(Edited by Dionysius Lardner, 1855)

Proposition II. Problem.

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

(59) | From a given point (A) to draw a right line equal to a given finite right line (B C). |

Let a right line be drawn from the given point A to either extremity B of the given finite right line B C (39). On the line A B let an equilateral triangle A D B be constructed (I). With the centre B and the radius B C let a circle be described (41). Let D B be produced to meet the circumference of this circle in F (40), and with the centre D and the radius D F let another circle F L K be described. Let the line D A be produced to meet the circumference of this circle in L. The line A L is then the required line.

The lines D L and D F are equal, being radii of the same circle F L K (17). Also the lines D A and D B are equal, being sides of the equilateral triangle B D A. Taking the latter from the former, the remainders A L and B F are equal (45). But B F and B C are equal, being radii of the same circle F C H (17), and since A L and B C are both equal to B F, they are equal to each other (43), Hence A L is equal to B C, and is drawn from the given point A, and therefore solves the problem.

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

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This proposition in other editions:

^{★}★^{★}The different positions which the given right line and given point may have with respect to each other, are apt to occasion such changes in the diagram as to lead the student into error in the execution of the construction for the solution of this problem.Hence it is necessary that in solving this problem, the student should be guided by certain

generaldirections, which are independent of any particular arrangement which the several lines concerned in the solution may assume. If the student is governed by the following general directions, no change which the diagram can undergo will mislead him.1° The given point is to be joined with

eitherextremity of the given right line. (Let us call the extremity with which it is connected, theconnected extremityof the given right line; and the line so connecting them, thejoining line.)2° The centre of the first circle is the

connected extremityof the given right line; and its radius, the given right line.3° The equilateral triangle may be constructed on

either sideof the joining line.4° The side of the equilateral triangle which is produced to meet the circle, is that side which is opposite to the given point, and it is produced through the centre of the first circle till it meets its circumference.

5° The centre of the second circle is that vertex of the triangle which is opposite to the joining line, and its radius is made up of that side of the triangle which is opposite to the given point, and its production which is the radius of the first circle. So that the radius of the second circle is the sum of the side of the triangle and the radius of the first circle.

6° The side of the equilateral triangle which is produced through the given point to meet the second circle, is that side which is opposite to the connected extremity of the given right line, and the production of this side is the line which solves the problem; for the sum of this line and the side of the triangle is the radius of the second circle, but also the sum of the given right line (which is the radius of the first circle) and the side of the triangle is equal to the radius of the second circle. The side of the triangle being taken away the remainders are equal.

As the given point may be joined with either extremity, there may be two different joining lines, and as the triangle may be constructed on either side of each of these, there may be four different triangles; so the right line and the point being given, there are four different constructions by which the problem may be solved.

If the student inquires further, he will perceive that the solution may be effected also by producing the side of the triangle opposite the given point, not through the extremity of the right line but through the vertex of the triangle. The various consequences of this variety in the construction we leave to the student to trace.

(60) By the second proposition a right line of a given length can be inflected from a given point P upon any given line A B. For from the point P draw a right line of the given length (II), and with P as centre, and that line as radius, describe a circle. A line drawn from P to any point C, where this circle meets the given line A B, will solve the problem.

By this proposition the first may be generalized; for an

isoscelestriangle may be constructed on a given line as base, and having its side of a given length. The construction will remain unaltered, except that the radius of each of the circles will be equal to the length of the side of the proposed triangle. If this length be not greater than half the base, the two circles will not intersect, and no triangle can be constructed, as will appear hereafter.