The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.

Let *AB* be the straight line which produces with a rational area a medial whole, and *CD* a rational straight line, and to *CD* let *CE* be applied equal to the square on *AB* and producing *CF* as breadth.

I say that *CF* is a fifth apotome.

Let *BG* be the annex to *AB*. Then *AG* and *GB* are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational.

To *CD* apply *CH* equal to the square on *AG*, and *KL* equal to the square on *GB*. Then the whole *CL* equals the sum of the squares on *AG* and *GB*.

But the sum of the squares on *AG* and *GB* together is medial, therefore *CL* is medial.

And it is applied to the rational straight line *CD* producing *CM* as breadth, therefore *CM* is rational and incommensurable with *CD*.

Since the whole *CL* equals the sum of the squares on *AG* and *GB*, and, in these, *CE* equals the square on *AB*, therefore the remainder *FL* equals twice the rectangle *AG* by *GB*.

Bisect *FM* at *N*, and draw *NO* through *N* parallel to either of the straight lines *CD* or *ML*. Then each of the rectangles *FO* and *NL* equals the rectangle *AG* by *GB*.

And, since twice the rectangle *AG* by *GB* is rational and equal to *FL*, therefore *FL* is rational.

And it is applied to the rational straight line *EF* producing *FM* as breadth, therefore *FM* is rational and commensurable in length with *CD*.

Now, since *CL* is medial, and *FL* rational, therefore *CL* is incommensurable with *FL*.

But *CL* is to *FL* as *CM* is to *MF*, therefore *CM* is incommensurable in length with *MF*.

And both are rational, therefore *CM* and *MF* are rational straight lines commensurable in square only. Therefore *CF* is an apotome.

I say next that it is also a fifth apotome.

We can prove similarly that the rectangle *CK* by *KM* equals the square on *NM*, that is, the fourth part of the square on *FM*.

And, since the square on *AG* is incommensurable with the square on *GB*, while the square on *AG* equals *CH*, and the square on *GB* equals *KL*, therefore *CH* is incommensurable with *KL*.

But *CH* is to *KL* as *CK* is to *KM*, therefore *CK* is incommensurable in length with *KM*.

Since *CM* and *MF* are two unequal straight lines, and a parallelogram equal to the fourth part of the square on *FM* and deficient by a square figure has been applied to *CM*, and divides it into incommensurable parts, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line incommensurable with *CM*.

And the annex *FM* is commensurable with the rational straight line *CD* set out, therefore *CF* is a fifth apotome.

Therefore, *the square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.*

Q.E.D.