If $a\mid b$ then $\gcd(a,c) \leq \gcd(b,c)$

Question

I need to show that:

If $a\mid b$ then $\gcd(a,c) \leq \gcd(b,c)$ where $a,b,c$ are positive integers.

I've come up with this, but I'm not 100% sure that it's correct:

Assume $a\mid b$, then $a \leq b$. Multiply both sides by an integer, $x$, such that $ax \leq bx$. Then add $cy$ for positive integer, $c$, and some integer, $y$, to both sides such that $ax + cy \leq bx + cy$. Then by EEA, $ax + cy = \gcd(a,c)$ and $bx + cy = \gcd(b,c)$. Therefore, $\gcd(a,c) \leq \gcd(b,c)$

Answer 1

This doesn’t work, I’m afraid. The extended Euclidean algorithm gives you some pair of integers $x$ and $y$ such that $ax+cy=\gcd(a,c)$, but there’s no guarantee that $bx+cy=\gcd(b,c)$ for that same pair of integers.

HINT: Since $a\mid b$, you know that for every $d$, if $d\mid a$, then $d\mid b$.

Answer 2

Let $\,b = an.\,$ If $\,d\mid a,c\,$ then $\,d\mid an,c,\,$ so the set $S = $ common divisors of $\,a,c\,$ is a subset of $T = $ common divisors of $\,an,c,\ $ so $\ S\subseteq T\Rightarrow \max S \le \max T,\ $ i.e $\ (a,c)\le (an,c).$

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Alternatively: $\,\ (a,c)\mid an,c\,\Rightarrow\,(a,c)\mid(an,c),\,$ therefore $\ (a,c)\le (an,c).$