(related to Proposition: Prop. 10.014: Commensurability of Squares on Proportional Straight Lines)

- 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 $F$.

- For since as $A$ is to $B$, so $C$ (is) to $D$, thus as the (square) on $A$ is to the (square) on $B$, so the (square) on $C$ (is) to the (square) on $D$ [Prop. 6.22].
- But the (sum of the squares) on $E$ and $B$ is equal to the (square) on $A$, and the (sum of the squares) on $D$ and $F$ is equal to the (square) on $C$.
- Thus, as the (sum of the squares) on $E$ and $B$ is to the (square) on $B$, so the (sum of the squares) on $D$ and $F$ (is) to the (square) on $D$.
- Thus, via separation, as the (square) on $E$ is to the (square) on $B$, so the (square) on $F$ (is) to the (square) on $D$ [Prop. 5.17].
- Thus, also, as $E$ is to $B$, so $F$ (is) to $D$ [Prop. 6.22].
- Thus, inversely, as $B$ is to $E$, so $D$ (is) to $F$ [Prop. 5.7 corr.] 1.
- But, as $A$ is to $B$, so $C$ also (is) to $D$.
- Thus, via equality, as $A$ is to $E$, so $C$ (is) to $F$ [Prop. 5.22].
- Therefore, $A$ 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 $F$ [Prop. 10.11].
- Thus, if, and so on ....∎

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