If a number is part of a number, and another (number) is the same part of another, also, alternately, which(ever) part, or parts, the first (number) is of the third, the second (number) will also be the same part, or the same parts, of the fourth. * For let a number $A$ be part of a number $BC$, and another (number) $D$ (be) the same part of another $EF$ that $A$ (is) of $BC$. * I say that, also, alternately, which(ever) part, or parts, $A$ is of $D$, $BC$ is also the same part, or parts, of $EF$.
In modern notation, this proposition states that if $A=\frac{BC}{n}$ and $D=\frac{EF}{n}$ for some integers $n \ge 1$ and $0 < BC < EF,$ then, using the division with quotient and remainder, $$\begin{array}{rclccl} D&=&m\cdot A&+&r&0\le r < A\\ &\Updownarrow&\\ EF&=&m\cdot BC&+&nr&0\le nr < BC. \end{array}$$
Proofs: 1
