Euclid's Elements: Book I: Proposition 2

Proposition 2: 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, 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.

Log in or register to write something here or to contact authors.