a/a-b = c/c-d
Here is an attempted proof, although it is probably wrong, I thought I'd have a go. It relies on some unwritten obvious proof, but you can prove that yourself.
For $\displaystyle a, b, c, d \in R$, and $\displaystyle a \neq b$ and $\displaystyle c \neq d$:
Then $\displaystyle \frac{a}{a-b}\cdot(a-b)\cdot(c-d) = \frac{c}{c-d}\cdot(a-b)\cdot(c-d)$
$\displaystyle a(c-d) = c(a-b)$
$\displaystyle \frac{c-d}{c} = \frac{a-b}{a}$
If $\displaystyle x = \frac{1}{x}$ then $\displaystyle x^2 = 1$, and thus $\displaystyle x = 1$.
Therefore, as shown above, $\displaystyle \frac{a}{a-b} = \frac{a-b}{a} = 1$, as does $\displaystyle \frac{c}{c-d} = \frac{c-d}{c} = 1$.
$\displaystyle 1=1$, QED.
You assume here that $\displaystyle \frac{a}{a-b}=\frac{c}{c-d}$ which is actually what you're trying to prove...
For this step to be correct, you must assume that a,c are non-zero (if they are, the equation holds trivially, although I don't really know what we're trying to "prove" here...).$\displaystyle a(c-d) = c(a-b)$
$\displaystyle \frac{c-d}{c} = \frac{a-b}{a}$
What about $\displaystyle x=-1$?If $\displaystyle x = \frac{1}{x}$ then $\displaystyle x^2 = 1$, and thus $\displaystyle x = 1$.
You basically said, "assume the equation holds, then it holds". This is obviously not a valid proof!Therefore, as shown above, $\displaystyle \frac{a}{a-b} = \frac{a-b}{a} = 1$, as does $\displaystyle \frac{c}{c-d} = \frac{c-d}{c} = 1$.
$\displaystyle 1=1$, QED.
Not to mention that it is quite obvious at a glance that this equation will definitely not hold for any four real numbers... what if we choose
$\displaystyle a=1, ~ b=2, ~ c=2, ~ d=1$ ?