Comparing P1 and P2

We start by noticing that property

\(\mathcal{P'}_2:\) If \(\theta \in\,\,[0,\pi]\) is a rational multiple of \(\pi\) and it has a rational cosine, then \(\theta\in \{0,\pi,\pi/2,\pi/3,2\pi/3\}\)

is equivalent to the previously deduced (\(\mathcal{P}_2\)).

In fact, since \[\cos(\{0,\pi,\pi/2\})=\{1, -1, 0\}\] and \[\forall \,\theta \in \mathbb{R},\quad \quad \cos(\theta) = -\cos(\pi-\theta)\] we may restrict our analysis to angles in the interval \(]0,\pi/2[\).

Each of the properties

\(\mathcal{P}_1:\) If in a triangle the ratios between the lengths of the three sides and between the amplitudes of the three angles are rational, then the triangle is equilateral.

and \(\mathcal{P'}_2\) (or \(\mathcal{P}_2\)) can be shown directly one from the other.

Let us see why this equivalence holds. Let \(\theta\,\in \,\,]0,\pi/2[\) be a rational multiple of \(\pi\) which has a rational cosine. We built a triangle with sides \(1,\,1,\,2\cos\,\theta\) and angles \(\theta, \,\theta\) and \(\pi-2\,\theta\), as shown in the figure, which it is possible because \(0<\theta <\pi/2\).

Since \(\theta\) is a rational multiple of \(\pi\), these three angles are rational multiples of \(\pi\). We also know, from the hypothesis that \(\cos\,\theta\) is a rational number, that the quotients between the lengths of the sides of this triangle are rational. Then, we may apply \(\mathcal{P}_1\) and conclude that the triangle is equilateral, and thus that \(\theta =\pi/3\).