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21. (8分)如图，在等腰三角形ABC中，AB = CB，点F在AB边上，连接CF并延长，交AD于点E，BD = BE，∠DBE = ∠ABC.
(1)求证：AD = CE.
(2)若∠ABC = 30°，∠AFC = 45°，求∠EAC的度数.
(1)求证：AD = CE.
(2)若∠ABC = 30°，∠AFC = 45°，求∠EAC的度数.
答案：
21.
(1)证明：$\because \angle DBE = \angle ABC$，
$\therefore \angle DBE + \angle ABE = \angle ABC + \angle ABE$，
即$\angle ABD = \angle CBE$.
在$\triangle ADB$和$\triangle CEB$中，$\begin{cases} AB = CB, \\ \angle ABD = \angle CBE, \\ BD = BE, \end{cases}$
$\therefore \triangle ADB \cong \triangle CEB (SAS)$.$\therefore AD = CE$.
(2)解：$\because BA = BC$，$\angle ABC = 30^{\circ}$，
$\therefore \angle BAC = \angle BCA = \frac{1}{2}(180^{\circ} - \angle ABC) = 75^{\circ}$.
$\because \angle AFC = 45^{\circ}$，$\therefore \angle BCE = \angle AFC - \angle ABC = 15^{\circ}$.
由
(1)得$\triangle ADB \cong \triangle CEB$，$\therefore \angle BAD = \angle BCE = 15^{\circ}$.
$\therefore \angle EAC = \angle BAD + \angle BAC = 90^{\circ}$.
(1)证明：$\because \angle DBE = \angle ABC$，
$\therefore \angle DBE + \angle ABE = \angle ABC + \angle ABE$，
即$\angle ABD = \angle CBE$.
在$\triangle ADB$和$\triangle CEB$中，$\begin{cases} AB = CB, \\ \angle ABD = \angle CBE, \\ BD = BE, \end{cases}$
$\therefore \triangle ADB \cong \triangle CEB (SAS)$.$\therefore AD = CE$.
(2)解：$\because BA = BC$，$\angle ABC = 30^{\circ}$，
$\therefore \angle BAC = \angle BCA = \frac{1}{2}(180^{\circ} - \angle ABC) = 75^{\circ}$.
$\because \angle AFC = 45^{\circ}$，$\therefore \angle BCE = \angle AFC - \angle ABC = 15^{\circ}$.
由
(1)得$\triangle ADB \cong \triangle CEB$，$\therefore \angle BAD = \angle BCE = 15^{\circ}$.
$\therefore \angle EAC = \angle BAD + \angle BAC = 90^{\circ}$.
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