Mathematical Collections (Commandino 1588)

...he said that by the contemplation of planes, given two straight lines, he could find two mean proportionals in continuous analogy. And that man wanted us, once we had diligently examined the construction made by him, to respond to it. Which indeed is as follows.

A geometric diagram shows a series of rectangles and triangles labeled with letters. The diagram illustrates a method for finding mean proportionals between two lines, featuring vertical and horizontal lines intersecting with diagonals.

Let there be two straight lines AB and AC at right angles to each other, and from point B let BD be drawn parallel to AC. Let BD be set equal to AB, and let DC be joined, which meets BA at E. From point E, however, let EH be drawn parallel to AC, and let BD be produced; and from point D let DG be drawn parallel to BE; and let DN, NL, LX, and XK be set equal to BD. Then, through points NL and XK, let NO, LM, XP, and KH be drawn parallel to BE; and let KR be set equal to AB, and be cut in half at point S; and as KH is to HS, so let SH be made to HT. And as SH is to HT, so let TH be to HΦ; and from the straight line XP let QX be removed equal to AB. And let QK and QΦ be joined. And from point S, let SΨ be drawn parallel to QΦ. From point Ψ, however, let ΨΩ be drawn parallel to KX.

A & B

And let it be as LM is to MΩ, so let ΩM be to Mα, and as ΩM is to Mα, so let αM be to Mβ, and from ON let Nγ be removed equal to AB: and let γL and γβ be joined. Then from point Ω let Ωδ be drawn parallel to βγ, and from δ let Aε be drawn parallel to LN, and as DG is to Gε, so let εG be to Gζ. And as εG is to Cζ, so let ζG be to Gθ. And let θC be joined, and parallel to θC let ζκελ be drawn. Finally, let κμ and λν be drawn from points κ and λ parallel to AC and BD. It must be shown that the mean proportionals of AC and BD are μκ and νλ.

These things, therefore, he wrote down and handed to me, not containing a demonstration of the proposed problem. But because Hieronymus the philosopher, and several others of his friends who were known to me, wanted me to respond in the meantime regarding the proposed construction, since he had promised to make a demonstration, I have these things to say now. Namely, that he did not use the construction rightly, but wrongly. For in cutting the straight line RK in two at S, and making SH to HT as KH is to HS, he establishes TH to HΦ in the same proportion. But it is absolutely necessary that neither he nor we find the point of section in the third proportion, such as Φ. By this doubt, therefore, following from its cause, he shows that he does not understand this consequence. For since the point of section cannot be determined as Φ in the third proportion, unless the proportion that KH has to HR is first posited, this...