To find two rational straight lines commensurable in square only such that the square on the greater is greater than the square on the less by the square on a straight line incommensurable in length with the greater.

Set out a rational straight line *AB*, and two square numbers *CE* and *ED* such that their sum *CD* is not square. Describe the semicircle *AFB* on *AB*. Let it be contrived that *DC* is to *CE* as the square on *BA* is to the square on *AF*, and join *FB*.

Then, in a similar manner to the preceding, we can prove that *BA* and *AF* are rational straight lines commensurable in square only.

Since *DC* is to *CE* as the square on *BA* is to the square on *AF*, therefore, in conversion, *CD* is to *DE* as the square on *AB* is to the square on *BF*.

But *CD* does not have to *DE* the ratio which a square number has to a square number, therefore neither has the square on *AB* to the square on *BF* the ratio which a square number has to a square number. Therefore *AB* is incommensurable in length with *BF*.

And the square on *AB* is greater than the square on *AF* by the square on *FB* incommensurable with *AB*.

Therefore *AB* and *AF* are rational straight lines commensurable in square only, and the square on *AB* is greater than the square on *AF* by the square on *FB* incommensurable in length with *AB*.

Q.E.D.

This proposition is used in the next three propositions.