Return to Elements I
Note Blue text is not part of Euclid's text.
On the given finite straight-line to construct an equilateral triangle.
(diagram 1)Let the given finite straight line be AB.
It is required to construct on straight-line AB an equilateral triangle.
With center A and distance AB let a circle be inscribed BGD, (diagram 3)
and again with center B and distance BA let a circle be inscribed AGE, and (diagram 4)
from point G, where the circles cut one another, to points A, B, (diagram 5)
let straight-lines be joined, GA, GB.
(diagram 6) or use (general diagram)
And since point A is center of circle GDB, AG is equal to AB. Again since point B is center of circle GAE, BG is equal to BA. But GA was also proved equal to AB. Therefore, each of GA, GB is equal to AB. But things equal to the same are equal to one another. Therefore GA too is equal to GB. Therefore, the three GA, AB, BG are equal to one another. Therefore triangle ABG is equilateral and is constructed on the given finite straight-line AB[. Therefore on the given finite straight-line an equilateral triangle is constructed], just what it was required to do.