Prove that $\gcd (−a, −b) = \gcd (a, b)$ where $a$ and $b$ are not both zero.
Suppose $g$ is $\gcd (−a, −b)$. Then, for some integers $m,n$
$$\begin{align} (-a) &= mg \\ \\ (-b) &= ng \end{align}$$
Multiplying these equations by (-1) gives us
$$\begin{align} a &= (-m)g \\ \\ b &= (-n)g \end{align}$$
So $g$, which is positive by definition, is divisor of both $a$ and $b$.
It is also the largest common divisor of $a$ and $b$ because it is also the largest common divisor of $(-1)a$ and $(-1)b$.
Put another way, if there were a larger common divisor $h$ of $a$ and $b$, then for some integers $p,q$
$$\begin{align} a &= ph \\ \\ b &= qh \end{align}$$
which gives us
$$\begin{align} (-a) &= (-p)h \\ \\ (-b) &= (-q)h \end{align}$$
then that $h$ and not $g$ would be the $\gcd (−a, −b)$.
And so
$$\gcd (−a, −b) = \gcd (a, b)$$