Side of Remaining Area from Rational Area from which Medial Area Subtracted

Proof

 * Euclid-X-108.png

Let $BC$ be a rational area.

Let the medial area $BD$ be subtracted from $BC$.

It needs to be demonstrated that the "side" of the remainder $EC$ is either an apotome or a minor straight line.

Let $FG$ be a rational straight line.

Let the rectangle $GH$ be applied to $FG$ equal to $BC$ producing $FH$ as breadth.

Let the area $GK$ equal to $BC$ be subtracted from $GH$.

Then the remainder $EC$ is equal to $LH$.

We have that $BC$ is rational and $BD$ is medial.

We also have:
 * $BC = GH$

and:
 * $BD = GK$

Therefore $GH$ is rational and $GK$ is medial.

But $GH$ and $GK$ are applied to a rational straight line $FG$.

Therefore from :
 * $FH$ is rational and commensurable in length with $FG$

while from :
 * $FK$ is rational and incommensurable in length with $FG$.

Therefore by :
 * $FH$ is incommensurable in length with $FK$.

Therefore $FH$ and $FK$ are rational straight lines which are commensurable in square only.

Therefore $KH$ is an apotome and $KF$ is the annex to $KH$.

We have that:
 * $HF^2 = FK^2 + \lambda^2$

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

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

First suppose $\lambda$ is commensurable in length with $HF$.

Then $HF$ is commensurable in length with the rational straight line $FG$.

Therefore $KH$ is a first apotome.

But from :
 * the "side" of $LH$ is an apotome.

Next suppose $\lambda$ is incommensurable in length with $HF$.

Then $HF$ is incommensurable in length with the rational straight line $FG$.

Therefore $KH$ is a fourth apotome.

But from :
 * the "side" of $LH$ is minor.