Square on Rational Straight Line applied to Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $BD$ be an apotome.
Let $BD \cdot KH$ be the rectangle on $BD$ equal to $A^2$.
It is to be demonstrated that:
:$KH$ is a binomial straight line whose terms are commensurable with the terms of $BD$ and in the same ratio
and:
:the order of $KH$ is the same as the order of $BD$.
Let $DC$ be the annex of $BD$.
By the definition of apotome, $BC$ and $CD$ are rational straight lines which are commensurable in square only.
Let $BC \cdot G$ be the rectangle on $BC$ equal to $A^2$.
But $A^2$ is rational.
Therefore $BC \cdot G$ is also rational.
We have that $BC \cdot G$ has been applied to a rational straight line $BC$.
Thus from :
:$G$ is rational and commensurable in length with $BC$.
We have that:
:$BC \cdot G = BD \cdot KH$
Therefore by :
:$CD \cdot BD = KH \cdot G$
But:
:$BC > BD$
and so from:
:
and:
:
it follows that:
:$KH > G$
Let $KE = G$.
Then $KE$ is commensurable in length with $BC$.
We have that:
:$CB : BD = HK : KE$
So from :
:$CD : BD = HF : FE$
Let it be contrived that $KH : HE = HF : FE$.
Therefore from :
:$KF : FH = KH : HE$
That is:
:$KF : FH = BC : CD$
But $BC$ and $CD$ are commensurable in square only.
Therefore from :
:$KF$ and $FH$ are commensurable in square only.
We have that:
:$KH : HE = KF : FH$
and:
:$KH : HE = HF : FE$
Therefore from :
:$KF : HE = HF : FE$
So from :
:$KF : FE = KF^2 : FH^2$
But $KF$ and $FH$ are commensurable in square.
So $KF^2$ is commensurable with $FH^2$.
Therefore from :
:$KF$ is commensurable in length with $FE$.
So from :
:$KF$ is commensurable in length with $KE$.
But $KE$ is rational and commensurable in length with $BC$.
Therefore from :
:$KF$ is rational and commensurable in length with $BC$.
We have that:
:$BC : CD = KF : FH$
Thus from :
:$BC : KF = DC : FH$
But $BC$ is commensurable in length with $KF$.
Therefore from :
:$FH$ is commensurable in length with $CD$.
But $BC$ and $CD$ are rational straight lines which are commensurable in square only.
Therefore $KF$ and $FH$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $KH$ is binomial.
We have that:
:$BC^2 = CD^2 + \lambda^2$
where either:
:$\lambda$ is commensurable in length with $BC$
or:
:$\lambda$ is incommensurable in length with $BC$.
First suppose $\lambda$ is commensurable in length with $BC$.
Then by :
: $KF^2 = FH^2 + \mu^2$
where $\mu$ is commensurable in length with $KF$.
Let $BC$ be commensurable in length with a rational straight line $\alpha$ which has been set out.
Then by:
:
and:
:
it follows that:
:$KF$ is commensurable in length with $\alpha$.
Let $CD$ be commensurable in length with $\alpha$.
Then by :
:$FH$ is commensurable in length with $\alpha$.
Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$.
Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$.
Next suppose $\lambda$ is incommensurable in length with $BC$.
Then by :
:$KF^2 = FH^2 + \mu^2$
where $\mu$ is incommensurable in length with $KF$.
Let $BC$ be commensurable in length with a rational straight line $\alpha$ which has been set out.
Then by:
:
and:
:
it follows that:
:$KF$ is commensurable in length with $\alpha$.
Let $CD$ be commensurable in length with $\alpha$.
Then by :
:$FH$ is commensurable in length with $\alpha$.
Let neither $BC$ nor $CD$ be commensurable in length with $\alpha$.
Then neither $KF$ nor $FH$ is commensurable in length with $\alpha$.
It follows that:
:the terms of $KH$ are commensurable with the terms of $BD$ and in the same ratio
and:
:the order of $KH$ is the same as the order of $BD$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory