Square on Second Apotome of Medial Straight Line applied to Rational Straight Line

Theorem

In the words of Euclid:

The square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.

Proof

Let $AB$ be a second apotome of a medial straight line.

Let $CD$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.

It is to be demonstrated that $CF$ is a third apotome.

Let $BG$ be the annex to $AB$.

Therefore, by definition, $AG$ and $GB$ are medial straight lines which are commensurable in square only which contain a medial rectangle.

Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.

Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.

Then the whole $CL$ is equal to the squares on $AG$ and $GB$.

From:

Proposition $15$ of Book $\text{X}$: Commensurability of Sum of Commensurable Magnitudes

and:

Porism to Proposition $23$ of Book $\text{X}$: Straight Line Commensurable with Medial Straight Line is Medial

it follows that:

$CL$ is medial.

We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.

$CM$ is rational and incommensurable in length with $CD$.

We have that:

$CL = AG^2 + GB^2$

and:

$AB^2 = CE$
$2 \cdot AG \cdot GB = FL$

Let $FM$ be bisected at the point $N$.

Let $NO$ be drawn through $N$ parallel to $CD$.

Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.

We have that the to the rectangle contained by $AB$ and $GB$ is medial.

Therefore $FL$ is medial.

Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.

$FM$ is rational and incommensurable in length with $CD$.

We have that $AG$ and $GB$ are commensurable in square only.

Therefore $AG$ is incommensurable in length with $GB$.

So from:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes:

we have that:

$AG^2$ is incommensurable with the rectangle contained by $AB$ and $GB$.

But the squares on $AG$ and $GB$ are commensurable with the square on $AG$.

Also $2 \cdot AG \cdot GB$ is commensurable with $AG \cdot GB$.

the squares on $AG$ and $GB$ are incommensurable with $2 \cdot AG \cdot GB$.

We have that:

$CL = AG^2 + GB^2$

and:

$FL = 2 \cdot AG \cdot GB$

Therefore $CL$ is incommensurable with $FL$.

$CL : FL = CM : FM$
$CM$ is incommensurable in length with $FM$.

But both $CM$ and $FM$ are rational.

Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $CF$ is an apotome.

It remains to be shown that $CF$ is a third apotome.

We have that the squares on $AG$ and $GB$ are commensurable with the square on $AG$.

Therefore $CH$ is commensurable with $KL$.

So from:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes:

we have that:

$CK$ is commensurable with $KM$.

We have that:

the rectangle contained by $AG$ and $GB$ is a mean proportional between the squares on $AG$ and $GB$.

and:

$CH = AG^2$

and:

$KL = BG^2$

and:

$NL = AG \cdot GB$

Therefore $NL$ is a mean proportional between $CH$ and $KL$.

Therefore:

$CH : NL = NL : KL$

But we also have:

$CH : NL = CK : NM$
$NL : KL = NM : KM$
$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$

We have that:

$CM$ and $MF$ are unequal straight lines

and:

the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure

while:

$CK$ is commensurable with $KM$.
$CM^2$ is greater than $MF^2$ by the square on a straight line which is commensurable in length with $CM$.

Also neither $CM$ nor $FM$ is commensurable in length with the rational straight line $CD$.

Therefore, by definition, $CF$ is a third apotome.

$\blacksquare$