Kitap 5
17
Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2]
17
And, since, as AB is to BE, so is CD to DF, and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP, therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less.
17
Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO.
17
But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP, and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP.
17
Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less.
17
And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD.
17
Therefore etc. Q. E. D.
PROPOSITION 18.
18
If magnitudes be proportional separando, they will also be proportional componendo.
18
Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD; I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD.
18
For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater.
18
First, let it be in that ratio to a less magnitude DG.
18
Then, since, as AB is to BE, so is CD to DG, they are magnitudes proportional componendo; so that they will also be proportional separando. [V. 17]
18
Therefore, as AE is to EB, so is CG to GD.
18
But also, by hypothesis, as AE is to EB, so is CF to FD.
18
Therefore also, as CG is to GD, so is CF to FD. [V. 11]
18
But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14]
18
But it is also less: which is impossible.
