Sum of Components of Equal Ratios

Theorem
That is:
 * $a_1 : b_1 = a_2 : b_2 = a_3 : b_3 = \cdots \implies \left({a_1 + a_2 + a_3 + \cdots}\right) : \left({b_1 + b_2 + b_3 + \cdots}\right)$

Proof
Let any number of magnitudes $A, B, C, D, E, F$ be proportional, so that:
 * $A : B = C : D = E : F$

etc.


 * Euclid-V-12.png

Of $A, C, E$ let equimultiples $G, H, K$ be taken, and of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.

We have that $A : B = C : D = E : F$.

Therefore:
 * $G > L \implies H > M, K > N$
 * $G = L \implies H = M, K = N$
 * $G < L \implies H < M, K < N$

So, in addition:


 * $G > L \implies G + H + K > L + M + N$
 * $G = L \implies G + H + K = L + M + N$
 * $G < L \implies G + H + K < L + M + N$

It follows from Multiplication of Numbers is Left Distributive over Addition that $G$ and $G + H + K$ are equimultiples of $A$ and $A + C + E$.

For the same reason, $L$ and $L + M + N$ are equimultiples of $B$ and $B + D + F$.

The result follows from.