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]
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.
甲戊、戊乙、之倍丙己、庚丙。其數等。卽其兩幷、甲乙之倍庚己。亦若甲戊之倍丙己也。本篇一而甲乙之倍丙丁。元若甲戊之倍丙己。則丙丁與庚己等也。次每減同用之丙己。卽庚丙與己丁、亦等。而戊乙之倍己丁。亦若戊乙之倍庚丙矣。夫戊乙之倍庚丙。旣若甲戊之倍丙己。則戊乙、為甲戊之分餘。所倍於己丁、為丙己之分餘者。亦若甲乙之倍丙丁也。