I need to prove the following:

1. Let $\displaystyle a,b,d\in\mathbb{Z}$

If $\displaystyle d|a$ and $\displaystyle d|b$, then:

$\displaystyle d=gcd(a,b) \Leftrightarrow (\frac{a}{d},\frac{b}{d})=1$

Here I was able to prove only one direction:

$\displaystyle d=gcd(a,b) \Rightarrow (\frac{a}{d},\frac{b}{d})=1$

Since $\displaystyle d=gcd(a,b)$, there exist $\displaystyle s,t\in\mathbb{Z}$ such that $\displaystyle d=sa+tb$.

I divide the whole thing by d and get:

$\displaystyle s\frac{a}{d}+t\frac{b}{d}=1$

Now since $\displaystyle gcd(x,y)=\min_{u,v\in\mathbb{Z}}}\left \{ xu+yv\in\mathbb{N}} \right \}$, and 1 is the minimal natural number, then $\displaystyle gcd(\frac{a}{d},\frac{b}{d})=1$.

2. $\displaystyle gcd(ab,ac)=a\cdot gcd(b,c)$

Here I don't have a clue what to do.

any help would be greatly appreciated!