Construction of Second Apotome of Medial is Unique

Proof

 * Euclid-X-81.png

Let $AB$ be the first apotome of a medial straight line.

Let $BC$ be added to $AB$ such that:
 * $AC$ and $CB$ are medial straight lines
 * $AC$ and $CB$ are commensurable in square only
 * $AC \cdot CB$ is a medial rectangle.

It is to be proved that no other medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a medial rectangle.

Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated.

Then by definition of the first apotome of a medial straight line, $AD$ and $DB$ are such that:
 * $AD$ and $DB$ are medial straight lines
 * $AD$ and $DB$ are commensurable in square only
 * $AD \cdot DB$ is a medial rectangle.

Let $EF$ be a rational straight line.

Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth.

Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth.

Therefore from :
 * the remainder $EL$ equals $AB^2$.

Thus $AB$ equals the "side" of $EL$.

Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth.

But $EL = AB^2$.

Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$.

We have that $AC$ and $CB$ are medial straight lines.

Therefore $AC^2$ and $CB^2$ are also medial.

Also $AC^2 + CB^2 = EG$.

So by:

and:

it follows that:
 * $EG$ is medial.

But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth.

Therefore from :
 * $EM$ is rational and incommensurable in length with $EF$.

We have that $AC \cdot CB$ is a medial rectangle.

So from :
 * $2 \cdot AC \cdot CB$ is a medial rectangle.

But $2 \cdot AC \cdot CB = HG$.

Therefore $HG$ is medial.

Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth.

Therefore by :
 * $HM$ is rational and incommensurable in length with $EF$.

Also, $AC$ and $CB$ are commensurable in square only.

Therefore $AC$ and $CB$ are incommensurable in length.

But:
 * $AC : CB = AC^2 : AC \cdot CB$

Therefore from :
 * $AC^2$ is incommensurable with $AC \cdot CB$.

But $AC^2 + CB^2$ is commensurable with $AC^2$.

Also from :
 * $2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$.

Therefore by :
 * $AC^2 + CB^2$ incommensurable with $AC \cdot CB$.

We have that:
 * $EG = AC^2 + CB^2$

and:
 * $GH = 2 \cdot AC \cdot CB$

Therefore:
 * $EG$ is incommensurable with $HG$.

But from :
 * $EG : HG = EM : HM$

Therefore from :
 * $EM$ is incommensurable in length with $MH$.

But both $EM$ and $MH$ are rational straight lines.

Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only.

Therefore $EH$ is an apotome, and $HM$ an annex to it.

Similarly it can be shown that $HN$ is also an annex to it.

Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole.

From, this is impossible.

The result follows.