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

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## 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