Square on First Bimedial Straight Line applied to Rational Straight Line

Theorem

In the words of Euclid:

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

Proof

Let $AB$ be a first bimedial straight line divided into its medials 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 second 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 first bimedial which has been divided into its medials at $C$.

Therefore, by definition, $AC$ and $CB$ are medial straight lines which are commensurable in square only such that $AC \cdot CB$ is a rational rectangle.

Thus, by definition, $AC^2$ and $CB^2$ are also medial.

From:

Proposition $15$ of Book $\text{X}$: Commensurability of Sum of Commensurable Magnitudes

and:

Porism to Proposition $23$ of Book $\text{X}$: Straight Line Commensurable with Medial Straight Line is Medial

it follows that:

$DL$ is medial.

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

$MD$ is rational and incommensurable in length with $DE$.

As $2 \cdot AC \cdot CB$ is rational, so is $MF$.

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

$MG$ is rational straight line and commensurable in length with $ML$, which equals $DE$.
$DM$ is incommensurable in length with $MG$.

But $DM$ and $MG$ are both rational.

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

Therefore by definition $DG$ is binomial.

It remains to be proved that $DG$ is a second binomial straight line.

$AC^2 + CB^2 > 2 \cdot AC \cdot CB$

Therefore $DL > MF$.

$DM > MG$

Since:

$AC^2$ is commensurable with $CB^2$

it follows that:

$DH$ 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 $KM$.

Also $DK \cdot KM = MN^2$.

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

Also:

$MG$ is commensurable in length with $DE$.

Therefore $DG$ is a second binomial straight line.

$\blacksquare$