Side of Area Contained by Rational Straight Line and Fourth Apotome

== Theorem ==
== Proof ==
:
Let the area $AB$ be contained by the rational straight line $AC$ and the fourth apotome $AD$.
It is to be proved that the "side" of $AB$ is minor.
Let $DG$ be the annex of the fourth apotome $AD$.
Then, by definition:
:$AG$ and $GD$ are rational straight lines which are commensurable in square only
:$AG$ is commensurable with the rational straight line $AC$
:the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$.
Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure.
From :
:that parallelogram divides $AG$ into incommensurable 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 $AG$ is rational and commensurable in length with $AC$.
Therefore from : 
:the rectangle $AK$ is rational.
We have that $DG$ isincommensurable in length with $AC$, while both are rational.
Therefore from : 
:the rectangles $DK$ is medial.
We have that $AF$ is incommensurable in length with $FG$.
Therefore from:
:
and:
::
:$AI$ is incommensurable with $FK$.
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 the irrational straight line called minor.
We have that $AK$ is rational and equal to $LP^2 + PN^2$.
Therefore $LP^2 + PN^2$ is rational.
We have that the rectangle $DK$ is medial.
But $DK$ equals twice the rectangle contained by $LP$ and $PN$.
Therefore twice the rectangle contained by $LP$ and $PN$ is medial.
We have that $AI$ is incommensurable with $FK$.
Therefore $LP^2$ is incommensurable with $PN^2$.
Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is rational and such that $2 \cdot LP \cdot PN$ is medial.
Therefore by definition $LN$ is minor.
But $LN$ is the "side" of the area $AB$.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory