First proposition.

d, K a b f, c

Upon a given straight line let the 3 species of triangles be set. Let the given line be a b, upon which first I wish to make the equilateral triangle, which I will make thus: I will open the compass as much as is the length of the said line, and according to that quantity, by the second postulate, I will make the circle, fixing one of the feet at the point a, which circle let be called a, named from its centre. Then, according to the same quantity, I will describe another circle, fixing the immobile foot at the point b, so that it is the circle b, named from its centre. These two circles will of necessity intersect in two points, which let be c d; one of which, let it be d, I will join with the extremity of the said line, by the first postulate, drawing the straights a b and b d. I say the proposition is concluded, that is, that the triangle a b d is equilateral, as was required, which I thus prove to you. The proof. Because the straights a d and a b are equal, by the definition of the circle, since they move from a same centre and terminate on a same circumference, hence the lines a d and a b are equal. Therefore, by the first conception, among them all are equal, because each of them is equal to the given line a b.