如图, 点 $\mathrm{O}$ 在直线 $\mathrm{AB}$ 上, $C O \perp A B, \mathrm{OE}$ 是 $\angle B O D$ 的平分线, $O F \perp O E$

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(1) 找出图中与 $\angle B O F$ 相等的角, 并说明理由;

(2) 若 $\angle 2-\angle 1=20^{\circ}$, 求 $\angle B O E$ 的度数.
(1) 解: $\because \mathrm{CO} \perp \mathrm{AB}, \mathrm{OF} \perp \mathrm{OE}$,

$\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}=\angle \mathrm{BOE}+\angle \mathrm{BOF}=90^{\circ}$

$\therefore \angle \mathrm{COE}=\angle \mathrm{BOF}$

(2) 解: $\because \mathrm{CO} \perp \mathrm{AB}$,

$\therefore \angle \mathrm{COA}=90^{\circ}$, 即 $\angle 2=90^{\circ}-\angle 1$,

又 $\because \angle 2-\angle 1=20^{\circ}$,

$\therefore \angle 2=20^{\circ}+\angle 1$,

$\therefore 90^{\circ}-\angle 1=20^{\circ}+\angle 1$,

解得: $\angle 1=35^{\circ}$,

$\therefore \angle 2=55^{\circ}$,

$\therefore \angle \mathrm{BOD}=180^{\circ}-\angle 2=125^{\circ}$,

$\because \mathrm{OE}$ 是 $\angle \mathrm{BOD}$ 的平分线,

$\therefore \angle \mathrm{BOE}=\angle \mathrm{DOE}=\frac{1}{2} \angle \mathrm{BOD}=62.5^{\circ}$.