The square on an apotome of a medial straight line applied to a rational straight line produces as breadth a first apotome.

Let *AB* be an apotome, and *CD* rational, and to *CD* let there be applied *CE* equal to the square on *AB* and producing *CF* as breadth.

I say that *CF* is a first apotome.

Let *BG* be the annex to *AB*. Then *AG* and *GB* are rational straight lines commensurable in square only.

To *CD* apply *CH*, equal to the square on *AG*, and *KL*, equal to the square on *BG*.

Then 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 *CD*. Then each of the rectangles *FO* and *LN* equals the rectangle *AG* by *GB*.

Now, since the sum of the squares on *AG* and *GB* is rational, and *DM* equals the sum of the squares on *AG* and *GB*, therefore *DM* is rational.

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

Again, since twice the rectangle *AG* by *GB* is medial, and *FL* equals twice the rectangle *AG* by *GB*, therefore *FL* is medial. And it is applied to the rational straight line *CD* producing *FM* as breadth, therefore *FM* is rational and incommensurable in length with *CD*.

Since the squares on *AG* and *GB* are 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*.

And *CL* equals the sum of the squares on *AG* and *GB*, and *FL* equals twice the rectangle *AG* by *GB*, therefore *DM* is incommensurable with *FL*.

But *DM* is to *FL* as *CM* is to *FM*, therefore *CM* is incommensurable in length with *FM*.

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 first apotome.

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 *BG*, 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*.

But *CH* is to *NL* as *CK* is to *NM*, and *NL* is to *KL* as *NM* is to *KM*, therefore the rectangle *CK* by *KM* equals the square on *NM*, that is, the fourth part of the square on *FM*.

Since the square on *AG* is commensurable with the square on *GB*, therefore *CH* is also commensurable with *KL*.

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

Since *CM* and *MF* are two unequal straight lines, and to *CM* there has been applied the rectangle *CK* by *KM* equal to the fourth part of the square on *FM* and deficient by a square figure, while *CK* is commensurable with *KM*, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line commensurable in length with *CM*.

And *CM* is commensurable in length with the rational straight line *CD* set out, therefore *CF* is a first apotome.

Therefore, *the square on an apotome of a medial straight line applied to a rational straight line produces as breadth a first apotome.*

Q.E.D.