Construction of Components of Side of Sum of Medial Areas

Proof

 * Euclid-X-34.png

From :

Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that:
 * $(1): \quad$ $AB$ and $BC$ contain a medial rectangle
 * $(2): \quad AB^2 = BC^2 + \rho^2$

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 $FB$.

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

Join $AD$ and $DB$.

From :
 * $AF$ 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.

... 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 :
 * $AB \cdot FD$ is a medial area.

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

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

Since:
 * $AB$ is incommensurable in length with $BC$

and
 * $CB$ is commensurable in length with $BE$

from :
 * $AB$ is incommensurable in length with $BE$.

By :
 * $AB^2 = AB \cdot BE$

From Pythagoras's Theorem:
 * $AD^2 + DB^2 = AB^2$

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

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.