Show that if $\gcd(a,c)=d$, $a\mid b$ and $c\mid b$, then $ac\mid bd$ 
I'm trying to prove: if $\gcd(a,c)=d$, $a\mid b$ and $c\mid b$, then $ac\mid bd$.

I know that $d\mid b$ by transitivity of $d\mid a$ and $a\mid b$, $d\mid c$ and $c\mid b$, but I don't know how to advance from here.
 A: I propose two solutions. 
First solution: We use the hint given by rogerl in the comments. Since $a\mid b$ and $c\mid b$, then by the universal definition of $\text{lcm}$ we have that $\text{lcm}(a,c)\mid b$, then $\text{lcm}(a,c)\gcd(a,c)\mid b\gcd(a,c)=bd$. Now, as $ac=\text{lcm}(a,c)\gcd(a,c)$, thus $ac\mid bd$.
Second solution: In this proof we won't use the identity $ab=\gcd(a,b)\text{lcm}(a,b)$. Since $a\mid b$ and $c\mid b$, then there are $x,y$ such that $b=ax$ and $b=cy$. On the other hand, as $d=\gcd(a,c)$, then $a=a'd$ and $c=c'd$, with $\gcd(a',c')=1$. Now, we have $ax=cy$, which can be written in the form $$a'dx=c'dy \implies a'x=c'y.$$
Therefore, $c'\mid a'x$ and since $\gcd(a',c')=1$ we deduce that $c'\mid x$ (see here for a proof of this theorem), thus $c'd\mid xd$, i.e., $c\mid xd$, then we can set $xd=cz$, and hence $$bd=axd=a(xd)=acz=(ac)z.$$
So we conclude that $ac\mid bd$. 
A: $a,c\mid b\,\Rightarrow\,ac\mid ba,bc\,\Rightarrow\,ac\mid(ba,bc) = b(a,c)=bd$ 
A: Using prime factorization:
$$\gcd(a,c) = d ,~~ a|b ,~~ c|b \implies ac|bd$$
$$\gcd(\Pi p_k^{a_k}, \Pi p_k^{c_k}) = \Pi p_k^{d_k} ,~~ \Pi p_k^{a_k}|\Pi p_k^{b_k} ,~~ \Pi p_k^{c_k}|\Pi p_k^{b_k} \implies 
\Pi p_k^{a_k} \Pi p_k^{c_k} | \Pi p_k^{b_k} \Pi p_k^{d_k}$$
$$\forall k ~:~\max(a_k, c_k) = d_k ,~~ a_k \le b_k ,~~ c_k \le b_k \implies a_k + c_k \le b_k + d_k$$
which can be establishabled from $a_k + c_k \le 2d_k \le b_k + d_k$.
A: First, $gcd(a,c)=d \implies a=d·a' \wedge c=d·c'$ with $gcd(a',c')=1$
Moreover, $a|b \implies b=a·b'=d·a'·b' \wedge c|b \implies b=c·b''=d·c'·b''$
And then $d|b \implies b= d·\delta$
Now we only have to compute the product of $a$ and $c$ and $b$ and $d$.
$$
a·c=d·a'·d·c' = d^2·a'·c'\wedge b·d= d·\delta·d = d^2·\delta \implies a'·c'=\delta
$$
Note that $a' | \delta \wedge c' | \delta$ and $gcd(a',c')=1$, then, by the fundamental theorem of arithmetic we can see that we haven't common factors and for this reason: 
$$
a' | \delta \implies c' | \frac{\delta}{a'} \implies b·d = a·c·\frac{\delta}{a'·c'} \implies ac| bd
$$
