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