Euclid, Proposition 1
Let AB be the given finite straight line.
Thus is required to construct an equilateral
triangle ont the straight line AB.
With The centre A and distance AB let the circle
BCD be described; [Post. 3]
again, with the 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 cirlce 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.