Since, as AB is to CD, so is E to F, and E is equal to AG, and F to CH,
therefore, as AB is to CD, so is AG to CH.
And since, as the whole AB is to the whole CD,
so is the part AG subtracted to the part CH subtracted,
the remainder GB will also be to the remainder HD as the whole AB is to the whole CD. [V. 19]
But AB is greater than CD;
therefore GB is also greater than HD.
And, since AG is equal to E, and CH to F,
therefore AG, F are equal to CH, E.
And if, GB, HD being unequal, and GB greater, AG, F be added to GB and CH, E be added to HD,
it follows that AB, F are greater than CD, E.