### Regular Decagon

Consider a regular decagon for which the points \(A\), \(B\), \(C\) and \(D\) are consecutive vertices, \(d\) is the length of its second shortest diagonal and \(l\) is the length of the side. If we assume that \(d\) and \(l\) are commensurable, we have \(d=mx\) and \(l=nx\), where \(m\) and \(n\) are positive integers and \(x\) is a common measure for the second shortest diagonal and the side of the decagon. Note that each side is a chord of the circle circumscribing the polygon, corresponding to an arc measuring \(\frac{1}{10}.360\,^{\circ}=36\,^{\circ}\). Choose the point \(E\) on the diagonal \([AD]\) such that \(\overline{AE}=\overline{AB}=l\). Then, as \([AEB]\) is isosceles and \[E\hat{A}B=D\hat{A}B=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ},\] we get \[A\hat{E}B=A\hat{B}E=\frac{1}{2}.(180\,^{\circ}-36\,^{\circ})=72\,^{\circ}.\] On the other hand, \[C\hat{B}E=C\hat{B}A-A\hat{B}E=\frac{1}{2}.8.36\,^{\circ}-72\,^{\circ}=144\,^{\circ}-72\,^{\circ}=72\,^{\circ}=A\hat{E}B,\] so the diagonal \([AD]\) is parallel to the side \([BC]\). Now choose the point \(F\) on the diagonal \([AD]\) such that \(\overline{EF}=\overline{BC}=l\). Then, since \([EF]\) is parallel to \([BC]\), \([EBCF]\) is a parallelogram, so \[D\hat{F}C=D\hat{E}B=180\,^{\circ}-A\hat{E}B=108\,^{\circ}.\] Since \[A\hat{C}D=\frac{1}{2}.7.36\,^{\circ}=126\,^{\circ}>108\,^{\circ}=D\hat{F}C,\] we choose the point \(G\) in the half-line \(DC\) such that \(D\hat{F}G=126\,^{\circ}\). As \(A\hat{D}C=G\hat{D}F\) and \(A\hat{C}D=D\hat{F}G\), we get that the triangles \([ACD]\) and \([DFG]\) are similar, so \([DG]\) is the second shortest diagonal of a regular decagon with side \([DF]\). On the other hand, we have \[C\hat{G}F=D\hat{G}F=D\hat{A}C=\frac{1}{2}.36\,^{\circ}=18\,^{\circ}\] and \[C\hat{F}G=D\hat{F}G-D\hat{F}C=126\,^{\circ}-108\,^{\circ}=18\,^{\circ},\] so \([FCG]\) is isosceles, with \(\overline{CG}=\overline{CF}\). We also have \[F\hat{D}C=A\hat{D}C=\frac{1}{2}.2.36\,^{\circ}=36\,^{\circ}\] and \[F\hat{C}D=C\hat{G}F+C\hat{F}G=36\,^{\circ},\] so \([DFC]\) is also isosceles, with \(\overline{CF}=\overline{DF}\). Therefore, we can construct a regular decagon passing through the points \(D\), \(F\) and \(G\), and whose side and second shortest diagonal have lengths given by: \[l_{1}=\overline{DF}=\overline{AD}-\overline{AE}-\overline{EF}=d-2l\] \[d_{1}=\overline{DG}=\overline{DC}+\overline{CG}=\overline{DC}+\overline{DF}=l+(d-2l)=d-l\]

As \(d=mx\) and \(l=nx\), we have: \[l_{1}=mx-2nx=(m-2n)x=n_{1}x\] \[d_{1}=mx-nx=(m-n)x=m_{1}x,\] where \(n_{1}=m-2n\) and \(m_{1}=m-n\) are two positive integer numbers, with \(m_{1}<m\). Proceeding in the same way, we obtain a new regular decagon whose second shortest diagonal is \(d_{2}=m_{2}x\) and whose side is \(l_{2}=n_{2}x\), where \(n_{2}=m_{1}-2n_{1}\) and \(m_{2}=m_{1}-n_{1}\) are two positive integer numbers, with \(m_{2}<m_{1}<m\).

Since we can go on indefinitely constructing new regular
decagons, we shall obtain a strictly decreasing sequence of
positive integer numbers:
\[m_{1}>m_{2}>m_{3}>m_{4}>...\] such that
\(m_{i}<m,\:\forall i\in\mathbb{N},\) which is absurd, since
there cannot be an infinite number of distinct positive integers
less than \(m\) (there exist exactly \(m-1\) such numbers: \(1\),
\(2\), \(3\),...,\(m-2\) and \(m-1\)). Therefore, the second
shortest diagonal and the side of the regular decagon are
incommensurable magnitudes.