If as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole. * For 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 $EB$ to the remainder $FD$ will also be as the whole $AB$ (is) to the whole $CD$.
In modern notation, this proposition reads that if \[\frac\alpha\beta=\frac\gamma\delta,\] then \[\frac\alpha\beta=\frac{\alpha-\gamma}{\beta-\delta},\] for all positive real numbers \(\alpha,\beta,\gamma,\delta\) with \(\alpha > \gamma\) and \(\beta > \delta\).