If as the whole (of a number) is to the whole (of another), so a (part) taken away (is) to a (part) taken away, then the remainder will also be to the remainder as the whole (is) to the whole. * Let the whole $AB$ be to the whole $CD$ as the (part) taken away $AE$ (is) to the (part) taken away $CF$. * I say that the remainder $EB$ is to the remainder $FD$ as the whole $AB$ (is) to the whole $CD$.
In modern notation, this proposition states that if \[\frac{AB}{CD}=\frac{AE+EB}{CF+FD}=\frac{AE}{CF},\] then \[\frac{EB}{FD}=\frac{AB}{CD}.\]
Proofs: 1
Proofs: 1
