(Edited by Dionysius Lardner, 1855)

Proposition I. Problem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(57) | On a given finite right line (A B) to construct an equilateral triangle. |

With the centre A and the radius A B let a circle B C D be described (41), and with the centre B and the radius B A let another circle A C E be described. From a point of intersection C of these circles let right lines be drawn to the extremities A and B of the given right line (39). The triangle A C B will be that which is required.

It is evident that the triangle A C B is constructed on the given right line A B. But it is also equilateral; for the lines A C and A B, being radii of the same circle B C D, are equal (17), and also B C and B A, being radii of the same circle A C E, are equal. Hence the lines B C and A C, being equal to the same line A B, are equal to each other (43). The three sides of the triangle A B C are therefore equal, and it is an equilateral triangle (28).

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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Previous: Euclid, Book I, Axioms (ed. Dionysius Lardner, 11th Edition, 1855)

This proposition in other editions:

(58) In the solution of this problem it is assumed that the two circles intersect, inasmuch as the vertex of the equilateral triangle is a point of intersection. This, however, is sufficiently evident if it be considered that a circle is a continued line which includes space, and that in the present instance each circle passing through the centre of the other must have a part of its circumference within that other, and a part without it, and must therefore intersect it.

It follows from the solution, that as many different equilateral triangles can be constructed on the same right line as there are points in which the two circles intersect. It will hereafter be proved that two circles cannot intersect in more than two points, but for the present this may be taken for granted.

Since there are but two points of intersection of the circles, there can be but two equilateral triangles constructed on the same finite right line, and these are placed on opposite sides of it, their vertices being at the points C and F.

After having read the first book of the elements, the student will find no difficulty in proving that the triangles C F E and C D F are equilateral. These lines are not in the diagram, but may easily be supplied.