Root of Area contained by Rational Straight Line and First Binomial

Proof

 * Euclid-X-54.png

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

Let $E$ divide $AD$ into its terms.

From this is possible at only one place.

Let $AE$ be its greater term.

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, $AE$ is commensurable in length with $AB$.

Let $ED$ be bisected at $F$.

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

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

By, it divides it into commensurable parts.

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

Then $AG$ 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 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 $MO$.

It remains to be demonstrated that $MO$ is binomial.

We have that $AG$ is commensurable in length with $GE$.

Therefore by :
 * $AE$ is commensurable in length with both $AG$ and $GE$.

But by hypothesis $AE$ is also commensurable in length with $AB$.

Therefore by :
 * $AB$ is commensurable in length with both $AG$ and $GE$.

We have that $AB$ is a rational straight line.

Therefore $AG$ and $GE$ are rational straight lines.

Therefore each of the rectangles $AH$ and $GK$ are rational.

Therefore by :
 * each of the rectangles $AH$ and $GK$ are rational.

and:
 * $AH$ is commensurable with $GK$.

But:
 * $AH = SN$

and:
 * $GK = NQ$

Therefore $SN$ and $NQ$, the squares on $MN$ and $NO$, are rational and commensurable.

Since:
 * $AE$ is incommensurable in length with $ED$

while:
 * $AE$ is commensurable in length with $AG$

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

it follows from that:
 * $AG$ is incommensurable in length with $EF$.

From:

and:

it follows that:
 * $AH$ is incommensurable with $EL$.

Therefore $SN$ is also incommensurable with $MR$.

But from :
 * $SN : MR = PN : NR$

and so from :
 * $PN$ is incommensurable with $NR$.

But:
 * $PN = MN$ and $NR = NO$

Therefore $MN$ is incommensurable with $NO$.

Also:
 * $MN^2$ is commensurable with $NO^2$

and:
 * $MN^2$ and $NO^2$ are both rational.

Therefore $MN$ and $NO$ are rational straight lines which are commensurable in square only.

Therefore by definition $MO$ is binomial and the "side" of the area $AC$.