《2025年暑假衔接七年级数学人教版延边人民出版社》

9. 如图,在$△ABC$中,$CD$平分$∠ACB$,$DE// AC$,$∠B = 72^{\circ}$,$∠EDC = 36^{\circ}$,求:
(1)$∠A$的度数为36°;
(2)$∠ADC$的度数为108°.

答案：
(1) $\because \angle EDC = 36^{\circ}, DE // AC$, $\therefore \angle ACD = 36^{\circ}$. $\because CD$平分$\angle ACE$, $\therefore \angle ACB = 72^{\circ}$. $\because \angle B = 72^{\circ}$, $\therefore \angle A = 36^{\circ}$.
(2) $\because \angle ACD = 36^{\circ}$, $CD$平分$\angle ACE$, $\therefore \angle DCB = 36^{\circ}$, $\therefore \angle ADC = \angle DBC + \angle DCB = 108^{\circ}$.

10. 如图,$CE$是$△ABC$的外角$∠ACD$的平分线,且$CE$交$BA$的延长线于点$E$.
(1)若$∠B = 30^{\circ}$,$∠ACB = 40^{\circ}$,求$∠E$的度数;40°
(2)求证:$∠BAC = ∠B + 2∠E$.

答案：
(1) $\because \angle ACB = 40^{\circ}$, $\therefore \angle ACD = 180^{\circ} - 40^{\circ} = 140^{\circ}$, $\because \angle B = 30^{\circ}$, $\therefore \angle EAC = \angle B + \angle ACB = 70^{\circ}$, $\because CE$是$\triangle ABC$的外角$\angle ACD$的平分线，$\therefore \angle ACE = 70^{\circ}$, $\therefore \angle E = 180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}$.
(2) $\because CE$平分$\angle ACD$, $\therefore \angle ACE = \angle DCE$, $\because \angle DCE = \angle B + \angle E$, $\therefore \angle ACE = \angle B + \angle E$, $\because \angle BAC = \angle ACE + \angle E$, $\therefore \angle BAC = \angle B + \angle E + \angle E = \angle B + 2\angle E$.

11. (1)如图1所示,把$△ABC$纸片沿$DE$折叠,使点$A$落在四边形$BCED$内部点$A'$处,试说明:$2∠A = ∠1 + ∠2$;
(2)如图2所示,若把$△ABC$纸片沿$DE$折叠,使点$A$落在四边形$BCED$外部点$A'$处,此时$∠A$与$∠1$,$∠2$之间的数量关系是______$2∠A = ∠1 - ∠2$______;
(3)如图3所示,若把四边形$ABCD$沿$EF$折叠,使点$A$,$D$落在四边形$BCFE$的内部点$A'$,$D'$处,请你探索$∠A$,$∠D$,$∠1$,$∠2$之间的数量关系.写出你的结论并说明理由.

答案：
(1) 根据翻折的性质，可得$\angle ADE = \frac{1}{2}(180^{\circ} - \angle 1)$, $\angle AED = \frac{1}{2}(180^{\circ} - \angle 2)$. $\because \angle A + \angle ADE + \angle AED = 180^{\circ}$, $\therefore \angle A + \frac{1}{2}(180^{\circ} - \angle 1) + \frac{1}{2}(180^{\circ} - \angle 2) = 180^{\circ}$. $\therefore 2\angle A = \angle 1 + \angle 2$.
(2) $2\angle A = \angle 1 - \angle 2$
(3) $2(\angle A + \angle D) = \angle 1 + \angle 2 + 360^{\circ}$. 理由如下：根据翻折的性质，可得$\angle AEF = \frac{1}{2}(180^{\circ} - \angle 1)$, $\angle DFE = \frac{1}{2}(180^{\circ} - \angle 2)$. $\because \angle A + \angle D + \angle AEF + \angle DFE = 360^{\circ}$, $\therefore \angle A + \angle D + \frac{1}{2}(180^{\circ} - \angle 1) + \frac{1}{2}(180^{\circ} - \angle 2) = 360^{\circ}$. $\therefore 2(\angle A + \angle D) = \angle 1 + \angle 2 + 360^{\circ}$.