《2025年学霸题中题八年级数学上册人教版》

9. 如图,在$\triangle ABC$中,$AB = AC = 6$,点D在边AC上,AD的垂直平分线交BC于点E.若$\angle AED = \angle B$,$CE = 3BE$,则CD等于 (B)

A. $\frac{3}{2}$
B. 2
C. $\frac{8}{3}$
D. 3

答案： B 解析：$ \because AB = AC = 6 $，$ \therefore \angle B = \angle C $。$ \because \angle AED = \angle B $，$ \angle BAE = 180^{\circ} - \angle B - \angle AEB $，$ \angle CED = 180^{\circ} - \angle AED - \angle AEB $，$ \therefore \angle BAE = \angle CED $。$ \because AD $ 的垂直平分线交 BC 于点 E，$ \therefore AE = DE $，由 AAS 得 $ \triangle ABE \cong \triangle ECD $，$ \therefore CE = AB = 6 $，$ BE = CD $。$ \because CE = 3BE $，$ \therefore CD = BE = 2 $。

10. (2024·北京校级月考)如图,四边形ABCD中,$AB = AD$,点B关于AC的对称点$B'$恰好落在CD上,若$\angle BAD = \alpha$,则$\angle ACB =$ ( )

A. $45^{\circ}$
B. $\alpha - 45^{\circ}$
C. $\frac{1}{2}\alpha$
D. $90^{\circ} - \frac{1}{2}\alpha$

答案：
D 解析：如图，连接 $ AB' $，$ BB' $，过点 A 作 $ AE \perp CD $ 于点 E，由轴对称的性质可知，AC 垂直平分 $ BB' $，$ AB = AB' $，$ \angle BAC = \angle B'AC $。$ \because AB = AD $，$ \therefore AD = AB' $。又 $ \because AE \perp CD $，$ \therefore \angle DAE = \angle B'AE $，$ \therefore \angle CAE = \frac{1}{2}\angle BAD = \frac{1}{2}\alpha $。又 $ \because \angle AEB' = 90^{\circ} $，$ \therefore \angle ACB' = 180^{\circ} - \angle AEB' - \angle CAE = 90^{\circ} - \frac{1}{2}\alpha $，$ \therefore \angle ACB = \angle ACB' = 90^{\circ} - \frac{1}{2}\alpha $。

11. (2024·内江中考)如图,在$\triangle ABC$中,$\angle DCE = 40^{\circ}$,$AE = AC$,$BC = BD$,则$\angle ACB$的度数为____$100^{\circ}$.

答案： $ 100^{\circ} $ 解析：$ \because \angle DCE = 40^{\circ} $，$ \therefore \angle CDE + \angle CED = 180^{\circ} - \angle DCE = 140^{\circ} $。$ \because AE = AC $，$ BC = BD $，$ \therefore \angle ACE = \angle AEC $，$ \angle BCD = \angle BDC $，$ \therefore \angle ACE + \angle BCD = \angle CDE + \angle CED = 140^{\circ} $，$ \therefore \angle ACB = \angle ACE + \angle BCE = \angle ACE + \angle BCD - \angle DCE = 140^{\circ} - 40^{\circ} = 100^{\circ} $。

12. (1)(2023·西宁中考)在$\triangle ABC$中,$AB = AC$,$\angle BAC = 100^{\circ}$,点D在BC边上,连接AD,若$\triangle ABD$为直角三角形,则$\angle ADB$的度数是____.
(2)在$\triangle ABC$中,$\angle B = 30^{\circ}$,点D在BC边上,点E在AC边上,$AD = BD$,$DE = CE$,若$\triangle ADE$为等腰三角形,则$\angle C$的度数为____.

答案：

(1) $ 50^{\circ} $或 $ 90^{\circ} $ 解析：$ \because AB = AC $，$ \angle BAC = 100^{\circ} $，$ \therefore \angle B = \angle C = \frac{180^{\circ} - \angle A}{2} = 40^{\circ} $。$ \because \triangle ABD $ 为直角三角形，$ \therefore $ 可分类讨论：①当 $ \angle BAD = 90^{\circ} $ 时，如图①，$ \therefore \angle ADB = 180^{\circ} - \angle BAD - \angle B = 50^{\circ} $；②当 $ \angle ADB = 90^{\circ} $ 时，如图②。综上可知，$ \angle ADB $ 的度数是 $ 50^{\circ} $或 $ 90^{\circ} $。


(2) $ 20^{\circ} $或 $ 40^{\circ} $ 解析：如图③，$ \because AD = BD $，$ \angle B = 30^{\circ} $，$ \therefore \angle ADC = 60^{\circ} $。$ \because DE = CE $，$ \therefore $ 可设 $ \angle C = \angle EDC = \alpha $，则 $ \angle ADE = 60^{\circ} - \alpha $，$ \angle AED = 2\alpha $，根据三角形内角和定理可得，$ \angle DAE = 120^{\circ} - \alpha $。分三种情况：①当 $ AD = AE $ 时，有 $ 60^{\circ} - \alpha = 2\alpha $，解得 $ \alpha = 20^{\circ} $；②当 $ DA = DE $ 时，有 $ 120^{\circ} - \alpha = 2\alpha $，解得 $ \alpha = 40^{\circ} $；③当 $ EA = ED $ 时，有 $ 120^{\circ} - \alpha = 60^{\circ} - \alpha $，方程无解。综上所述，$ \angle C $ 的度数为 $ 20^{\circ} $或 $ 40^{\circ} $。

13. 如图,$\triangle ABC$中,$AC = BC$,$\angle ACB = 90^{\circ}$,点D,E分别在AB,BC上,且$AD = BE$,$BD = AC$,过E作$EF \perp AB$于F.
(1)求证:$\angle FED = \angle CED$;
(2)若$BF = \frac{5}{2}$,则CE的长为____.

答案：

(1) 如图，连接 CD，$ \because AC = BC $，$ \angle ACB = 90^{\circ} $，$ \therefore \angle A = \angle B = 45^{\circ} $。在 $ \triangle ADC $ 和 $ \triangle BED $ 中，$ \left\{ \begin{array} { l } { A D = B E, } \\ { \angle A = \angle B, } \\ { A C = B D, } \end{array} \right. $$ \therefore \triangle ADC \cong \triangle BED (SAS) $，$ \therefore DC = DE $，$ \angle DCA = \angle EDB $，$ \therefore \angle ECD = \angle CED $，$ \angle DCA + \angle ECD = \angle EDB + \angle FED = 90^{\circ} $，$ \therefore \angle FED = \angle ECD $，$ \therefore \angle FED = \angle CED $。
(2) 5 解析：如图，过 D 作 $ DH \perp EC $ 于 H，$ \because DC = DE $，$ DH \perp EC $，$ \therefore EH = HC = \frac{1}{2}EC $，$ \angle EDH = \angle CDH $，$ \therefore DH // AC $，$ \therefore \angle CDH = \angle ACD $，$ \therefore \angle FDE = \angle EDH $。又 $ EF \perp AB $，$ EH \perp DH $，$ \therefore EF = EH = \frac{1}{2}CE $。$ \because \angle BFE = 90^{\circ} $，$ \angle B = 45^{\circ} $，$ \therefore EF = BF = \frac{5}{2} $，$ \therefore CE = 5 $。

14. (绍兴中考)如图,钢架中焊上等长的13根钢条来加固钢架,若$AP_1 = P_1P_2 = P_2P_3 = \cdots = P_{13}P_{14} = P_{14}A$,则$\angle A$的度数是____$12^{\circ}$.

答案： $ 12^{\circ} $ 解析：$ \because AP_1 = P_1P_2 = P_2P_3 = \cdots = P_{13}P_{14} = P_{14}A $，$ \therefore $ 设 $ \angle A = \angle AP_2P_1 = \angle AP_{13}P_{14} = x $，则 $ \angle P_1P_3P_2 = \angle P_2P_1P_3 = \angle P_{13}P_{14}P_{12} = \angle P_{13}P_{12}P_{14} = 2x $，$ \therefore \angle P_3P_2P_4 = \angle P_{12}P_{13}P_{11} = 3x $，$ \cdots $，$ \angle P_7P_6P_8 = \angle P_8P_9P_7 = 7x $，$ \therefore \angle AP_7P_8 = 7x $，$ \angle AP_8P_7 = 7x $。在 $ \triangle AP_7P_8 $ 中，$ \angle A + \angle AP_7P_8 + \angle AP_8P_7 = 180^{\circ} $，即 $ x + 7x + 7x = 180^{\circ} $，解得 $ x = 12^{\circ} $，即 $ \angle A = 12^{\circ} $。

15. (1)有一张三角形纸片ABC,$\angle A = 80^{\circ}$,点D是AC边上一点,沿BD方向剪开三角形纸片后,发现所得两纸片均为等腰三角形,则$\angle C$的度数可以是____;
(2)一个三角形可被剪成两个等腰三角形,原三角形的一个内角为$36^{\circ}$,求原三角形的最大内角的所有可能值.

答案：

(1) $ 40^{\circ} $或 $ 25^{\circ} $或 $ 10^{\circ} $ 解析：由题意知 $ \triangle ABD $ 与 $ \triangle DBC $ 均为等腰三角形，对于 $ \triangle ABD $ 可能有 ① $ AB = BD $，此时 $ \angle ADB = \angle A = 80^{\circ} $，$ \therefore \angle BDC = 180^{\circ} - \angle ADB = 100^{\circ} $，$ \angle C = \frac{1}{2} \times (180^{\circ} - 100^{\circ}) = 40^{\circ} $；② $ AB = AD $，此时 $ \angle ADB = \frac{1}{2} \times (180^{\circ} - \angle A) = 50^{\circ} $，$ \therefore \angle BDC = 180^{\circ} - \angle ADB = 130^{\circ} $，$ \angle C = \frac{1}{2} \times (180^{\circ} - 130^{\circ}) = 25^{\circ} $；③ $ AD = BD $，此时 $ \angle ADB = 180^{\circ} - 2 \times 80^{\circ} = 20^{\circ} $，$ \therefore \angle BDC = 180^{\circ} - \angle ADB = 160^{\circ} $，$ \angle C = \frac{1}{2} \times (180^{\circ} - 160^{\circ}) = 10^{\circ} $。综上所述，$ \angle C $ 的度数可以为 $ 40^{\circ} $或 $ 25^{\circ} $或 $ 10^{\circ} $。
(2) ①原三角形是锐角三角形，最大角是 $ 72^{\circ} $ 的情况：如图①，$ \angle ABC = \angle ACB = 72^{\circ} $，$ \angle A = 36^{\circ} $，$ AD = BD = BC $；
②原三角形是直角三角形，最大角是 $ 90^{\circ} $ 的情况：如图②，$ \angle ABC = 90^{\circ} $，$ \angle A = 36^{\circ} $，$ AD = BD = CD $；③原三角形是钝角三角形，最大角是 $ 108^{\circ} $ 的情况：如图③，$ \angle BAC = 108^{\circ} $，$ \angle B = 36^{\circ} $，$ BD = AB $，$ AD = DC $；④原三角形是钝角三角形，最大角是 $ 132^{\circ} $ 的情况：如图④，$ \angle C = 132^{\circ} $，$ \angle ABC = 36^{\circ} $，$ AD = BD $，$ CD = BC $；⑤原三角形是钝角三角形，最大角是 $ 126^{\circ} $ 的情况：如图⑤，$ \angle BAC = 126^{\circ} $，$ \angle C = 36^{\circ} $，$ BD = AD = AC $。
综上，原三角形的最大内角的所有可能值为 $ 72^{\circ} $，$ 90^{\circ} $，$ 108^{\circ} $，$ 132^{\circ} $或 $ 126^{\circ} $。