Straight Line Commensurable with that which produces Medial Whole with Medial Area

Proof

 * Euclid-X-103.png

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

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

It is to be demonstrated that $CD$ is a straight line which produces with a medial area a medial whole.

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

Therefore by definition of a straight line which produces with a rational area a medial whole:
 * $AE$ and $EB$ are incommensurable in square
 * $AE^2 + EB^2$ is medial
 * $2 \cdot AE \cdot EB$ is a medial rectangle.

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

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

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

We have that $AE$ and $EB$ are incommensurable in square.

Therefore from :
 * $CF$ and $FD$ are incommensurable in square.

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

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

So from :
 * $AE^2 : EB^2 = CF^2 : FD^2$

Therefore from :
 * $AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$

But $EB^2$ is commensurable with $FD^2$.

So from:

and:

we have that:
 * $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.

But by definition $AE^2 + EB^2$ is medial.

Therefore by $CF^2 + FD^2$ is medial.

We have that:
 * $AE : EB = CF : FD$

Therefore:
 * $AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$

Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.

Therefore $CF \cdot FD$ is medial.

Thus $CF$ and $FD$ are such that:
 * $CF$ and $FD$ are incommensurable in square
 * $CF^2 + FD^2$ is medial
 * $2 \cdot CF \cdot FD$ is a medial rectangle.

Hence the result.