如图，在$\mathit{\Delta ABC}$中，$\mathit{AB}=\mathit{AC}$，以$\mathit{AB}$为直径的$\odot O$交$\mathit{BC}$于点$D$，连接$\mathit{AD}$，过点$D$作$\mathit{DM}\perp \mathit{AC}$，垂足为$M$，$\mathit{AB}$、$\mathit{MD}$的延长线交于点$N$．

（1）求证：$\mathit{MN}$是$\odot O$的切线；

（2）求证：$D{N}^2=\mathit{BN}·(\mathit{BN}+\mathit{AC})$；

（3）若$\mathit{BC}=6$，$\cos C=\frac{3}{5}$，求$\mathit{DN}$的长．

证明：（1）如图，连接$\mathit{OD}$，

$\because \mathit{AB}$是直径，

$\therefore \angle \mathit{ADB}=90°$，

又$\because \mathit{AB}=\mathit{AC}$，

$\therefore \mathit{BD}=\mathit{CD}$，$\angle \mathit{BAD}=\angle \mathit{CAD}$，

$\because \mathit{AO}=\mathit{BO}$，$\mathit{BD}=\mathit{CD}$，

$\therefore \mathit{OD}//\mathit{AC}$，

$\because \mathit{DM}\perp \mathit{AC}$，

$\therefore \mathit{OD}\perp \mathit{MN}$，

又$\because \mathit{OD}$是半径，

$\therefore \mathit{MN}$是$\odot O$的切线；

（2）$\because \mathit{AB}=\mathit{AC}$，

$\therefore \angle \mathit{ABC}=\angle \mathit{ACB}$，

$\because \angle \mathit{ABC}+\angle \mathit{BAD}=90°$，$\angle \mathit{ACB}+\angle \mathit{CDM}=90°$，

$\therefore \angle \mathit{BAD}=\angle \mathit{CDM}$，

$\because \angle \mathit{BDN}=\angle \mathit{CDM}$，

$\therefore \angle \mathit{BAD}=\angle \mathit{BDN}$，

又$\because \angle N=\angle N$，

$\therefore \mathit{\Delta BDN}\backsim \mathit{\Delta DAN}$，

$\therefore \frac{\mathit{BN}}{\mathit{DN}}=\frac{\mathit{DN}}{\mathit{AN}}$，

$\therefore D{N}^2=\mathit{BN}·\mathit{AN}=\mathit{BN}·(\mathit{BN}+\mathit{AB})=\mathit{BN}·(\mathit{BN}+\mathit{AC})$；

（3）$\because \mathit{BC}=6$，$\mathit{BD}=\mathit{CD}$，

$\therefore \mathit{BD}=\mathit{CD}=3$，

$\because \cos C=\frac{3}{5}=\frac{\mathit{CD}}{\mathit{AC}}$，

$\therefore \mathit{AC}=5$，

$\therefore \mathit{AB}=5$，

$\therefore \mathit{AD}=\sqrt{A{B}^2-B{D}^2}=\sqrt{25-9}=4$，

$\because \mathit{\Delta BDN}\backsim \mathit{\Delta DAN}$，

$\therefore \frac{\mathit{BN}}{\mathit{DN}}=\frac{\mathit{DN}}{\mathit{AN}}=\frac{\mathit{BD}}{\mathit{AD}}=\frac{3}{4}$，

$\therefore \mathit{BN}=\frac{3}{4}\mathit{DN}$，$\mathit{DN}=\frac{3}{4}\mathit{AN}$，

$\therefore \mathit{BN}=\frac{3}{4}\left(\frac{3}{4}\mathit{AN}\right)=\frac{9}{16}\mathit{AN}$，

$\because \mathit{BN}+\mathit{AB}=\mathit{AN}$，

$\therefore \frac{9}{16}\mathit{AN}+5=\mathit{AN}$

$\therefore \mathit{AN}=\frac{80}{7}$，

$\therefore \mathit{DN}=\frac{3}{4}\mathit{AN}=\frac{60}{7}$．