# Square on Binomial Straight Line applied to Rational Straight Line

## Theorem

In the words of Euclid:

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

### Lemma

In the words of Euclid:

If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

## Proof

Let $AB$ be a binomial straight line divided into its terms at $C$.

Let $AC > CB$.

Let $DE$ be a rational straight line.

Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.

It is to be demonstrated that $DG$ is a first binomial straight line.

$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.

Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.

Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.

Let $MG$ be bisected at $N$.

Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).

Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.

We have that $AB$ is a binomial which has been divided into its terms at $C$.

Therefore, by definition, $AC$ and $CB$ are rational straight lines which are commensurable in square only.

Therefore $AC^2$ and $CB^2$ are commensurable and rational.

$AC^2 + CB^2$ is a rational area.

But $AC^2 + CB^2 = DL$.

Therefore $DL$ is a rational area.

We have that $DL$ is applied to the rational straight line $DE$.

$DM$ is rational and commensurable in length with $DE$.

We have that $AC$ and $CB$ are rational straight lines which are commensurable in square only.

$2 \cdot AC \cdot CB = MF$ is medial.

We have that $MF$ is applied to the rational straight line $ML$.

$MG$ is also rational and incommensurable in length with $DE$.

But $MD$ is also rational and commensurable in length with $DE$.

$DM$ is incommensurable in length with $MG$.

But $DM$ and $MG$ are both rational straight lines which are commensurable in square only.

Therefore, by definition, $DG$ is binomial.

It remains to be proved that $DG$ is first binomial.

$AC \cdot CB$ is a mean proportional between $AC^2$ and $CB^2$.

Therefore $MO$ is a mean proportional between $DH$ and $KL$.

$DK : MN = MN : MK$
$DK \cdot KM = MN^2$

We have that $AC^2$ is commensurable with $CB^2$.

Therefore $DH$ is commensurable with $KL$.

So from:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes

it follows that:

$DK$ is commensurable in length with $KM$.
$AC^2 + CB^2 > 2 \cdot AC \cdot CB$

Therefore:

$DL > MF$
$DM > MG$

And:

$DK \cdot KM = MN^2 = \dfrac {MG^2} 4$

and:

$DK$ is commensurable in length with $KM$.
$DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$.

We also have that:

$DM$ and $MG$ are rational

and:

the greater term $DM$ is commensurable in length with the rational straight line $DE$.

Therefore by definition $DG$ is a first binomial straight line.

$\blacksquare$