《2025年多维导学案八年级数学上册人教版》

三角形的内角和定理：
三角形的内角和等于$180^{\circ}$.
几何语言：
如图，已知$\triangle ABC$，
则有$\angle A+\angle B+\angle C=$$180$$^{\circ}$.

答案： $180^{\circ}$ $180$ 

如图，已知$\triangle ABC$，求证：$\angle BAC+\angle B+\angle C=180^{\circ}$.

证明：如图，过点$A$作直线$DE// BC$.
$\therefore \angle B=\angle 2$，$\angle C=$____.
$\because \angle 1+\angle 2+\angle 3=$____$^{\circ}$，
$\therefore \angle BAC+\angle B+\angle C=$____$^{\circ}$.

答案： $\angle 1$ $180$ $180$

【例1】在$\triangle ABC$中：
(1)若$\angle A=70^{\circ}$，$\angle B=30^{\circ}$，则$\angle C=$80$^{\circ}$；
(2)若$\angle A=65^{\circ}$，则$\angle B+\angle C=$115$^{\circ}$；
(3)若$\angle A:\angle B:\angle C=1:2:3$，则$\angle A=$30$^{\circ}$；
(4)若$\angle A=45^{\circ}$，$\angle B$比$\angle C$大$25^{\circ}$，求$\angle B$的度数.
解:$\because \angle B$比$\angle C$大$25^{\circ},$
$\therefore$设$\angle B=x$,则$\angle C=x-25^{\circ}.$
$\because \angle A+\angle B+\angle C=180^{\circ},$
$\therefore 45^{\circ}+x+x-25^{\circ}=180^{\circ}.$
解得$x=80^{\circ}.$
$\therefore \angle B$的度数为$80^{\circ}.$

答案：
(1)$80$
(2)$115$
(3)$30$
(4)解:$\because \angle B$比$\angle C$大$25^{\circ},$
$\therefore$设$\angle B=x$,则$\angle C=x-25^{\circ}.$
$\because \angle A+\angle B+\angle C=180^{\circ},$
$\therefore 45^{\circ}+x+x-25^{\circ}=180^{\circ}.$
解得$x=80^{\circ}.$
$\therefore \angle B$的度数为$80^{\circ}.$

【变式1】(1)求出下列各图形中$x$的值：

$x=$40 $x=$60
(2)若$\angle A=20^{\circ}$，$\angle B=3\angle C$，求$\angle C$的度数.
解:$\because \angle B=3\angle C,$
$\therefore$设$\angle C=x$,则$\angle B=3x.$
$\because \angle A+\angle B+\angle C=180^{\circ},$
$\therefore 20^{\circ}+x+3x=180^{\circ}.$
解得$x=40^{\circ}.$
$\therefore \angle C$的度数为$40^{\circ}.$

答案：
(1)$40$ $60$
(2)解:$\because \angle B=3\angle C,$
$\therefore$设$\angle C=x$,则$\angle B=3x.$
$\because \angle A+\angle B+\angle C=180^{\circ},$
$\therefore 20^{\circ}+x+3x=180^{\circ}.$
解得$x=40^{\circ}.$
$\therefore \angle C$的度数为$40^{\circ}.$

【例2】(人教教材P12例1改编)如图，在$\triangle ABC$中，$\angle BAC=40^{\circ}$，$\angle B=65^{\circ}$，$AD$是$\triangle ABC$的角平分线，求$\angle ADB$的度数.

答案： 解:$\because AD$平分$\angle BAC$,且$\angle BAC=40^{\circ},$
$\therefore \angle BAD=\frac{1}{2}\angle BAC=20^{\circ}.$
$\because \angle BAD+\angle B+\angle ADB=180^{\circ},$
$\angle B=65^{\circ},$
$\therefore \angle ADB=180^{\circ}-65^{\circ}-20^{\circ}=95^{\circ}.$

【变式2】(多维原创)如图，$AE$是$\triangle ABC$的角平分线，$AD$是高，$\angle B=30^{\circ}$，$\angle C=65^{\circ}$.
(1)求$\angle CAE$度数；
(2)求$\angle DAE$的度数.

答案： 解:
(1)$\because$在$\triangle ABC$中,
$\angle B=30^{\circ},\angle C=65^{\circ},$
$\therefore \angle BAC=180^{\circ}-\angle B-\angle C=180^{\circ}-$
$30^{\circ}-65^{\circ}=85^{\circ}.$
$\because AE$平分$\angle BAC,$
$\therefore \angle CAE=\frac{1}{2}\angle BAC=\frac{1}{2}\times85^{\circ}=42.5^{\circ}.$
(2)$\because AD$是高,$\therefore \angle ADC=90^{\circ}.$
$\therefore \angle CAD=180^{\circ}-90^{\circ}-65^{\circ}=25^{\circ}.$
$\therefore \angle DAE=\angle CAE-\angle CAD=42.5^{\circ}-$
$25^{\circ}=17.5^{\circ}.$