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

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

I say that *CF* is a third apotome.

Let *BG* be the annex to *AB*, therefore *AG* and *GB* are medial straight lines commensurable in square only which contains a medial rectangle.

Apply *CH*, equal to the square on *AG*, to *CD* producing *CK* as breadth, and apply *KL*, equal to the square on *BG*, to *KH* producing *KM* as breadth. 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*.

Now, 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 *LF* equals twice the rectangle *AG* by *GB*.

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

But the rectangle *AG* by *GB* is medial, therefore *FL* is also medial.

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

Since *AG* and *GB* are commensurable in square only, therefore *AG* is incommensurable in length with *GB*. Therefore the square on *AG* is also incommensurable with the rectangle *AG* by *GB*.

But the sum of the squares on *AG* and *GB* is commensurable with the square on *AG*, and twice the rectangle *AG* by *GB* with the rectangle *AG* by *GB*, 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 also 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 third apotome.

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

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, then, *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 commensurable parts, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line commensurable with *CM*.

And neither of the straight lines *CM* nor *MF* is commensurable in length with the rational straight line *CD* set out, therefore *CF* is a third apotome.

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

Q.E.D.