If $c \mid a, c \mid b$ and $d = \gcd(a, b)$, then $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$. what I've already done is I showed 
$d \mid ck_{1}$ and $d \mid ck_{2}$ where $ck_1$ and $ck_2$ are $a$ and $b$
since $d$ is the gcd, then $d\geq c$
and if $d = c$, then  $\gcd(\frac{a}{c},\frac{b}{c}) = \gcd(k_1,k_2)= 1 = \frac{d}{c}$.
but now I don't know how to prove when $c<d$ 
Maybe I am in wrong direction? Could anyone show me how to do this step?
 A: It's very good that you looked at a special case.
Have you learned about Bezout's Theorem?
This theorem tells us that since $\gcd (a, b)=d$, there exist integers $x, y$ such that $ax+by=d$. Now dividing both sides by $c$ we get $\dfrac ac x + \dfrac bc y = \dfrac dc$. All of the fractions are integers (why?). Applying Bezout's Theorem again we conclude that  $\gcd \left(\dfrac ac, \dfrac bc \right)$ divides $\dfrac dc$. Now if we can show the other way around, namely that $\dfrac dc$ divides $\gcd \left(\dfrac ac, \dfrac bc\right)$, we would be done. How can we show this? By showing that $\dfrac dc$ divides both $\dfrac ac$ and $\dfrac bc$. And indeed, since $d|a$, $a=dk$ for some $k \in \mathbb{Z}$ so $\dfrac ac = \dfrac dck$; and again the fractions are integers. The exact same argument can be used for $\dfrac bc$, so we are done. 
A: Let
$a
=\prod_{p \in P} p^{a_p}
$,
$b
=\prod_{p \in P} p^{b_p}
$,
$c
=\prod_{p \in P} p^{c_p}
$,
$d
=\prod_{p \in P} p^{d_p}
$.
Since
$c|a, c|b$,
$c_p \le a_p$
and
$c_p \le b_p$
so
$c_p \le \min(a_p, b_p)$.
Since
 $d = \gcd(a, b)$,
$d_p = \min(a_p, b_p)$.
$a/c
=\prod_{p \in P} p^{a_p-c_p}
$,
$b/c
=\prod_{p \in P} p^{b_p-c_p}
$,
so
$\gcd(a/c, b/c)
=\prod_{p \in P} p^{\min(a_p-c_p, b_p-c_p)}
$.
$d/c
=\prod_{p \in P} p^{d_p-c_p}
$,
so if
$d_p-c_p
=\min(a_p-c_p, b_p-c_p)
$
then
we are done.
But
$\min(a_p-c_p, b_p-c_p)
=\min(a_p, b_p)-c_p
=d_p-c_p
$
and we are done.
A: $\exists r \text{ and } s\in\mathbb Z: ra+sb=d \implies r\frac ac +s \frac bc=\frac dc\implies gcd(\frac ac, \frac bc) | \frac dc$...
But,  conversely, $\frac dc | \frac ac$ since the ratio is $\frac ad$...
Similarly, $\frac dc | \frac bc$...  
Now using that $t \frac ac+u\frac bc=gcd(\frac ac, \frac bc)$, for some $t,u\in \mathbb Z$, we get $\frac dc| gcd(\frac ac, \frac bc)$...
