Mathematical Collections (Commandino 1588)

LIBER TERTIVS.

But this not being given, what is proposed by him cannot be found by means of plane geometry; just as, by the numbers themselves, it will be permissible to persuade those who wish, consistently with the resolution, using the table of Ptolemy concerning straight lines given in a circle. But it would have been better to doubt this along with others than to find it in this way. We, however, will now demonstrate the things that were posited above.

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A

And let it be as LM to MΩ, so ΩM to Mκ] The Greek codex had it incorrectly. kai estō hōs lm pros mō, houtōs hē ōm pros mk and let it be as lm to mō, thus om to mk when it should be read pros mk to mk.

B

And let parallels to θC be drawn, ζκ, ελ. Finally, let parallels to AC, BD be drawn from points κ, λ, namely κμ, λν. It must be shown that μκ, νλ are mean proportionals of AC, BD,] The Greek codex says kai ēchthōsan tē thg parallēloi hai, zk el, kai apo tōn k a tōis ag ba parallēloi hai km a n leixai, etc. hai mk na and let be drawn parallel to θΓ the lines ζκ ελ, and from κ α parallels to αγ βα the lines κμ αν, to show... the μκ να. But it should be read kai ēchthōsan tē thg parallēloi hai, zk en, kai apo tōn kl tais ag bd parallēloi hai km, ln deixai hai mk, ln and let be drawn parallel to θγ the lines ζκ εν, and from κλ parallels to αγ βδ the lines κμ, λν, to show the mk, ln.

C

Unless the ratio which KH has to HR, that is BE to EA, is first posited.] Greek codex, mē proteron hupotethentos ton logon hon echei hē kh pros kr, toutesti ton hon echei hē h pros tēn ba without first having posited the ratio which kh has to kr, that is which h has to ba. But I think it should be read from those things which follow subsequently: hon echei hē kh pros th r toutesti ton hon echei hē be pros tēn ea which kh has to th r, that is which be has to ea.

D

And as 3 to 2 1/4, so 2 1/4 to another certain value.] Greek codex hōs de kai ta g pros ta duo kai d, houtōs auta ta bd pros allēla and as 3 to 2 and d, so these 2 d to each other. But it should be read pros allēn tina to another certain one.

E

And the ratio of KH to HS, which 10 has to 6.] Greek codex kai tēs kh e ara pros tēn ths logos esti, hon ta e pros ta st and of kh therefore to ths the ratio is, which 5 to 6. But it should be read hon ta i pros ta st which 10 to 6.

F

He is refuted by himself, assuming the requirement as conceded.] Greek codex autothen elegchetai to zētoumenon hōs homologoumenon labōn the requirement is refuted from itself having been taken as agreed. Read to zētoumenon hōs homologoumenon labōn having taken the requirement as agreed.

G

And by drawing parallels to Kϖ through the points STR, what is sought will be done.] The Greek codex was corrupt in this place and below, which we have restored.

H

For it was also as Kϖ to ςg, so ςg to Tσ.] This follows from 4 of the sixth book of elements due to the similarity of triangles.

K

Since, therefore, the ratio of KH to HR is given.] Greek codex epei toinun dotheisēs esti ho tēs kh pros th r since therefore being given is the ratio of kh to th r read epei toinun dotheis esti ho tēs kh pros th r since therefore the ratio of kh to th r is given.

L

Therefore the angle RΦQ is given, which is equal to the exterior angle KSψ.] Greek codex dotheisa ara hē hupo r ph gōnia kai isē estin tē hupo ks ph ektos gōnia given therefore the angle r ph and equal is to the angle ks ph outside. But it should be read dotheisa ara hē hupo r ph ch gōnia kai isē estin tē hupo ks ps ektos gōnia given therefore the angle r ph ch and equal is to the exterior angle ks ps.

M

Therefore, when Aψ is extended to F, the right-angled triangle SFψ will be given in species and size.] Greek codex ekblētheisēs ara etc. tō eidei houtōs having been extended therefore etc. in species thus. I think it should be read tō eidei kai megethei in species and size, which clearly appears from what follows.

N

If N is necessarily the point of section of the third proportion.] Greek codex n ei to st estin ex anankēs to tēs tomēs sēmeion ton triton logon n if the st is by necessity the point of the section the third ratio. But for st st I would gladly put b b. That is V as we translated.

FIRST THEOREM, PROPOSITION I.

Let a certain straight line AG be divided into equal parts at points BCDEF. I say that as AC is to CB, so is BC to half of CB; and as AD is to DB, so is BD to DC and the third part of CB; and as AE is to EB, so is BE to EC and the fourth part of CB; and as AF is to FB, so is BF to FC and the fifth part of CB; finally, as AG is to GB, so is BG to GC and the sixth part of CB.