Sum of Antecedents of Proportion

Theorem

 * If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to a fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth.

Proof
Let a first magnitude $AB$ have to a second $C$ the same ratio as a third $DE$ has to a fourth $F$.

Let also a fifth $BG$ have to the second $C$ the same ratio as a sixth $EF$ has to a fourth $F$.


 * Euclid-V-24.png

We need to show that $AG : C = DH : F$.

We have that $BG : C = EH : F$ and so $C : BG = F : EH$.

We also have that $AB : C = DE : F$.

So from Equality of Ratios Ex Aequali $AB : BG = DE : EH$.

From Magnitudes Proportional Separated are Proportional Compounded it follows that $AG : GB = DH : HE$.

But $BG : C = EH : F$.

So from Equality of Ratios Ex Aequali, $AG : C = DH : F$.