If \(a ≡ b \bmod m\) and \(c ≡ d \bmod m\):
$$\begin{gathered} a = b + mk \\ c = d + ml\end{gathered}$$
If we add them:
$$\begin{gathered} ac = (b + mk)(d + ml) \\ ac = bd + bml + dmk + mkml = bd + m(b+d+mkl) \end{gathered}$$
This proves that \(ac ≡ bd \bmod m\). As a corollary, we can also claim \(a^k ≡ b^k \bmod m\).