《2025年全国重点高中提前招生同步强化全真试卷八年级数学上册》

1. (2023·广州大学附中教育集团自主招生)如图 8 - 1,$\angle AOB = 30^{\circ}$. 按下列步骤作图:①在射线$OA$上取一点$C$,以点$O$为圆心,$OC$长为半径作圆弧$DE$,交射线$OB$于点$F$,连接$CF$;②以点$F$为圆心,$CF$长为半径作圆弧,交弧$DE$于点$G$;③连接$FG$,$CG$,作射线$OG$. 根据以上作图过程及所作图形,下列结论中错误的是(D)

A.$\angle AOG = 60^{\circ}$
B.$OF$垂直平分$CG$
C.$OG = CG$
D.$OC = 2FG$

答案： 1.D 提示：由作法得$OC = OF = OG$，$FG = FC$，则$OF$垂直平分$CG$，所以$B$选项的结论正确；$\because C$点与$G$点关于$OF$对称，$\therefore \angle FOG = \angle FOC = 30^{\circ}$，$\therefore \angle AOG = 60^{\circ}$，所以$A$选项的结论正确；$\therefore \triangle OCG$为等边三角形，$\therefore OG = CG$，所以$C$选项的结论正确；在$Rt\triangle OCM$中，$\because \angle COM = 30^{\circ}$，$\therefore OC = 2CM$，$\because CF \gt CM$，$FC = FG$，$\therefore OC \neq 2FG$，所以$D$选项的结论错误.故选$D$.

2. (浙江自主招生)在$\triangle ABC$中,$\angle A = 30^{\circ}$,$\angle B = 70^{\circ}$,直线将$\triangle ABC$分成两个三角形,如果其中一个三角形是等腰三角形,这样的直线有(C)

A.5 条
B.7 条
C.9 条
D.10 条

答案：
2.C 提示：如图8−1，最多画9条.
　　

3. (2023 秋·凤山县期末)如图 8 - 2,在$2× 2$的方格纸中有一个以格点为顶点的$\triangle ABC$,则与$\triangle ABC$成轴对称且以格点为顶点的三角形共有(C)

A.3 个
B.4 个
C.5 个
D.6 个

答案：
3.C 提示：如图8−2，与△ABC成轴对称且以格点为顶点三角形有△ABG,△CDF,△AEF,△DBH,△BCG共5个，故选C.
　　

4. (2024·西安碑林区月考)如图 8 - 3,$\triangle ABC$的三边都不相等,点$P$是三条内角平分线的交点,点$O$是三边垂直平分线的交点,当点$P$,$O$同时在$\triangle ABC$的内部时,$\angle BOC = 124^{\circ}$,则$\angle BPC$的度数为(A)

A.$121^{\circ}$
B.$122^{\circ}$
C.$123^{\circ}$
D.$124^{\circ}$

答案： 4.A 提示：连接$AO$，$\because$点$O$是这个三角形三边垂直平分线的交点，$\therefore OA = OB = OC$，$\therefore \angle OAB = \angle OBA$，$\angle OAC = \angle OCA$，$\angle OBC = \angle OCB$，$\therefore \angle AOB = 180^{\circ} - 2\angle OAB$，$\angle AOC = 180^{\circ} - 2\angle OAC$，$\angle BOC = 360^{\circ} - (\angle AOB + \angle AOC) = 360^{\circ} - (180^{\circ} - 2\angle OAB + 180^{\circ} - 2\angle OAC) = 2\angle OAB + 2\angle OAC = 2\angle BAC$，$\because \angle BOC = 124^{\circ}$，$\therefore \angle BAC = 62^{\circ}$，$\because BP$平分$\angle ABC$，$CP$平分$\angle ACB$，$\therefore \angle PBC = \frac{1}{2}\angle ABC$，$\angle PCB = \frac{1}{2}\angle ACB$，$\therefore \angle BPC = 180^{\circ} - (\angle PBC + \angle PCB) = 180^{\circ} - \frac{1}{2}(\angle ABC + \angle ACB) = 180^{\circ} - \frac{1}{2}(180^{\circ} - \angle BAC) = 90^{\circ} + \frac{1}{2}\angle BAC = 90^{\circ} + \frac{1}{2} × 62^{\circ} = 121^{\circ}$，故选$A$.

5. (青岛市南区自主招生)如图 8 - 4,在$\triangle ABC$中,$\angle BAC$的平分线与$BC$边的中垂线交于点$D$,$DE\perp AB$于$E$,连接$CD$. 若$CD = 2$,$DE = \sqrt{3}$,则$\angle ACD = $(C)

A.$150^{\circ}$
B.$135^{\circ}$
C.$120^{\circ}$
D.$110^{\circ}$

答案：
5.C 提示：如图8−3，连接$BD$，作$DF \perp AC$交$AC$的延长线于$F$，$\because$点$D$在$BC$边的中垂线上，$\therefore DB = DC = 2$，在$Rt\triangle DEB$中，$\frac{DE}{DB}=\frac{\sqrt{3}}{2}$，$\therefore \angle DBE = 60^{\circ}$，$\because AD$是$\angle BAC$的平分线，$DE \perp AB$，$DF \perp AC$，$\therefore DE = DF$，可证$Rt\triangle DEB \cong Rt\triangle DFC(HL)$，$\therefore \angle DCF = \angle DBE = 60^{\circ}$，$\therefore \angle ACD = 180^{\circ} - 60^{\circ} = 120^{\circ}$，故选$C$.
　　

6. (2023 秋·哈尔滨香坊区期末)如图 8 - 5,在$\triangle ABC$中,$\angle BAC = 120^{\circ}$,$AC ＞ AB$,点$P$是$\triangle ABC$外一点,$PM$垂直平分$AC$于$M$,交$BC$于点$E$,$PN$垂直平分$AB$于$N$,交$BC$于点$F$,连接$AE$,$AF$. 则以下各说法:①$\angle P = 60^{\circ}$;②$\angle EAF = 60^{\circ}$;③$PE = PF$;④点$P$到点$B$和点$C$的距离相等. 其中正确的个数是(C)

A.1
B.2
C.3
D.4

答案：
6.C 提示：如图8−4，连接$PC$，$PA$，$PB$，$\because PM$垂直平分$AC$，$PN$垂直平分$AB$，$\therefore CE = EA$，$FA = FB$，$PC = PA$，$PB = PA$，$\therefore \angle EAC = \angle C$，$\angle FAB = \angle B$，$PC = PB$，$\therefore \angle EAC + \angle FAB = \angle B + \angle C = 180^{\circ} - \angle BAC = 60^{\circ}$，$\therefore \angle EAF = \angle BAC - (\angle EAC + \angle FAB) = 120^{\circ} - 60^{\circ} = 60^{\circ}$，故②④符合题意；$\because AC \gt AB$，$\therefore \angle C \neq \angle B$，$\because \angle CEM + \angle C = \angle BFN + \angle B = 90^{\circ}$，$\therefore \angle CEM \neq \angle BFN$，$\because \angle PEF = \angle CEM$，$\angle PFE = \angle BFN$，$\therefore \angle PEF \neq \angle PFE$，$\therefore PE \neq PF$，故③不符合题意；$\because PM \perp AC$，$PN \perp AB$，$\therefore \angle PMA = \angle PNA = 90^{\circ}$，$\because \angle BAC = 120^{\circ}$，$\therefore \angle P = 360^{\circ} - 90^{\circ} - 90^{\circ} - 120^{\circ} = 60^{\circ}$，故①符合题意.所以正确的个数有3个，故选$C$.