# Side of Sum of Two Medial Areas is Divisible Uniquely

## Theorem

In the words of Euclid:

*The side of the sum of two medial areas is divided at one point only.*

(*The Elements*: Book $\text{X}$: Proposition $47$)

## Proof

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

Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:

- $AC$ and $CB$ are incommensurable in square
- $AC^2 + CB^2$ is medial
- $AC$ and $CB$ contain a medial rectangle
- $AC \cdot CB$ is incommensurable with $AC^2 + CB^2$.

Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.

Let $AC > BD$.

From Proposition $4$ of Book $\text{II} $: Square of Sum:

- $AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

and:

- $AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$

Let $EF$ be a rational straight line.

- Let $EG$ be a rectangle set out on $EF$ equal to $AC^2 + CB^2$

and

- Let $HK$ be a rectangle set out on $EF$ equal to $2 \cdot AC \cdot CB$.

Therefore $EK$ is a rectangle equal to $AB^2$.

- Let $EL$ be a rectangle set out on $EF$ equal to $AD^2 + DB^2$

The remainder $MK$ is a rectangle set out on $EF$ equal to $2 \cdot AD \cdot DB$.

By hypothesis, $AC^2 + CB^2$ is medial.

We have that $AC^2 + CB^2$ is applied to a rational straight line $EF$.

So from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:

- $HE$ is rational and incommensurable in length with $EF$.

For the same reason:

- $HN$ is rational and incommensurable in length with $EF$.

Since $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$:

- $EG$ is incommensurable with $GN$.

So by:

and:

it follows that:

- $EH$ is incommensurable with $HN$.

Also, $EH$ and $HN$ are rational.

Therefore $EH$ and $HN$ are rational straight lines commensurable in square only.

Therefore from Proposition $36$ of Book $\text{X} $: Binomial is Irrational:

- $EN$ is a binomial straight line divided at $H$.

Using the same argument we can deduce:

- $EN$ is a binomial straight line divided at $M$.

But this contradicts Proposition $42$ of Book $\text{X} $: Binomial Straight Line is Divisible into Terms Uniquely.

So there can be no such $D$.

$\blacksquare$

## Historical Note

This proof is Proposition $47$ of Book $\text{X}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions