Next let GB be a multiple of E: HD shall be the same multiple of F.
Make CK the same multiple of F that GB is of E. Then, because AG is the same multiple of E that CH is of F, [Hypothesis.
and GB is the same multiple of E that CK is of F [Constr.
therefore AB is the same multiple of E that KM is of F. [V. 2.
But AB is the same multiple of E that CD is of F; [Hyp.
therefore KH is the same multiple of F that CD is of F; [Hyp
therefore KH is equal to CD. [V. Axiom 1.
From each of these take the common magnitude CH; then the remainder CK is equal to the remainder HD.
And because CK is the same multiple of F that GB is of E, [Construction.
and that CK is equal to HD;
therefore HD is the same multiple of F that GB is of E.
Wherefore, if two magnitudes &c. q.e.d.
PROPOSITION A. THEOREM.
If the first of four magnitudes have the same ratio to the second that the third has to the fourth, then, if the first be greater than the second, the third shall also be greater than the fourth, and if equal equal, and if less less.
Take any equimultiples of each of them, as the doubles of each.
Then if the double of the first be greater than the double of the second, the double of the third is greater than the double of the fourth. [V. Definition 5.
But if the first be greater than the second, the double of the first is greater than the double of the second
