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Proposition 1

Tue, 2006-10-10 07:11 — ashley

On a given finite straight line to construct an equilateral triangle.

Let $ AB $ be the given finite straight line.

Thus it is required to construct an equilateral triangle on the straight line $ AB $.

With centre $ A $ and distance $ AB $ let the circle $ BCD $ be described; [Post. 3]

again, with centre $ B $ and distance $ BA $ let the circle $ ACE $ be described; [Post. 3]

and from the point $ C $, in which the circles cut one another, to the points $ A $, $ B $ let the straight lines $ CA $, $ CB $ be joined. [Post. 1]

Now, since the point $ A $ is the centre of the circle $ CDB $, $ AC $ is equal to $ AB $. [Def. 15]

Again, since the point $ B $ is the centre of the circle $ CAE $, $ BC $ is equal to $ BA $. [Def. 15]

But $ CA $ was also proved equal to $ AB $;

therefore each of the straight lines $ CA $, $ CB $ is equal to $ AB $.

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

therefore $ CA $ is also equal to $ CB $.

Therefore the three straight lines $ CA $, $ AB $, $ BC $ are equal to one another.

Therefore the triangle $ ABC $ is equilateral; and it has been constructed on the given finite straight line $ AB $.

(Being) what it was required to do.