《2025年学霸题中题八年级数学上册人教版》

5. 如图①,$\triangle ABD$和$\triangle AEC$均为等边三角形,连接BE,CD.
(1)请判断:线段BE与CD的大小关系是____BE=CD.
(2)观察图②,当$\triangle ABD$和$\triangle AEC$分别绕点A旋转时,BE,CD之间的大小关系是否会改变?线段BE与CD的大小关系不会改变
(3)观察图③和图④,若四边形ABCD和四边形DEFG都是正方形,猜想类似的结论是____AE=CG,并证明你的猜想.

答案：
(1) $ BE = CD $ 解析: $ \because \triangle ABD, \triangle ACE $ 均为等边三角形, $ \therefore AB = AD, AE = AC, \angle BAD = \angle EAC = 60 ^ { \circ }, \therefore \angle BAD + \angle DAE = \angle EAC + \angle DAE, \therefore \angle BAE = \angle DAC $. 在 $ \triangle ABE $ 与 $ \triangle ADC $ 中, $ \left\{ \begin{array} { l } { AB = AD }, \\ { \angle BAE = \angle DAC }, \\ { AE = AC }, \end{array} \right. $ $ \therefore \triangle ABE \cong \triangle ADC $, $ \therefore BE = CD $.
(2) 线段 $ BE $ 与 $ CD $ 的大小关系不会改变. $ \because \triangle ABD, \triangle ACE $ 均为等边三角形, $ \therefore AB = AD, AE = AC, \angle BAD = \angle EAC = 60 ^ { \circ } $, $ \therefore \angle BAD + \angle BAC = \angle EAC + \angle BAC, \therefore \angle DAC = \angle BAE $. 在 $ \triangle ABE $ 与 $ \triangle ADC $ 中, $ \left\{ \begin{array} { l } { AB = AD }, \\ { \angle BAE = \angle DAC }, \\ { AE = AC }, \end{array} \right. $ $ \therefore \triangle ABE \cong \triangle ADC, \therefore BE = CD $.
(3) $ AE = CG $ 证明如下: 正方形 $ ABCD $ 与正方形 $ DEFG $ 中, $ \because AD = CD, DE = DG, \angle ADE = \angle CDG $, $ \therefore \triangle ADE \cong \triangle CDG, \therefore AE = CG $.

6. (2024·重庆期中)在$\triangle ABC$中,$AB=AC$,点D是直线BC上一点(不与点B,C重合),把线段AD绕着点A逆时针旋转至AE(即$AD=AE$),使得$∠DAE=∠BAC$,连接DB,CE.
(1)如图①,点D在线段BC上,若$∠BAC=90^{\circ }$,则$∠BCE=$____$^{\circ }$.
(2)如图②,当点D在线段BC上时,若$∠BAC=60^{\circ }$,则$∠BCE=$____$^{\circ }$.
(3)如图③,设$∠BAC=α,∠BCE=β$,当点D在线段BC上移动时,α,β的数量关系是什么?请说明理由.
(4)设$∠BAC=α,∠BCE=β$,当点D在直线BC上移动时,请直接写出α,β的数量关系,不用证明.

答案：

(1) $ 90 $
(2) $ 120 $
(3) $ \alpha + \beta = 180 ^ { \circ } $, 理由如下: $ \because AB = AC, AD = AE, \angle DAE = \angle BAC $, $ \therefore \angle BAD = \angle CAE $. 在 $ \triangle BAD $ 和 $ \triangle CAE $ 中, $ \left\{ \begin{array} { l } { AB = AC }, \\ { \angle BAD = \angle CAE }, \\ { AD = AE }, \end{array} \right. $ $ \therefore \triangle BAD \cong \triangle CAE ( SAS ), \therefore \angle ACE = \angle B, \therefore \angle ACE + \angle ACB = \angle B + \angle ACB $. $ \because \angle BCE = \angle ACB + \angle ACE = \beta, \therefore \angle B + \angle ACB = \beta $. $ \because \angle BAC = \alpha, \angle BAC + \angle B + \angle ACB = 180 ^ { \circ }, \therefore \alpha + \beta = 180 ^ { \circ } $.
(4) $ \alpha + \beta = 180 ^ { \circ } $ 或 $ \alpha = \beta $. 解析: 如图①, 当点 $ D $ 在 $ BC $ 的延长线上时, $ \alpha + \beta = 180 ^ { \circ } $, 证明方法同
(3);

如图②, 当点 $ D $ 在 $ CB $ 的延长线上时, $ \alpha = \beta $, 理由如下: $ \because \angle DAE = \angle BAC, \therefore \angle DAB + \angle BAE = \angle EAC + \angle BAE, \therefore \angle DAB = \angle EAC $. 在 $ \triangle BAD $ 和 $ \triangle CAE $ 中, $ \left\{ \begin{array} { l } { AB = AC }, \\ { \angle BAD = \angle CAE }, \\ { AD = AE }, \end{array} \right. $ $ \therefore \triangle BAD \cong \triangle CAE ( SAS ) $, $ \therefore \angle ABD = \angle ACE $. $ \because \angle ABD = \angle BAC + \angle ACB, \angle ACE = \angle BCE + \angle ACB, \therefore \angle BAC = \angle BCE $. $ \because \angle BAC = \alpha, \angle BCE = \beta, \therefore \alpha = \beta $. 综上, $ \alpha + \beta = 180 ^ { \circ } $ 或 $ \alpha = \beta $.