If four straight lines are proportional, and the square on the first is greater than (the square on) the second by the (square) on (some straight line) commensurable [in length] with the first, then the square on the third will also be greater than (the square on) the fourth by the (square) on (some straight line) commensurable [in length] with the third. And if the square on the first is greater than (the square on) the second by the (square) on (some straight line) incommensurable [in length] with the first, then the square on the third will also be greater than (the square on) the fourth by the (square) on (some straight line) incommensurable [in length] with the third.

- Let $A$, $B$, $C$, $D$ be four proportional straight lines, (such that) as $A$ (is) to $B$, so $C$ (is) to $D$.
- And let the square on $A$ be greater than (the square on) $B$ by the (square) on $E$, and let the square on $C$ be greater than (the square on) $D$ by the (square) on $F$.
- I say that $A$ is either commensurable (in length) with $E$, and $C$ is also commensurable with $F$, or $A$ is incommensurable (in length) with $E$, and $C$ is also incommensurable with $F$.

(not yet contributed)

Proofs: 1

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"

**Prime.mover and others**: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016