Straight Line Commensurable with Side of Rational plus Medial Area

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Theorem

In the words of Euclid:

A straight line commensurable with the side of a rational plus a medial area is itself also the side of a rational plus a medial area.

(The Elements: Book $\text{X}$: Proposition $69$)


Proof

Euclid-X-66.png

Let $AB$ be the side of a rational plus a medial area.

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

It is to be shown that $CD$ is also the side of a rational plus a medial area.


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 rational.


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 $CF \cdot FD$.



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

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

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

Therefore $CF \cdot FD$ is rational.

Hence:

$CF$ and $FD$ are incommensurable in square
$CF^2 + FD^2$ is medial
$CF \cdot FD$ is rational.

Thus by definition $CD$ is the side of a rational plus a medial area.

$\blacksquare$


Historical Note

This proof is Proposition $69$ of Book $\text{X}$ of Euclid's The Elements.


Sources