Straight Line Commensurable with that which produces Medial Whole with Medial Area
Theorem
In the words of Euclid:
- A straight line commensurable with that which produces with a medial area a medial whole is itself also a straight line which produces with a medial area a medial whole.
(The Elements: Book $\text{X}$: Proposition $107$)
Proof
Let $AB$ be a straight line which produces with a medial area a medial whole.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that $CD$ is a straight line which produces with a medial area a medial whole.
Let $BE$ be the annex of $CD$.
Therefore by definition of a straight line which produces with a rational area a medial whole:
- $AE$ and $EB$ are incommensurable in square
- $AE^2 + EB^2$ is medial
- $2 \cdot AE \cdot EB$ is a medial rectangle.
From Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line, let it be contrived that:
- $BE : DF = AB : CD$
From Proposition $12$ of Book $\text{V} $: Sum of Components of Equal Ratios:
- $AE : CF = AB : CD$
and so from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $AE : CF = BE : DF$
We have that $AE$ and $EB$ are incommensurable in square.
Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:
- $CF$ and $FD$ are incommensurable in square.
We have that:
- $AE : CF = BE : DF$
So from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:
- $AE : EB = CF : FD$
So from Proposition $22$ of Book $\text{VI} $: Similar Figures on Proportional Straight Lines:
- $AE^2 : EB^2 = CF^2 : FD^2$
Therefore from Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded:
- $AE^2 + EB^2 : EB^2 = CF^2 + FD^2 : FD^2$
But $EB^2$ is commensurable with $FD^2$.
So from:
and:
we have that:
- $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.
But by definition $AE^2 + EB^2$ is medial.
Therefore by $CF^2 + FD^2$ is medial.
We have that:
- $AE : EB = CF : FD$
Therefore:
- $AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Therefore $CF \cdot FD$ is medial.
Thus $CF$ and $FD$ are such that:
- $CF$ and $FD$ are incommensurable in square
- $CF^2 + FD^2$ is medial
- $2 \cdot CF \cdot FD$ is a medial rectangle.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $107$ 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