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

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

I say that the side of the area *AB* is a second 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 neither of the straight lines *AG* and *GD* is commensurable in length with the rational straight line *AC* set out, while the square on the whole *AG* is greater than the square on the annex *DG* by the square on a straight line commensurable with *AG*.

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

Then *AF* and *FG* are commensurable. Therefore *AI* is also commensurable with *FK*.

Since *AF* and *FG* are commensurable in length, 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*, so that *AF* and *FG* are so also.

Therefore each of the rectangles *AI* and *FK* is medial.

Again, 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 *GD* 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.

Since *AG* and *GD* are commensurable in square only, therefore *AG* is incommensurable in length with *GD*.

But *AG* is commensurable in length with *AF*, and *DG* with *EG*, therefore *AF* is incommensurable in length with *EG*.

But *AF* is to *EG* as *AI* is to *EK*, therefore *AI* is incommensurable with *EK*.

Now construct the square *LM* equal to *AI*, and subtract *NO*, equal to *FK*, about the same angle with *LM*. Then *LM* and *NO* are about the same diameter.

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

Now, 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 *FK*, therefore *AI* is to *EK* 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*, and *FK* equals *NO*, therefore *EK* also equals *MN*.

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

But *AK* equals the sum of *LM* and *NO*, therefore the remainder *AB* equals *ST*, that is, to the square on *LN*. Therefore *LN* is the side of the area *AB*.

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

Since *AI* and *FK* were proved medial, and equal the squares on *LP*, therefore each of the squares on *LP* and *PN* is also medial. Therefore each of the straight lines *LP* and *PN* is medial.

Since *AI* is commensurable with *FK*, therefore the square on *LP* is also commensurable with the square on *PN*.

Again, since *AI* was proved incommensurable with *EK*, therefore *LM* is also incommensurable with *MN*, that is, the square on *LP* with the rectangle *LP* by *PN*, so that *LP* is also incommensurable in length with *PN*.

Therefore *LP* and *PN* are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle.

Since *EK* was proved medial, and equals the rectangle *LP* by *PN*, therefore the rectangle *LP* by *PN* is also medial, so that *LP* and *PN* are medial straight lines commensurable in square only which contain a medial rectangle.

Therefore *LN* is a second apotome of a medial straight line, and it is the side of the area *AB*.

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

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

Q.E.D.