Suppose that $(a,b)\sim(c,d)$; you want to show that $(c,d)\sim(a,b)$.
Translate this into something more basic: by definition there is some non-zero $\lambda$ such that
$$\begin{align*}a&=\lambda c\\
b&=\lambda d\;,
\end{align*}\tag{1}$$ and you want to show that there is a non-zero $\mu$ such that
$$\begin{align*}c&=\mu a\\
d&=\mu b\;.
\end{align*}\tag{2}$$
Knowing that $\lambda\ne 0$, you should be able to use $(1)$ to solve $(2)$ for $\mu$ in terms of $\lambda$ without any real trouble.
You should approach transitivity the same way. Suppose that $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$. Then you know that there are non-zero $\lambda$ and $\mu$ such that
$$\begin{align*}a&=\lambda c\\
b&=\lambda d\\
c&=\mu e\\
d&=\mu f\;,
\end{align*}\tag{3}$$
and you want to show that there is a non-zero $\alpha$ such that
$$\begin{align*}
a&=\alpha e\\
b&=\alpha f\;.
\end{align*}$$
Can you see how to combine the information in $(3)$ to get this $\alpha$ in terms of $\lambda$ and $\mu$?