How to proof the reason? I have this statement:

If $\frac{a}{b} = \frac{c}{d},$ prove that
  $\frac{a+b}{a-b}=\frac{c+d}{c-d}$

I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised to $^{-1}$)
So far I have arrived, without much success.
How can I prove it? Thanks in advance.
 A: It is given
$$\frac{c}{d}=\frac{a}{b}$$ from here we get
$$2bc=2ad$$ and then
$$ac+bc-ad-bd=ac+ad-bc-bd$$
and this is $$(a+b)(c-d)=(c+d)(a-b)$$
Can you finish? 
A: $$\frac{a}{b} = \frac{c}{d}$$
write  $$\frac{a}{b}+1 = \frac{c}{d}+1$$
$$\frac{a+b}{b} = \frac{c+d}{d}$$
also if $$\frac{a}{b} = \frac{c}{d}$$
then $$\frac{b}{a} = \frac{d}{c}$$
write $$1-\frac{b}{a} =1- \frac{d}{c}$$
 $$\frac{a-b}{a} = \frac{c-d}{c}$$
we first obtained that $$\frac{a+b}{b} = \frac{c+d}{d}$$
$$\frac{(a+b)a}{(a-b)b} = \frac{(c+d)c}{(c-d)d}$$
also $$\frac{a}{b} = \frac{c}{d}$$
thus 
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$
A: Actually the result is not true if $a = b$ and $\frac ab =\frac cd =  1$ and $c = d$.  Then $\frac {a+b}{a-b}$ and $\frac {c+d}{c-d} $ are both undefined and not equal to each other.
.......
Three ideas:  
1) Proportions:
$\frac ab = \frac cd$ means there is some ratio so that $a = cr$ and $b = dr$.  (That's clear, isn't it?  $\frac {cr}{dr} = \frac ab$ for all $r$ and if [assuming $a\ne 0; c\ne 0$] then if $r = \frac ac$ means $\frac ab = \frac {cr}{dr} =  \frac {c\frac ac}{dr} = \frac a{dr}$.  So $dr = b$.)
So $\frac {a+b}{a-b} = \frac {cr +dr}{cr - dr} = \frac {r(c+d)}{r(c-d)}=\frac {c+d}{c-d}$.
[Note:  we should take $a=0$ as a possiblity.  $\frac 0b = \frac cd \implies c = 0$ so $\frac {a+b}{a-b} = \frac {b}{-b} = - 1$ and $\frac {c+d}{c-d}=\frac d{-d} = -1$.]
2) Substitution:
$\frac ab = \frac cd$ so $a = \frac {bc}d$ (or any of other variable)
So $\frac {a+ b}{a-b} = \frac {\frac {bc}d + b}{\frac {bc}d - b} ==\frac {b(\frac cd +1)}{b(\frac cd - 1)}= \frac {\frac cd +1}{\frac cd - 1}==\frac {d(\frac cd +1)}{d(\frac cd - 1)} = \frac {c + d}{c-d}$.
3)  Consequences:
$\frac ab =\frac cd \implies ad = bc$.
And $\frac {a+b}{a-b} = \frac{c+d}{c-d} \iff$
$(a+b)(c-d) = (c+d)(a-b) \iff$
$ac + bc -ad - bd = ac +ad - bc = bd \iff$
$2bc = 2ad \iff$
$bc = ad$
And we determined that that is true.
(NOTE:  It is !!!!!!VITALLLLLLLY!!!!!! important the we note that every step in that argument is an if AND ONLY if statement.  Otherwise we can NOT EVER start with what we want to prove.  If you EVER try to to do a proof where you start with your conclusion and get a true result and conclude your conclusion was true and your statements were not if and only if statements, I will reach through my computer screen and pound your head to the table.)
(Less important but still important.  We can not end with $bc = ad \iff \frac ab = \frac cd$ as it is possible that $b=0$ or $d=0$ and it will not be true that $\frac ab = \frac cd$.  In particular we coulde have $a=0; d=5; b=0;c = 6$ and it certainly isnt true that $\frac 00 = \frac 65$.  We could say $bc = ad \iff \frac ab = \frac cd$ or $bc = ad = 0$.)
... or less dramatic than threatening to pound heads into table:
Once we figured this, we work in the real direction:
$\frac ab = cd \implies$
$ad = bc \implies$
$2ad = 2bc \implies$
$ad -bc = bc - ad \implies$
$ac + ad - bc - bd = ac +bc -ad -bd \implies$
$a(c+d) - b(c+d) = c(a+b) - d(a+b) \implies$
$(a-b)(c+d) = (c-d)(a+b)$
Assumming $a - b \ne 0$ and $c-d \ne 0$ then
$\frac {a+b}{a-b} = \frac {c+d}{c-d}$.
However if $a - b = 0$ then $a = b$ and $\frac ab = 1$ and $\frac cd = 1$ and $c = d$ and then.... well, the result simply isn't true.
A: Any identity similar to yours -- involving 'direct sums/differences' of fractions, can be proved by multiplying out the denominator (of the identity to be proved), rearranging and cancelling terms and getting $ad-bc=0$. All of the operations above are easily done the opposite way, which builds up the proof. (Although it will look like a proof out of the blue!)
For yours, write the following sequence of equations backwards.
$$
\begin{align}
\frac{a+b}{a-b}&=\frac{c+d}{c-d}\\
(a+b)(c-d)&=(c+d)(a-b)\\
ac+bc-ad-bd&=ac+ad-bc-bd\\
2bc&=2ad\\
ad&=bc\\
\end{align}
$$
... with the original equation on the top.
A: Let, $\frac{a}{b}=\frac{c}{d}=k$ or, $a=bk,c=dk$. Now consider the LHS:$$\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{k+1}{k-1}$$
Also, $$\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{k+1}{k-1}~~~~~~~~~\boxed{}$$
