If $\gcd(a,b)= 1$ and $a$ divides $bc$ then $a$ divides $c\ $ [Euclid's Lemma] Well I thought this is obvious. since $\gcd(a,b)=1$,  then we have that $a\not\mid b$ AND $a\mid bc$. This implies that $a$ divides $c$. But apparently this is wrong. Help explain why this way is wrong please. 
The question tells you give me two relatively prime numbers $a$ and $b$ such that $a$ divides the product $bc$, prove that $a$ divides $c$. how is this not obvious? Explain to me how my "proof" is not correct.

Your reasoning is wrong for the following reason:
You say that

we have that $a$  does not divide $b$  AND $a$  divides $bc$ .

From here you conclude that $a|c$. This part is wrong. It is possible to have $a$ does not divide $b$  AND $a$  divides $bc$, but $a$ does not divide $c$.
For example, $a=4, b=6$ and $c=2$.
