Mathematical Collections (Commandino 1588)

to the section CE, let it be made equal to that which is drawn from the same point G to the section AF. And so, let it be already done, and let the position of the ruler be EBF, and let EG be equal to GF. I say that the straight lines AF, CE are mean proportionals of the lines AB, BC.
A For since the parallelogram ABCD is a rectangle, the four straight lines DG, GA, BG, GC are equal to one another. And since DG and GA are equal, and GF has been drawn, the rectangle DFA together with the square from AG will be equal to the square from GF. And by the same reasoning, the rectangle DEC together with the square from CG is equal to the square from GE. And FG and GE are equal. Therefore, the rectangle DFA together with the square from AG will be equal to the rectangle DEC together with the square from CG. Of these, the square from CG is equal to the square from GA. Therefore, the remaining rectangle DEC is equal to the rectangle DFA.
14 of Book VI Therefore, as ED is to DF, so is FA to CE. But as ED is to DF, so also is BA to AF, and EC to CB. Wherefore, also as BA is to AF, so is FA to CE, and EC to CB. Therefore, AF and CE are mean proportionals of the straight lines AB and BC.

A "For since the parallelogram ABCD is a rectangle, the four straight lines DG, GA, BG, GC are equal to one another": The text provides a geometric proof: The two sides CB and BA of triangle ABC are equal to the two sides BC and CD of triangle DCB, and the right angle at B is equal to the right angle at C. Therefore, the base AC will be equal to the base BD. But angle AGB is equal to angle DGC because they are vertical angles; and angle ABG is equal to angle CDG, and angle BAG is equal to angle DCG. Triangle ABG is therefore similar to triangle CDG; and as AB is to BG, so is CD to DG: and by alternating, as AB is to CD, so is BG to GD. Therefore, BG and GD are equal. And in the same way, AG and GC will be shown to be equal. Therefore, as BG is to GD, so is AG to GC. And by compounding, as BD is to DG, so is AC to CG; and again by alternating, as BD is to AC, so is DG to GC. But BD and AC are equal, as was shown. Therefore, DG and GC are also equal, and consequently BG and GA, and all are equal to one another.

B "And since DG and GA are equal, and GF has been drawn; the rectangle DFA together with the square from AG will be equal to the square from GF": Let GH be drawn from point G perpendicular to AD. Triangle AGH will be similar to triangle DGH: and as GH is to HA, so is GH to HD. Therefore, AH is equal to HD. And since AD has been cut in half at H, and AF is added to it, the rectangle DFA together with the square from AH is equal to the square from HF. Let the square from GH be added in common; therefore, the rectangle DFA together with the squares from AH and HG—that is, together with the square from AG—will be equal to the squares from FH and HG, namely the square from FG. And similarly, if a perpendicular is drawn from point G to DC, the rectangle DEC together with the square from CG will be shown equal to the square from GE.

But the cube of a cube is found not only to be double through the provided instrument, which we have devised, but also generally to possess any mandated proportion.