<s>[INST] 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? :)
 [/INST] 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
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