Magnitudes Proportional Separated are Proportional Compounded

Theorem

 * If magnitudes be proportional separando, they will also be proportional componendo.

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

Proof
Let $AE, EB, CF, FD$ be magnitudes which are proportional separando, that is:
 * $AE : EB = CF : FD$

We need to show that they are also proportional componendo, that is:
 * $AB : BE = CD : FD$


 * Euclid-V-18.png

Suppose $CD : DF \ne AB : BE$.

Then as $AB : BE$ so will $CD$ be to some magnitude less than $DF$ or greater.

Suppose that it be in that ratio to a less magnitude $DG$.

Then since $AB : BE = CD : DG$ it follows from Magnitudes Proportional Compounded are Proportional Separated that:
 * $AE : EB = CG : GD$

But by hypothesis:
 * $AE : EB = CF : FD$

So by Equality of Ratios is Transitive we have that:
 * $CG : GD = CF : FD$

But $CG > CF$.

Therefore $GD > FD$ from Relative Sizes of Components of Ratios.

But it is also less, which is impossible.

Therefore as $AB$ is to $BE$, so is not $CD$ to a lesser magnitude than $FD$.

Similarly we can show that neither is it in that ratio to a greater.

Hence the result.