$\operatorname{lcm}(a,b) = c$ and $\gcd(a,b) = d$ => $\operatorname{lcm}(\frac{a}{d},\frac{b}{d}) = \frac{c}{d}$ in a Euclidean domain or PID

I know that in an integral domain $c=\operatorname{lcm}(a,b)$ if and only if $a\mid c, b\mid c$ and if there exists $c'$ such that $a\mid c', b\mid c'$ then this implies that $c\mid c'$.

And $d=\gcd(a,b)$ if and only if $d\mid a$ and $d\mid b$ and if there exists $d'\mid a, d'\mid b$ then $d'\mid d$

Now if $\operatorname{lcm}(a,b) = c$ then there exists integers $m,n$ such that $am=c$ and $bn = c$ then

And if $\gcd(a,b) = d$ then there exists integers $s,t$ such that $ds =a$ and $dt = b$

so now, $\frac{a}{d} = s$ and $\frac{b}{d} = t$ and so clearly $\frac{a}{d} \mid \frac{c}{d}$ since $c=am$ and so $ \frac{a}{d}m = \frac{c}{d}$ and also $\frac{b}{d} \mid \frac{c}{d}$ since $c= bn$ and so $\frac{b}{d}n = \frac{c}{d}$. I now need to prove that if there is another integer $c'$ that is a multiple of both $\frac{a}{d}$ and $\frac{b}{d}$ then $\frac{c}{d}$ divides $c'$ to conclude that $\operatorname{lcm}(\frac{a}{d},\frac{b}{d}) = \frac{c}{d}$.

So I began by saying that if $\frac{a}{d} \mid c'$ and $\frac{b}{d} \mid c'$, and so there exits integers $l$ and $k$ such that $\frac{a}{d}l = c'$ and $\frac{b}{d}k = c'$ but I can't show that $\frac{c}{d} | c'$ , how can I Proceed from here ? Any suggestions

It's an immediate consequence of the LCM distributive law $\ {\rm lcm}(na,nb) = n\,{\rm lcm}(a,b).\,$ Proof:

$\, {\rm lcm}(na,nb)\mid k\!\!\overset{\rm\color{#c00}U\!\!}\iff na,nb\mid k\iff a,b\mid k/n $ $ \overset{\rm\color{#c00}U\!\!}\iff {\rm lcm}(a,b)\mid k/n \iff n\,{\rm lcm}(a,b)\mid k $

where we twice applied $\,\rm\color{#c00}U = $ LCM Universal Property: $\ x,y\mid z\iff {\rm lcm}(x,y)\mid z$

Thus in your case $\ d\ {\rm lcm}(a/d,b/d) = {\rm lcm}(a,b) = c\ $ holds for all $\,d\neq 0$