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.
 A: 

I hope those hints will help you.
A: $CE/CD=CF/CB=2$
$\angle ECF=\angle DCB$
These triangles are similar. Including BD parallel to FE => $BD/FE=2$ => $BD=20$ Diagonals in a rhombus are always perpendicular. $X=FE \cap AC$, but FE perpendicular to the AC. => AXF right-angled triangle. $AF=13$ $FX=10/2=5$ from the similarity. => $AX=(169-25)^{1/2}=12=3/4*AC$ from the similarity. => $S=12/(3/4)*20/2=160$
A: "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 $S=|AB||BC|\sin \angle DAB = AB^2\sqrt{1-\cos^2 \angle DAB} = \sqrt{AB^4 - (AB,BC)^2}=\sqrt{164^2-36^2}=160$.
A: I am assuming that the midpoint of $CD$ is $E$ and not $F$. Let $BF=CF=CE=a$. We then have $AB=2a$. If $\angle{ECF} = \theta$, we have $\angle{ABF} = \pi-\theta$.
Applying cosine rule for triangles $ABF$ and $CEF$, we obtain
\begin{align}
\cos(\theta) & = \dfrac{a^2+a^2-10^2}{2\cdot a \cdot a}\\
\cos(\pi-\theta) & = \dfrac{a^2+(2a)^2-13^2}{2\cdot a \cdot (2a)}
\end{align}
Adding both the equations, we obtain
$$\dfrac{2a^2-10^2}{2a^2} + \dfrac{5a^2-13^2}{4a^2} = 0 \implies 1-\dfrac{50}{a^2} + \dfrac54 - \dfrac{169}{4a^2} = 0 \implies \dfrac94 - \dfrac{369}{4a^2} = 0$$
This gives us the side to be $$2a = 2\sqrt{41}$$
