Then, since K is less than N first, therefore K is not less than M.
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]
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.
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.
But K is not less than M;
therefore neither is GH less than M.
And FG is greater than D;
therefore the whole FH is greater than D, M together.
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.
But FH is greater than M, D;
therefore FH is in excess of N,
while K is not in excess of N.
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]