A geometric proof about decagons. Let us have a regular decagon. Prove that the radius of the circle that can be drawn around it (if I have drawn correctly, that's the blue line) equals $|ad|-|ab|$.

How can I prove this statement, any ideas? :)
 A: Geometrically:

Angles $ a,b,c $ of triangle 
$\begin{align}
a &= (\pi - 3\pi/5)/2 = \pi/5 \\
b & =  2\pi/5\\
\therefore c &= \pi - (a+b) =  2\pi/5\\
\end{align}$
so $\Delta a b c$ is isosceles and so is the similar triangle above, giving the result.
A: WLOG you can take the radius to be unity. Using the Cosine Law we can easily say that $AD=2\sin\dfrac{3\pi}{10}$ and $AB=2\sin\dfrac{\pi}{10}$ . Then the difference $AD-AB:$
$$2\left(\sin\dfrac{3\pi}{10}-\sin\dfrac{\pi}{10}\right)=4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}={4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}\cos\dfrac{\pi}{10}\over\cos\dfrac{\pi}{10}}$$
$$={2\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{5}\over\cos\dfrac{\pi}{10}}={\sin\dfrac{2\pi}{5}\over \cos\dfrac{\pi}{10}}=1$$
QED
A: We know that the angle subtended at the centre by $ab$ is $\sin{\pi/5}$ so its length is  $2r\sin{\pi/10}$. Similarly  the length of $ad$ is $2r\sin{3\pi/10}$. 
Define $k = \pi/10$, and note that $\sin(x) = \cos(5k-x)$.
Then 
$$\begin{align}|ad|-|ab| &= 2r (\sin{3k}-\sin{k}) \\ 
&= 2r (\sin(2k+k)-\sin(2k-k))\\
&= 2r (2\cos 2k \sin k)\\
&= 2r (\cos 2k \cdot 2\sin k\cos k)/(\cos k)\\
&= r (2\cos 2k \sin 2k)/(\cos k)\\
&= r (\sin 4k)/(\cos k)\\
&= r \\
\square
\end{align}$$
