Square on Rational Straight Line applied to Binomial Straight Line

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