1.(教材变式)如图,$\angle 1+\angle 2= 4\angle A$,求$\angle A$的度数.

解: $\because \angle 1=\angle A E D+\angle A$,
$\angle 2=\angle A D E+\angle A$,
$\therefore \angle 1+\angle 2=\angle A E D+\angle A+$
$=\angle A D E+\angle A$
$=180^{\circ}+\angle A$.
$\because \angle 1+\angle 2=4 \angle A$,
$\therefore 4 \angle A=180^{\circ}+\angle A, \therefore \angle A=60^{\circ}$.

2.(教材变式)如图,$\angle 1= \angle 2$,$\angle 3= \angle 4$,$\angle A= 110^{\circ }$,求$x$的值.

解: $\because \angle A=110^{\circ}$,
$\therefore \angle A B C+\angle A C B=180^{\circ}-\angle A=$
$70^{\circ}$,
即 $\angle 1+\angle 2+\angle 3+\angle 4=70^{\circ}$.
$\because \angle 1=\angle 2, \angle 3=\angle 4$,
$\therefore 2 \angle 2+2 \angle 4=70^{\circ}$,
$\therefore \angle 2+\angle 4=35^{\circ}$,
$\because x^{\circ}+\angle 2+\angle 4=180^{\circ}$,
解得 $x=145$,
$\therefore x$ 的值为 145.

3.(教材变式)如图,求$\angle A+\angle B+\angle C+\angle D+\angle E$的度数.

解: $\because \angle 1=\angle B+\angle C$,
$\angle 2=\angle D+\angle E$,
$\therefore \angle A+\angle B+\angle C+\angle D+\angle E=$
$\angle A+\angle 1+\angle 2=180^{\circ}$.

4.如图,在$\triangle ABC$中,$\angle B= \angle C$,$D为BC$边上一点,点$E在CA$的延长线上,连接$AD$,$DE$,若$\angle E= \angle ADE$,$\angle BAD= 26^{\circ }$,求$\angle BDE$的度数.

解: 设 $\angle E=\angle A D E=\alpha$,
$\therefore \angle C A D=\angle E+\angle A D E=2 \alpha$.
$\because \angle B A D=26^{\circ}$,
$\therefore \angle B A C=2 \alpha+26^{\circ}$.
$\because \angle B=\angle C$,
$\therefore \angle C=\frac{180^{\circ}-\angle B A C}{2}=77^{\circ}-\alpha$.
$\because \angle B D E=\angle C+\angle E$,
![img alt=4]
$\therefore \angle B D E=77^{\circ}$.