Square on Side of Sum of two Medial Area applied to Rational Straight Line
Theorem
In the words of Euclid:
- The square on the side of the sum of two medial areas applied to a rational straight line produces as breadth the sixth binomial.
(The Elements: Book $\text{X}$: Proposition $65$)
Proof
Let $AB$ be the side of the sum of two medial areas divided at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a sixth binomial straight line.
From Proposition $4$ of Book $\text{II} $: Square of Sum:
- $AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is the side of the sum of two medial areas divided at $C$.
Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that:
- $AC^2 + CB^2$ is medial
- $AC \cdot CB$ is a medial rectangle
- $AC^2 + CB^2$ is incommensurable with $AC \cdot CB$.
Since $AC^2 + CB^2$ is medial, $DL$ is medial.
Therefore from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $DM$ is rational and incommensurable in length with $DE$.
We have that:
- $2 \cdot AC \cdot CB$, which equals $MF$, is medial.
We also have that $MF$ is applied to the rational straight line $ML$.
Therefore from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $MG$ is rational and incommensurable in length with $DE$.
Therefore $DM$ and $MG$ are rational straight lines which are incommensurable in length with $DE$.
We have that $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
Thus $DL$ is incommensurable with $MF$.
So from:
and:
it follows that:
- $DM$ is incommensurable in length with $MG$.
That is, $DM$ and $MG$ are rational straight lines which are commensurable in square only.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a sixth binomial straight line.
- $AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:
- $DM > MG$
Also $DK \cdot KM = MN^2$.
Because:
- $AC^2$ is incommensurable with $CB^2$
it follows that:
- $DH$ is incommensurable with $KL$.
So from:
and:
it follows that:
- $DK$ is incommensurable in length with $KM$.
- $DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$.
Also, neither $DM$ nor $MG$ is commensurable in length with the rational straight line $DE$.
Therefore $DG$ is a sixth binomial straight line.
$\blacksquare$
Historical Note
This proof is Proposition $65$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $59$: Root of Area contained by Rational Straight Line and Sixth Binomial.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions