 (diagram 1) For let a first, A, have the same ratio to second, B, as third, G, has to fourth, D, (diagram 2) and let equally-times multiples be taken of A, G, i.e., E, Z, and other equally-times multiples of B, D, as it happens, H, Q. I say that it is: as E to H, so Z to Q. (diagram 3) For let equally-times multiples of E, Z, be taken, K, L, and other equally-times multiples of H, Q, as it happens, M, N. (diagram 4) And since E is equally-times a multiple of A with Z of G, and K, L, are taken as equally-times multiples of E, Z, therefore, K is equally-times a multiple of A as L of G. (v 3) (diagram 5) For the same reasons, in fact, M is equally-times a multiple of B as N of L. (diagram 6) And since it is: as A to B so G to D, and K, L have been taken as equal-times multiples of A, G, but others, whatever happens, M, N, as equal-times multiples of B, D, therefore, if K exceeds M, L also exceeds N, and if equal, equal, and if smaller, smaller. And K, L are equal-times multiples of E, Z and M, N are other, whatever happens, equal-times multiples of H, Q. Therefore, it is: as E to H so Z to Q. Therefore, if first to second has the same ratio as third to fourth, the equal-times multiples of the first and third to the equal-times multiples of the second and fourth, according to any multiplication, will have the same ratio taken to one another, just what it was required to show.