Mathematical Collections (Commandino 1588) (1588) - Page 27

PROBLEM IV. PROPOSITION VIII.

Given straight lines FB and BC, to find the major extreme.

A geometric diagram shows a triangle with vertices A, H, E. A vertical line HC drops to the base. Points F and C are marked on the base line. A line connects A to H and H to E.

Let CH be drawn at right angles, which the circumference described around center B through F should intersect at H, and let HA be drawn at right angles to the joined BH. Therefore, AB is the third proportional of the lines CB and BF, for this is clearly evident from what was demonstrated before.

PROBLEM V. PROPOSITION IX.

Given straight lines AB and BC, to find the minor extreme in harmonic mean.

4. of the sixth [book].

A geometric diagram shows a horizontal line AB with points C and G. A vertical line DA is drawn from A. Lines connect D to B and D to C. A perpendicular line FG is drawn to the base.

Again, let there be two straight lines AB and BC; and let DAE be drawn at right angles to AB, such that DA is equal to AE. Let BD and ECF be joined, and from point F, let FG be drawn perpendicular to CB. I say that as AB is to BG, so is AC to CG, the excess of BC to the excess of CB and BG. For since it is as AB to BG, so is DA to FG; that is, AE to FG: for AE is equal to AD. It will be as AB to BG, so is AE to FG. But as AE is to FG, so is AC to CG: because the triangles ACE and CFG are equiangular. Therefore, as AB is to BG, so is AC to CG. And AC is the excess of the straight lines AB and BC, and CG is the excess of CB and BG. Therefore, as AB is to BG, so is the excess of AB and BC to the excess of CB and BG. This theorem is useful for the harmonic mean. For the first is AB, the second is BC, the third is BG.

COMMENTARY

Given straight lines AB and BC, to find the minor extreme in harmonic mean.] We have added this proposition for the sake of clarity, just as in those which follow.

A geometric diagram similar to the one above but with additional lines and points K, L, M, N illustrates the commentary's proof regarding the intersection of lines.

It must be noted, however, that Pappus does not determine the magnitude of the straight lines DA and AE, because however they are taken, whether larger or smaller, it is necessary that the same thing happens. For let two other lines KA and AL be taken, yet such that they are equal to each other; and let KB be joined, which intersects the straight line FG at point M; let LC be drawn. I say that LC, when produced, meets FG at M. For if it is possible, let it meet at another point, namely N. Since therefore we have demonstrated