### Introduction

Given a plane triangle \(\triangle ABC\), the equality between the angles \(\angle B=\angle A\) is equivalent to a simple relation between opposite sides to these angles, \(|AC| = |BC|\). What if \(\angle B = n \angle A\) for some natural number \(n\)?

This question was considered by Euler in 1765 in a 35 pages paper [1], which is cataloged in the Eneström index as E324. Euler begins by studying some particular cases, from \(n=1\) to \(n=5\), and realized the existence of a pattern in the equations obtained. However he only presents the cases for \(n\) up to \(13\) and doesn’t present a complete proof of the equivalences that he claims to be true. Therefore, it is worth returning to this problem.

First suppose \(\angle B = \angle A\) and consider the line segment \(CP\) perpendicular to the line \(AB\). This way we have two similar right triangles, \(\triangle APC\) and \(\triangle BPC\), with angles \(\angle A,\frac{\pi}{2}\) and \(\frac{\pi}{2} - \angle A\). Besides that, these triangles have a common side \(CP\), which is corresponding by similarity. So, these triangles are congruent and \(a=b\). Conversely, if \(a=b\), let \(M\) be the midpoint of \(AB\) (figure 1). The triangles \(\triangle AMC\) and \(\triangle BMC\) are congruent, so, in particular, \(\angle B = \angle A\).

\[\frac{|AC|}{|BC|}=\frac{|AB|}{|BQ|}=\frac{|BC|}{|CQ|}\] that is,\[\frac{b}{a}=\frac{c}{|BQ|}=\frac{a}{|CQ|}\] and, consequently, \(|BQ|=\frac{ac}{b}\) and \(|CQ|=\frac{a^{2}}{b}\). Besides that, by construction, we have \(|AC| = |AQ| + |CQ|\), so, \(|AQ|=b-\frac{a^{2}}{b}\). Since in the triangle \(\triangle ABQ\), we have \(\angle B = \angle A\), we also have \(|AQ| = |BQ|\). Replacing the above values in this equality, we obtain \(b-\frac{a^{2}}{b}=\frac{ac}{b}\) which is equivalent to \(b^{2}-a^{2} = ac\). Conversely, suppose that for the triangle \(\triangle ABC\), with corresponding sides \(a\), \(b\) and \(c\), the equality \(b^{2}-a^{2} = ac \) holds. Then \(b>a\) and we may fix a point \(Q\) on the side \(AC\) such that the angle \(\angle QBC\) is equal to \(\angle A\). This way we created a triangle \(\triangle BQC\) similar to \(\triangle ABC\), and, as we saw before, \[|BQ|=\frac{ac}{b},|CQ|=\frac{a^{2}}{b} \mbox{ and } |AQ|=b-\frac{a^{2}}{b}.\]Since by hypothesis \(b-\frac{a^{2}}{b}=\frac{ac}{b}\), that is, \(|AQ| = |BQ|\), we know that the triangle \(\triangle AQB\) is isosceles and \(\angle ABQ = \angle A\). It follows that, in the triangle \(\triangle ABC\), we have \(\angle B = 2 \angle A\).