If a straight line is cut into unequal parts, then the sum of the squares on the unequal parts is greater than twice the rectangle contained by the unequal parts.

Let *AB* be a straight line, and let it be cut into unequal parts at *C*, and let *AC* be the greater.

I say that the sum of the squares on *AC* and *CB* is greater than twice the rectangle *AC* by *CB*.

Bisect *AB* at *D*.

Since a straight line is cut into equal parts at *D* and into unequal parts at *C*, therefore the rectangle *AC* by *CB* together with the square on *CD* equals the square on *AD*, so that the rectangle *AC* by *CB* is less than the square on *AD*. Therefore twice the rectangle *AC* by *CB* is less than double the square on *AD*.

But the sum of the squares on *AC* and *CB* is double the sum of the squares on *AD* and *DC*, therefore the sum of the squares on *AC* and *CB* is greater than twice the rectangle *AC* by *CB*.

Q.E.D.

The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.

Let *AB* be a binomial straight line divided into its terms at *C*, so that *AC* is the greater term, let a rational straight line *DE* be set out, and let *DEFG* equal the square on *AB* be applied to *DE* producing *DG* as its breadth.

I say that *DG* is a first binomial straight line.

Apply to *DE* the rectangle *DH* equal to the square on *AC*, and *KL* equal to the square on *BC*. Then the remainder, twice the rectangle *AC* by *CB*, equals *MF*.

Bisect *MG* at *N*, and draw *NO* parallel to *ML* or *GF*. Then each of the rectangles *MO* and *NF* equals once the rectangle *AC* by *CB*.

Now, since *AB* is a binomial divided into its terms at *C*, therefore *AC* and *CB* are rational straight lines commensurable in square only.

Therefore the squares on *AC* and *CB* are rational and commensurable with one an other, so that the sum of the squares on *AC* and *CB* is also rational. And it equals *DL*, therefore *DL* is rational.

And it is applied to the rational straight line *DE*, therefore *DM* is rational and commensurable in length with *DE*.

Again, since *AC* and *CB* are rational straight lines commensurable in square only, therefore twice the rectangle *AC* by *CB*, that is *MF*, is medial.

And it is applied to the rational straight line *ML*, therefore *MG* is also rational and incommensurable in length with *ML*, that is, *DE*.

But *MD* is also rational and is commensurable in length with *DE*, therefore *DM* is incommensurable in length with *MG*.

And they are rational, therefore *DM* and *MG* are rational straight lines commensurable in square only. Therefore *DG* is binomial.

It is next to be proved that it is also a first binomial straight line.

Since the rectangle *AC* by *CB* is a mean proportional between the squares on *AC* and *CB*, therefore *MO* is also a mean proportional between *DH* and *KL*.

Therefore *DH* is to *MO* as *MO* is to *KL*, that is *DK* is to *MN* as *MN* is to *MK*. Therefore the rectangle *DK* by *KM* equals the square on *MN*.

Since the square on *AC* is commensurable with the square on *CB*, therefore *DH* is also commensurable with *KL*, so that *DK* is also commensurable with *KM*.

Since the sum of the squares on *AC* and *CB* is greater than twice the rectangle *AC* by *CB*, therefore *DL* is also greater than *MF*, so that *DM* is also greater than *MG*.

And the rectangle *DK* by *KM* equals the square on *MN*, that is, to the fourth part of the square on *MG*, and *DK* is commensurable with *KM*.

But, if there are two unequal straight lines, and to the greater there is applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divides it into commensurable parts, then the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater. Therefore the square on *DM* is greater than the square on *MG* by the square on a straight line commensurable with *DM*.

And *DM* and *MG* are rational, and *DM*, which is the greater term, is commensurable in length with the rational straight line *DE* set out.

Therefore *DG* is a first binomial straight line.

Therefore, *the square on the binomial straight line applied to a rational straight line produces as breadth the first binomial.*

Q.E.D.