23. 如图10①,AD平分$∠BAC$,$AE⊥BC$,$∠B= 40^{\circ }$,$∠C= 70^{\circ }$。
(1)求$∠DAE$的度数。
(2)如图10②,若把“$AE⊥BC$”变成“点F在DA的延长线上,$FE⊥BC$”,其他条件不变,求$∠DFE$的度数。
(3)如图10③,若把“$AE⊥BC$”变成“AE平分$∠BEC$”,其他条件不变,$∠DAE$的度数大小是否有变化？请说明理由。

(1)$\because \angle B = 40^{\circ}$,$\angle C = 70^{\circ}$,
$\therefore \angle BAC = 70^{\circ}$.
$\because AD$平分$\angle BAC$,
$\therefore \angle BAD = \angle CAD = 35^{\circ}$.
$\therefore \angle ADE = \angle B + \angle BAD = 75^{\circ}$.
$\because AE \perp BC$,$\therefore \angle AEB = 90^{\circ}$.
$\therefore \angle DAE = 90^{\circ} - \angle ADE = 15^{\circ}$.
(2)同(1),可得$\angle ADE = 75^{\circ}$.
$\because FE \perp BC$,$\therefore \angle FEB = 90^{\circ}$.
$\therefore \angle DFE = 90^{\circ} - \angle ADE = 15^{\circ}$.
(3)结论:$\angle DAE$的度数大小不变.理由如下:
$\because AE$平分$\angle BEC$,$\therefore \angle AEB = \angle AEC$.
$\therefore \angle C + \angle CAE = \angle B + \angle BAE$.
$\because \angle CAE = \angle CAD - \angle DAE$,$\angle BAE = \angle BAD + \angle DAE$,
$\therefore \angle C + \angle CAD - \angle DAE = \angle B + \angle BAD + \angle DAE$.
$\because AD$平分$\angle BAC$,$\therefore \angle BAD = \angle CAD$.
$\therefore 2\angle DAE = \angle C - \angle B = 30^{\circ}$.
$\therefore \angle DAE = 15^{\circ}$.