Binomial Straight Line is Divisible into Terms Uniquely/Lemma

Proof

 * Euclid-X-42-Lemma.png

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

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$