《2025年新课程问题解决导学方案八年级数学上册华师大版》

12. 问题：如图，在$\triangle ABD$中，$BA = BD$，在$BD$的延长线上取点$E$、$C$，作$\triangle AEC$，使$EA = EC$.若$\angle BAE = 90^{\circ}$，$\angle B = 45^{\circ}$，求$\angle DAC$的度数.
答案：$\angle DAC = 45^{\circ}$.
思考：(1)如果把以上“问题”中的条件“$\angle B = 45^{\circ}$”去掉，其余条件不变，那么$\angle DAC$的度数会改变吗？请说明理由.
(2)如果把以上“问题”中的条件“$\angle B = 45^{\circ}$”去掉，再将“$\angle BAE = 90^{\circ}$”改为“$\angle BAE = n^{\circ}$”，其余条件不变，求$\angle DAC$的度数.

答案： 12.
(1) $\angle DAC$的度数不会改变.
理由：$\because EA = EC$，
$\therefore \angle AED = 2\angle C$. ①
$\because \angle BAE = 90^{\circ}$，
$\therefore \angle BAD = \frac{1}{2}[180^{\circ} - (90^{\circ} - 2\angle C)]$
$= 45^{\circ} + \angle C$.
$\therefore \angle DAE = 90^{\circ} - \angle BAD$
$= 90^{\circ} - (45^{\circ} + \angle C)$
$= 45^{\circ} - \angle C$.
由①②，得
$\angle DAC = \angle DAE + \angle CAE = 45^{\circ}$.
(2) 设$\angle B = m^{\circ}$，
则$\angle BAD = \frac{1}{2}(180^{\circ} - m^{\circ})$
$= 90^{\circ} - \frac{1}{2}m^{\circ}$，
$\angle AEB = 180^{\circ} - n^{\circ} - m^{\circ}$.
$\therefore \angle DAE = n^{\circ} - \angle BAD$
$= n^{\circ} - 90^{\circ} + \frac{1}{2}m^{\circ}$.
$\because EA = EC$，
$\therefore \angle CAE = \frac{1}{2}\angle AEB$
$= 90^{\circ} - \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ}$.
$\therefore \angle DAC = \angle DAE + \angle CAE$
$= n^{\circ} - 90^{\circ} + \frac{1}{2}m^{\circ} + 90^{\circ} - \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ}$
$= \frac{1}{2}n^{\circ}$.