# Square on Side of Sum of two Medial Area applied to Rational Straight Line

## Theorem

In the words of Euclid:

The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.

## Proof

Let $AB$ be the side of the sum of two medial areas divided 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 sixth 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 the side of the sum of two medial areas divided at $C$.

Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that:

$AC^2 + CB^2$ is medial
$AC \cdot CB$ is a medial rectangle
$AC^2 + CB^2$ is incommensurable with $AC \cdot CB$.

Since $AC^2 + CB^2$ is medial, $DL$ is medial.

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

We have that:

$2 \cdot AC \cdot CB$, which equals $MF$, is medial.

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

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

Therefore $DM$ and $MG$ are rational straight lines which are incommensurable in length with $DE$.

We have that $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.

Thus $DL$ is incommensurable with $MF$.

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:

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

That is, $DM$ 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 sixth binomial straight line.

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

Therefore $DL > MF$.

$DM > MG$

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

Because:

$AC^2$ is incommensurable with $CB^2$

it follows that:

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

Also, neither $DM$ nor $MG$ is commensurable in length with the rational straight line $DE$.

Therefore $DG$ is a sixth binomial straight line.

$\blacksquare$