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

Let *AB* be the straight line which produces with a medial 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 sixth 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, twice the rectangle *AG* by *GB* medial, and the sum of the squares on *AG* and *GB* incommensurable with twice the rectangle *AG* by *GB*.

Now to *CD* apply *CH* equal to the square on *AG* and producing *CK* as breadth, and *KL* equal to the square on *BG*. Then the whole *CL* equals the sum of the squares on *AG* and *GB*. Therefore *CL* is also medial.

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

Since *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*. And twice the rectangle *AG* by *GB* is medial, therefore *FL* is also medial.

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

Since the sum of the squares on *AG* and *GB* is incommensurable with twice the rectangle *AG* by *GB*, *CL* equals the sum of the squares on *AG* and *GB*, and *FL* equals twice the rectangle *AG* by *GB*, 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 sixth apotome.

Since *FL* equals twice the rectangle *AG* by *GB*, bisect *FM* at *N*, and draw *NO* through *N* parallel to *CD*, therefore each of the rectangles *FO* and *NL* equals the rectangle *AG* by *GB*.

And, since *AG* and *GB* are incommensurable in square, therefore the square on *AG* is incommensurable with the square on *GB*.

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

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

Since the rectangle *AG* by *GB* is a mean proportional between the squares on *AG* and *GB*, *CH* equals the square on *AG*, *KL* equals the square on *GB*, and *NL* equals the rectangle *AG* by *GB*, therefore *NL* is also a mean proportional between *CH* and *KL*. Therefore *CH* is to *NL* as *NL* is to *KL*.

And for the same reason as before the square on *CM* is greater than the square on *MF* by the square on a straight line incommensurable with *CM*.

And neither of them is commensurable with the rational straight line *CD* set out, therefore *CF* is a sixth apotome.

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

Q.E.D.