If a number is part of a number, and another (number) is the same part of another, then the sum (of the leading numbers) will also be the same part of the sum (of the following numbers) that one (number) is of another. * For let a number $A$ be part of a [number] $BC$, and another (number) $D$ (be) the same part of another (number) $EF$ that $A$ (is) of $BC$. * I say that the sum $A$, $D$ is also the same part of the sum $BC$, $EF$ that $A$ (is) of $BC$.
This proposition states $$\begin{array}{rclc}BC&=&n\cdot A&\wedge\\ EF&=&n\cdot D\\ &\Downarrow&\\ BC+EF&=&n(A+D). \end{array}$$ In particular, $$n\mid BC\wedge n\mid EF\Rightarrow n\mid (BC+EF).$$
Proofs: 1
