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
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
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
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;
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.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
Next: Proposition 2
Previous: Euclid, Book I, Common Notions (ed. Sir Thomas L. Heath, 1st Edition, 1908)
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