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

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

I say that the side of the area *AB* is minor.

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

Since the square on *AG* is greater than the square on *GD* by the square on a straight line incommensurable 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 incommensurable 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 incommensurable in length with *FG*.

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

Since *AG* is rational and commensurable in length with *AC*, therefore the whole *AK* is rational.

Again, since *DG* is incommensurable in length with *AC*, and both are rational, therefore *DK* is medial.

Again, since *AF* is incommensurable in length with *FG*, therefore *AI* is incommensurable with *FK*.

Now construct the square *LM* equal to *AI*, and subtract *NO*, equal to *FK*, about the same angle, the angle *LPM*.

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

Since, then, the whole *AK* equals the sum of the squares *LM* and *NO*, and, in these, *DK* equals the gnomon *UVW* and the square *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 the irrational straight line called minor.

Since *AK* is rational and equals the sum of the squares on *LP* and *PN*, therefore the sum of the squares on *LP* and *PN* is rational.

Again, since *DK* is medial, and *DK* equals twice the rectangle *LP* by *PN*, therefore twice the rectangle *LP* by *PN* is medial.

And, since *AI* was proved incommensurable with *FK*, therefore the square on *LP* is also incommensurable with the square on *PN*.

Therefore *LP* and *PN* are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.

Therefore *LN* is the irrational straight line called minor, and it is the side of the area *AB*.

Therefore the side of the area *AB* is minor.

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

Q.E.D.