如图，$\angle \mathit{ABD}=\angle \mathit{BCD}=90°$，$\mathit{DB}$平分$\angle \mathit{ADC}$，过点$B$作$\mathit{BM}//\mathit{CD}$交$\mathit{AD}$于$M$．连接$\mathit{CM}$交$\mathit{DB}$于$N$．

（1）求证：$B{D}^2=\mathit{AD}·\mathit{CD}$；

（2）若$\mathit{CD}=6$，$\mathit{AD}=8$，求$\mathit{MN}$的长．