如图所示, 直线 $A B 、 C D$ 相交于点 $O, O M \perp A B$.

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(1) 若 $\angle 1=\angle 2$, 判断 $\mathrm{ON}$ 与 $\mathrm{CD}$ 的位置关系, 并说明理由;

(2) 若 $\angle 1=\frac{1}{5} \angle \mathrm{BOC}$, 求 $\angle \mathrm{MOD}$ 的度数.
(1) 解: $\mathrm{ON} \perp \mathrm{CD}$. 理由如下:

## $\because \mathrm{OM} \perp \mathrm{AB}$

$\therefore \angle \mathrm{AOM}=90^{\circ}$,

$\therefore \angle 1+\angle \mathrm{AOC}=90^{\circ}$

又 $\because \angle 1=\angle 2$,

$\therefore \angle 2+\angle \mathrm{AOC}=90^{\circ}$,

即 $\angle \mathrm{CON}=90^{\circ}$,

$\therefore \mathrm{ON} \perp \mathrm{CD}$.

(2) 解: $\because \mathrm{OM} \perp \mathrm{AB}, \quad \therefore \angle \mathrm{MOB}=90^{\circ}$,

$\therefore \angle \mathrm{BOC}=\angle \mathrm{MOB}+\angle 1=90^{\circ}+\angle 1$

$\because \angle 1=\frac{1}{5} \angle \mathrm{BOC}=\frac{1}{5}\left(\angle 1+90^{\circ}\right)$,

$\therefore \angle 1=22.5^{\circ}$,

又 $\because \angle 1+\angle \mathrm{MOD}=180^{\circ}$,
$\therefore \angle \mathrm{MOD}=180^{\circ}-\angle 1=157.5^{\circ}$.