# Sum of two Incommensurable Medial Areas give rise to two Irrational Straight Lines

## Theorem

In the words of Euclid:

*If two medial areas incommensurable with one another be added together, the remaining two irrational straight lines arise, namely either a second bimedial or a side of the sum of two medial areas.*

(*The Elements*: Book $\text{X}$: Proposition $72$)

## Proof

Let $AB$ and $CD$ be medial areas.

It is to be demonstrated that the "side" of their combined area $AD$ is either:

- $(1): \quad$ a second bimedial

or:

- $(2): \quad$ a side of the sum of two medial areas.

WLOG let $AB > CD$.

Let a rational straight line $EF$ be set out.

Let the rectangle $EG = AB$ be applied to $EF$, producing $EH$ as breadth.

Let the rectangle $HI = DC$ be applied to $EF$, producing $HK$ as breadth.

We have that $AB$ and $CD$ are medial areas and equals $EG$ and $HI$ respectively.

Therefore $EG$ and $HI$ are medial areas.

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

- $EH$ and $HK$ are rational straight lines which are both incommensurable in length with $EF$.

We have that:

- $AB$ is incommensurable with $CD$

and:

- $AB = EG, CD = HI$

Therefore:

- $EG$ is incommensurable with $HI$.

But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

- $EG : HI = EH : HK$

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

- $EH$ is incommensurable in Length with $HK$.

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

Therefore, by definition, $EK$ is a binomial straight line which is divided at $H$.

We have that:

- $AB > CD$

and:

- $AB = EG, CD = HI$

Therefore:

- $EH > HK$

Thus $EH^2$ is greater than $HK^2$ by either:

- the square on a straight line which is commensurable in length with $EH$

or:

- the square on a straight line which is incommensurable in length with $EH$.

Let $EH^2 = HK^2 + \lambda^2$.

First, let $\lambda$ be commensurable in length with $EH$.

Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.

Therefore $EK$ is a third binomial.

But $EF$ is rational.

- the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is second bimedial.

Therefore the "side" of $EI$ is second bimedial.

That is, the "side" of $AD$ is second bimedial.

$\Box$

Next, let $\lambda$ be incommensurable in length with $EH$.

Neither of $EH$ or $HK$ is commensurable in length with the rational straight line $EF$.

Therefore $EK$ is a sixth binomial.

But $EF$ is rational.

- the "side" of the square equal to the rectangle contained by a rational straight line and a first binomial is the side of the sum of two medial areas.

Therefore the "side" of $EI$ is the side of the sum of two medial areas.

That is, the "side" of $AD$ is the side of the sum of two medial areas.

$\blacksquare$

## Historical Note

This proof is Proposition $72$ 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