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

## Theorem

In the words of Euclid:

A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.

## Proof

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 Proposition $12$ of Book $\text{VI}$: Construction of Fourth Proportional Straight Line, let it be contrived that:

$BE : DF = AB : CD$
$AE : CF = AB : CD$
$AE : CF = BE : DF$

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

$CF$ and $FD$ are incommensurable in square.

We have that:

$AE : CF = BE : DF$
$AE : EB = CF : FD$
$AE^2 : EB^2 = CF^2 : FD^2$
$AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$

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

So from:

Proposition $16$ of Book $\text{V}$: Proportional Magnitudes are Proportional Alternately:

and:

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

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.

$\blacksquare$

## Historical Note

This proof is Proposition $107$ of Book $\text{X}$ of Euclid's The Elements.