Construction of that which produces Medial Whole with Medial Area is Unique

Theorem

In the words of Euclid:

To a straight line which produces with a medial area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial and twice the rectangle contained by them both medial and also incommensurable with the sum of the squares on them.

(The Elements: Book $\text{X}$: Proposition $84$)

Proof

Let $AB$ be a straight line which produces with a medial area a medial whole.

Let $BC$ be added to $AB$ such that:

$AC$ and $CB$ are incommensurable in square

$AC^2 + CB^2$ is medial

$2 \cdot AC \cdot CB$ is a medial rectangle

$2 \cdot AC \cdot CB$ is a incommensurable with $AC^2 + CB^2$.

It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions.

Suppose $BD$, different from $BC$, can be added to $AB$ such that:

$AD$ and $DB$ are incommensurable in square

$AD^2 + DB^2$ is medial

$2 \cdot AD \cdot DB$ is a medial rectangle

$2 \cdot AD \cdot DB$ is a incommensurable with $AD^2 + DB^2$.

Let $EF$ be a rational straight line.

Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth.

Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth.

Therefore from Proposition $7$ of Book $\text{II} $: Square of Difference:

the remainder $EL$ equals $AB^2$.

Thus $AB$ equals the "side" of $EL$.

Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth.

But $EL = AB^2$.

Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$.

Therefore $AC^2 + CB^2$ are medial.

Also $AC^2 + CB^2 = EG$.

Therefore $EG$ is medial.

But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth.

Therefore from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:

$EM$ is rational and incommensurable in length with $EF$.

We have that $2 \cdot AC \cdot CB$ is a medial rectangle.

But $2 \cdot AC \cdot CB = HG$.

Therefore $HG$ is medial.

Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth.

Therefore by Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:

$HM$ is rational and incommensurable in length with $EF$.

We have that:

$AC^2 + CB^2$ incommensurable with $2 \cdot AC \cdot CB$.

We also have that:

$EG = AC^2 + CB^2$

and:

$GH = 2 \cdot AC \cdot CB$

Therefore:

$EG$ is incommensurable with $HG$.

But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

$EG : HG = EM : HM$

Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

$EM$ is incommensurable in length with $MH$.

But both $EM$ and $MH$ are rational straight lines.

Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only.

Therefore $EH$ is an apotome, and $HM$ an annex to it.

Similarly it can be shown that $HN$ is also an annex to it.

Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole.

From Proposition $79$ of Book $\text{X} $: Construction of Apotome is Unique, this is impossible.

The result follows.

$\blacksquare$

Historical Note

This proof is Proposition $84$ 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