How can I prove that $a|bc$ if and only if $\frac{a}{(a,b)}|c$?

Note that $x|y$ if and only if $(x,y) = x$; then use the properties of the gcd.

\begin{align*} a|bc &\Longleftrightarrow (a,bc) = a\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&&\mbox{(by Euclid's Lemma)}\\\ &\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c, \end{align*} using Euclid's Lemma ("If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$") and $$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$

The manipulations are consequences of $x(y,z) = (xy,xz)$; see Bill Dubuque's proof here.

Follows quickly by employing basic gcd laws, namely the gcd (or lcm) Universal Property $\, x\mid y,z\!\!\!\overset{\rm\color{#c00}{U}\!}\iff\! x\mid (y,z)_{\phantom{|_{|_|}}} $ (this is the gcd definition in general domains), and also using the ubiquitous $\,\rm\color{#0a0}D\,$= GCD $ $ distributive law $\, \ (ac,bc)\, =\, (a,b)\,c,\,$ proved here,. With such

it is easy: $\ \ \ a\mid bc\iff a\mid ac,bc\smash{\overset{\rm\color{#c00}{U}\!}\iff} a\mid (ac,bc) \overset{\rm\color{#0a0}D}= (a,b)\,c\iff a/(a,b)\mid c\quad$ QED

Or dually: $\ a\mid bc\iff a,b\mid bc\smash{\overset{\rm\color{#c00}{U}\!}\iff} [a,b]\mid bc\iff [a,b]/b\ |\ c,\ $ $\, [m,n]\, :=\ {\rm lcm}(m,n)$

Combining the two yields the GCD $\cdot$ LCM law: $\ [a,b]/b\, =\, a/(a,b),\ $ i.e. $\ (a,b)\ [a,b] =\, ab\,.$

Alternatively, cancelling $\ (a,b)\,$ reduces it to Euclid's Lemma $\ (a,b)= 1,\ a\,|\,bc\ \Rightarrow\ a|c\,.$

Generally the result is the special case $\ a\,|\,bc\ $ of $\ (a,bc) = (a,(a,b)\,c),\,$ which holds true for both GCDs and ideals. As you can see these basic properties are all intimately connected, so it's useful to master these properties to become proficient with GCD arithmetic. For another example of applying these properties to obtain simple proofs see this proof of the Freshman's Dream $\ (A,B)^n = (A^n,B^n)\ $ for GCDs and invertible ideals.