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

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

I say that *CF* is a second apotome.

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

To *CD* apply *CH*, equal to the square on *AG*, producing *CK* as breadth, and *KL*, equal to the square on *GB*, producing *KM* as breadth.

Therefore the whole *CL* equals the sum of the squares on *AG*. 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*.

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

But twice the rectangle *AG* by *GB* is rational, therefore *FL* is rational.

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

Now, since the sum of the squares on *AG* and *GB*, that is, *CL*, is medial, while twice the rectangle *AG* by *GB*, that is, *FL*, is rational, therefore *CL* is incommensurable with *FL*.

But *CL* 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 second apotome.

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

Now, since the rectangle *AG* by *GB* is a mean proportional between the squares on *AG* and *GB*, the square on *AG* equals *CH*, the rectangle *AG* by *GB* equals *NL*, and the square on *BG* equals *KL*, 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 *MK*, therefore *CK* is to *NM* 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 *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 the greater, *CM*, and divides it into commensurable parts, 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 the annex *FM* is commensurable in length with the rational straight line *CD* set out, therefore *CF* is a second apotome.

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

Q.E.D.