If four magnitudes are proportional then the (sum of the) largest and the smallest [of them] is greater than the (sum of the) remaining two (magnitudes). * Let $AB$, $CD$, $E$, and $F$ be four proportional magnitudes, (such that) as $AB$ (is) to $CD$, so $E$ (is) to $F$. * And let $AB$ be the greatest of them, and $F$ the least. * I say that $AB$ and $F$ is greater than $CD$ and $E$.
In modern notation, this proposition reads that if $\alpha > \beta,\gamma >\delta > 0$ and if \[\frac{\alpha}{\beta}=\frac{\gamma}{\delta},\] then \[\alpha+\delta > \beta+\gamma.\]
Proofs: 1
Sections: 1