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

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

I say that the side of the area *AB* is a first apotome of a medial straight line.

Let *DG* be the annex to *AD*. Then *AG* and *GD* are rational straight lines commensurable in square only, and the annex *DG* is commensurable with the rational straight line *AC* set out, while the square on the whole *AG* is greater than the square on the annex *GD* by the square on a straight line commensurable in length with *AG*.

Since the square on *AG* is greater than the square on *GD* by the square on a straight line commensurable with *AG*, therefore, if there is applied to *AG* a parallelogram equal to the fourth part of the square on *GD* and deficient by a square figure, then it divides it into commensurable parts.

Bisect, then, *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 in length with *FG*.

Therefore *AG* is also commensurable in length with each of the straight lines *AF* and *FG*.

But *AG* is rational and incommensurable in length with *AC*, therefore each of the straight lines *AF* and *FG* is also rational and incommensurable in length with *AC*. Therefore each of the rectangle *AI* by *FK* is medial.

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

But *DG* is commensurable in length with *AC*.

Therefore each of the rectangles *DH* and *EK* is rational.

Construct the square *LM* equal to *AI*, and subtract *NO*, equal to *FK*, about the same angle with *LM*, namely the angle *LPM*. Then the squares *LM* and *NO* are about the same diameter.

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

Since *AI* and *FK* are medial and equal the squares on *LP* and *PN*, the squares on *LP* and *PN* are also medial, therefore *LP* and *PN* are also medial straight lines commensurable in square only.

Since the rectangle *AF* by *FG* equals the square on *EG*, therefore *AF* is to *EG* as *EG* is to *FG*, while *AF* is to *EG* as *AI* is to *EK*, and *EG* is to *FG* as *EK* is to *FK*. Therefore *EK* is a mean proportional between *AI* and *FK*.

But *MN* is also a mean proportional between the squares *LM* and *NO*, and *AI* equals *LM* while *FK* equals *NO*, therefore *MN* also equals *EK*.

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

Since, then, the whole *AK* equals *LM* and *NO*, and, in these, *DK* equals the gnomon *UVW* and *NO*, therefore the remainder *AB* equals *TS*.

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

I say that *LN* is a first apotome of a medial straight line.

Since *EK* is rational and equals *LO*, therefore *LO*, that is, the rectangle *LP* by *PN*, is rational.

But *NO* was proved medial, therefore *LO* is incommensurable with *NO*.

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

Therefore *LP* and *PN* are medial straight lines commensurable in square only, which contain a rational rectangle. Therefore *LN* is a first apotome of a medial straight line.

And it is the side of the area *AB*.

Therefore the side of the area *AB* is a first apotome of a medial straight line.

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

Q.E.D.