Kitap 1
2
Enunciation To place at a given point (as an extremity) a straight line equal to a given straight line.
2
Proof. Let A be the given point, and BC the given straight line.
2
Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC.
2
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]
2
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]
2
Then, since the point B is the centre of the circle CGH, BC is equal to BG.
2
Again, since the point D is the centre of the circle GKL, DL is equal to DG.
2
And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C.N. 3]
2
But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG.
2
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.
2
Therefore at the given point A the straight line AL is placed equal to the given straight line BC.
2
QED. (Being) what it was required to do.
Proposition 3.
3
Enunciation Given two unequal straight lines, to cut off from the greater a straight line equal to the less.
3
Proof. Let AB, C be the-two given unequal straight lines, and let AB be the greater of them.
3
Thus it is required to cut off from AB the greater a straight line equal to C the less.
3
At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15] But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]
