The square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.

Let *AB* be a minor and *CD* a rational straight line, and to the rational straight line *CD* let *CE* be applied equal to the square on *AB* and producing *CF* as breadth.

I say that *CF* is a fourth 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 *AG* and *GB* rational, but twice the rectangle *AG* by *GB* medial.

To *CD* apply *CH*, equal to the square on *AG*, producing *CK* as breadth, and *KL*, equal to the square on *BG*, producing *KM* as breadth. Then the whole *CL* equals the sum of the squares on *AG* and *GB*.

And the sum of the squares on *AG* and *GB* is rational, therefore *CL* is also rational.

And it is applied to the rational straight line *CD* producing *CM* as breadth, therefore *CM* is also rational and commensurable in length 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 the point *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 medial and equals *FL*, 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 rational, while twice the rectangle *AG* by *GB* is medial, therefore the sum of the squares on *AG* and *GB* is incommensurable with twice the rectangle *AG* by *GB*.

But *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 that it is also a fourth apotome.

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

But *CH* is to *NL* as *CK* is to *NM*, and *NL* is to *KL* as *NM* is to *KM*, therefore *CK* is to *MN* as *MN* is to *KM*.

Therefore the rectangle *CK* by *KM* equals the square on *MN*, that is, to the fourth part of the square on *FM*.

Since *CM* and *MF* are two unequal straight lines, and the rectangle *CK* by *KM*, equal to the fourth part of the square on *MF* 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 whole *CM* is commensurable in length with the rational straight line *CD* set out, therefore *CF* is a fourth apotome.

Therefore, *the square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.*

Q.E.D.