Let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F.
I say that the first A also has to the second B a greater ratio than the fifth E to the sixth F.
Since there are some equimultiples of C and E, and of D and F other equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, let them be taken. Let G and H be equimultiples of C and E, and K and L other, arbitrary, equimultiples of D and F, so that G is in excess of K, but H is not in excess of L. Whatever multiple G is of C, let M also be that multiple of A, and, whatever multiple K is of D, let N also be that multiple of B.
Now, since A is to B as C is to D, and of A and C equimultiples M and G have been taken, and of B and D other, arbitrary, equimultiples N and K, therefore, if M is in excess of N, G is also in excess of K; if equal, equal; and if less, less.
But G is in excess of K, therefore M is also in excess of N.
But H is not in excess of L, and M and H are equimultiples of A and E, and N and L other, arbitrary, equimultiples of B and F, therefore A has to B a greater ratio than E has to F.
Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth has to a sixth, then the first also has to the second a greater ratio than the fifth to the sixth.
The magnitudes may be of three different kinds with a and b of one kind, c and d of a second kind, and e and f of a third kind.
The analogous statement for lesser ratios isn’t stated, but, of course, it holds as well. Euclid uses it as well as this proposition, in V.20.
Transitivity of the greater than relation also holds:
The proof isn’t difficult, but without symbolic algebra it becomes unwieldy. Euclid would have required 20 lines in his diagram.
This proposition is used in the next one as well as V.20 and V.21.