# Construction of Components of Side of Rational plus Medial Area

## Theorem

In the words of Euclid:

To find two straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

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

$AB^2 = BC^2 + \rho^2$

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 $E$.

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 $FB$.

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

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

Join $AD$ and $DB$.

We have that $AF$ is incommensurable in length with $FB$.

$BA \cdot AF$ is incommensurable with $AB \cdot BF$.
$BA \cdot AF = AD^2$

and:

$AB \cdot BF = DB^2$

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

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 $AF + FB$ is medial.

As $BC = 2 DF$:

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

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

$AB \cdot FD$ is a rational area.
$AB \cdot FD = AD \cdot DB$

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

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

$\blacksquare$

## Historical Note

This proof is Proposition $34$ of Book $\text{X}$ of Euclid's The Elements.