consider a vector basis $AB$,$AD$ and $A$ to have coordinates $(0,0)$,
$AE=AB+BC/2$ , $EF=1/2AD - 1/2BC$, we're given
$AE^2=(AB+BC/2)^2=AB^2+(AB,BC)+BC^2/4=13^2$,
$EF^2=1/4AB^2 - 1/2(AB,BC) + 1/4 BC^2 = 10^2$.
These are linear equations over $AB^2=BC^2$ and $(AB,BC)$ so we obtain $AB^2$ and $(AB,BC)$ (namely, $AB^2=164$ and $(AB,BC)=-36$) and then $S=|AB||BC|\sin \angle DAB = AB^2\sqrt{1-\cos^2 \angle DAB} = \sqrt{AB^4 - (AB,BC)^2}=\sqrt{164^2-36^2}=160$.
