Construction of Components of Side of Rational plus Medial Area

Proof

 * Euclid-X-34.png

From :

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 $E$.

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 $FB$.

From :
 * $AF$ 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 $FB$.

So from :
 * $BA \cdot AF$ is incommensurable with $AB \cdot BF$.

From :
 * $BA \cdot AF = AD^2$

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.

Therefore from :
 * $AB \cdot FD$ is a rational area.

But from :
 * $AB \cdot FD = AD \cdot DB$

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.

Also see

 * Definition:Side of Rational plus Medial Area