《2025年轻松作业本八年级数学上册人教版》

20. (2024·泰州姜堰区二模)如图,在等边三角形ABC中,D是边AC上的一点,点E在边BC的延长线上.
(1) 若______,______,求证:CD= CE(请从信息“① BD= ED,② D为AC的中点$,③ BD= \sqrt{3}CE”$中选择两个分别填入两条横线中,将题目补充完整,并完成证明);
(2) 过点D作DM⊥BC于点M,在(1)的条件下,当MC= 1时,求BE的长.

答案：
解：
(1) 若 $BD = ED$，$D$ 为 $AC$ 的中点，求证：$CD = CE$. 证明：$\because \triangle ABC$ 是等边三角形，$\therefore \angle ACB = \angle ABC = 60^{\circ}$，$\because D$ 为 $AC$ 的中点，$\therefore \angle DBC = \frac{1}{2}\angle ABC = 30^{\circ}$，$\because BD = DE$，$\therefore \angle DBE = \angle DEB = 30^{\circ}$，$\because \angle ACB = \angle CED + \angle CDE$，$\therefore \angle CED = \angle CDE = 30^{\circ}$，$\therefore CD = CE$；
(2) $\because DM \perp BC$，$\therefore \angle DMC = 90^{\circ}$，$\because \angle DCM = 60^{\circ}$，$\therefore CD = 2CM = 2$，$\because CE = CD = 2$，$\therefore EM = 3$，$\because BD = DE$，$\therefore BE = 2ME = 6$.

21. (1) 问题发现
如图①,△ACB和△DCE均为等边三角形,点A、D、E在同一直线上,连接BE.
① ∠AEB的度数为60°;
② 线段AD、BE之间的数量关系是AD=BE.
(2) 拓展探究
如图②,△ACB和△DCE均为等腰直角三角形,∠ACB= ∠DCE= 90°,点A、D、E在同一直线上,CM为△DCE中DE边上的高,连接BE.
① 求∠AEB的度数;
② 判断线段CM、AE、BE之间的数量关系,并说明理由.

答案： 解：
(1) ① $60^{\circ}$ ② $AD = BE$
(2) ① $\because \triangle ACB$ 和 $\triangle DCE$ 均为等腰直角三角形，$\angle ACB = \angle DCE = 90^{\circ}$，$\therefore AC = BC$，$CD = CE$，$\therefore \angle ACB - \angle DCB = \angle DCE - \angle DCB$，即 $\angle ACD = \angle BCE$，$\therefore \triangle ACD \cong \triangle BCE$，$\therefore AD = BE$，$\angle BEC = \angle ADC = 135^{\circ}$，$\therefore \angle AEB = \angle BEC - \angle CED = 135^{\circ} - 45^{\circ} = 90^{\circ}$；② $AE = 2CM + BE$. 理由：$\because$ 在等腰直角三角形 $DCE$ 中，$CM$ 为斜边 $DE$ 上的高，$\therefore CM = DM = ME$，$\therefore DE = 2CM$，$\therefore AE = DE + AD = 2CM + BE$.