# Commensurability of Squares on Proportional Straight Lines

## Theorem

In the words of Euclid:

If two straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line commensurable with the third.

And, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line incommensurable with the third.

### Lemma

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.

## Proof

Let $A$, $B$, $C$ and $D$ be four straight lines in proportion, such that:

$A : B = C : D$

Using the lemma let the straight lines $E$ and $F$ be found such that:

$A^2 = B^2 + E^2$
$C^2 = D^2 + F^2$

As $A : B = C : D$ it follows from Similar Figures on Proportional Straight Lines that:

$A^2 : B^2 = C^2 : D^2$

But:

$E^2 + B^2 = A^2$
$F^2 + D^2 = C^2$

Therefore:

$E^2 + B^2 : B^2 = F^2 + D^2 : D^2$
$E^2 : B^2 = F^2 : D^2$
$B : E = D : F$

But:

$A : B = C : D$

Therefore from Equality of Ratios Ex Aequali:

$A : E = C : F$

From Commensurability of Elements of Proportional Magnitudes it follows that:

if $A$ is commensurable with $E$ then $C$ is also commensurable with $F$

and:

if $A$ is incommensurable with $E$ then $C$ is also incommensurable with $F$.

$\blacksquare$

## Historical Note

This proof is Proposition $14$ of Book $\text{X}$ of Euclid's The Elements.