To construct an equilateral triangle on a given finite straight-line. Let $AB$ be the given finite straight-line. So it is required to construct an equilateral triangle on the straight-line $AB$. 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, 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.