18. 如图①,直线$AB与CD交于点O$,且$∠EOA = 90^{\circ}$.
(1)若点$B在点O$的正东方向上,点$D在点O的北偏东20^{\circ}$方向上,则点$C在点O$的______方向上.
(2)判断$∠AOC与∠EOD$的数量关系并说明理由.
(3)如图②,$OM是∠BOC$的平分线,设$∠EOD = \alpha(\alpha ＜ 90^{\circ})$.
i. 求$∠BOM$的度数(用含$\alpha$的代数式表示)；
ii. 直线$CD$由如图②位置开始,绕$O点以每秒\frac{1}{10}\alpha的速度顺时针旋转t s$(旋转角度始终小于$180^{\circ}$),请你直接写出$∠BOM$的度数(用含$\alpha$,$t$的代数式表示).

解：(1) 南偏西$20^{\circ}$
(2) $\angle AOC + \angle EOD = 90^{\circ}$，理由略.
(3) i. $\angle BOM = 45^{\circ} + \frac{1}{2}\alpha$.
ii. $\angle DOE = \alpha + \frac{1}{10}\alpha t$.
当$\angle DOE \leq 90^{\circ}$时，即$t \leq \frac{900^{\circ} - 10\alpha}{\alpha}$，
由i知，$\angle BOM = 45^{\circ} + \frac{1}{2}(\alpha + \frac{1}{10}\alpha t) = 45^{\circ} + \frac{1}{2}\alpha + \frac{1}{20}\alpha t$；
当$90^{\circ} ＜ \angle DOE ＜ 180^{\circ}$时，
即$\frac{900^{\circ} - 10\alpha}{\alpha} ＜ t ＜ \frac{1800^{\circ} - 10\alpha}{\alpha}$，如图①，
$\therefore \angle AOC = \angle DOE - 90^{\circ} = \alpha + \frac{1}{10}\alpha t - 90^{\circ}$，
$\therefore \angle BOM = \frac{1}{2}(180^{\circ} - \angle AOC) = 135^{\circ} - \frac{1}{2}\alpha - \frac{1}{20}\alpha t$；
当旋转角大于或等于$180^{\circ} - \alpha$时，
即$\frac{1800^{\circ} - 10\alpha}{\alpha} \leq t ＜ \frac{1800^{\circ}}{\alpha}$，如图②，
$\therefore \angle DOE = 360^{\circ} - \alpha - \frac{1}{10}\alpha t$，
$\therefore \angle AOD = \angle DOE - 90^{\circ} = 270^{\circ} - \alpha - \frac{1}{10}\alpha t$，
$\therefore \angle BOC = \angle AOD = 270^{\circ} - \alpha - \frac{1}{10}\alpha t$.
$\because OM$为$\angle BOC$的平分线，
$\therefore \angle BOM = \frac{1}{2}\angle BOC = 135^{\circ} - \frac{1}{2}\alpha - \frac{1}{20}\alpha t$.
综上所述，$\angle BOM =\begin{cases}45^{\circ} + \frac{1}{2}\alpha + \frac{1}{20}\alpha t, & 0 \leq t \leq \frac{900^{\circ} - 10\alpha}{\alpha}, \\135^{\circ} - \frac{1}{2}\alpha - \frac{1}{20}\alpha t, & \frac{900^{\circ} - 10\alpha}{\alpha} ＜ t ＜ \frac{1800^{\circ}}{\alpha}.\end{cases}$