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Proposition 2

To place at a given point [as an extremity] a straight line equal to a given straight line.

This proposition provides a way of extending from a given point a line of equal length of a given distinct line. This proposition requires constructing an equilateral triangle - by way of Proposition 1, constructing circles, finding the intersection points, and comparing lengths. If we wanted to accomplish the same goal in analytic geometry we would simply determine the exact length of the given line, perhaps using a ruler, and extend a line of equal length from the desired point. This is an instance of where synthetic geometry proves to be a little cumbersome.

This proposition is criticized because it doesn't consider any other cases, such as the cases where A=B or B=C. In these cases the proof would be trivial because the line would already be extended from the given point itself. This can be visualized using the interactive diagram below.

Points A, B, and C can be moved in this diagram. D depends on the length of AB. Lines E and F are then extended from DA and DB respectively. A circle H with center B and radius BC is constructed, then a circle K at center D with radius DG is constructed, where G is the intercept of line F and circle H. So DB+BG=DG, and BC=BG since G is a point on the circle H. The intercept of line E and circle K is taken as point L, and AL is proved to be equal to BC, and therefore we have drawn a line of length BC that extends from point A. Whether directly or indirectly, this construction was built from just A, B, and C, which is why they are the only movable parts.

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]To draw a straight line from any point to any point. and on it let the equilateral triangle DAB be constructed. [1.1]On a given finite straight line to construct an equilateral triangle.
Let the straight lines AE, BF be produced in a straight line with DA, DB. [Post. 2]To produce a finite straight line continuously in a straight line.

With centre B and distance BC let the circle CGH be described; [Post. 3]To describe a circle with any centre and distance.

And again, with centre D and distance DG let the circle GKL be described. [Post. 3]To describe a circle with any centre and distance.

Then, since the point B is the centre of the circle CGH, BC is equal to BF.

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]If equals be subtracted from equals, the remainders are equal.

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 equal to one another; [C.N. 1]Things which are equal to the same thing are also equal to one another. 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.

Proof in Nuprl

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