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2025年假期冲冠黑龙江教育出版社八年级数学
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12. 如图,已知$D为\triangle ABC边BC$延长线上一点,$DF\perp AB于F$,且交$AC于E$,$\angle A = 30^{\circ}$,$\angle D = 55^{\circ}$
(1) 求$\angle ACD$的度数；
(2) 求$\angle FEC$的度数.
(1) 求$\angle ACD$的度数；
(2) 求$\angle FEC$的度数.
答案：
解：
(1) $\because DF\perp AB$，
$\therefore \angle BFD = 90^{\circ}$，
$\therefore \angle B = 90^{\circ}-\angle D = 35^{\circ}$，
$\because \angle ACD = \angle B+\angle A,\angle A = 30^{\circ}$，
$\therefore \angle ACD = 65^{\circ}$。
(2) $\because \angle FEC = \angle ECD+\angle D,\angle ECD = 65^{\circ},\angle D = 55^{\circ}$，
$\therefore \angle FEC = 55^{\circ}+65^{\circ}=120^{\circ}$。
(1) $\because DF\perp AB$，
$\therefore \angle BFD = 90^{\circ}$，
$\therefore \angle B = 90^{\circ}-\angle D = 35^{\circ}$，
$\because \angle ACD = \angle B+\angle A,\angle A = 30^{\circ}$，
$\therefore \angle ACD = 65^{\circ}$。
(2) $\because \angle FEC = \angle ECD+\angle D,\angle ECD = 65^{\circ},\angle D = 55^{\circ}$，
$\therefore \angle FEC = 55^{\circ}+65^{\circ}=120^{\circ}$。
13. 如图,在四边形$ABCD$中,$\angle B = 50^{\circ}$,$\angle C = 110^{\circ}$,$\angle D = 90^{\circ}$,$AE\perp BC$,$AF是\angle BAD$的平分线,与边$BC交于点F$. 求$\angle EAF$的度数.
答案：
解：在四边形 $ABCD$ 中，$\angle B = 50^{\circ},\angle C = 110^{\circ},\angle D = 90^{\circ}$，
$\therefore \angle BAD = 360^{\circ}-\angle B-\angle C-\angle D = 360^{\circ}-50^{\circ}-110^{\circ}-90^{\circ}=110^{\circ}$，
$\because AF$ 是 $\angle BAD$ 的平分线，
$\therefore \angle BAF=\frac{1}{2}\angle BAD=\frac{1}{2}×110^{\circ}=55^{\circ}$，
$\because AE\perp BC,\angle B = 50^{\circ}$，
$\therefore \angle BAE = 90^{\circ}-\angle B = 90^{\circ}-50^{\circ}=40^{\circ}$，
$\therefore \angle EAF = \angle BAF-\angle BAE = 55^{\circ}-40^{\circ}=15^{\circ}$。
$\therefore \angle BAD = 360^{\circ}-\angle B-\angle C-\angle D = 360^{\circ}-50^{\circ}-110^{\circ}-90^{\circ}=110^{\circ}$，
$\because AF$ 是 $\angle BAD$ 的平分线，
$\therefore \angle BAF=\frac{1}{2}\angle BAD=\frac{1}{2}×110^{\circ}=55^{\circ}$，
$\because AE\perp BC,\angle B = 50^{\circ}$，
$\therefore \angle BAE = 90^{\circ}-\angle B = 90^{\circ}-50^{\circ}=40^{\circ}$，
$\therefore \angle EAF = \angle BAF-\angle BAE = 55^{\circ}-40^{\circ}=15^{\circ}$。
14. 如图,在$\triangle ABC$中,$\angle 1 = \angle 2$,$\angle C ＞ \angle B$,$E为射线AD$上一点,且$EF\perp BC于F$.
(1) 若$\angle B = 40^{\circ}$,$\angle C = 60^{\circ}$,试求$\angle DEF$的度数；
(2) 由解答(1)的经历,试探索$\angle DEF与\angle B$、$\angle C$的数量关系,并说明理由.
(1) 若$\angle B = 40^{\circ}$,$\angle C = 60^{\circ}$,试求$\angle DEF$的度数；
(2) 由解答(1)的经历,试探索$\angle DEF与\angle B$、$\angle C$的数量关系,并说明理由.
答案：
解：
(1) $\because \angle B = 40^{\circ},\angle C = 60^{\circ},\therefore \angle BAC = 80^{\circ}$，
$\therefore \angle 1=\angle 2=\frac{1}{2}\angle BAC = 40^{\circ}$，
$\therefore \angle FDE = \angle ADC = 180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}$，
$\because EF\perp BC,\therefore \angle DEF = 90^{\circ}-80^{\circ}=10^{\circ}$。
(2) $\angle DEF=\frac{1}{2}(\angle C-\angle B)$。理由如下：
$\because \angle BAC = 180^{\circ}-\angle B-\angle C,\angle 1=\angle 2$，
$\therefore \angle 2=\frac{1}{2}(180^{\circ}-\angle B-\angle C)$，
$\therefore \angle ADC = 180^{\circ}-\angle C-\angle 2 = 90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B$，
$\therefore \angle EDF = 90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B$，
$\because EF\perp BC,\angle DEF = 90^{\circ}-(90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B)=\frac{1}{2}\angle C-\frac{1}{2}\angle B=\frac{1}{2}(\angle C-\angle B)$。
(1) $\because \angle B = 40^{\circ},\angle C = 60^{\circ},\therefore \angle BAC = 80^{\circ}$，
$\therefore \angle 1=\angle 2=\frac{1}{2}\angle BAC = 40^{\circ}$，
$\therefore \angle FDE = \angle ADC = 180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}$，
$\because EF\perp BC,\therefore \angle DEF = 90^{\circ}-80^{\circ}=10^{\circ}$。
(2) $\angle DEF=\frac{1}{2}(\angle C-\angle B)$。理由如下：
$\because \angle BAC = 180^{\circ}-\angle B-\angle C,\angle 1=\angle 2$，
$\therefore \angle 2=\frac{1}{2}(180^{\circ}-\angle B-\angle C)$，
$\therefore \angle ADC = 180^{\circ}-\angle C-\angle 2 = 90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B$，
$\therefore \angle EDF = 90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B$，
$\because EF\perp BC,\angle DEF = 90^{\circ}-(90^{\circ}-\frac{1}{2}\angle C+\frac{1}{2}\angle B)=\frac{1}{2}\angle C-\frac{1}{2}\angle B=\frac{1}{2}(\angle C-\angle B)$。
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