This follows easily from basic laws of GCD arithmetic. First I will present the proof using the standard notation $\rm\: (a,b)\:$ for $\rm\: gcd(a,b),\: $ immediately followed by a proof employing a more suggestive arithmetical notation, namely denoting $\rm\:\gcd(a,b)\:$ by $\rm\ a \dot+ b\:.\:$ Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes more intuitive using a notation that highlights this common arithmetical structure.
The proof below employs only said basic GCD laws, namely the $ $ associative law, $ $ and the commutative law, combined with the fundamental distributive law $\rm\, (ac,bc)\ =\ (a,b)\:c\, $ and the GCD-specific law $\rm\: \color{#C00}{(c,1) = 1}.\: $
$$\begin{eqnarray} &\rm (ac,\:b)\! &=&\rm\: (ac,b\:\color{#C00}{(c,1)})\!\! &=&\rm\: (ac,\:bc,\:b)\!\! &=&\rm\ ((a,b)\:c,b)\! &[\!\!\rm&=&\rm\: (c,\:b)\ &\rm if\ \ \ (a,b)\!\! &=&\rm 1\:] \\[0.2em] &\rm ac \dot+ b &=&\rm\ ac \dot+ b\:\color{#C00}{(c \dot+ 1)}\!\! &=&\rm\,\ ac\dot+bc\dot+b\!\! &=&\rm\ (a\dot+b)\:c \dot+ b\ &[\!\! &=&\rm\ \: c \dot+ b\ \:\ &\rm if\rm\quad a\dot+b\!\! &=&\rm 1\:] \end{eqnarray}$$
Notice how the suggestive notation in the second proof invites us to exploit our well-honed arithmetical intuition regarding the associative, commutative and distributive laws of integer arithmetic during analogous GCD arithmetic proofs. $ $ For a less trivial example see a similar proof of the Freshman's Dream $\rm\ (A,B)^n\, =\ (A^n,B^n)\ $ for GCDs and cancellable ideals.
The motivation behind this powerful abstract axiomatic approach becomes much clearer should you go on to study ideal theory and/or divisor theory (and you should, for there lies much beauty).
For further discussion and generalizations see this post and see my menagerie of posts on GCDs.