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