Side of Area Contained by Rational Straight Line and First Apotome

Proof

 * Euclid-X-91.png

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

It is to be proved that $AB$ is an apotome.

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

Then, by definition:
 * $AG$ and $GD$ are rational straight lines which are commensurable in square only
 * the whole $AG$ is commensurable with the rational straight line $AC$
 * the square on $AG$ is greater than the square on $GD$ by the square on a straight line which is commensurable in length with $AG$.

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

From :
 * that parallelogram divides $AG$ into commensurable 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 commensurable with $FG$.

Through $E$, $F$ and $G$ let $EH$, $FI$ and $GK$ be drawn parallel to $AC$.