Let AE, EB, CF, FD be proportionals; that is, let AE be to EB as CF is to FD: they shall also be proportionals when taken jointly; that is, AB shall be to BE as CD is to DF.
Take of AB, BE, CD, DF any equimultiples whatever GH, HK, LM, MN;
and, again, of BE, DF take any equimultiples whatever KO, NP.
Then, because KO and NP are equimultiples of BE and DF, and that KH and NM are also equimultiples of BE and DF; [Construction.
therefore if KO, the multiple of BE, be greater than KH, which is a multiple of the same BE, then NP the multiple of DF is also greater than NM the multiple of the same DF; and if KO be equal to KH, NP is equal to NM; and if less, less.
First, let KO be not greater than KH;
therefore NP is not greater than NM.
And because GH and HK are equimultiples of AB and BE, [Construction.
and that AB is greater than BE,
therefore GH is greater than HK; [V. Axiom 3.
but KO is not greater than KH; [Hypothesis.
therefore GH is greater than KO.
In like manner it may be shewn that LM is greater than NP.
Thus if KO be not greater than KH, then GH, the multiple of AB, is always greater than KO, the multiple of BE;
and likewise LM, the multiple of CD, is greater than NP, the multiple of DF.
