解: (1) 如圖，直線$MN$即為所求.
(2) 如圖，$\because MN$垂直平分線段$AB，$$\therefore DA = DB，$設(shè)$DA = DB = x.$ 在$Rt\triangle ACD$中，$\because AD^{2}=AC^{2}+CD^{2}，$$\therefore x^{2}=4^{2}+(8 - x)^{2}，$
$\begin{aligned}x^{2}&=16 + 64-16x+x^{2}\\16x&=80\\x&= 5\end{aligned}$
$\therefore BD = 5.$
