Proportional Magnitudes have Proportional Remainders

Theorem

 * 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.

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 $b$.

Proof
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 $EB : FD = AB : CD$.


 * Euclid-V-19.png

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 $BA : EA = DF : CF$.

Alternatively, from Proportional Magnitudes are Proportional Alternately: $BE : DF = EA : FC$.

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

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