Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio/Lemma

Lemma to Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio

In the words of Euclid:

That the double of $AC$ is greater than $BC$ is to be proved thus.

(The Elements: Book $\text{XIII}$: Proposition $2$ : Lemma)

Proof

Suppose that $BC = 2 \cdot CA$.

Therefore:

$BC^2 = 4 \cdot CA^2$

Therefore:

$BC^2 + CA^2 = 5 CA^2$

But by hypothesis:

$BA^2 = 5 CA^2$

Therefore by Proposition $4$ of Book $\text{II} $: Areas of Triangles and Parallelograms Proportional to Base:

$BA^2 = BC^2 + CA^2$

which is impossible.

Therefore:

$BC \ne 2 \cdot CA$

Similarly it can be shown that it is not the case that:

$BC < 2 \cdot CA$

as it leads to a much greater absurdity.

Therefore:

$BC > 2 \cdot CA$

$\blacksquare$

Historical Note

This proof is Proposition $2$ of Book $\text{XIII}$ of Euclid's The Elements.
Heiberg has reason to believe that this lemma is spurious.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XIII}$. Propositions