# Construction of Components of Side of Sum of Medial Areas

## Theorem

In the words of Euclid:

*To find two straight lines incommensurable in square which make the sum of the squares on them medial and the rectangle contained by them medial and moreover incommensurable with the sum of the squares on them.*

(*The Elements*: Book $\text{X}$: Proposition $35$)

## Proof

Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that:

such that $\rho$ is incommensurable in length with $AB$.

Let the semicircle $ADB$ be drawn with $AB$ as the diameter.

Let $BC$ be bisected at $E$.

From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram:

Let a parallelogram be applied to $AB$ equal to the square on either of $BE$ or $EC$, and deficient by a square.

Let this parallelogram be the rectangle contained by $AF$ and $FB$.

Let $FD$ be drawn perpendicular to $AB$.

Join $AD$ and $DB$.

- $AF$ is incommensurable in length with $FB$.

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

- $BA \cdot AF$ is incommensurable with $AB \cdot BF$.

From Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:

- $BA \cdot AF = AD^2$

and:

- $AB \cdot BF = DB^2$

Therefore $AD^2$ and $DB^2$ are incommensurable.

... this far

As $AB$ is medial, it follows by definition that $AB^2$ is a medial area.

From Pythagoras's Theorem:

- $AB^2 = \left({AD + DB}\right)^2$

Thus $\left({AD + DB}\right)^2$ is also a medial area.

Therefore $AD + DB$ is medial.

Since $AF \cdot FB = BE^2$ and also $AF \cdot FB = DF^2$:

- $BE^2 = DF^2$

As $BC = 2 DF$:

- $AB \cdot BC = 2 AB \cdot FD$

But $AB \cdot BC$ is a medial area.

Therefore from Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:

- $AB \cdot FD$ is a medial area.

But from Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:

- $AB \cdot FD = AD \cdot DB$

Thus $AD \cdot DB$ is a medial area.

Since:

- $AB$ is incommensurable in length with $BC$

and

- $CB$ is commensurable in length with $BE$

- $AB$ is incommensurable in length with $BE$.

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

- $AB^2 = AB \cdot BE$

From Pythagoras's Theorem:

- $AD^2 + DB^2 = AB^2$

But from Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:

- $AB \cdot FD = AD \cdot DB$

and:

- $AD \cdot DB = AD \cdot BE$

Therefore $AD^2 + DB^2$ is incommensurable with $AD \cdot DB$.

Therefore we have found two straight lines $AD$ and $DB$ which are incommensurable in square whose sum of squares is medial, but such that the rectangle contained by them is medial.

$\blacksquare$

## Historical Note

This proof is Proposition $35$ 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