Proportional Magnitudes have Proportional Remainders

From ProofWiki
Jump to navigation Jump to search


In the words of Euclid:

If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole.

(The Elements: Book $\text{V}$: Proposition $19$)

That is:

$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$

where $a : b$ denotes the ratio of $a$ to $b$.


In the words of Euclid:

From this it is manifest that, if magnitudes be proportional componendo, they will also be proportional convertendo.

(The Elements: Book $\text{V}$: Proposition $19$ : Porism)


As the whole $AB$ is to the whole $CD$, so let the part $AE$ subtracted be to the part $CF$ subtracted.

That is:

$AB : CD = AE : CF$

We need to show that $EB : FD = AB : CD$.


We have that :$AB : CD = AE : CF$.

So from Proportional Magnitudes are Proportional Alternately we have that $BA : AE = DC : CF$.

From Magnitudes Proportional Compounded are Proportional Separated we have that $BA : EA = DF : CF$.

From Proportional Magnitudes are Proportional Alternately: $BE : DF = EA : FC$.

But by hypothesis $AE : CF = AB : CD$.

So by Equality of Ratios is Transitive $EB : FD = AB : CD$.


Historical Note

This proof is Proposition $19$ of Book $\text{V}$ of Euclid's The Elements.