Square on Second Bimedial Straight Line applied to Rational Straight Line

Proof

 * Euclid-X-60.png

Let $AB$ be a second 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 third binomial straight line.

From :
 * $AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

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 second 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 medial rectangle.

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

From:

and:

it follows that:
 * $DL$ is medial.

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

Therefore from :
 * $MD$ is rational and incommensurable in length with $DE$.

For the same reason:
 * $MG$ is also rational straight line and incommensurable in length with $ML$, which equals $DE$.

Therefore each of $DM$ and $MG$ is rational straight line and incommensurable in length with $DE$.

We have that $AC$ is incommensurable in length with $CB$.

Also:
 * $AC : CB = AC^2 : AC \cdot CB$

Therefore by :
 * $AC^2$ is incommensurable with $AC \cdot CB$.

Hence by:

and:

it follows that:
 * $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.

That is:
 * $DL$ is incommensurable with $MF$.

So from:

and:

it follows that:
 * $DM$ is incommensurable in length with $MG$.

But $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other.

Therefore by definition $DG$ is binomial.

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

From :
 * $AC^2 + CB^2 > 2 \cdot AC \cdot CB$

Therefore $DL > MF$.

From :
 * $DM > MG$

Since:
 * $AC^2$ is commensurable with $CB^2$

it follows that:
 * $DH$ is commensurable with $KL$.

So from:

and:

it follows that:
 * $DK$ is commensurable in length with $KM$.

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

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

Also:
 * Neither $DM$ nor $MG$ is commensurable in length with $DE$.

Therefore $DG$ is a third binomial straight line.