Mathematical Collections (Commandino 1588) (1588) - Page 21

BOOK THREE.

square. Therefore, the rectangle BKC together with the squares from CE and EF—that is, together with the square from CF—is equal to the squares from KE and EF, which is the square from FK. And since as MA is to AB, so is MD to DK; but as MD is to DK, so is BC to CK; it will be as MA is to AB, so is BC to CK. And AL is half of AB, and CG is double BC. Therefore, as MA is to AL, so is GC to CK; but as GC is to CK, so is FH to HK, on account of the parallel lines GF and CH. Wherefore, also by compounding, as ML is to LA, so is FK to KH. But AL is posited as equal to HK, since it is also equal to CF. Therefore, ML will also be equal to FK, and the square from ML will be equal to the square from FK. But the square from ML is equal to the rectangle BMA together with the square from AL; and it is shown that the square from FK is equal to the rectangle BKC together with the square from CF. Of these, the square from AL is equal to the square from CF, for AL is posited equal to CF. Therefore, the remaining rectangle BMA is equal to the remaining rectangle BKC. As therefore MB is to BK, so is CK to MA. But as MB is to BK, so is DC to CK. Wherefore as DC is to CK, so is CK to MA, and as MA is to AD. Therefore, as DC is to CK, so is CK to MA, and MA to AD.

B C D 6 Book II, 14 Book VI, 4 Book VI, 4 Book VI.

C O M M E N T A R Y.

"For it has been shown that this can be done via the conchoid line": See how that is done via the conchoid line in the fourth book, proposition 24, and in Eutocius in his commentaries on the second book of Archimedes' On the Sphere and Cylinder.

"And CG is double BC": Because of the similarity of the triangles DGC and LGB, as DC is to LB, so is CG to GB. But AB—that is, DC—is double LB. Therefore CG will also be double GB, and for that reason also double BC.

"As therefore MA is to AL, so is GC to CK": For since it is as MA is to AB, so is BC to CK, and as BA is to AL, so is CG to BC; it will be by equality in a perturbed ratio, as MA is to AL, so is GC to CK.

"Wherefore also by compounding, as ML is to LA, so is FK to KH; but AL is posited equal to HK, since it is also equal to CF. Therefore ML will be equal to FK": For from the aforementioned and the eleventh of the fifth book of the Elements, it follows that as MA is to AL, so is FH to HK. Therefore, also by compounding, as ML is to LA, so is FK to KH, and by permuting, as ML is to FK, so is LA to KH. But LA is equal to KH; wherefore ML will be equal to FK.

AS HERO.

A geometric diagram shows a rectangle ABCD with extended lines and a diagonal. A line segment labeled G, F, H, E intersects the extended lines, illustrating Hero's construction.

We shall show in what way we are able to find two mean proportionals for two straight lines mechanically; for this problem, as Hero also says, is solid. We shall therefore set forth the demonstration, which is most suitable for manual operations.

Let there be given two straight lines AB and BC, set at right angles to each other, for which it is necessary to find two mean proportionals. Let parallelogram ABCD be completed, and let DC and DA be produced; and let DB and AC be joined. And let a ruler be fitted to point B, which, when moved, cuts the lines CE and AF, such that that which is drawn from point G...