# Binomial Straight Line is Divisible into Terms Uniquely/Lemma

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## Lemma to Binomial Straight Line is Divisible into Terms Uniquely

In the words of Euclid:

*Let the straight line $AB$ be set out, let the whole be cut into unequal parts at each of the points $C, D$, and let $AC$ be supposed greater than $DB$;*

I say that the squares on $AC, CB$ are greater than the squares on $AD, DB$.

(*The Elements*: Book $\text{X}$: Proposition $42$ : Lemma)

## Proof

Let $AB$ be a straight line.

Let $AB$ be divided into unequal parts at $C$ and $D$ such that $AC > DB$.

Let $AB$ be bisected at $E$.

Let $DC$ be subtracted from both $AC$ and $DB$.

Then as $AC > DB$ it follows that $AD > CB$.

But as $AE = EB$ it follows that $DE < EC$.

Therefore $C$ and $D$ are not equidistant from the point of bisection.

From Proposition $5$ of Book $\text{II} $: Difference of Two Squares:

- $AC \cdot CB = EB^2 - EC^2 and: :$AD \cdot DB = EB^2 - ED^2

so:

- $AC \cdot CB + EC^2 = AD \cdot DB + ED^2$

We have that $ED^2 < EC^2$ and so:

- $AC \cdot CB < AD \cdot DB$

and so:

- $2 \cdot AC \cdot CB < 2 \cdot AD \cdot DB$

But from Proposition $4$ of Book $\text{II} $: Square of Sum:

- $AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

and:

- $AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$

and so it follows that:

- $ AC^2 + CB^2 > AD^2 + DB^2$

$\blacksquare$

## Historical Note

This proof is Proposition $42$ 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