Root of Area contained by Rational Straight Line and Third Binomial

Proof

 * Euclid-X-54.png

Let the rectangular area $ABCD$ be contained by the rational straight line $AB$ and the third binomial $AD$.

Let $E$ divide $AD$ into its terms such that $AE > ED$.

From this is possible at only one place.

By definition of binomial, $AE$ and $ED$ are rational straight lines which are commensurable in square only.

Thus $AE^2$ is greater than $ED^2$ by the square on a straight line which is commensurable with $AE$.

By, neither of the terms $AE$ and $ED$ is commensurable in length with $AB$.

Let $ED$ be bisected at $F$.

Using :
 * Let a parallelogram be applied to $AE$ equal to $EF^2$ and deficient by a square.

Let the rectangle contained by $AG$ and $GE$ equal to $EF^2$ be applied to $AE$.

By :
 * $AG$ is commensurable in length with $EG$.

Let $GH$, $EK$ and $FL$ be drawn from $G$, $E$ and $F$ parallel to $AB$ and $CD$.

Using :
 * let the square $SN$ be constructed equal to the parallelogram $AH$

and:
 * let the square $NQ$ be constructed equal to the parallelogram $GK$.

Let $SN$ and $NQ$ be placed so that $MN$ is in a straight line with $NO$.

Therefore $RN$ is also in a straight line with $NP$.

Let the parallelogram $SQ$ be completed.

From :
 * $SQ$ is a square.

We have that $AD$ is a third binomial straight line.

Therefore $AE$ and $ED$ are rational straight lines which are commensurable in square only.

We have that the rectangle contained by $AG$ and $GE$ is equal to $EF^2$.

Then by :
 * $AG : EF = EF : EG$

and so by :
 * $AH : EL = EL : KG$

Therefore $EL$ is a mean proportional between $AH$ and $GK$.

But:
 * $AH = SN$

and:
 * $GK = NQ$

therefore $EL$ is a mean proportional between $SN$ and $NQ$.

But from :
 * $MR$ is a mean proportional between $SN$ and $NQ$.

Therefore:
 * $EL = MR$

and so:
 * $EL = PO$

But:
 * $AH + GK = SN + NQ$

Therefore:
 * $AC = SQ$

That is:
 * $AC$ is an area contained by a rational straight line $AB$ and the first binomial $AD$, while the side of the area $SQ$ is $MO$.

We have that:
 * $AE$ is incommensurable in length with $ED$

and:
 * $ED$ is commensurable in length with $AB$

it follows from :
 * $AE$ is incommensurable in length with $AB$.

As:
 * $AG$ is commensurable in length with $EG$

from :
 * $AE$ is commensurable in length with each of $AG$ and $GE$.

But $AE$ is incommensurable in length with $AB$.

So from :
 * $AG$ and $GE$ are incommensurable in length with $AB$.

Therefore:
 * $BA$ and $AG$ are rational straight lines which are commensurable in square only

and:
 * $BA$ and $GE$ are rational straight lines which are commensurable in square only.

Therefore by definition rectangles $AH$ and $GK$ are both medial.

Therefore squares $SN$ and $NQ$ are both medial.

Therefore $MN$ and $NO$ are both medial straight lines.

It remains to be demonstrated that $MO$ is second bimedial.

We have that:
 * $DE$ is incommensurable in length with $AB = EK$

and:
 * $DE$ is commensurable with $EF$

so from :
 * $EF$ is incommensurable in length with $EK$.

Both $EF$ and $EK$ are rational straight lines.

Therefore $EF$ and $EK$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $EL = MR$ are medial.

But $MR$ is the rectangle contained by $MN$ and $MO$.

But $MN$ and $NO$ have been shown to be medial and commensurable in square.

Therefore, by definition, $MN$ and $NO$ are medial and commensurable in square only.

By definition, $MO$ is a second bimedial straight line.