# Straight Line Commensurable with Medial Straight Line is Medial

## Theorem

In the words of Euclid:

*A straight line commensurable with a medial straight line is medial.*

(*The Elements*: Book $\text{X}$: Proposition $23$)

### Porism

In the words of Euclid:

*From this it is manifest that an area commensurable with a medial area is medial.*

(*The Elements*: Book $\text{X}$: Proposition $23$ : Porism)

## Proof

Let $A$ be a medial straight line.

Let $B$ be a straight line which is commensurable in length with $A$.

Let $CD$ be a rational straight line.

Let a rectangle $CE$ be set up on $CD$ equal to the square on $A$.

Let the breadth of $CE$ be $DE$.

Then from Square on Medial Straight Line:

- $ED$ is a rational straight line which is commensurable in length with CD.

Let a rectangle $CF$ be set up on $CD$ equal to the square on $B$.

Let the breadth of $CF$ be $DF$.

As $A$ is commensurable in length with $B$, the square on $A$ is also commensurable with the square on $B$.

But as $CE = A^2$ and $CF = B^2$, $CE$ is commensurable with $CF$.

From Areas of Triangles and Parallelograms Proportional to Base:

- $EC : CF = ED : DF$

From Commensurability of Elements of Proportional Magnitudes:

- $ED$ is commensurable with $DF$.

But $ED$ is a rational straight line which is commensurable in length with CD.

By Book $\text{X}$ Definition $3$: Rational Line Segment, $DF$ is a rational straight line.

By Commensurable Magnitudes are Incommensurable with Same Magnitude:

- $DF$ is incommensurable in length with $DC$.

Therefore $CD$ and $DF$ are rational and commensurable in square only.

But from Medial is Irrational, the side of the square equal to therectangle contained by $CD$ and $DF$ is medial.

Therefore $B$ is medial.

$\blacksquare$

## Historical Note

This proof is Proposition $23$ 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