Square on Straight Line which produces Medial Whole with Medial Area applied to Rational Straight Line

Proof

 * Euclid-X-97.png

Let $AB$ be a straight line which produces with a medial area a medial whole.

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 sixth apotome.

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

Therefore, by definition, $AG$ and $GB$ are straight lines which are incommensurable in square which make $AG^2 + GB^2$ medial and $2 \cdot AG \cdot GB$ medial, and $AG^2 + GB^2$ incommensurable with $2 \cdot AG \cdot GB$.

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 $GB$.

But $AG^2 + GB^2$ medial.

Therefore $CL$ is medial.

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

Therefore by :
 * $CM$ is rational and incommensurable in length with $CD$.

We have that:
 * $CL = AG^2 + GB^2$

and:
 * $AB^2 = CE$

Therefore by :
 * $2 \cdot AG \cdot GB = FL$

But $2 \cdot AG \cdot GB$ is medial.

Therefore $FL$ is medial.

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

Therefore by :
 * $FM$ is rational and incommensurable in length with $CD$.

We have that $AG^2 + GB^2$ incommensurable with $2 \cdot AG \cdot GB$.

But:
 * $CL = AG^2 + GB^2$

and:
 * $FL = 2 \cdot AG \cdot GB$

Therefore $CL$ incommensurable with $FL$.

But from :
 * $CL : FL = CM : FM$

Therefore by :
 * $CM$ 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 sixth apotome.

We have that $FL = 2 \cdot AG \cdot GB$.

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 $AG$ and $GB$ are incommensurable in square.

Therefore $AG^2$ is incommensurable with $GB^2$.

But:
 * $CH = AG^2$

and:
 * $KL = BG^2$

Therefore $CH$ is incommensurable with $KL$.

But from :
 * $CH : KL = CK : KM$

Therefore by :
 * $CK$ is incommensurable 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:
 * $CH : NL = CK : NM$

and from :
 * $NL : KL = NM : KM$

Therefore by :
 * $CK : MN = MN : KM$

Therefore by :
 * $CK \cdot KM = NM^2 = \dfrac {FM^2} 4$

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 incommensurable with $KM$.

Therefore from :
 * $CM^2$ is greater than $MF^2$ by the square on a straight line which is incommensurable in length with $CM$.

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

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