Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = \gcd(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$. 
Let $a$ and $b$ be non-zero integers, and $c$ be an integer. Let $d = hcf(a, b)$. Prove that if $a|c$ and $b|c$ then $ab|cd$.

We know that if $a|c$ and $b|c$ then $a\cdot b\cdot s=c$ (for some positive integer $s$). $(ab|c)$
Then doesn't $ab|dc$ since $ab|c$?
I feel like I'm misunderstanding my givens.  
Can we say $\operatorname{lcm}(a,b)=c$,  $\operatorname{hcf}(a,b)=d$, and $\operatorname{lcd}(a,b) \operatorname{hcf}(a*b)=a*b$?
Thus, $ab|dc$  as $dc = ab$.

Let us put it this way: Define a', b' so that a=a'd and b=b'd, where d=GCD(a,b).  Then a|c means c = a'ds for some s, and b|c means c = b'dt for some t.  But that doesn't mean ab|c because we could have a'|t and b'|s, i.e c = a'b'du for some u.  I.e in dividing c, a and b are sharing d between them.
Of course, if c = a'b'du, cd = a'db'du = abu.
