Kitap 5
5
Therefore etc. Q. E. D. let EB be made that multiple of CG, τοσαυταπλάσιον γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. From this way of stating the construction one might suppose that CG was given and EB had to be found equal to a certain multiple of it. But in fact EB is what is given and CG has to be found, i.e. CG has to be constructed as a certain submultiple of EB.
PROPOSITION 6.
6
If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same or equimultiples of them.
6
For let two magnitudes AB, CD be equimultiples of two magnitudes E, F, and let AG, CH subtracted from them be equimultiples of the same two E, F; I say that the remainders also, GB, HD, are either equal to E, F or equimultiples of them.
6
For, first, let GB be equal to E; I say that HD is also equal to F.
6
For let CK be made equal to F.
6
Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F, therefore AB is the same multiple of E that KH is of F. [V. 2]
6
But, by hypothesis, AB is the same multiple of E that CD is of F; therefore KH is the same multiple of F that CD is of F.
6
Since then each of the magnitudes KH, CD is the same multiple of F, therefore KH is equal to CD.
6
Let CH be subtracted from each; therefore the remainder KC is equal to the remainder HD.
6
But F is equal to KC; therefore HD is also equal to F.
6
Hence, if GB is equal to E, HD is also equal to F.
6
Similarly we can prove that, even if GB be a multiple of E, HD is also the same multiple of F.
6
Therefore etc. Q. E. D.
PROPOSITION 7.
7
Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes.
7
Let A, B be equal magnitudes and C any other, chance, magnitude; I say that each of the magnitudes A, B has the same ratio to C, and C has the same ratio to each of the magnitudes A, B.
7
For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F.
