Straight Line Commensurable with Side of Sum of two Medial Areas

Theorem

In the words of Euclid:

A straight line commensurable with the side of the sum of two medial areas is the side of the sum of two medial areas.

(The Elements: Book $\text{X}$: Proposition $70$)

Proof

Let $AB$ be the side of the sum of two medial areas.

Let $CD$ be commensurable in length with $AB$.

It is to be shown that $CD$ is also the side of the sum of two medial areas.

Let $AB$ be divided into its terms by $E$.

Let $AE > EB$.

By definition, $AE$ and $EB$ are straight lines such that:

$AE$ and $EB$ are incommensurable in square

$AE^2 + EB^2$ is medial

$AE \cdot EB$ is medial.

Using Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line, let it be contrived that:

$AB : CD = AE : CF$

Therefore by Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

$EB : FD = AB : CD$

Therefore by Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

$AE : CF = EB : FD$

But $AB$ is commensurable in length with $CD$.

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

$AE$ is commensurable in length with $CF$

and:

$EB$ is commensurable in length with $FD$.

We have that:

$AE : CF = EB : FD$

Therefore by Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

$AE : EB = CF : FD$

and by by Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded:

$AB : BE = CD : DF$

Using a similar line of reasoning to Proposition $68$ of Book $\text{X} $: Straight Line Commensurable with Major Straight Line is Major:

$CF$ and $FD$ are incommensurable in square

$AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$

$AE \cdot EB$ is commensurable with $CF \cdot FD$.

This article, or a section of it, needs explaining. In particular: The above needs to be worked through in detail. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

We have that $AE^2 + EB^2$ is medial.

Therefore $CF^2 + FD^2$ is medial.

Similarly we have that $AE \cdot EB$ is medial.

Therefore $CF \cdot FD$ is medial.

Hence:

$CF$ and $FD$ are incommensurable in square

$CF^2 + FD^2$ is medial

$CF \cdot FD$ is medial.

Thus by definition $CD$ is the side of the sum of two medial areas.

$\blacksquare$

Historical Note

This proof is Proposition $70$ 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