If an area is contained by a rational straight line and a first apotome, then the side of the area is an apotome.

Let the area *AB* be contained by the rational straight line *AC* and the first apotome *AD*.

I say that the side of the area *AB* is an apotome.

Since *AD* is a first apotome, let *DG* be its annex, therefore *AG* and *GD* are rational straight lines commensurable in square only. Also, the whole *AG* is commensurable with the rational straight line *AC* set out, and the square on *AG* is greater than the square on *GD* by the square on a straight line commensurable in length with *AG*.

Therefore if there is applied to *AG* a parallelogram equal to the fourth part of the square on *DG* and deficient by a square figure, then it divides it into commensurable parts.

Bisect *DG* at *E*, apply to *AG* a parallelogram equal to the square on *EG* and deficient by a square figure, and let it be the rectangle *AF* by *FG*. Then *AF* is commensurable with *FG*.

Draw *EH*, *FI*, and *GK* through the points *E*, *F*, and *G* parallel to *AC*.

Now, since *AF* is commensurable in length with *FG*, therefore *AG* is also commensurable in length with each of the straight lines *AF* and *FG*.

But *AG* is commensurable with *AC*, therefore each of the straight lines *AF* and *FG* is commensurable in length with *AC*.

And *AC* is rational, therefore each of the straight lines *AF* and *FG* is also rational, so that each of the rectangles *AI* and *FK* is also rational.

Now, since *DE* is commensurable in length with *EG*, therefore *DG* is also commensurable in length with each of the straight lines *DE* and *EG*.

But *DG* is rational and incommensurable in length with *AC*, therefore each of the straight lines *DE* and *EG* is also rational and incommensurable in length with *AC*. Therefore each of the rectangles *DH* and *EK* is medial.

Now make the square *LM* equal to *AI*, and subtract the square *NO* having a common angle with it, the angle *LPM*, and equal to *FK*. Then the squares *LM* and *NO* are about the same diameter.

Let *PR* be their diameter, and draw the figure.

Since the rectangle *AF* by *FG* equals the square on *EG*, therefore *AF* is to *EG* as *EG* is to *FG*.

But *AF* is to *EG* as *AI* is to *EK*, and *EG* is to *FG* as *EK* is to *KF*, therefore *EK* is a mean proportional between *AI* and *KF*.

But it was proved before that *MN* is also a mean proportional between *LM* and *NO*, and *AI* equals the square *LM*, and *KF* equals *NO*, therefore *MN* also equals *EK*.

But *EK* equals *DH*, and *MN* equals *LO*, therefore *DK* equals the gnomon *UVW* and *NO*.

But *AK* also equals the sum of the squares *LM* and *NO*, therefore the remainder *AB* equals *ST*.

But *ST* is the square on *LN*, therefore the square on *LN* equals *AB*. Therefore *LN* is the side of *AB*.

I say next that *LN* is an apotome.

Since each of the rectangles *AI* and *FK* is rational, and they equal *LM* and *NO*, therefore each of the squares *LM* and *NO*, that is, the squares on *LP* and *PN* respectively, is also rational. Therefore each of the straight lines *LP* and *PN* is also rational.

Again, since *DH* is medial and equals *LO*, therefore *LO* is also medial.

Since, then, *LO* is medial, while *NO* is rational, therefore *LO* is incommensurable with *NO*.

But *LO* is to *NO* as *LP* is to *PN*, therefore *LP* is incommensurable in length with *PN*.

And both are rational, therefore *LP* and *PN* are rational straight lines commensurable in square only. Therefore *LN* is an apotome.

And it is the side of the area *AB*, therefore the side of the area *AB* is an apotome.

Therefore, *if an area is contained by a rational straight line and a first apotome, then the side of the area is an apotome.*

Q.E.D.