If there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less but falling short by a square, then it divides it into incommensurable parts.

Let *A* and *BC* be two unequal straight lines, of which *BC* is the greater, and to *BC* let there be applied a parallelogram equal to the fourth part of the square on the less, *A*, but falling short by a square. Let this be the rectangle *BD* by *DC*, and let *BD* be incommensurable in length with *DC*.

I say that the square on *BC* is greater than the square on *A* by the square on a straight line incommensurable with *BC*.

With the same construction as before, we can prove similarly that the square on *BC* is greater than the square on *A* by the square on *FD*.

It is to be proved that *BC* is incommensurable in length with *DF*.

Since *BD* is incommensurable in length with *DC*, therefore *BC* is also incommensurable in length with *CD*.

But *DC* is commensurable with the sum of *BF* and *DC*, therefore *BC* is incommensurable with the sum of *BF* and *DC*, so that *BC* is also incommensurable in length with the remainder *FD*.

And the square on *BC* is greater than the square on *A* by the square on *FD*, therefore the square on *BC* is greater than the square on *A* by the square on a straight line incommensurable with *BC*.

Next, let the square on *BC* be greater than the square on *A* by the square on a straight line incommensurable with *BC*. Apply to *BC* a parallelogram equal to the fourth part of the square on *A* but falling short by a square. Let this be the rectangle *BD* by *DC*.

It is to be proved that *BD* is incommensurable in length with *DC*.

With the same construction, we can prove similarly that the square on *BC* is greater than the square on *A* by the square on *FD*.

But the square on *BC* is greater than the square on *A* by the square on a straight line incommensurable with *BC*, therefore *BC* is incommensurable in length with *FD*, so that *BC* is also incommensurable with the remainder, the sum of *BF* and *DC*.

But the sum of *BF* and *DC* is commensurable in length with *DC*, therefore *BC* is also incommensurable in length with *DC*, so that, taken separately, *BD* is also incommensurable in length with *DC*.

Therefore, *if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less but falling short by a square, and if it divides it into incommensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater. And if the square on the greater is greater than the square on the less by the square on a straight line incommensurable with the greater, and if there is applied to the greater a parallelogram equal to the fourth part of the square on the less but falling short by a square, then it divides it into incommensurable parts.*

Q.E.D.