Square on Straight Line which produces Medial Whole with Medial Area applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a [[[Definition:That which produces Medial Whole with Medial Area|straight line which produces with a medial area a medial whole]].
Let $CD$ be a rational straight line.
Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.
It is to be demonstrated that $CF$ is a sixth apotome.
Let $BG$ be the annex to $AB$.
Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial and $2 \cdot AG \cdot GB$ medial, and $AG^2 + GB^2$ incommensurable with $2 \cdot AG \cdot GB$.
Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.
Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.
Then the whole $CL$ is equal to the squares on $AG$ and $GB$.
But $AG^2 + GB^2$ medial.
Therefore $CL$ is medial.
We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.
Therefore by :
:$CM$ is rational and incommensurable in length with $CD$.
We have that:
:$CL = AG^2 + GB^2$
and:
:$AB^2 = CE$
Therefore by :
:$2 \cdot AG \cdot GB = FL$
But $2 \cdot AG \cdot GB$ is medial.
Therefore $FL$ is medial.
We have that $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.
Therefore by :
:$FM$ is rational and incommensurable in length with $CD$.
We have that $AG^2 + GB^2$ incommensurable with $2 \cdot AG \cdot GB$.
But:
:$CL = AG^2 + GB^2$
and:
:$FL = 2 \cdot AG \cdot GB$
Therefore $CL$ incommensurable with $FL$.
But from :
:$CL : FL = CM : FM$
Therefore by :
:$CM$ is incommensurable in length with $FM$.
But both $CM$ and $FM$ are rational.
Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $CF$ is an apotome.
It remains to be shown that $CF$ is a sixth apotome.
We have that $FL = 2 \cdot AG \cdot GB$.
Let $FM$ be bisected at the point $N$.
Let $NO$ be drawn through $N$ parallel to $CD$.
Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.
We have that $AG$ and $GB$ are incommensurable in square.
Therefore $AG^2$ is incommensurable with $GB^2$.
But:
:$CH = AG^2$
and:
:$KL = BG^2$
Therefore $CH$ is incommensurable with $KL$.
But from :
:$CH : KL = CK : KM$
Therefore by :
:$CK$ is incommensurable with $KM$.
We have that:
:the rectangle contained by $AG$ and $GB$ is a mean proportional between the squares on $AG$ and $GB$.
and:
:$CH = AG^2$
and:
:$KL = BG^2$
and:
:$NL = AG \cdot GB$
Therefore $NL$ is a mean proportional between $CH$ and $KL$.
Therefore:
:$CH : NL = NL : KL$
But we also have:
:$CH : NL = CK : NM$
and from :
:$NL : KL = NM : KM$
Therefore by :
:$CK : MN = MN : KM$
Therefore by :
:$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$
We have that:
:$CM$ and $MF$ are unequal straight lines
and:
:the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure
while:
:$CK$ is incommensurable with $KM$.
Therefore from :
:$CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.
Also, neither $MF$ nor $CM$ is commensurable in length with the rational straight line $CD$.
Therefore, by definition, $CF$ is a sixth apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory