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
