Side of Area Contained by Rational Straight Line and First Apotome

== Theorem ==
== Proof ==
:
Let the area $AB$ be contained by the rational straight line $AC$ and the first apotome $AD$.
It is to be proved that the "side" of $AB$ 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 $AC$.
We have that $AF$ is commensurable with $FG$.
Therefore from :
:$AG$ is commensurable with each of $AF$ and $FG$.
But $AG$ is commensurable with $AC$.
Therefore from :
:each of the straight lines $AF$ and $FG$ is commensurable in length with $AC$.
We have that $AC$ is rational.
Therefore each of the straight lines $AF$ and $FG$ is rational.
Therefore from :
:each of the rectangles $AI$ and $FK$ is rational.
We have that $DE$ is commensurable in length with $EG$.
Therefore from :
:$DG$ is also commensurable in length with each of the straight lines $DE$ and $EG$.
But $DG$ is rational and incommensurable in length with $AC$.
Therefore from : 
:each of the straight lines $DE$ and $ED$ is rational and incommensurable in length with $AC$.
Therefore from : 
:each of the rectangles $DH$ and $EK$ is medial.
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 $NO$.
We have that the rectangle contained by $AF$ and $FG$ equals the square on $EG$.
Therefore from :
:$AF : EG = EG : FG$
But we also have:
:$AF : EG = AI : EK$
And from :
:$EG : FG = EK : KF$
Therefore from :
:$AI : EK = EK : FK$
Therefore $EK$ is a mean proportional between $AI$ and $FK$.
But from :
:$MN$ is a mean proportional between $LM$ and $NO$.
We have that:
:$AI$ equals the square $LM$
and:
:$KF$ equals the square $NO$.
Therefore:
:$MN = EK$
But:
:$EK = DH$
and:
:$MN = LO$
Therefore $DK$ equals the gnomon $UVW$ and $NO$.
But:
:$AK$ equals the squares $LM$ and $NO$.
Therefore the remainder $AB$ equals $ST$.
But $ST$ is the square on $LN$.
Therefore the square on $LN$ equals $AB$.
Therefore $LN$ is the "side" of $AB$.
Now it is to be shown that $LN$ is an apotome.
We have that each of the rectangles $AI$ and $FK$ is rational.
Also:
:$AI = LM$
and:
:$FK = NO$
Therefore each of the squares $LM$ and $NO$, that is the squares on $LP$ and $PN$, is rational.
Therefore $LP$ and $PN$ are both rational.
We have that $DH$ is medial and equals $LO$.
Therefore $LO$ is medial.
But $NO$ is rational.
Therefore $LO$ is incommensurable with $NO$.
But from :
:$LO : NO = LP : PN$
Therefore from :
:$LP$ is incommensurable in length with $PN$.
We have that both $LP$ and $PN$ are rational.
Thus $LP$ and $PN$ are rational straight lines which are commensurable in square only.
Therefore by definition $LN$ is an apotome.
But $LN$ is the "side" of the area $AB$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory