Proof: By Euclid

Let the circle $BCD$ with center $A$ and radius $AB$ have been drawn [Post. 3] , and again let the circle $ACE$ with center $B$ and radius $BA$ have been drawn [Post. 3] .
And let the straight lines $CA$ and $CB$ have been joined from the point $C$, where the circles cut one another,¹ to the points $A$ and $B$ (respectively) [Post. 1] .
And since the point $A$ is the center of the circle $CDB$, $AC$ is equal to $AB$ [Def. 1.15] .
Again, since the point $B$ is the center of the circle $CAE$, $BC$ is equal to $BA$ [Def. 1.15] .
But $CA$ was also shown (to be) equal to $AB$.
Thus, $CA$ and $CB$ are each equal to $AB$.
But things equal to the same thing are also equal to one another [C.N. 1] .
Thus, $CA$ is also equal to $CB$.
Thus, the three (straight lines) $CA$, $AB$, and $BC$ are equal to one another.
Thus, the triangle $ABC$ is equilateral, and has been constructed on the given finite straight line $AB$.
(Which is) the very thing it was required to do.

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The assumption that the circles do indeed cut one another should be counted as an additional postulate. There is also an implicit assumption that two straight lines cannot share a common segment (translator's note) ↩