abstract algebra - How can I show that in a UFD $operatorname{gcd} (ra,rb) sim r operatorname{gcd} (a,b)$?

Let $R$ be a UFD. Let $r,a,b \in R$. Then how can I show that $\operatorname{gcd} (ra,rb) \sim r \operatorname{gcd} (a,b)$?

If $\operatorname{gcd} (a,b) = d$ and $\operatorname{gcd} (ra,rb) = d'$.Then $d|a$ and $d|b$ $\implies$ $rd|ra$ and $rd|rb$ i.e. $rd$ becomes a common divisor of $ra$ and $rb$. Hence we must have $rd|d'$. But I fail to show the converse part i.e. $d'|rd$.

How can I show this? Please help me.

Thank you in advance.