### Regular Decagon

Let \([ABCDEFGHIJ]\) be a regular decagon with centre \(O\) such that the lengths of the side and the longest diagonal are \(l\) and \(d\), respectively. Let \([AF]\), \([CH]\) and \([EJ]\) be three diagonals passing through the centre of the decagon and \([AH]\) and \([BE]\) two of the second shortest diagonals. Consider the points \(K\) and \(L\), produced by the intersection between the diagonals \([AH]\) and \([BE]\) with the diagonals \([EJ]\) and \([CH]\), respectively. We have \[J\hat{A}L=J\hat{A}H=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},\] \[L\hat{J}A=E\hat{J}A=\frac{1}{2}.4.36\,^{\circ}=72\,^{\circ}\] and \[J\hat{L}A=180\,^{\circ}-J\hat{A}L-L\hat{J}A=72\,^{\circ}=L\hat{J}A,\] so \([AJL]\) is an isosceles triangle, with \(\overline{AL}=\overline{AJ}=l\). Analogously, \([BCK]\) is also isosceles, with \(\overline{BK}=\overline{BC}=l\). Additionally, since these triangles are congruent, we have \(\overline{JL}=\overline{CK}\). The triangle \([LAO]\) is also isosceles, since \[L\hat{A}O=H\hat{A}F=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}\] and \[L\hat{O}A=J\hat{O}A=36\,^{\circ}=L\hat{A}O,\] so \(\overline{LO}=\overline{AL}=l\) and \[\overline{KO}=\overline{CO}-\overline{CK}=\overline{JO}-\overline{JL}=\overline{LO}=l.\] Therefore, \[\overline{AB}=\overline{BK}=\overline{KO}=\overline{OL}=\overline{LA}=l.\] On the other hand, we have \[L\hat{A}B=H\hat{A}B=\frac{1}{2}.6.36\,^{\circ}=108\,^{\circ},\] \[A\hat{B}K=A\hat{B}E=\frac{1}{2}.6.36\,^{\circ}=108\,^{\circ},\] \[B\hat{K}O=180\,^{\circ}-B\hat{K}C=108\,^{\circ},\] \[K\hat{O}L=C\hat{O}J=3.36\,^{\circ}=108\,^{\circ}\] and \[O\hat{L}A=180\,^{\circ}-J\hat{L}A=108\,^{\circ}.\] Therefore, the polygon \([ABKLO]\) has all its sides equal and all its interior angles equal, so it is a regular pentagon with side \(\overline{AB}=l\) and whose diagonal is \(\overline{AO}=\frac{1}{2}\overline{AF}=\frac{1}{2}d\). If the longest diagonal of \([ABCDEFGHIJ]\) and its side were commensurable, that is, if the ratio \(\frac{d}{l}\) were a rational number, then the ratio \(\frac{\frac{1}{2}d}{l}=\frac{1}{2}.\frac{d}{l}\) would also be rational and the diagonal of \([ABKLO]\) would be commensurable with its side, which we have already seen is not the case. Therefore, the longest diagonal and the side of the decagon \([ABCDEFGHIJ]\) are incommensurable magnitudes.

**Remark:** actually, it can be proved that the
ratio between the longest diagonal and the side of a \(2n\)-sides
regular polygon, with \(n\) an odd number larger than \(3\), is
always double the ratio between the longest diagonal and the side
of a \(n\)-sides regular polygon. The above case corresponds to
taking \(n=5\). To visualize the next case, consult this *app*.