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.

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 $$2a = 2\sqrt{41}$$
