Finding the area of a triangle, given the distance between center of incircle and circumscribed circle Consider the following depiction:

$ABC$ is an isosceles triangle ($AB=AC$), where the two angles opposite the equal sides are equal $\beta$ ($\beta>60$), and $AD$ perpendicular to $BC$.
$O$ is the center of the circumscribed circle of $ABC$, and $M$ is the center of the incircle of $ABC$.
I want to express the area of the triangle $BOM$ and the radius of the incircle of $ABC$, in terms of $k$ and $\beta$.  
This is what I tried: I denoted the radius of the circumscribed circle as $R$, and the radius of the incircle as $r$.
The area of  $BOM$ can be obtained by: $\frac{k\cdot BD}{2}$, and $BD=\frac{BC}{2}$, I get that the area is actually:  $\frac{k\cdot BC}{4}$.
Now from Law of sines I get $BC=\sin(180-2\beta)\cdot 2 R$.
So this reduces to problem to expressing $R$ (which is $OA$) in terms of $\beta$ and $k$. This is where I'm stuck. I couldn't find a way to do that.
I know that $OA=OB=R$, and that $AOB$ is also isosceles triangle, and I have a feeling it is important, but I can't figure out what am I missing...
Hints and guidelines would be very appreciated. 

Assuming $BD=1$, we have $AB=AC=\frac{1}{\cos\beta}$, $[ABC]=\tan\beta$, 
$$ R = \frac{AB}{2\sin\beta} = \frac{1}{\sin(2\beta)}$$
and:
$$ r = \frac{2[ABC]}{AB+AC+BC} = \frac{\tan\beta}{1+\frac{1}{\cos\beta}}=\frac{\sin\beta}{1+\cos\beta}=\tan\frac{\beta}{2}. $$
Moreover, by Euler's theorem we have:
$$ k^2 = OM^2 = R^2-2Rr = \frac{1}{4\sin^2\beta\cos^2\beta}-\frac{1}{\cos\beta(1+\cos\beta)} $$
so:
$$ k^2 = \frac{(1-2\cos\beta)^2}{4\cos^2\beta(1-\cos^2\beta)} $$
and:
$$ k = \frac{1-2\cos\beta}{\sin(2\beta)}. $$
This gives, in general,
$$ BD = \frac{\sin(2\beta)}{1-2\cos\beta} k, $$
so:
$$ [BOM]=\frac{k}{2}BD = \frac{\sin\beta\cos\beta}{1-2\cos\beta} k^2,\tag{1} $$
and:
$$ r = \frac{\sin\beta}{1+\cos\beta} BD = \frac{2(1-\cos\beta)\cos\beta}{1-2\cos\beta} k.\tag{2}$$
