Denominator Identity Verification

Question

Here is the given information for the following proof I'm about to begin:

Suppose a, b, c, d, e and f are non-zero elements of field such that $$ \frac{a}{b} = \frac {c}{d} = \frac{e}{f}$$

I'm suppose to show the following identity is true, whenever the denominator in question is non-zero:

$$\frac{a}{b} = \frac{(a+c+e)}{(b+d+f)}$$

Here my thinking process: In order for this identity to hold, $c + e$ and $d + f$ must add up to $0$, or so I believe. Otherwise, the equality wouldn't hold true at all. Another way of thinking about this is that perhaps the letters a, c,e are multiples of one another (same with b, d, f).

Answer 1

We have $ad=bc$, $af=be$ and $cf=de$. Now let's compute

$$a(b+d+f)-b(a+c+e)=ab+ad+af-ab-bc-be=0$$

Therefore whenever the denominators are non zero

$${a\over b}={a+c+e\over b+d+f}$$

Answer 2

$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}\rightarrow c=ax,d=bx,e=ay,f=by\\\frac{a+c+e}{b+d+f}=\frac{a+ax+ay}{b+bx+by}=\frac{a(1+x+y)}{b(1+x+y)}=\frac{a}{b}$$

Answer 3

$$\begin{align} b\,x\, &=\, a\\ d\,x\, &=\, c\\ f\,x\, &=\, e \\ \smash[t]{\overset{\rm add}\Longrightarrow}\,\ (b\!+\!d\!+\!f)\,x\, &=\, a\!+\!c\!+\!e \end{align}$$

Said geometrically: the vectors $\,(a,b),(c,d),(e,f)\,$ have the same slope hence so does their sum, i.e. the solutions of $\, ay = bx\,$ are a submodule of $\,\Bbb Z^2,\,$ just like in the vectors space case.