Mathematical Collections (Commandino 1588)

A geometric diagram shows a semicircle with points A, B, C. A series of intersecting lines and chords are labeled with letters A, F, N, B, K, G, H, E, D, C, M, L. A rule or line segment AGHK intersects various points within and on the boundary of the semicircle.

For let a semicircle ABC be described: and from the center D, let DB be drawn at right angles; and let a ruler be moved around point A, so that one of its ends is placed upon point A with some sort of pin, and the remaining part is moved around the pin, as if around a center, between BC. With these things constructed in this way, the proposal is to find two cubes that have a given proportion between them. Let the proportion of BD to DE be made the same as the given proportion; and let CE be joined and produced to F. Let the ruler be moved between BC, until the part of it intercepted between the lines FE and EB is equal to that which is intercepted between the straight line BE and the circumference BKC. For by attempting this and moving the ruler, we shall easily achieve it. Let it therefore be already done, and let the ruler have the position AGHK, so that GH and HK are equal to one another. I say that the cube made from the line BD to the cube from the line DH has the given proportion, namely that which BD has to DE. For let the complete circle be understood: and having joined KD, let it be produced 2 of Book VI to L, and let LG be joined. Therefore, LG is parallel to BD, because KH is equal to HG, and KD is equal to DL. Let AL and LC also be joined. And since angle GAL in the semicircle is a right angle; and AM is a perpendicular; it will be Corollary 8 as the square from LM is to the square from MA—that is, as the straight line CM is to the straight line MA—so is the square from AM to the square from MG: for as LM is to MA, so is MA to MG. Therefore, as the square from LM is to the square from MA, so is A the square from AM to the square from MG, and the straight line CM is to MA. Let the proportion of AM to MG be applied in common. Therefore, the proportion composed of the B proportion of GM to MA, and the proportion of AM to MG—namely the proportion of C CM to MG—is the same as that which is composed of the proportion of the square from AM to the square from MG; and of the proportion of the straight line AM to MG. But the proportion composed of the proportion of the square from AM to the square from MG, and of the proportion of the straight line AM to MG, is the same as that which the cube made from AM has to the cube which is from MG. Therefore, the proportion of CM to MG will be the same 4 of Book VI as the cube from AM has to the cube from MG. But as CM is to MG, so is CD to DE, that is BD to DE. And as AM is to MG, so is AD to DH, that is BD to DH. Therefore, as BD is to DE, which is the given proportion, so is the cube from BD to that which is made from the DH cube: If, therefore, it is made so that as BD is to DH, so is DH to some other line, as DX; then DH and DX will be two mean proportionals of the straight lines BD and DE.

COMMENTARY.

A "Therefore, as the square from LM is to the square from MA, so is the square from AM to the square from MG": From the 22nd of Book VI.
B "And the straight line CM to MA": Again, because angle ALC in the semicircle is a right angle and LM is a perpendicular, there will be three straight lines CM, ML, MA in continuous proportion.