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And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF,
therefore AE is the same multiple of CF that EB is of FD.

But, by hypothesis,

AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD.

That is, the remainder EB will be the same multiple of
the remainder FD that the whole AB is of the whole CD.

Therefore etc. Q. E. D. 1

PROPOSITION 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.

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.

For, first, let GB be equal to E; I say that HD is also equal to F.

For let CK be made equal to F.

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]

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.

Since then each of the magnitudes KH, CD is the same multiple of F,

therefore KH is equal to CD.

1 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.

Euclid. Euclid's Elements. Sir Thomas Little Heath. New York. Dover. 1956.

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