Root of Area contained by Rational Straight Line and Third Binomial

== Theorem ==
== Proof ==
:
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.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory