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Proposition 2

Wed, 2006-10-11 09:20 — ashley

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. [Prop. 1.1]

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]

threfore $ 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.