《2025年假期面对面南方出版社八年级数学沪科版》

13.(泰州中考)如图所示的网格由边长相同的小正方形组成,点A,B,C,D,E,F,G在小正方形的顶点上,则$\triangle ABC$的重心是 (A)

A. 点D
B. 点E
C. 点F
D. 点G

答案： A

14.(成都中考)等腰三角形的一个底角为$50^{\circ}$,则它的顶角的度数为$80^{\circ}$.

答案： $80^{\circ}$

15.(齐齐哈尔中考)等腰三角形的两条边长分别为3和4,则这个等腰三角形的周长是10 或 11.

答案： 10 或 11

16.(安徽中考)命题“如果$a + b = 0$,那么a,b互为相反数”的逆命题为如果 $a$，$b$ 互为相反数，那么 $a + b = 0$.

答案： 如果 $a$，$b$ 互为相反数，那么 $a + b = 0$

17.(成都中考)在$\triangle ABC$中,$\angle A:\angle B:\angle C = 2:3:4$,则$\angle A$的度数为______$40^{\circ}$.

答案： $40^{\circ}$

18.(陕西中考)如图,在$\triangle ABC$中,BD和CE是$\triangle ABC$的两条角平分线.若$\angle A = 52^{\circ}$,则$\angle 1 + \angle 2$的度数为$64^{\circ}$.

答案： $64^{\circ}$

19.(泰州中考)如图,将分别含有$30^{\circ}$,$45^{\circ}$角的一副三角板重叠,使直角顶点重合,若两直角重叠形成的角为$65^{\circ}$,则图中角$\alpha$的度数为______$140^{\circ}$.

答案： $140^{\circ}$

20.(6分)(淄博中考)已知:如图,$\triangle ABC$是任意一个三角形,求证:$\angle A + \angle B + \angle C = 180^{\circ}$.

答案：
证明：如图，过点 $A$ 作 $EF // BC$。$\because EF // BC$， $\therefore \angle 1 = \angle B$，$\angle 2 = \angle C$。$\because \angle 1 + \angle 2 + \angle BAC = 180^{\circ}$，$\therefore \angle BAC + \angle B + \angle C = 180^{\circ}$，即 $\angle A + \angle B + \angle C = 180^{\circ}$。

21.(8分)(宜昌中考)如图,在$Rt\triangle ABC$中,$\angle ACB = 90^{\circ}$,$\angle A = 40^{\circ}$,$\triangle ABC$的外角$\angle CBD$的平分线BE交AC的延长线于点E.
(1)求$\angle CBE$的度数;65°
(2)过点D作$DF// BE$,交AC的延长线于点F,求$\angle F$的度数.25°

答案： 解：
(1) $\because \angle ACB = 90^{\circ}$，$\angle A = 40^{\circ}$，$\therefore \angle ABC = 90^{\circ} - \angle A = 50^{\circ}$，$\therefore \angle CBD = 130^{\circ}$。$\because BE$ 是 $\angle CBD$ 的平分线，$\therefore \angle CBE = \frac{1}{2} \angle CBD = 65^{\circ}$。
(2) $\because \angle ACB = 90^{\circ}$，$\angle CBE = 65^{\circ}$，$\therefore \angle CEB = 90^{\circ} - 65^{\circ} = 25^{\circ}$。$\because DF // BE$，$\therefore \angle F = \angle CEB = 25^{\circ}$。

22.(10分)(绍兴中考)问题:如图,在$\triangle ABD$中,$BA = BD$.在BD的延长线上取点E,C,作$\triangle AEC$,使$EA = EC$.若$\angle BAE = 90^{\circ}$,$\angle B = 45^{\circ}$,求$\angle DAC$的度数.
答案:$\angle DAC = $$45^{\circ}$.
思考:(1)如果把以上“问题”中的条件“$\angle B = 45^{\circ}$”去掉,其余条件不变,那么$\angle DAC$的度数会改变吗? 说明理由;
(2)如果把以上“问题”中的条件“$\angle B = 45^{\circ}$”去掉,再将“$\angle BAE = 90^{\circ}$”改为“$\angle BAE = n^{\circ}$”,其余条件不变,求$\angle DAC$的度数.

答案： 解：
(1) $\angle DAC$ 的度数不会改变。$\because EA = EC$，$\therefore \angle AED = 2 \angle C$。$\because \angle BAE = 90^{\circ}$，$\therefore \angle BAD = \frac{1}{2}[180^{\circ} - (90^{\circ} - 2 \angle C)] = 45^{\circ} + \angle C$，$\therefore \angle DAE = 90^{\circ} - \angle BAD = 90^{\circ} - (45^{\circ} + \angle C) = 45^{\circ} - \angle C$，$\therefore \angle DAC = \angle DAE + \angle CAE = 45^{\circ}$。
(2) 设 $\angle ABC = m^{\circ}$，则 $\angle BAD = \frac{1}{2}(180^{\circ} - m^{\circ}) = 90^{\circ} - \frac{1}{2}m^{\circ}$，$\angle AEB = 180^{\circ} - n^{\circ} - m^{\circ}$，$\therefore \angle DAE = n^{\circ} - \angle BAD = n^{\circ} - 90^{\circ} + \frac{1}{2}m^{\circ}$。$\because EA = EC$，$\therefore \angle CAE = \frac{1}{2} \angle AEB = 90^{\circ} - \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ}$，$\therefore \angle DAC = \angle DAE + \angle CAE = n^{\circ} - 90^{\circ} + \frac{1}{2}m^{\circ} + 90^{\circ} - \frac{1}{2}n^{\circ} - \frac{1}{2}m^{\circ} = \frac{1}{2}n^{\circ}$。