Binomial Straight Line is Divisible into Terms Uniquely/Lemma

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