# Construction of Medial Straight Lines Commensurable in Square Only containing Medial Rectangle whose Square Differences Commensurable with Greater

## Theorem

In the words of Euclid:

To find two medial straight lines commensurable in square only, containing a medial rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater.

## Proof

Let $A$, $B$ and $C$ be rational straight lines which are commensurable in square only.

Let $A^2 > C^2$ such that:

$A^2 = C^2 + \rho^2$

where $\rho$ is a straight line which is commensurable in length with $A$.

Let $D$ be a straight line such that $D^2 = A B$.

From Medial is Irrational, $A B$ is medial.

Therefore $D^2$ is medial.

So by definition $D$ is medial.

Let $D E = B C$.

Since:

$A B : B C = A : C$

and:

$D^2 = A B$

and:

$D E = B C$

then:

$A : C = D^2 : D E$

So as:

$D^2 : D E = D : E$

it follows that:

$A : C = D : E$

But $A$ is commensurable in square only with $C$.

$D$ is commensurable in square only with $E$.

We have that $D$ is medial.

Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $E$ is medial.

We have that:

$A : C = D : E$

and:

$A^2 = C^2 + \rho^2$

where $\rho$ is a straight line which is commensurable in length with $A$.

$D^2 = E^2 + \sigma^2$

where $\sigma$ is a straight line which is commensurable in length with $D$.

We have that:

$B C = D E$

while from Medial is Irrational:

$B C$ is medial.

Therefore $D E$ is also medial.

Therefore two medial straight lines $D$ and $E$ have been found which are commensurable in square only and which contain a medial rectangle.

Further, $D^2$ is greater than $E^2$ by the square on a straight line commensurable in length with $D$.

Similarly it can be proved that the square on $D$ exceeds the square on $E$ by the square on a straight line incommensurable in length with $D$, when the square on $A$ is greater than the square on $C$ by the square on a straight line incommensurable in length with $A$.

$\blacksquare$

## Historical Note

This proof is Proposition $32$ of Book $\text{X}$ of Euclid's The Elements.