Explaination of step in proof that If $d|a$ and $d|b$ and $\displaystyle\gcd(\frac{a}{d}, \frac{b}{d})=1$ then $\gcd(a,b)=d$ My professor presented the following proof and I am wondering if it makes sense. There is one piece of it that I do not quite grasp. 

If $d|a$ and $d|b$ and $\displaystyle\gcd(\frac{a}{d}, \frac{b}{d})=1$ then $\gcd(a,b)=d$ 

Proof: We know that there are $x$ and $y$ such that $a=dx$ and $b=dy$. But since we are given $\displaystyle\gcd(\frac{a}{d}, \frac{b}{d})=1$ we have that $\gcd(x,y)=1$. Any common divisor of $a$ and $b$ that is greater than $d$ would have to be a common divisor of $x$ and $y$. But $x$ and $y$ have no common divisors greater than 1. So $\gcd(a,b)=d$. 
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It is not clear to me that the bolded statement above is true. If it is true then could someone verify that it is true and possibly elaborate on why it must be true. 
 A: I agree with the part of Greg Martin's comment which says

The bolded statement doesn't make sense.

I obviously can't be exactly sure what your professor was trying to convey, but I believe it was something along the line of

Any common divisor of $a$ and $b$ that is greater than $d$ and is of the form $de$ would have $e$ be a common divisor of $x$ and $y$.

The rest of the stated proof would then follow. Using this same basic approach, here is how I would write a formal proof. Let
$$f = \gcd(a,b), \; a = fg, \; b = fh, \; \gcd(g,h) = 1 \tag{1}\label{eq1A}$$
Since $d \mid a$ and $d \mid b$, then $d \mid f$. Then for some integer $e \ge 1$, you have
$$f = ed \tag{2}\label{eq2A}$$
This gives
$$\begin{equation}\begin{aligned}
\gcd\left(\frac{a}{d},\frac{b}{d}\right) & = \gcd\left(\frac{(ed)g}{d},\frac{(ed)h}{d}\right) \\
& = \gcd(eg,eh) \\
& = e(\gcd(g,h)) \\
& = e \\
& = 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Note going from the second to third lines used, as Bill Dubuque's comment states, the GCD distributive law, with multiple proofs for this such as in his answer. You thus have $e = 1$, so using \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} gives
$$\gcd(a,b) = f = ed = d \tag{4}\label{eq4A}$$
