Side of Area Contained by Rational Straight Line and Sixth Apotome

Proof

 * Euclid-X-93.png

Let the area $AB$ be contained by the rational straight line $AC$ and the sixth apotome $AD$.

It is to be proved that the "side" of $AB$ is a straight line which produces with a medial area a medial whole.

Let $DG$ be the annex of the sixth apotome $AD$.

Then, by definition:
 * $AG$ and $GD$ are rational straight lines which are commensurable in square only
 * neither the whole $AG$ nor the annex $GD$ is commensurable with the rational straight line $AC$
 * the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$.

Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure.

From :
 * that parallelogram divides $AG$ into incommensurable parts.

Let $DG$ be bisected at $E$.

Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure.

Therefore $AF$ is incommensurable with $FG$.

We have that $AF$ is incommensurable in length with $FG$.

But from :
 * $AF : FG = AI : FK$

Therefore from :
 * $AI$ is incommensurable with $FK$.

We have that $AG$ commensurable in length with $AC$, and both are rational.

Therefore from :
 * the rectangle $AK$ is medial.

We have that $DG$ is incommensurable in length with $AC$, while both are rational.

Therefore from :
 * the rectangles $DK$ is medial.

But $AG$ and $GD$ are rational straight lines which are commensurable in square only.

Therefore $AG$ is incommensurable in length with $GD$.

But:
 * $AG : GD = AK : KD$

Therefore from:

and:
 * $AK$ is incommensurable with $KD$.
 * $AK$ is incommensurable with $KD$.

Let the square $LM$ be constructed equal to $AI$.

Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$.

Therefore from :
 * the squares $LM$ and $NO$ are about the same diameter.

Let $PR$ be the diameter of $LM$ and $NO$.

It is to be shown that $LN$ is the "side" of the area $AB$.

We have that $AK$ is medial and equals $LP^2 + PN^2$.

Therefore $LP^2 + PN^2$ is medial.

We have that the rectangle $DK$ is medial.

But $DK$ equals twice the rectangle contained by $LP$ and $PN$.

Therefore twice the rectangle contained by $LP$ and $PN$ is medial.

We have that $AK$ is incommensurable with $DK$.

Therefore $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$.

We have that $AI$ is incommensurable with $FK$.

Therefore $LP^2$ is incommensurable with $PN^2$.

Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is medial and such that $2 \cdot LP \cdot PN$ is also medial.

Further, $LP^2 + PN^2$ is incommensurable with $2 \cdot LP \cdot PN$.

Therefore by definition $LN$ is a straight line which produces with a medial area a medial whole.

But $LN$ is the "side" of the area $AB$.

Hence the result.