Let line AB be the given straight line.

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

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

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

But line AB equals line 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 line AB and line 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.

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