PROPOSITION 2. PROBLEM.
From a given point to draw a straight line equal to a given straight line.
Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.
From the point A to B draw the straight line AB;
and on it describe the equilateral triangle DAB, [I. 1.
and produce the straight lines DA, DB to E and F. [Post. 2.
From the centre B, at the distance BC, describe the circle CGH, meeting DF at G. [Post. 3.
From the centre D, at the distance DG, describe the circle GKL, meeting DE at L. [Post. 3.
AL shall be equal to BC.
Because the point B is the centre of the circle CGH, BC is equal to BG. [Definition 15.
And because the point D is the centre of the circle GKL, DL is equal to DG; [Definition 15.
and DA, DB parts of them are equal; [Definition 24.
therefore the remainder AL is equal to the remainder BG. [Axiom 3.
But it has been shewn that BC is equal to BG;
therefore AL and BC are each of them equal to BG.
But things which are equal to the same thing are equal to one another. [Axiom 1.
Therefore AL is equal to BC.
Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. q.e.f.
PROPOSITION 3. PROBLEM.
From the greater of two given straight lines to cut off a part equal to the less.
Let AB and C be the two given straight lines, of which