To describe a square on a given straight line.

Let *AB* be the given straight line.

It is required to describe a square on the straight line *AB.*

Draw *AC* at right angles to the straight line *AB* from the point *A* on it. Make *AD* equal to *AB.* Draw *DE* through the point *D* parallel to *AB,* and draw *BE* through the point *B* parallel to *AD.*

Then *ADEB* is a parallelogram. Therefore *AB* equals *DE,* and *AD* equals *BE.*

But *AB* equals *AD,* therefore the four straight lines *BA, AD, DE,* and *EB* equal one another. Therefore the parallelogram *ADEB* is equilateral.

I say next that it is also right-angled.

Since the straight line *AD* falls upon the parallels *AB* and *DE,* therefore the sum of the angles *BAD* and *ADE* equals two right angles.

But the angle *BAD* is right, therefore the angle *ADE* is also right.

And in parallelogrammic areas the opposite sides and angles equal one another, therefore each of the opposite angles *ABE* and *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.

There are quite a few steps needed to construct a square on AB. In order to construct the perpendicular AC, first AB has to be extended in the direction of A and a point F on the far side the same distance from A as B is, then two more circles centered at B and F to get a perpendicular line, and then it needs to be cut off at length C, but fortunately, the needed circle has already been drawn.
Next, This abbreviation of Euclid’s construction requires six circles and four lines. There are alternate constructions that are a bit shorter. |