To place at a given point (as an extremity) a straight line equal to a given straight line.
Let A be the given point, and BC the given straight line. Thus it is required to place at the point A (as an extremity)
a straight line equal to the given straight line BC. From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangle
DAB be constructed. [I. 1] Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let the
circle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3] Then, since the point B is the centre of the circle CGH,
Again, since the point D is the centre of the circle GKL,
But BC was also proved equal to BG;
(Being) what it was required to do. 1 2 3