1. 如图,O 是$\triangle ABC$内一点,$AO = BO = CO$,$\angle BAC = 40^{\circ}$。求$\angle BOC$的度数。

解: $\because OA = OB = OC$,
$\therefore \angle OAB = \angle OBA$, $\angle OAC = \angle OCA$,
$\therefore \angle OAB + \angle OAC = \angle OBA + \angle OCA$,
$\therefore \angle OBA + \angle OCA = \angle BAC = 40^{\circ}$,
$\angle ABC + \angle ACB = 180^{\circ} - \angle BAC = 140^{\circ}$,
$\therefore \angle OBC + \angle OCB = 140^{\circ} - 40^{\circ} = 100^{\circ}$,
$\therefore \angle BOC = 180^{\circ} - (\angle OBC + \angle OCB)$
$= 80^{\circ}$.

2. (2024 内江中考)如图,在$\triangle ABC$中,$\angle DCE = 40^{\circ}$,$AE = AC$,$BC = BD$。求$\angle ACB$的度数。

解: $\because AC = AE$, $BC = BD$,
$\therefore$ 设 $\angle AEC = \angle ACE = x$,
$\angle BDC = \angle BCD = y$,
$\therefore \angle A = 180^{\circ} - 2x$, $\angle B = 180^{\circ} - 2y$.
$\because \angle ACB + \angle A + \angle B = 180^{\circ}$,
$\therefore \angle ACB = 2(x + y) - 180^{\circ}$.
$\because \angle BDC + \angle AEC + \angle DCE = 180^{\circ}$,
$\therefore x + y + 40^{\circ} = 180^{\circ}$,
$\therefore x + y = 140^{\circ}$,
$\therefore \angle ACB = 2 \times 140^{\circ} - 180^{\circ}$
$= 100^{\circ}$.

3. 如图,D,E,F 分别为$\triangle ABC$边上一点,且$AD = AF$,$BD = BE$。
(1)若$\angle C = 90^{\circ}$,求$\angle EDF$的度数；
(2)若$\angle C = \alpha$,直接写出$\angle EDF$的度数为______。

解: (1) $45^{\circ}$;
(2) $90^{\circ} - \frac{1}{2}\alpha$.
$\because \angle A + \angle B = 180^{\circ} - \alpha$,
$\therefore \angle AFD + \angle ADF + \angle BDE + \angle BED = 360^{\circ} - (180^{\circ} - \alpha) = 180^{\circ} + \alpha$.
$\because AD = AF$, $BD = BE$,
$\therefore \angle AFD = \angle ADF$,
$\angle BDE = \angle BED$,
$\therefore \angle EDF = 180^{\circ} - (\angle ADF + \angle BDE)$
$= 180^{\circ} - \frac{180^{\circ} + \alpha}{2}$
$= 90^{\circ} - \frac{1}{2}\alpha$.

4. 如图,在$\triangle ABC$中,$CA = CB$,D,E,F 分别在$AB$,$BC$,$AC$上,且$BD = AF$,$AD = BE$。
(1)若$\angle C = 40^{\circ}$,求$\angle EDF$的度数；
(2)若$\angle C = \alpha$,直接写出$\angle EDF$的度数为______。

解: (1) $\because CA = CB$,
$\therefore \angle A = \angle B$.
又 $\because AF = BD$, $AD = BE$,
$\therefore \triangle ADF \cong \triangle BED$,
$\therefore \angle ADF = \angle DEB$.
$\because CA = CB$, $\angle C = 40^{\circ}$,
$\therefore \angle A = \angle B = 70^{\circ}$,
$\therefore \angle DEB + \angle EDB = 110^{\circ}$,
$\therefore \angle ADF + \angle EDB = 110^{\circ}$,
$\therefore \angle EDF = 180^{\circ} - 110^{\circ} = 70^{\circ}$;
(2) 同(1)可得 $\angle EDF = 90^{\circ} - \frac{1}{2}\alpha$.