### Regular Decagon

Let \([ABCDEFGHIJ]\) be a regular decagon with centre \(O\) such that the lengths of the side, the longest diagonal and the second shortest diagonal are \(l\), \(d\) and \(d'\), respectively. Let \([AF]\) and \([CH]\) be two diagonals passing through the centre of the decagon, \([AD]\) the second shortest diagonal joining the vertices \(A\) and \(D\), and \(K\) the point of intersection of the diagonal \([AD]\) with the diagonal \([CH]\). Then, \[C\hat{D}K=C\hat{D}A=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}\] and \[K\hat{C}D=H\hat{C}D=\frac{1}{2}.4.36\,^{\circ}=72\,^{\circ},\] so that \[C\hat{K}D=180\,^{\circ}-C\hat{D}K-K\hat{C}D=72\,^{\circ}\] and \([KCD]\) is an isosceles triangle, with \(\overline{KD}=\overline{CD}=l\). In addition, \[O\hat{A}K=F\hat{A}D=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},\] \[A\hat{K}O=C\hat{K}D=72\,^{\circ}\] and \[A\hat{O}K=180\,^{\circ}-O\hat{A}K-A\hat{K}O=72\,^{\circ},\] so that \([AOK]\) is an isosceles triangle, with \(\overline{AK}=\overline{AO}\). So, the length of the longest diagonal is given by: \[d=\overline{AF}=2\overline{AO}=2\overline{AK}=2\left(\overline{AD}-\overline{KD}\right)=2(d'-l)\]

Hence we have: \[\frac{d}{l}=2\left(\frac{d'}{l}-1\right)\]

Equivalently, we have: \[\frac{d'}{l}=\frac{1}{2}.\frac{d}{l}+1\]

Therefore, \(\frac{d}{l}\) is a rational number if and only if \(\frac{d'}{l}\) is also a rational number, which we already saw is not the case. So, the longest diagonal and the side of the decagon \([ABCDEFGHIJ]\) are incommensurable magnitudes.