Let there be two squares *AB* and *BC*, and let them be placed so that *DB* is in a straight line with *BE*. Then *FB* is also in a straight line with *BG*. Complete the parallelogram *AC*.

I say that *AC* is a square, that *DG* is a mean proportional between *AB* and *BC*, and further that *DC* is a mean proportional between *AC* and *CB*.

Since *DB* equals *BF*, and *BE* is to *BG*, therefore the whole *DE* equals the whole *FG*.

But *DE* equals each of the straight lines *AH* and *KC*, and *FG* equals each of the straight lines *AK* and *HC*, therefore each of the straight lines *AH* and *KC* also equals each of the straight lines *AK* and *HC*.

Therefore the parallelogram *AC* is equilateral. And it is also rectangular, therefore *AC* is a square.

Since *FB* is to *BG* as *DB* is to *BE*, while *FB* is to *BG* as *AB* is to *DG*, and *DB* is to *BE* as *DG* is to *BC*, therefore *AB* is to *DG* as *DG* is to *BC*.

Therefore *DG* is a mean proportional between *AB* and *BC*.

I say next that *DC* is also a mean proportional between *AC* and *CB*.

Since *AD* is to *DK* as *KG* is to *GC*, for they are equal respectively, and, taken jointly, *AK* is to *KD* as *KC* is to *CG*, while *AK* is to *KD* as *AC* is to *CD*, and *AC* is to *CG* as *DC* is to *CB*, therefore *AC* is to *DC* as *DC* is to *BC*.

Therefore *DC* is a mean proportional between *AC* and *CB*.

Q.E.D.

If an area is contained by a rational straight line and the first binomial, then the side of the area is the irrational straight line which is called binomial.

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

I say that the side of the area *AC* is the irrational straight line which is called binomial.

Since *AD* is a first binomial straight line, divide it into its terms at *E*, and let *AE* be the greater term.

It is then manifest that *AE* and *ED* are rational straight lines commensurable in square only, the square on *AE* is greater than the square on *ED* by the square on a straight line commensurable with *AE*, and *AE* is commensurable in length with the rational straight line *AB* set out.

Bisect *ED* at the point *F*.

Then, since the square on *AE* is greater than the square on *ED* by the square on a straight line commensurable with *AE*, therefore, if there is applied to the greater *AE* a parallelogram equal to the fourth part of the square on the less, that is, to the square on *EF*, and deficient by a square figure, then it divides it into commensurable parts.

Apply to *AE* the rectangle *AG* by *GE* equal to the square on *EF*. Then *AG* is commensurable in length with *EG*.

Draw *GH*, *EK*, and *FL* from *G*, *E*, and *F* parallel to either of the straight lines *AB* and *CD*. Construct the square *SN* equal to the parallelogram *AH*, and the square *NQ* equal to *GK*, and place them so that *MN* is in a straight line with *NO*. Then *RN* is also in a straight line with *NP*. Complete the parallelogram *SQ*. Then *SQ* is a square.

Now, since the rectangle *AG* by *GE* equals the square on *EF*, therefore *AG* is to *EF* as *FE* is to *EG*.

Therefore *AH* is to *EL* as *EL* is to *KG*. Therefore *EL* is a mean proportional between *AH* and *GK*.

But *AH* equals *SN*, and *GK* equals *NQ*, therefore *EL* is a mean proportional between *SN* and *NQ*. But *MR* is also a mean proportional between the same *SN* and *NQ*, therefore *EL* equals *MR*, so that it also equals *PO*.

But *AH* and *GK* also equal *SN* and *NQ*, therefore the whole *AC* equals the whole *SQ*, that is, it equals the square on *MO*, Therefore *MO* is the side of *AC*.

I say next that *MO* is binomial.

Since *AG* is commensurable with *GE*, therefore *AE* is also commensurable with each of the straight lines *AG* and *GE*.

But *AE* is also, by hypothesis, commensurable with *AB*, therefore *AG* and *GE* are also commensurable with *AB*.

And *AB* is rational, therefore each of the straight lines *AG* and *GE* is also rational. Therefore each of the rectangles *AH* and *GK* is rational, and *AH* is commensurable with *GK*.

But *AH* equals *SN*, and *GK* equals *NQ*, therefore the sum of *SN* and *NQ*, that is the squares on *MN* and *NO*, are rational and commensurable.

Since *AE* is incommensurable in length with *ED*, while *AE* is commensurable with *AG*, and *DE* is commensurable with *EF*, therefore *AG* is also incommensurable with *EF*, so that *AH* is also incommensurable with *EL*.

But *AH* equals *SN*, and *EL* equals *MR*, therefore *SN* is also incommensurable with *MR*. But *SN* is to *MR* as *PN* is to *NR*, therefore *PN* is incommensurable with *NR*.

But *PN* equals *MN*, and *NR* equals *NO*, therefore *MN* is incommensurable with *NO*. And the square on *MN* is commensurable with the square on *NO*, and each is rational, therefore *MN* and *NO* are rational straight lines commensurable in square only.

Therefore *MO* is binomial and the side of *AC*.

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

Q.E.D.