解: (1) 如图,直线$MN$即为所求.
(2) 如图,$\because MN$垂直平分线段$AB,$$\therefore DA = DB,$设$DA = DB = x.$ 在$Rt\triangle ACD$中,$\because AD^{2}=AC^{2}+CD^{2},$$\therefore x^{2}=4^{2}+(8 - x)^{2},$
$\begin{aligned}x^{2}&=16 + 64-16x+x^{2}\\16x&=80\\x&= 5\end{aligned}$
$\therefore BD = 5.$