Book 5
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And, since FG is the same multiple of AE that GH is of EB, therefore FG is the same multiple of AE that FH is of AB. [V. 1]
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But FG is the same multiple of AE that K is of C; therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C.
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Again, since GH is the same multiple of EB that K is of C, and EB is equal to C, therefore GH is equal to K.
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But K is not less than M; therefore neither is GH less than M.
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And FG is greater than D; therefore the whole FH is greater than D, M together.
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But D, M together are equal to N, inasmuch as M is triple of D, and M, D together are quadruple of D, while N is also quadruple of D; whence M, D together are equal to N.
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But FH is greater than M, D; therefore FH is in excess of N, while K is not in excess of N.
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And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D; therefore AB has to D a greater ratio than C has to D. [V. Def. 7]
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I say next, that D also has to C a greater ratio than D has to AB.
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For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH.
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And N is a multiple of D, while FH, K are other, chance, equimultiples of AB, C; therefore D has to C a greater ratio than D has to AB. [V. Def. 7]
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[Case 2.]
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Again, let AE be greater than EB.
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Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4]
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Let it be multiplied, and let GH be a multiple of EB and greater than D; and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K of C.
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Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M.
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But GH is greater than D; therefore the whole FH is in excess of D, M, that is, of N.
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Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N.
