Area of rhombus given circumradii of its contained triangles Here is a question I came across

Find the area of rhombus ABCD given that the radii of the circles
  circumscribed around triangles ABD and ACD are 12.5 and 25,
  respectively.

With the diagonals of the rhombus perpendicularly bisecting each other, let $a$ be half of diagonal BD and $b$ half of diagonal AC. Thus, the sides of the rhombus are $\sqrt{a^2 + b^2}$ long. 
The area of any triangle can be expressed as $\frac{\text{product of side lengths}}{\text{4 multiplied by the circumradius}}$. Thus, the area $\Delta ABC$ is $\frac{2b (a^2 + b^2)}{4(12.5)}$, while the area of $\Delta ABD$ is $\frac{2a (a^2 + b^2)}{4(25)}$. These two areas are equal since both of the said triangles are half of the rhombus. Setting these two expressions equal to each other and simplifying gives $b = 2a$. 
I've been stuck here for the past few hours. I'm blatantly missing something here. What's the next step?
 A: we denote the Areas of the triangles by $A_1$ and $A_2$ and the diagonals by $e$ and $f$: then we have $$A_1=\frac{a^2f}{50}$$ and $$A_2=\frac{a^2e}{100}$$ also we get
$$A_1=\frac{1}{2}a^2\sin(\alpha)$$ and $$A_2=\frac{1}{2}a^2\sin(180^{\circ}-\alpha)$$
from here we get $$f=\frac{1}{2}e$$ with $$e^2+f^2=4a^2$$ we get $$e=\frac{4}{\sqrt{5}}a$$ and $$f=\frac{2}{\sqrt{5}}a$$ and then we have $$\tan\left(\frac{\alpha}{2}\right)=\frac{1}{2}$$ can you finish now?
A: 
Let $\angle DAB=\alpha$, $R_1=12.5$, $R_2=2R_1=25$.
Using the general formula for the area of triangle $ABC$
in terms of its curcumradius $R$ and sinuses of its angles $A,B,C$
\begin{align} 
S_{ABC}&=2R^2\sin A\sin B\sin C
\tag{1}\label{1}
,
\end{align} 
we have
\begin{align} 
S_{ABCD}&=2S_{ABD}=2S_{ACD}
\tag{2}\label{2}
,\\
S_{ABD}&=2R_1^2\sin\alpha\sin^2(90^\circ-\tfrac\alpha2)
=2R_1^2\sin\alpha\cos^2\tfrac\alpha2
\tag{3}\label{3}
,\\
S_{ACD}&=2R_2^2\sin(180^\circ-\alpha)\sin^2\tfrac\alpha2
=2R_2^2\sin\alpha\sin^2\tfrac\alpha2
\tag{4}\label{4}
,
\end{align} 
hence we must have
\begin{align}
\tan\tfrac\alpha2&=\frac{R_1}{R_2}=\frac{12.5}{25}=\frac12
,
\end{align}
and the rest is straightforward:
\begin{align}
\sin\alpha&=\frac{2\tan\tfrac\alpha2}{1+\tan^2\tfrac\alpha2}=\frac45
. 
\end{align}
Since given $R_2=2R_1$, we can rewrite \eqref{4} as
\begin{align} 
S_{ACD}
&=
2(2R_1)^2\sin\alpha\sin^2\tfrac\alpha2
=
8R_1^2\sin\alpha\sin^2\tfrac\alpha2
\tag{6}\label{6}
,
\end{align} 
and avoid calculation of 
$\sin\tfrac\alpha2$
and
$\cos\tfrac\alpha2$
\begin{align}
4S_{ABD}+S_{ACD}&=5S_{ABD}
\\
&=8R_1^2\sin\alpha(\sin^2\tfrac\alpha2+\cos^2\tfrac\alpha2)
=8R_1^2\cdot\frac45
,\\
S_{ABD}&=\frac{32R_1^2}{25}=200
,
\end{align}
and the answer is
\begin{align}
S_{ABCD}&=400
.
\end{align}
