I.2

To place at a given point (as an extremity) a straight line equal to a given straight line.

Proof

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, 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.