# Straight Line Commensurable with Side of Sum of two Medial Areas

## Theorem

In the words of Euclid:

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

## Proof

Let $AB$ be the side of the sum of two medial areas.

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

It is to be shown that $CD$ is also the side of the sum of two medial areas.

Let $AB$ be divided into its terms by $E$.

Let $AE > EB$.

By definition, $AE$ and $EB$ are straight lines such that:

$AE$ and $EB$ are incommensurable in square
$AE^2 + EB^2$ is medial
$AE \cdot EB$ is medial.

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

$AB : CD = AE : CF$
$EB : FD = AB : CD$
$AE : CF = EB : FD$

But $AB$ is commensurable in length with $CD$.

$AE$ is commensurable in length with $CF$

and:

$EB$ is commensurable in length with $FD$.

We have that:

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

Using a similar line of reasoning to Proposition $68$ of Book $\text{X}$: Straight Line Commensurable with Major Straight Line is Major:

$CF$ and $FD$ are incommensurable in square
$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$
$AE \cdot EB$ is commensurable with $CF \cdot FD$.

We have that $AE^2 + EB^2$ is medial.

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

Similarly we have that $AE \cdot EB$ is medial.

Therefore $CF \cdot FD$ is medial.

Hence:

$CF$ and $FD$ are incommensurable in square
$CF^2 + FD^2$ is medial
$CF \cdot FD$ is medial.

Thus by definition $CD$ is the side of the sum of two medial areas.

$\blacksquare$

## Historical Note

This proof is Proposition $70$ of Book $\text{X}$ of Euclid's The Elements.