Proportional Proof 
Question: How would you prove this theorem:



If $$\frac ab=\frac cd=\frac ef=\&c\tag1$$
  Then$$\frac ab=\frac {a+c+e+\&c}{b+d+f+\&c}\tag{2}$$

And using that proof, is it possible to prove the following generalized theorem:

General Formula: If $\frac ab=\frac cd=\frac ef=\&c=k$, then we have $k$ as$$k=\frac {pa^n+qc^n+re^n+\&c}{pb^n+qd^n+rf^n+\&c}^{\frac 1n}\tag3$$
  For $p,q,r,\&c$ are any quantities.


I know about compenendo et dividendo where if $\frac ab=\frac cd$, then $\frac {a+b}{a-b}=\frac {c+d}{c-d}$. But I'm not too sure if that will help with proving both formulae.
Basically, I'm not sure where to even begin and what techniques I should employ. Anything is appreciated!
 A: You either have a typo or that is very false:
Let $a = c = e = \lambda = 2$ and $b = d = f = 1$.
Then you have $$\frac ab=\frac {a+c+e+\lambda}{b+d+f+\lambda} \iff \frac 21=\frac {2+2+2+2}{1+1+1+2} \iff 2 = \frac85$$
Assuming you didn't mean to put the $\lambda$ there, thus having
$$\frac ab=\frac {a+c+e}{b+d+f}$$ you can follow the hint in the comments and set
$$\begin{cases} a = \lambda b\\c = \lambda d\\ e = \lambda f\end{cases}$$
and have:
$$\lambda=\frac {\lambda b + \lambda d + \lambda f}{b+d+f} \iff \lambda = \lambda \frac {b + d + f}{b+d+f} \iff \lambda = \lambda$$
Similarly, with the same substitutions we can have 
$$\lambda=\frac {p(\lambda b)^n+q(\lambda d)^n+r(\lambda f)^n}{pb^n+qd^n+rf^n}^{\frac 1n} \iff \lambda = ((\lambda)^n\frac {pb^n+qd^n+rf^n}{pb^n+qd^n+rf^n})^{\frac 1n} \iff \lambda = {\lambda^n}^\frac1n$$
A: Just another way:

We have a theorem where given $\frac ab=\frac cd=\frac ef=\ldots=k$, we can write it as$$k=\frac {a+c+e\ldots}{b+d+f\ldots}\tag1$$

We can multiply the numerator and denominators by $p,q,r\ldots$ to keep the fraction the same and raise everything the $n$ power. Thus, we now have $(1)$ as$$k^n=\frac {pa^n+qc^n+re^n+\ldots}{pb^n+qd^n+rf^n+\ldots}\tag{2}$$
And taking the $n^{\text{th}}$ root, gives us $(3)$.
