To describe a square on a given straight-line. Let $AB$ be the given straight-line. So it is required to describe a square on the straight-line $AB$.  Let $AC$ have been drawn at right-angles to the straight-line $AB$ from the point $A$ on it [Prop.~1.11], and let $AD$ have been made equal to $AB$ [Prop.~1.3]. And let $DE$ have been drawn through point $D$ parallel to $AB$ [Prop.~1.31], and let $BE$ have been drawn through point $B$ parallel to $AD$ [Prop.~1.31]. Thus, $ADEB$ is a parallelogram. Therefore, $AB$ is equal to $DE$, and $AD$ to $BE$ [Prop.~1.34]. But, $AB$ is equal to $AD$. Thus, the four (sides) $BA$, $AD$, $DE$, and $EB$ are equal to one another. Thus, the parallelogram $ADEB$ is equilateral. So I say that (it is) also right-angled. For since the straight-line $AD$ falls across the parallels $AB$ and $DE$, the (sum of the) angles $BAD$ and $ADE$ is equal to two right-angles [Prop.~1.29]. But $BAD$ (is a) right-angle. Thus, $ADE$ (is) also a right-angle. And for parallelogrammic figures, the opposite sides and angles are equal to one another [Prop.~1.34]. Thus, each of the opposite angles $ABE$ and $BED$ (are) also right-angles. Thus, $ADEB$ is right-angled. And it was also shown (to be) equilateral.   Thus, ($ADEB$) is a square [Def.~1.22]. And it is described on the straight-line $AB$. (Which is) the very thing it was required to do.