Square on Second Apotome of Medial Straight Line applied to Rational Straight Line

Proof

 * Euclid-X-97.png

Let $AB$ be a second apotome of a medial straight line.

Let $CD$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.

It is to be demonstrated that $CF$ is a third apotome.

Let $BG$ be the annex to $AB$.

Therefore, by definition, $AG$ and $GB$ are medial straight lines which are commensurable in square only which contain a medial rectangle.

Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.

Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.

Then the whole $CL$ is equal to the squares on $AG$ and $GB$.

From:

and:

it follows that:
 * $CL$ is medial.

We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.

Therefore by :
 * $CM$ is rational and incommensurable in length with $CD$.

We have that:
 * $CL = AG^2 + GB^2$

and:
 * $AB^2 = CE$

Therefore by :
 * $2 \cdot AG \cdot GB = FL$

Let $FM$ be bisected at the point $N$.

Let $NO$ be drawn through $N$ parallel to $CD$.

Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.

We have that the to the rectangle contained by $AB$ and $GB$ is medial.