Proof: By Euclid

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 $F$.

fig014e

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 ....
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