《2025年假期面对面南方出版社八年级数学沪科版》

9.如图，在$\triangle ABC$中，$AB=BC$，$\angle ABC=120^{\circ}$，$D$是$AC$边上的点，$DA=DB=3$，则$AC$的长为____9。

答案： 9

10.如图，已知点$P$是射线$ON$上一动点(即$P$可在射线$ON$上运动)，$\angle AON=30^{\circ}$，当$\angle A=$$30^{\circ}$或$75^{\circ}$或$120^{\circ}$时，$\triangle AOP$为等腰三角形。

答案： $30^{\circ}$或$75^{\circ}$或$120^{\circ}$

11.(四川中考)直线上依次有$A$，$B$，$C$，$D$四个点，$AD=7$，$AB=2$。若$AB$，$BC$，$CD$可构成以$BC$为腰的等腰三角形，则$BC$的长为2或2.5。

答案： 2或2.5

12.如图，在$\triangle ABC$中，$AB=AC$，$AD$平分$\angle BAC$。
(1)求$\angle ADB$的度数；$90^{\circ}$
(2)若$\angle BAC=100^{\circ}$，求$\angle B$，$\angle C$的度数；$40^{\circ}$，$40^{\circ}$
(3)若$BC=3cm$，求$BD$的长。$1.5cm$

答案： 解:
(1) $\because AB = AC$，$AD$平分$\angle BAC$，$\therefore AD\perp BC$，$\therefore \angle ADB = 90^{\circ}$。
(2) $\because \angle BAC = 100^{\circ}$，$\therefore \angle B+\angle C = 80^{\circ}$。$\because AB = AC$，$\therefore \angle C=\angle B = 40^{\circ}$。
(3) $\because AB = AC$，$AD$平分$\angle BAC$，$\therefore BD=\frac{1}{2}BC=\frac{1}{2}\times3 = 1.5(\text{cm})$。

13.如图，已知$\triangle ABC$和$\triangle DBE$均为等腰直角三角形。
(1)求证：$AD=CE$；
(2)猜想：$AD$和$CE$是否垂直？若垂直，请说明理由；若不垂直，则只要写出结论，不用写理由。

答案：

(1) 证明: $\because \triangle ABC$和$\triangle DBE$均为等腰直角三角形，$\therefore AB = BC$，$BD = BE$，$\angle ABC=\angle DBE = 90^{\circ}$，$\therefore \angle ABC-\angle DBC=\angle DBE-\angle DBC$，即$\angle ABD=\angle CBE$，$\therefore \triangle ABD\cong\triangle CBE$，$\therefore AD = CE$。
(2) 解: 垂直. 理由如下: 如图，延长$AD$分别交$BC$和$CE$于点$G$和点$F$。$\because \triangle ABD\cong\triangle CBE$，$\therefore \angle BAD=\angle BCE$。$\because \angle BAD+\angle ABC+\angle BGA=\angle BCE+\angle AFC+\angle CGF = 180^{\circ}$，又$\because \angle BGA=\angle CGF$，$\therefore \angle AFC=\angle ABC = 90^{\circ}$，$\therefore AD\perp CE$。

14.如图，点$O$是等边$\triangle ABC$内一点，$\angle AOB=110^{\circ}$，$\angle BOC=\alpha$。以$OC$为一边作等边三角形$OCD$，连接$AD$。
(1)当$\alpha=150^{\circ}$时，试判断$\triangle AOD$的形状，并说明理由；
(2)当$\alpha$为多少度时，$\triangle AOD$是等腰三角形？
(1)$\triangle AOD$是直角三角形. 理由如下: $\because \triangle OCD$是等边三角形，$\therefore OC = CD$，$\angle OCD=\angle ODC = 60^{\circ}$。又$\because \triangle ABC$是等边三角形，$\therefore BC = AC$，$\angle ACB = 60^{\circ}$。$\because \angle ACB=\angle OCD = 60^{\circ}$，$\therefore \angle BCO=\angle ACD$。在$\triangle BOC$和$\triangle ADC$中，$\because \begin{cases}OC = DC,\\\angle BCO=\angle ACD,\\BC = AC,\end{cases}$ $\therefore \triangle BOC\cong\triangle ADC(SAS)$。$\therefore \angle BOC=\angle ADC = 150^{\circ}$，$\therefore \angle ADO = 150^{\circ}-60^{\circ}=90^{\circ}$。$\therefore \triangle AOD$是直角三角形.
(2) 由(1)得，$\angle ADO=\alpha - 60^{\circ}$，结合题图易知$\angle AOD = 360^{\circ}-110^{\circ}-60^{\circ}-\alpha = 190^{\circ}-\alpha$，$\therefore \angle OAD = 180^{\circ}-(\alpha - 60^{\circ})-(190^{\circ}-\alpha)=50^{\circ}$。①要使$AO = AD$，需$\angle AOD=\angle ADO$，$\therefore 190^{\circ}-\alpha=\alpha - 60^{\circ}$，$\therefore \alpha = 125^{\circ}$。②要使$OA = OD$，需$\angle OAD=\angle ADO$，$\therefore \alpha - 60^{\circ}=50^{\circ}$，$\therefore \alpha = 110^{\circ}$。③要使$OD = AD$，需$\angle OAD=\angle AOD$，$\therefore 190^{\circ}-\alpha = 50^{\circ}$，$\therefore \alpha = 140^{\circ}$。综上，当$\alpha$为$110^{\circ}$或$125^{\circ}$或$140^{\circ}$时，$\triangle AOD$是等腰三角形.

答案： 解:
(1) $\triangle AOD$是直角三角形. 理由如下: $\because \triangle OCD$是等边三角形，$\therefore OC = CD$，$\angle OCD=\angle ODC = 60^{\circ}$。又$\because \triangle ABC$是等边三角形，$\therefore BC = AC$，$\angle ACB = 60^{\circ}$。$\because \angle ACB=\angle OCD = 60^{\circ}$，$\therefore \angle BCO=\angle ACD$。在$\triangle BOC$和$\triangle ADC$中，$\because \begin{cases}OC = DC,\\\angle BCO=\angle ACD,\\BC = AC,\end{cases}$ $\therefore \triangle BOC\cong\triangle ADC(SAS)$。$\therefore \angle BOC=\angle ADC = 150^{\circ}$，$\therefore \angle ADO = 150^{\circ}-60^{\circ}=90^{\circ}$。$\therefore \triangle AOD$是直角三角形.
(2) 由
(1)得，$\angle ADO=\alpha - 60^{\circ}$，结合题图易知$\angle AOD = 360^{\circ}-110^{\circ}-60^{\circ}-\alpha = 190^{\circ}-\alpha$，$\therefore \angle OAD = 180^{\circ}-(\alpha - 60^{\circ})-(190^{\circ}-\alpha)=50^{\circ}$。①要使$AO = AD$，需$\angle AOD=\angle ADO$，$\therefore 190^{\circ}-\alpha=\alpha - 60^{\circ}$，$\therefore \alpha = 125^{\circ}$。②要使$OA = OD$，需$\angle OAD=\angle ADO$，$\therefore \alpha - 60^{\circ}=50^{\circ}$，$\therefore \alpha = 110^{\circ}$。③要使$OD = AD$，需$\angle OAD=\angle AOD$，$\therefore 190^{\circ}-\alpha = 50^{\circ}$，$\therefore \alpha = 140^{\circ}$。综上，当$\alpha$为$110^{\circ}$或$125^{\circ}$或$140^{\circ}$时，$\triangle AOD$是等腰三角形.