Let AB be the given straight line;
thus it is required to describe a square on the straight line AB.

Let AC be drawn at right angles to the straight line AB from the point A on it, [I. 11] and let AD be made equal to AB;
through the point D let DE be drawn parallel to AB,
and through the point B let BE be drawn parallel to AD. [I. 31]

Therefore ADEB is a parallelogram;
therefore AB is equal to DE, and AD to BE [I. 34] . But AB is equal to AD;
therefore the four straight lines BA, AD, DE, EB are equal to one another;
therefore the parallelogram ADEB is equilateral.

I say next that it is also right-angled.

For, since the straight line AD falls upon the parallels AB, DE,
the angles BAD, ADE are equal to two right angles. [I. 29]
But the angle BAD is right;
therefore the angle ADE is also right.
And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34]
therefore each of the opposite angles ABE, BED is also right.
Therefore ADEB is right-angled.

And it was also proved equilateral.

Therefore it is a square; and it is described on the straight line AB. Q.E.F.