《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》

18. (2023 秋·肇源县期中)如图 8 - 16,在$\triangle ABC$中,$DM$,$EN$分别垂直平分$AC$和$BC$,交$AB$于$M$,$N$两点,$DM$与$EN$相交于点$F$.
(1)若$\angle ACB = 120^{\circ}$,则$\angle MCN$的度数为60°;
(2)若$\angle MCN = \alpha$,则$\angle MFN$的度数为90°−α;(用含$\alpha$的代数式表示)
(3)连接$FA$,$FB$,$FC$,$\triangle CMN$的周长为$6cm$,$\triangle FAB$的周长为$14cm$,求$FC$的长.

答案： 18.解：
(1)在$\triangle ABC$中，$\because \angle ACB = 120^{\circ}$，$\therefore \angle A + \angle B = 180^{\circ} - 120^{\circ} = 60^{\circ}$. $\because DM$，$EN$分别垂直平分$AC$和$BC$，$\therefore MA = MC$，$NB = NC$，$\therefore \angle ACM = \angle A$，$\angle NCB = \angle B$，$\therefore \angle MCN = \angle ACB - (\angle ACM + \angle NCB)=120^{\circ} - 60^{\circ} = 60^{\circ}$.故答案为$60^{\circ}$；
(2)$\because DM$，$EN$分别垂直平分$AC$和$BC$，$\therefore MA = MC$，$NB = NC$，$\therefore \angle ACM = \angle A$，$\angle NCB = \angle B$. 又$\because$在$\triangle ABC$中，$\angle A + \angle B + \angle ACM + \angle NCB + \angle MCN = 180^{\circ}$，$\therefore 2(\angle A + \angle B) + \angle MCN = 180^{\circ}$，即$2(\angle A + \angle B) + \alpha = 180^{\circ}$，$\therefore \angle A + \angle B = \frac{1}{2}(180^{\circ} - \alpha)=90^{\circ}-\frac{\alpha}{2}$. 在$\triangle FMN$中，$\angle MFN = 180^{\circ} - \angle FMN - \angle FNM$，$\because \angle FMN = \angle AMD = 90^{\circ} - \angle A$，$\angle FNM = \angle BNE = 90^{\circ} - \angle B$，$\therefore \angle MFN = 180^{\circ}-(90^{\circ} - \angle A)-(90^{\circ} - \angle B)=\angle A + \angle B = 90^{\circ}-\alpha$.故答案为$90^{\circ}-\alpha$；
(3)$\because \triangle CMN$的周长$= 6$，$\because MC = MA$，$NC = NB$，$\therefore MA + MN + NB = 6cm$，即$AB = 6cm$. $\because \triangle FAB$的周长$= 14cm$，$\therefore FA + FB + AB = 14cm$，$\because DF$，$EF$分别垂直平分$AC$和$BC$，$\therefore FA = FC$，$FB = FC$，$\therefore 2FC = 8cm$，$\therefore FC = 4cm$.

19. (“希望杯”全国数学竞赛)如图 8 - 17,$\triangle ABC$是等边三角形,延长$BC$到$D$,延长$BA$到$E$,且$AE = BD$,若$CE = 12$,求$DE$的长.

答案： 19.解：延长$BD$到$F$，使$DF = AB$，连接$EF$. $\because \triangle ABC$是等边三角形，$\therefore \angle B = 60^{\circ}$，$AB = BC = DF$，$\because AE = BD$，$\therefore \triangle BEF$是等边三角形，$\therefore \angle F = 60^{\circ}$，$EB = EF$，在$\triangle EBC$和$\triangle EFD$中，$BC = DF$，$\angle B = \angle F = 60^{\circ}$，$BE = EF$，$\therefore \triangle EBC \cong \triangle EFD(SAS)$，$\therefore ED = EC = 12$.

20. (江苏省竞赛)如图 8 - 18,河岸$l$同侧的两个居民小区$A$,$B$到河岸的距离分别为$a\ m$,$b\ m$(即图中表示$AA' = a\ m$,$BB' = b\ m$),$A'B' = c\ m$. 现欲在河岸边建一个长度为$s\ m$的绿化带$CD$(宽度不计),使$C$到小区$A$的距离与$D$到小区$B$的距离之和最小.
(1)在图中画出绿化带的位置,并写出画图过程;
(2)求$AC + BD$的最小值.

答案：
20.解：
(1)如图8−10，作线段$AP // l$，使$AP = s$，且点$P$在点$A$右侧，取点$P$关于$l$的对称点$P'$，连$BP'$交$l$于点$C$，过点$C$作$CD // AP$，使$DC = s$，则$CD$即为所求绿化带的位置.设绿化带建于另一位置$C'D'$.连接$BD'$，$PD'$，$AC'$，$PD$，则由对称性及$AP // CD$，$AP // C'D'$，知$AC = PD$，$AC' = PD'$.但$P'D'+D'B \geq P'B = P'D + BD$，即$PD'+D'B \geq PD + BD$，所以，这样画出的$AC + BD$最小.
(2)$AC + BD$的最小值即为线段$P'B$的长度.延长$BB'$，作$P'H \perp BB'$，$BH = BB'+B'H = b + a$，$P'H = c - s$，$P'B = \sqrt{P'H^{2}+BH^{2}}=\sqrt{(c - s)^{2}+(b + a)^{2}}$，则$AC + BD$的最小值为$\sqrt{(c - s)^{2}+(b + a)^{2}}$.