Prove that $a \mid bc$ implies $a \mid \gcd(a,b)\times \gcd(a,c)$

Question

I want to show that $a \mid bc$ implies $a \mid \gcd(a,b)\times \gcd(a,c)$.

My answer: since $$ a\mid ac \quad \text{ and } \quad a\mid bc,$$ we get that $a \mid \gcd(ac,bc)$ which implies that $$ a \mid |c|\gcd(a,b).$$

However, I don't know the final step.

Answer 1

Recall that the $gcd$ is a linear combination. i.e. $(a,b)(a,c)$ can be written as

$$(a,b)(a,c)=(am+bn)(ak+cl)=a^2mk+amcl+bnak+bncl$$

$$=a(amk+mcl+bnk) + bc(nl)$$

If $a\mid bc$ then $a$ divides the right hand side.