To describe a square on a given straight-line.
ΑΒ 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 ΑΒ 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 ΑΒ, (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 rightangled. 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.