To place a straight-line equal to a given straight-line at a given point (as an extremity). Let $A$ be the given point, and $BC$ the given straight-line. So  it is required to place a straight-line at point $A$ equal to the given straight-line $BC$.  For  let the straight-line $AB$ have been joined from point $A$ to point $B$ [Post.~1], and let the equilateral triangle $DAB$ have been been constructed upon it [Prop.~1.1].  And let the straight-lines $AE$ and $BF$ have been produced in a straight-line with $DA$ and $DB$  (respectively) [Post.~2]. And let the circle $CGH$ with center $B$ and radius $BC$ have been drawn [Post.~3], and again let the circle $GKL$ with center $D$ and radius $DG$ have been drawn [Post.~3].      Therefore, since the point $B$ is the center of (the circle) $CGH$, $BC$ is equal to  $BG$ [Def.~1.15]. Again, since the point $D$ is the center of the circle $GKL$, $DL$ is equal to $DG$ [Def.~1.15]. And within these,  $DA$ is equal to $DB$. Thus, the remainder $AL$ is equal to the remainder $BG$ [C.N.~3]. But $BC$ was also shown (to be)  equal to $BG$. Thus,  $AL$ and $BC$ are each equal to $BG$. But things equal to the same thing are also equal to one another [C.N.~1]. Thus, $AL$ is also equal to $BC$.  Thus, the straight-line $AL$, equal to the given straight-line $BC$, has been placed at the given point $A$. (Which is) the very thing it was required to do.