PROPOSITION 1.¶
Let AB be the given finite straight line.[2]
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;[3] [Post. 3]
again, with centre B and distance BA let the circle ACE be described; [Post. 3]
and from the point C,[4] 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,
Again, since the point B is the centre of the circle CAE,
But CA was also proved equal to AB; therefore each of the straight lines CA, CB[5] is equal to AB.
And things which are equal to the same thing are also equal to one another; [C. N. 1]
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.
See Commentary on Proposition I.1.