《2025年学霸题中题八年级数学上册人教版》

1. (2024·重庆期中)(1)如图①，$\triangle ABC$是等边三角形，$D$是边$BC$下方一点，$\angle BDC = 120^{\circ}$，探索线段$DA$，$DB$，$DC$之间的数量关系。
(2)如图②，$\triangle ABC$是等边三角形，直线$a// AB$，$D$为边$BC$上一点，$\angle ADE$交直线$a$于点$E$，且$\angle ADE = 60^{\circ}$，求证：$CD + CE = CA$。

答案：

(1) $DA = DC + DB$，理由如下：
如图①，延长 $DC$ 到点 $E$，使 $CE = BD$，连接 $AE$。$\because \triangle ABC$ 是等边三角形，$\therefore AB = AC$，$\angle BAC = 60^{\circ}$。$\because \angle BDC = 120^{\circ}$，$\therefore \angle ABD + \angle ACD = 180^{\circ}$。又$\because \angle ACE + \angle ACD = 180^{\circ}$，$\therefore \angle ABD = \angle ACE$。
在$\triangle ABD$和$\triangle ACE$中，$\begin{cases}AB = AC\\\angle ABD = \angle ACE\\BD = CE\end{cases}$，$\therefore \triangle ABD \cong \triangle ACE (SAS)$，$\therefore AD = AE$，$\angle BAD = \angle CAE$。$\because \angle BAC = 60^{\circ}$，即$\angle BAD + \angle DAC = 60^{\circ}$，$\therefore \angle DAC + \angle CAE = 60^{\circ}$，即$\angle DAE = 60^{\circ}$，$\therefore \triangle ADE$是等边三角形，$\therefore DA = DE = DC + CE = DC + DB$，即$DA = DC + DB$。

　　
(2) 如图②，在 $AC$ 上截取 $CM = CD$，连接 $DM$。$\because \triangle ABC$ 是等边三角形，$\therefore \angle ACB = 60^{\circ}$，$\therefore \triangle CDM$ 是等边三角形，$\therefore MD = CD = CM$，$\angle CMD = \angle CDM = 60^{\circ}$，$\therefore \angle AMD = 120^{\circ}$。
$\because \angle ADE = 60^{\circ}$，$\therefore \angle ADE = \angle MDC$，$\therefore \angle ADM = \angle EDC$。
$\because$ 直线 $a // AB$，$\therefore \angle ACE = \angle BAC = 60^{\circ}$，$\therefore \angle DCE = 120^{\circ} = \angle AMD$。
在$\triangle ADM$和$\triangle EDC$中，$\begin{cases}\angle ADM = \angle EDC\\MD = CD\\\angle AMD = \angle ECD\end{cases}$，$\therefore \triangle ADM \cong \triangle EDC (ASA)$，$\therefore AM = EC$，$\therefore CA = CM + AM = CD + CE$，即 $CD + CE = CA$。

2. (2025·东营期末)【问题初探】
如图①，已知$\triangle ABC$为等边三角形，点$D$为边$BC$上一动点(点$D$不与点$B$，点$C$重合)。以$AD$为边向右侧作等边三角形$ADE$，连接$CE$。
(1)求证：$\triangle ABD\cong\triangle ACE$；
(2)猜想并证明：①$AB$与$CE$的位置关系$AB// CE$；②线段$CE$，$AC$，$CD$之间的数量关系$CE=AC-CD$。
【类比探究】
(3)如图②，若点$D$在边$BC$的延长线上，其他条件不变，随着动点$D$的位置不同，猜想(2)的两个结论是否还成立？若成立，说明理由；若不成立，请写出正确的结论，并说明理由。
①平行成立，②数量关系不成立，应为$CE=AC+CD$。

答案：
(1) $\because \triangle ABC$ 和 $\triangle ADE$ 是等边三角形，$\therefore AB = AC$，$AD = AE$，$\angle BAC = \angle DAE = 60^{\circ}$，$\therefore \angle BAC - \angle DAC = \angle DAE - \angle DAC$，即$\angle BAD = \angle CAE$。
在$\triangle ABD$和$\triangle ACE$中，$\begin{cases}AB = AC\\\angle BAD = \angle CAE\\AD = AE\end{cases}$，$\therefore \triangle ABD \cong \triangle ACE (SAS)$。
(2) ① $AB // CE$；② $CE = AC - CD$。
证明如下：由
(1) 得$\triangle ABD \cong \triangle ACE (SAS)$，$\therefore \angle B = \angle ACE = 60^{\circ}$，$CE = BD$，$\therefore \angle BAC = \angle ACE = 60^{\circ}$，$\therefore AB // CE$。$\because CE = BD$，$AC = BC$，$\therefore CE = BD = BC - CD = AC - CD$。
(3) ① 平行成立，② 数量关系不成立，应为 $CE = AC + CD$。
理由：$\because \triangle ABC$ 和 $\triangle ADE$ 是等边三角形，$\therefore AB = AC$，$AD = AE$，$\angle BAC = \angle DAE = 60^{\circ}$。$\because \angle BAC = \angle DAE$，$\therefore \angle BAC + \angle DAC = \angle DAE + \angle DAC$，即$\angle BAD = \angle CAE$。在$\triangle ABD$和$\triangle ACE$中，$\begin{cases}AB = AC\\\angle BAD = \angle CAE\\AD = AE\end{cases}$，$\therefore \triangle ABD \cong \triangle ACE (SAS)$，$\therefore \angle B = \angle ACE = 60^{\circ}$，$CE = BD$，$\therefore \angle BAC = \angle ACE = 60^{\circ}$，$\therefore AB // CE$。$\because CE = BD$，$AC = BC$，$\therefore CE = BD = BC + CD = AC + CD$。

3. (2024·赣州期末)等边三角形的三条边都相等，三个内角都相等，并且都等于$60^{\circ}$，由此可得等边三角形的每一个外角都等于$120^{\circ}$，等边三角形与$120^{\circ}$的角是否还有某些特殊关系，请你完成证明过程或解答过程。
(1)如图①，$\triangle ABC$是等边三角形，点$D$，$E$分别在$CB$和$BC$的延长线上，且$\angle DAE = 120^{\circ}$，当$\angle D$的度数确定时，$\angle E$的度数也随之确定。
①若$\angle D = 26^{\circ}$，则$\angle E$的度数为______$34^{\circ}$；
②求证：$\angle D = \angle EAC$。
$\because \triangle ABC$ 是等边三角形，$\therefore \angle ACB = \angle E + \angle EAC = 60^{\circ}$。$\because \angle DAE = 120^{\circ}$，$\therefore \angle D + \angle E = 60^{\circ}$，$\therefore \angle D = \angle EAC$。
(2)如图②，$\triangle ABC$是等边三角形，点$P$是三角形内一点，且$\angle APB = 120^{\circ}$，延长$AP$交$BC$于点$D$，延长$BP$交$AC$于点$E$，求证：$AD = BE$。
$\because \triangle ABC$ 是等边三角形，$\therefore AB = BC$，$\angle ABD = \angle BCE = 60^{\circ}$，$\therefore \angle EBC + \angle BEC = 120^{\circ}$。$\because \angle APB = 120^{\circ}$，$\therefore \angle EBC + \angle ADB = 120^{\circ}$，$\therefore \angle BEC = \angle ADB$。在$\triangle ABD$和$\triangle BCE$中，$\begin{cases}\angle ADB = \angle BEC\\\angle ABD = \angle BCE\\AB = BC\end{cases}$，$\therefore \triangle ABD \cong \triangle BCE (AAS)$，$\therefore AD = BE$。

答案：
(1) ① $34^{\circ}$
② $\because \triangle ABC$ 是等边三角形，$\therefore \angle ACB = \angle E + \angle EAC = 60^{\circ}$。$\because \angle DAE = 120^{\circ}$，$\therefore \angle D + \angle E = 60^{\circ}$，$\therefore \angle D = \angle EAC$。
(2) $\because \triangle ABC$ 是等边三角形，$\therefore AB = BC$，$\angle ABD = \angle BCE = 60^{\circ}$，$\therefore \angle EBC + \angle BEC = 120^{\circ}$。$\because \angle APB = 120^{\circ}$，$\therefore \angle EBC + \angle ADB = 120^{\circ}$，$\therefore \angle BEC = \angle ADB$。在$\triangle ABD$和$\triangle BCE$中，$\begin{cases}\angle ADB = \angle BEC\\\angle ABD = \angle BCE\\AB = BC\end{cases}$，$\therefore \triangle ABD \cong \triangle BCE (AAS)$，$\therefore AD = BE$。