Commentary on Proposition IV.2¶
Back to Proposition IV.2.
Here again, since any point on the circle may be taken as an angular point of the triangle, there are an infinite number of solutions. Even when a particular point has been chosen to form one angular point, the required triangle may be constructed in six ways. For any one of the three angles may be placed at the point; and, whichever is placed there, the positions of the two others relatively to it may be interchanged. The sides of the triangle will, in all the different solutions, be of the same length respectively; only their relative positions will be different.
This problem can of course be reduced (as it was by Borelli) to III. 34, namely the problem of cutting off from a given circle a segment containing an angle equal to a given angle. It can also be solved by the alternative method applicable to III. 34 of drawing “angles at the centre” equal to double the angles of the given triangle respectively; and by this method we can easily solve this problem, or III. 34, with the further condition that one side of the required triangle, or the base of the required segment, respectively, shall be parallel to a given straight line.
As a particular case, we can, by the method of this proposition, describe an equilateral triangle in any circle after we have first constructed any equilateral triangle by the aid of I. 1. The possibility of this is assumed in IV. 16. It is of course equivalent to dividing the circumference of a circle into three equal parts. As De Morgan says, the idea of dividing a revolution into equal parts should be kept prominent in considering Book IV.; this aspect of the construction of regular polygons is obvious enough, and the reason why the division of the circle into three equal parts is not given by Euclid is that it happens to be as easy to divide the circle into three parts which are in the ratio of the angles of any triangle as to divide it into three equal parts.