To place at a given point (as an extremity) a straight line equal to a given straight line

Let $A$ be the given point, and $BC$ the given straight line.Thus it is required to place at the point $A$ (as an extremity) a straight line equal to the given straight line $BC$.From the point $A$ to the point $B$ let the straight line $AB$ be joined; [Post. 1]and on it let the equilateral triangle $DAB$ be constructed. [Prop. 1.1]Let the straight lines $AE$, $BF$ be produced in a straight line with $DA$, $DB$; [Post. 2]with centre $B$ and distance $BC$ let the circle $CGH$ be described; [Post. 3]and again, with centre $D$ and distance $DG$ let the circle $GKL$ be described. [Post. 3]Then, since the point $B$ is the centre of the circle $CGH$,$BC$ is equal to $BG$.Again, since the point $D$ is the centre of the circle $GKL$,$DL$ is equal to $DG$.And in these $DA$ is equal to $DB$;therefore the remainder $AL$ is equal to the remainder $BG$. [C.N. 3]But $BC$ was also proved equal to $BG$;therefore each of the straight lines $AL$, $BC$ is equal to $BG$.And things which are equal to the same thing are also equal to one another; [C.N. 1]threfore $AL$ is also equal to $BC$.Therefore at the given point $A$ the straight line $AL$ is placed equal to the given straight line $BC$.(Being) what it was required to do.