If $a, b \mid c \text { and } \gcd(a, b) = d, \text { then } ab \mid cd $

Question

$a \mid c \implies c = ak \text { and } b \mid c \implies c = bj.$

$ak + bj = 2c = d \implies c \mid d.$

$d \mid a \implies a = dj.$

$c = ak = d(jk) \implies d \mid c.$

So, $c = d.$

$a \mid c \text { and } b \mid c \implies ab \mid cc \implies ab \mid cd.$

Does this make sense to you?

Answer 1

There is a mistake:

$$ak+bj = 2c = d$$

I agree, since $ak=bj=c$, that $ak+bj=2c$, but why would $2c$ be equal to $\gcd(a,b)?$