For, whatever multiple AE is of CF, let EB be made that multiple of CG.

Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. [[V. 1](elem.5.1)]

But, by the assumption, AE is the same multiple of CF that AB is of CD.

Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD.

Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD.

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.

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.