Next, let KO be greater than KH; therefore, as has been shewn, NP is greater than NM. And because the whole GH is the same multiple of the whole AB that HK is of BE [Construction.
therefore the remainder GK is the same multiple of the remainder AE that GH is of AB; [V. 5.
which is the same that LM is of CD. [Construction.
In like manner, because the whole LM is the same multiple of the whole CD that MN is of DF, [Construction. therefore the remainder LN is the same multiple of the remainder CF that LM is of CD. [V. 5.
But it was shewn that LM is the same multiple of CD that G is of CD.
Therefore GK is the same multiple of AE that LN is of CF;
that is, G and LN are equimultiples of BE and CF.
And because KO and NP are equimultiples of BE and DF; [Construction.
therefore, if from KO and NP there be taken KH and NM, which are also equimultiples of BE and DF, [Constr.
the remainders HO and MP are either equal to BE and DF, or are equimultiples of them.
Suppose that HO and MP are equal to BE and DF. Then, because AE is to EB as CF is to FD, [Hypothesis. and that GK and LN are equimultiples of AE and CF; therefore GK is to EB as LN is to FD. [V. 4, Cor.
But HO is equal to BE, and MP is equal to DF; [Hyp
therefore GK is to HO as LN is to MP.
