If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and the sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively). * For let a first (magnitude) $AB$ and a third $DE$ be equal multiples of a second $C$ and a fourth $F$ (respectively). * And let a fifth (magnitude) $BG$ and a sixth $EH$ also be (other) equal multiples of the second $C$ and the fourth $F$ (respectively). * I say that the first (magnitude) and the fifth, being added together, (to give) $AG$, and the third (magnitude) and the sixth, (being added together, to give) $DH$, will also be equal multiples of the second (magnitude) $C$ and the fourth $F$ (respectively).
If we are given two positive real numbers \(\alpha\), \(\beta\),^{1} and the following multiples of aliquot parts \(m > 1\), \(n > 1\): \[m\alpha=m\beta,\quad n\alpha=n\beta\quad\quad( * )\] then adding both equations gives us \[(m+n)\alpha=m\beta+n\beta.\]
See distributivity law for real numbers.
Proofs: 1
From a geometrical point of view, \(\alpha,\beta\) could e.g. mean the lengths of some segments, the areas of some plane figures or the volumes of some solids. ↩