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 $BCD$ be described; [Post. 3]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 circle $CDB$, $AC$ 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 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.