Magnitudes Proportional Compounded are Proportional Separated

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

Proof
Let $AB, BE, CD, DF$ be magnitudes which are proportional componendo, so that $AB : BE = CD : DF$.

We need to show that they are proportional separando, that is, that $AE : EB = CF : DF$.


 * Euclid-V-17.png

Let equimultiples $GH, HK, LM, MN$ be taken of $AE, EB, CF, FD$.

Let $KO, NP$ be other arbitrary equimultiples of $EB, FD$.

We have that $GH$ is the same multiple of $AE$ that $HK$ is of $EB$.

So from Multiplication of Numbers is Left Distributive over Addition $GH$ is the same multiple of $AE$ that $GK$ is of $AB$.

But $GH$ is the same multiple of $AE$ that $LM$ is of $CF$.

Therefore $GK$ is the same multiple of $AB$ that $LM$ is of $CF$.

Again, we have that $LM$ is the same multiple of $CF$ that $MN$ is of $FD$.

So from Multiplication of Numbers is Left Distributive over Addition $LM$ is the same multiple of $CF$ that $LN$ is of $CD$.

Therefore $GK, LN$ are equimultiples of $AB, CD$.

Again, we have that $HK$ is the same multiple of $EB$ that $MN$ is of $FD$.

Also, $KO$ is the same multiple of $EB$ that $NP$ is of $FD$.

So from Multiplication of Numbers is Right Distributive over Addition the sum $HO$ is also the same multiple of $EB$ that $MP$ is of $FD$.

We also have that $AB : BE = CD : DF$.

Also, we have that $GK, LN$ are equimultiples of $AB, CD$.

Also, we have that $EB, FD$ are equimultiples of $HO, MP$.

Therefore:
 * $GK > HO \implies LN > MP$
 * $GK = HO \implies LN = MP$
 * $GK < HO \implies LN < MP$

Suppose $GK > HO$.

If we subtract $HK$ from each, we have that $GH > KO$.

But $GK > HO \implies LN > MP$.

Therefore $LN > MP$.

So if $MN$ is subtracted from each, we have that $LM > NP$.

So:
 * $GH > KO \implies LM > NP$

Similarly we can prove that:
 * $GH = KO \implies LM = NP$
 * $GH < KO \implies LM < NP$

Also, we have that:
 * $GH, LM$ are equimultiples of $AE, CF$
 * $KO, NP$ are equimultiples of $EB, FD$

Therefore $AE : EB = CF : FD$.