《2025年拔尖特训八年级数学上册人教版》

1. 如图，在$\triangle ABE$中，$BA=BE$，$F$为$AE$的中点. 若$\angle ABC=34^{\circ}$，$\angle C=50^{\circ}$，则$\angle ADB$的度数为（C）

A. $60^{\circ}$
B. $63^{\circ}$
C. $67^{\circ}$
D. $70^{\circ}$

答案： C

2. 如图，在$\triangle ABC$中，$AB=AC$，$AD\perp BC$于点$D$，$DE\perp AB$于点$E$，$BF\perp AC$于点$F$. 若$DE=3\mathrm{cm}$，则$BF$的长为（B）

A. $4.8\mathrm{cm}$
B. $6\mathrm{cm}$
C. $5\mathrm{cm}$
D. $6.4\mathrm{cm}$

答案： B

3. 如图，$AB=AC$，$D$是$\triangle ABC$内一点，$\angle D=110^{\circ}$，$\angle 1=\angle 2$，则$\angle A$的度数为$40^{\circ}$.

答案： $40^{\circ}$ 解析：$\because \angle D = 110^{\circ}$，$\therefore \angle 1+\angle BCD = 180^{\circ}-\angle D = 70^{\circ}$。$\because \angle 1=\angle 2$，$\therefore \angle 2+\angle BCD=\angle ACB = 70^{\circ}$。$\because AB = AC$，$\therefore \angle ABC=\angle ACB = 70^{\circ}$。$\therefore \angle A = 180^{\circ}-70^{\circ}-70^{\circ}=40^{\circ}$。

4. （2025·渭南大荔期中）如图，在$\triangle ABC$中，点$D$在$AC$上，且$BD=BC=AD$，$\angle DBC=20^{\circ}$，求$\angle A$，$\angle C$，$\angle ABC$的度数.

答案： $\because BD = BC$，$\angle DBC = 20^{\circ}$，$\therefore \angle C=\angle BDC=\frac{1}{2}(180^{\circ}-\angle DBC)=80^{\circ}$。$\because AD = BD$，$\therefore \angle A=\angle ABD$。$\because \angle BDC=\angle A+\angle ABD$，$\therefore \angle A=\angle ABD=\frac{1}{2}\angle BDC = 40^{\circ}$。$\therefore \angle ABC=\angle ABD+\angle DBC = 60^{\circ}$。

5. 借助如图所示的三等分角仪能三等分任意一个角. 这个三等分角仪由两根有槽的棒$OA$，$OB$组成，两根棒在点$O$处相连并可绕点$O$转动，点$C$固定，$OC=CD=DE$，点$D$，$E$可在槽中滑动. 若$\angle BDE=78^{\circ}$，则$\angle CDE$的度数为（C）

A. $52^{\circ}$
B. $66^{\circ}$
C. $76^{\circ}$
D. $78^{\circ}$

答案： C 解析：$\because OC = CD = DE$，$\therefore \angle O=\angle ODC$，$\angle DCE=\angle DEC$。$\therefore \angle DCE=\angle O+\angle ODC = 2\angle ODC$。$\because \angle O+\angle OED = 3\angle ODC=\angle BDE = 78^{\circ}$，$\therefore \angle ODC = 26^{\circ}$。$\because \angle CDE+\angle ODC = 180^{\circ}-\angle BDE = 102^{\circ}$，$\therefore \angle CDE = 102^{\circ}-\angle ODC = 76^{\circ}$。

6. 如图，在$\triangle ABC$中，$\angle ACB=90^{\circ}$，$AC=AN$，$BC=BM$，则$\angle MCN$的度数为（B）

A. $30^{\circ}$
B. $45^{\circ}$
C. $60^{\circ}$
D. $55^{\circ}$

答案： B 解析：设$\angle BMC = x$，$\angle ANC = y$。$\because BC = BM$，$\therefore \angle BCM=\angle BMC = x$。$\therefore \angle B = 180^{\circ}-2x$。$\because AC = AN$，$\therefore \angle ACN=\angle ANC = y$。$\therefore \angle A = 180^{\circ}-2y$。$\because \triangle ABC$为直角三角形，$\angle ACB = 90^{\circ}$，$\therefore \angle A+\angle B = 90^{\circ}$，即$180^{\circ}-2y + 180^{\circ}-2x = 90^{\circ}$。$\therefore x + y = 135^{\circ}$。$\therefore \angle BCM+\angle ACN = 135^{\circ}$。$\therefore \angle MCN=\angle BCM+\angle ACN-\angle ACB = 135^{\circ}-90^{\circ}=45^{\circ}$。

7. 如图，在$\triangle ABC$中，$AB=AC$，$\angle BAC=40^{\circ}$，$AD$是边$BC$上的高. 线段$AC$的垂直平分线交$AD$于点$E$，交$AC$于点$F$，连接$BE$，则$\angle EBD$的度数为______.

答案：
$50^{\circ}$ 解析：如图，连接$CE$。$\because AB = AC$，$AD$是边$BC$上的高，$\angle BAC = 40^{\circ}$，$\therefore BD = CD$，$\angle ABC=\frac{1}{2}(180^{\circ}-\angle BAC)=70^{\circ}$，$\angle BAE=\frac{1}{2}\angle BAC = 20^{\circ}$。$\therefore AD$为$BC$的垂直平分线。$\because$点$E$在$AD$上，$\therefore BE = CE$。又$\because$线段$AC$的垂直平分线交$AD$于点$E$，交$AC$于点$F$，$\therefore AE = CE$。$\therefore AE = BE$。$\therefore \angle ABE=\angle BAE = 20^{\circ}$。$\therefore \angle EBD=\angle ABC-\angle ABE = 50^{\circ}$。

8. 小丽从一张等腰三角形纸片$ABC(AB=AC)$中恰好剪出五个如图所示的小等腰三角形，其中$BC=BD$，$EC=EF=FG=DG=DA$，则$\angle B$的度数为______$67.5^{\circ}$.

答案： $67.5^{\circ}$ 解析：设$\angle ECF = x$。$\because EC = EF$，$\therefore \angle EFC=\angle ECF = x$。$\therefore \angle GEF = 2x$。$\because EF = GF$，$\therefore \angle FGE=\angle GEF = 2x$。$\therefore \angle DFG=\angle FGC+\angle GCF = 3x$。$\because DG = GF$，$\therefore \angle GDF=\angle DFG = 3x$。$\therefore \angle AGD=\angle GDC+\angle GCD = 4x$。$\because DG = DA$，$\therefore \angle A=\angle AGD = 4x$。$\therefore \angle BDC=\angle A+\angle DCA = 5x$。$\because BC = BD$，$\therefore \angle BDC=\angle BCD = 5x$。$\therefore \angle ACB=\angle BCD+\angle DCA = 6x$。$\because AB = AC$，$\therefore \angle B=\angle ACB = 6x$。$\because \angle A+\angle B+\angle ACB = 180^{\circ}$，$\therefore 4x + 6x + 6x = 180^{\circ}$，解得$x = 11.25^{\circ}$。$\therefore \angle B = 67.5^{\circ}$。

9. 如图①，$C$为线段$AB$上一点，分别以$AC$，$BC$为底边，在$AB$的同侧作等腰三角形$ACD$和等腰三角形$BCE$，且$\angle A=\angle EBC$. 在线段$EC$上取一点$F$，使$EF=AD$，连接$BF$，$DE$.
（1）判断$DE$与$FB$之间的数量关系，并说明理由.
（2）如图②，若$\angle A=\alpha$，延长$BF$交$DE$于点$G$，探究$\angle BGE$与$\angle GBC$之间的数量关系，并说明理由.

答案：
(1)$DE = FB$。理由：$\because \triangle ACD$，$\triangle BCE$分别是以$AC$，$BC$为底边的等腰三角形，$\therefore \angle A=\angle DCA$，$\angle ECB=\angle EBC$，$CE = BE$，$AD = CD$。$\because EF = AD$，$\therefore EF = CD$。$\because \angle A=\angle EBC$，$\therefore \angle A=\angle ECB=\angle DCA=\angle EBC$。$\therefore AD// CE$，$DC// BE$。$\therefore \angle ADC=\angle DCE$，$\angle DCE=\angle CEB$。在$\triangle DCE$和$\triangle FEB$中，$\left\{\begin{array}{l}CD = EF\\\angle DCE=\angle FEB\\CE = EB\end{array}\right.$，$\therefore \triangle DCE\cong\triangle FEB$。$\therefore DE = FB$。
(2)$\angle BGE = 2\angle GBC$。理由：由
(1)，可知$\angle A=\angle ECB=\angle CBE=\alpha$，$\triangle DCE\cong\triangle FEB$。$\therefore \angle DEC=\angle GBE$。$\because \angle GBE=\angle CBE-\angle GBC=\alpha-\angle GBC$，$\therefore \angle DEC=\alpha-\angle GBC$。$\because \angle BGE+\angle DEC+\angle EFG = 180^{\circ}$，$\angle ECB+\angle GBC+\angle CFB = 180^{\circ}$，$\angle EFG=\angle CFB$，$\therefore \angle BGE+\angle DEC=\angle ECB+\angle GBC$。$\therefore \angle BGE+\alpha-\angle GBC=\alpha+\angle GBC$。$\therefore \angle BGE = 2\angle GBC$。