|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.||I.11
|Then ADEB is a parallelogram. Therefore AB equals DE, and AD equals BE.||I.34|
|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.||I.29|
|But the angle BAD is right, therefore the angle ADE is also right.||I.Post.4|
|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.||I.34|
|And it was also proved equilateral.|
|Therefore it is a square, and it is described on the straight line AB.||I.Def.22|
|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.|
This abbreviation of Euclid's construction requires six circles and four lines. There are alternate constructions that are a bit shorter. For instance, E may be found as the other intersection of the circles of radii BA and DA.
Next proposition: I.47