Straight Line Commensurable with Apotome

Proof

 * Euclid-X-103.png

Let $AB$ be an apotome.

Let $CD$ be commensurable in length with $AB$.

It is to be demonstrated that:
 * $CD$ is an apotome

and:
 * the order of $CD$ is the same as the order of $AB$.

Let $BE$ be the annex of $CD$.

This can be constructed uniquely by Construction of Apotome is Unique.

By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only.

From, let it be contrived that:
 * $BE : DF = AB : CD$

From :
 * $AE : CF = AB : CD$

and so from :
 * $AE : CF = BE : DF$

But $AB$ is commensurable in length with $CD$.

Therefore from :
 * $AE$ is commensurable in length with $CF$

and:
 * $BE$ is commensurable in length with $DF$.

We have that $AE$ and $EB$ are rational straight lines which are commensurable in square only.

Therefore from :
 * $CF$ and $FD$ are rational straight lines which are commensurable in square only.

Thus $CD$ is an apotome.

It remains to be shown that the order of $CD$ is the same as the order of $AB$.

We have that:
 * $AE : CF = BE : DF$

So from :
 * $AE : EB = CF : FD$

We have that:
 * $AE^2 = EB^2 + \lambda^2$

where either:
 * $\lambda$ is commensurable in length with $AE$

or:
 * $\lambda$ is incommensurable in length with $AE$.

Suppose $\lambda$ is commensurable in length with $AE$.

Then by :
 * $CF^2 > FD^2 + \mu^2$

where $\mu$ is commensurable in length with $CF$.

Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.

Then by :
 * $CD$ is commensurable in length with $\alpha$.

Similarly, let $BE$ be commensurable in length with $\alpha$.

Then by :
 * $DF$ is commensurable in length with $\alpha$.

Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.

Then by :
 * neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.

Let $\lambda$ be incommensurable in length with $AE$.

Then by :
 * $CF^2 > FD^2 + \mu^2$

where $\mu$ is incommensurable in length with $CF$.

Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.

Then by :
 * $CD$ is commensurable in length with $\alpha$.

Similarly, let $BE$ be commensurable in length with $\alpha$.

Then by :
 * $DF$ is commensurable in length with $\alpha$.

Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.

Then by :
 * neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.

It follows that $CD$ is an apotome of the same as the order as $AB$.