Square on Second Bimedial Straight Line applied to Rational Straight Line
Theorem
In the words of Euclid:
- The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.
(The Elements: Book $\text{X}$: Proposition $62$)
Proof
Let $AB$ be a second bimedial straight line divided into its medials 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 third 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 a second bimedial which has been divided into its medials at $C$.
Therefore, by definition, $AC$ and $CB$ are medial straight lines which are commensurable in square only such that $AC \cdot CB$ is a medial rectangle.
Thus, by definition, $AC^2$ and $CB^2$ are also medial.
From:
and:
it follows that:
- $DL$ is medial.
We have that $DL$ has been applied to the rational straight line $DE$.
Therefore from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $MD$ is rational and incommensurable in length with $DE$.
For the same reason:
- $MG$ is also rational straight line and incommensurable in length with $ML$, which equals $DE$.
Therefore each of $DM$ and $MG$ is rational straight line and incommensurable in length with $DE$.
We have that $AC$ is incommensurable in length with $CB$.
Also:
- $AC : CB = AC^2 : AC \cdot CB$
Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $AC^2$ is incommensurable with $AC \cdot CB$.
Hence by:
and:
it follows that:
- $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
That is:
- $DL$ is incommensurable with $MF$.
So from:
and:
it follows that:
- $DM$ is incommensurable in length with $MG$.
But $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a third 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:
- $DM > MG$
Since:
- $AC^2$ is commensurable with $CB^2$
it follows that:
- $DH$ is commensurable with $KL$.
So from:
and:
it follows that:
- $DK$ is commensurable in length with $KM$.
Also $DK \cdot KM = MN^2$.
- $DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$.
Also:
- Neither $DM$ nor $MG$ is commensurable in length with $DE$.
Therefore $DG$ is a third binomial straight line.
$\blacksquare$
Historical Note
This proof is Proposition $62$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $56$: Root of Area contained by Rational Straight Line and Third Binomial.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions