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;
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, BC is equal to BG.

Again, since the point D is the centre of the circle GKL, DL is equal to DG.

And in these DA is equal to DB;
therefore the remainder AL is equal to the remainder BG. [C.N. 3]

But BC was also proved equal to BG;
therefore each of the straight lines AL, BC is equal to BG.

And things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore AL is also equal to BC.

Therefore at the given point A the straight line AL is placed equal to the given straight line BC.
(Being) what it was required to do.