Square on Rational Straight Line applied to Apotome

Proof

 * Euclid-X-113.png

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 $BD$.