《2025年拔尖特训八年级数学上册人教版》

10. 如图,在$\triangle ABC$中,$AD$平分$\angle BAC$.
(1) 尺规作图:作线段$AD$的垂直平分线$EF$,垂足为$O$,分别交$AB$,$AC$于点$E$,$F$.连接$DF$(保留作图痕迹,不要求写作法).
(2) 在(1)的条件下,猜想线段$AE$与$DF$的关系,并说明理由.

答案：
（1）如图，直线$EF$，线段$DF$即为所求。
（2）线段$AE$与$DF$平行且相等。
理由：$\because$直线$EF$为线段$AD$的垂直平分线，$\therefore OA = OD$，$AF = DF$。$\therefore \angle DAF=\angle ADF$。$\because AD$平分$\angle BAC$，$\therefore \angle BAD=\angle FAD$。$\therefore \angle BAD=\angle ADF$。$\therefore AE// DF$。$\because \angle AOE=\angle DOF$，$\therefore \triangle AOE\cong\triangle DOF$。$\therefore AE = DF$。$\therefore$线段$AE$与$DF$平行且相等。

11. 如图,$\angle BAC$的平分线与$BC$的垂直平分线$DG$相交于点$D$,连接$BD$,过点$D$作$DE\perp AB$于点$E$,$DF\perp AC$交$AC$的延长线于点$F$.
(1) 求证:$BE = CF$.
(2) 若$AB = 15$,$AC = 9$,求$BE$的长.

答案：
（1）如图，连接$CD$。$\because DG$是$BC$的垂直平分线，$\therefore BD = CD$。$\because DE\perp AB$，$DF\perp AC$，$AD$平分$\angle BAC$，$\therefore DE = DF$，$\angle BED=\angle CFD = 90^{\circ}$。在$Rt\triangle BDE$和$Rt\triangle CDF$中，$\left\{\begin{array}{l}BD = CD,\\ DE = DF,\end{array}\right.$ $\therefore Rt\triangle BDE\cong Rt\triangle CDF$。$\therefore BE = CF$。
（2）由（1），得$BE = CF$，设$BE = CF = x$。$\because DE\perp AB$，$DF\perp AC$，$\therefore \angle DEA=\angle DFA = 90^{\circ}$。在$Rt\triangle ADE$和$Rt\triangle ADF$中，$\left\{\begin{array}{l}AD = AD,\\ DE = DF,\end{array}\right.$ $\therefore Rt\triangle ADE\cong Rt\triangle ADF$。$\therefore AE = AF$。$\because AB = 15$，$AC = 9$，$\therefore 15 - x = 9 + x$，解得$x = 3$。$\therefore BE = 3$。

12. (2024·雅安雨城段考)已知$\triangle ABC$是三边都不相等的三角形,$P$是三个内角的平分线的交点,$O$是三边的垂直平分线的交点,当点$P$,$O$同时在不等边三角形$ABC$的内部时,那么$\angle BOC$和$\angle BPC$的数量关系是$\angle BOC =$______.

答案：
$4\angle BPC - 360^{\circ}$ 解析：$\because BP$平分$\angle ABC$，$CP$平分$\angle ACB$，$\therefore \angle PBC=\frac{1}{2}\angle ABC$，$\angle PCB=\frac{1}{2}\angle ACB$。$\therefore \angle BPC = 180^{\circ}-(\angle PBC+\angle PCB)=180^{\circ}-(\frac{1}{2}\angle ABC+\frac{1}{2}\angle ACB)=180^{\circ}-\frac{1}{2}(\angle ABC+\angle ACB)=180^{\circ}-\frac{1}{2}(180^{\circ}-\angle BAC)=90^{\circ}+\frac{1}{2}\angle BAC$，即$\angle BAC = 2\angle BPC - 180^{\circ}$。如图，连接$AO$。$\because O$是这个三角形三边的垂直平分线的交点，$\therefore OA = OB = OC$。$\therefore \angle OAB=\angle OBA$，$\angle OAC=\angle OCA$，$\angle OBC=\angle OCB$。$\therefore \angle AOB = 180^{\circ}-2\angle OAB$，$\angle AOC = 180^{\circ}-2\angle OAC$。$\therefore \angle BOC = 360^{\circ}-(\angle AOB+\angle AOC)=360^{\circ}-(180^{\circ}-2\angle OAB + 180^{\circ}-2\angle OAC)=2\angle OAB + 2\angle OAC = 2\angle BAC = 2(2\angle BPC - 180^{\circ})=4\angle BPC - 360^{\circ}$。

13. 如图,在$\triangle ABC$中,$DM$,$EN$分别垂直平分$AC$,$BC$,交$AB$于$M$,$N$两点,$DM$与$EN$的延长线相交于点$F$,连接$CM$,$CN$.
(1) 若$\angle ACB = 120^{\circ}$,则$\angle MCN$的度数为$60^{\circ}$.
(2) 若$\angle MCN = \alpha$,则$\angle MFN$的度数为$90^{\circ}-\frac{1}{2}\alpha$(用含$\alpha$的式子表示).
(3) 连接$FA$,$FB$,$FC$.若$\triangle CMN$的周长为$6\mathrm{c}\mathrm{m}$,$\triangle FAB$的周长为$14\mathrm{c}\mathrm{m}$,求$FC$的长.
$\because \triangle CMN$的周长为$6cm$，$\therefore MC + MN + NC = 6cm$。又$\because MC = MA$，$NC = NB$，$\therefore MA + MN + NB = 6cm$，即$AB = 6cm$。$\because \triangle FAB$的周长为$14cm$，$\therefore FA + FB + AB = 14cm$。$\therefore FA + FB = 8cm$。$\because DF$，$EF$分别垂直平分$AC$，$BC$，$\therefore FA = FC$，$FB = FC$。$\therefore 2FC = 8cm$。$\therefore FC = 4cm$。

答案： （1）$60^{\circ}$。
（2）$90^{\circ}-\frac{1}{2}\alpha$。解析：$\because DM$，$EN$分别垂直平分$AC$，$BC$，$\therefore MA = MC$，$NB = NC$。$\therefore \angle ACM=\angle CAM$，$\angle NCB=\angle NBC$。又$\because$在$\triangle ABC$中，$\angle CAM+\angle NBC+\angle ACM+\angle NCB+\angle MCN = 180^{\circ}$，$\therefore 2(\angle CAM+\angle NBC)+\angle MCN = 180^{\circ}$，即$2(\angle CAM+\angle NBC)+\alpha = 180^{\circ}$。$\therefore \angle CAM+\angle NBC=\frac{1}{2}(180^{\circ}-\alpha)=90^{\circ}-\frac{1}{2}\alpha$。$\because$在$\triangle FMN$中，$\angle MFN = 180^{\circ}-\angle FMN-\angle FNM$，易得$\angle FMN=\angle AMD = 90^{\circ}-\angle CAM$，$\angle FNM=\angle BNE = 90^{\circ}-\angle NBC$，$\therefore \angle MFN = 180^{\circ}-(90^{\circ}-\angle CAM)-(90^{\circ}-\angle NBC)=\angle CAM+\angle NBC = 90^{\circ}-\frac{1}{2}\alpha$。
（3）$\because \triangle CMN$的周长为$6cm$，$\therefore MC + MN + NC = 6cm$。又$\because MC = MA$，$NC = NB$，$\therefore MA + MN + NB = 6cm$，即$AB = 6cm$。$\because \triangle FAB$的周长为$14cm$，$\therefore FA + FB + AB = 14cm$。$\therefore FA + FB = 8cm$。$\because DF$，$EF$分别垂直平分$AC$，$BC$，$\therefore FA = FC$，$FB = FC$。$\therefore 2FC = 8cm$。$\therefore FC = 4cm$。