Commensurability of Squares on Proportional Straight Lines/Lemma

Lemma to Commensurability of Squares on Proportional Straight Lines

In the words of Euclid:

Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.

(The Elements: Book $\text{X}$: Proposition $14$ : Lemma)

Proof

Let $AB$ and $C$ be two unequal straight lines.

Let $AB > C$.

Let the semicircle $ADB$ be described with $AB$ as the diameter.

Using Fitting Chord Into Circle, let $AD$ be fitted into $ADB$ equal to $C$.

Let $DB$ be joined.

From Relative Sizes of Angles in Segments, $\angle ADB$ is a right angle.

From Pythagoras's Theorem:

$AB^2 = AD^2 + DB^2$

and so $AB^2$ is greater than $AD^2$, that is, $C^2$, by $DB^2$.

Conversely, given two straight lines $A$ and $B$ the same technique can be used to find the straight line the square of whose length equals the sum of the squares on $A$ and $B$.

$\blacksquare$

Historical Note

This proof is Proposition $14$ 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