Q:

Proving relationship between lowest common denominators of common multiples

prove that $l.c.m.(ab,ad)=a[l.c.m.(b,d)]$. 
my work so far:
I know $l.c.m.(ab,ad)=a^2bd/g.c.d.(ab,ad)$  and ∃ $x,y\in \mathbb{Z}$  | $g.c.d.(ab,ad)=abx+ady$
∴ we now have $l.c.m.(ab,ad)=abd/(bx+dy)$ 
So we know that ∃ $x,y\in \mathbb{Z}$ | $bx+dy=c$ for some $c\in \mathbb{Z}$. If $c=g.c.d.(b,d)$ we are done.  But, how do we know that the $x$ and $y$'s are | $c=g.c.d.(b,d)$?  
here are the relevant corollaries:
1) if $d=g.c.d.(a,b)$, then ∃ $x,y\in \mathbb{Z}$ | $ax+by=d$
2)In order that ∃ $x,y\in \mathbb{Z}$ | $ax+by=c$ it is necessary and sufficient that d|c, where $d=g.c.d.(a,b)$.
Thank you in advance.    

A:

$ {\rm lcm}(ab,ad)\mid n\color{#c00}\iff ab,ad\mid n\iff\begin{align} a &\mid n\\ b,d &\mid n/a\end{align}$ $\color{#c00}\iff \begin{align} a &\mid n\\ {\rm lcm}(b,d) &\mid n/a\end{align}  \iff a\,{\rm lcm}(b,d) \mid n $
Remark $\ $ Above we used $\ x,y\mid z\color{#c00}\iff {\rm lcm}(x,y)\mid z,\,$ the universal property of $\,\rm lcm$
See here for a few proofs of the gcd distributive law.