《2025年天府前沿九年级数学上册北师大版》

11. (2022·长沙)如图,在平面直角坐标系中,已知点$A(2,0)$,$B(4,3)$,$D(5,0)$,$\triangle ABC与\triangle DEF$位似,原点$O$是位似中心,则点$E$的坐标是____。

答案： $(10,7.5)$

12. (2025·编写)如图,在平面直角坐标系中,点$A$,$B$,$C$,$D的坐标分别为(-2,5)$,$(0,5)$,$(0,-1)$,$(4,-1)$。若线段$AB和CD$是位似图形,位似中心在$y$轴上,则位似中心的坐标为____。

答案： $(0,3)$

13. (2025·编写)如图,在平面直角坐标系中,$\triangle ABC的顶点A$在第二象限,点$B的坐标为(-2,0)$,点$C的坐标为(-1,0)$,以点$C$为位似中心,在$x轴的下方作\triangle ABC的位似图形\triangle A'B'C$。若点$A的对应点A'的坐标为(2,-3)$,点$B的对应点B'的坐标为(1,0)$,则点$A$的坐标为____。

答案： $(-\frac{5}{2},\frac{3}{2})$

14. (2025·编写)如图,在矩形$ABCD$中,$AB = 9$,$BC = 6$,点$E$,$F分别在BC$,$CD$上。若$DF = 2$,$\angle EAF = 45^{\circ}$,求$BE$的长度。

答案：
如图，在$AB$，$CD$上截取$AM = DN = 6$，连接$MN$，交$AE$于点$H$，连接$FH$，延长$CD$至点$G$，使$DG = MH$，连接$AG$。

$\because AM = DN = 6$，$AM // DN$，
$\therefore$四边形$AMND$是平行四边形。
$\because \angle ADC = 90^{\circ}$，$AD = DN = 6$，
$\therefore$四边形$AMND$是正方形，
$\therefore MN = AD = 6$，$\angle AMN = \angle ADF = 90^{\circ}$。
$\because AM = AD = 6$，$\angle AMH = \angle ADG = 90^{\circ}$，$MH = DG$，
$\therefore \triangle ADG \cong \triangle AMH(SAS)$，
$\therefore \angle DAG = \angle MAH$，$AG = AH$。
$\because \angle EAF = 45^{\circ}$，
$\therefore \angle MAH + \angle DAF = 45^{\circ} = \angle DAF + \angle GAD$，
$\therefore \angle GAF = 45^{\circ} = \angle EAF$，且$AG = AH$，$AF = AF$，
$\therefore \triangle AFH \cong \triangle AFG(SAS)$，
$\therefore FH = FG$。
$\because DF = 2$，
$\therefore FN = 4$，$GF = 2 + DG = 2 + MH$。
$\because FH^{2} = HN^{2} + FN^{2}$，
$\therefore (2 + MH)^{2} = (6 - MH)^{2} + 16$，
$\therefore MH = 3$。
$\because \angle BAE = \angle MAH$，$\angle AMN = \angle ABC = 90^{\circ}$，
$\therefore \triangle AMH \sim \triangle ABE$，
$\therefore \frac{AM}{AB} = \frac{MH}{BE}$，
$\therefore \frac{6}{9} = \frac{3}{BE}$，
$\therefore BE = \frac{9}{2}$。

15. (2025·彭州)【观察与猜想】
(1) 如图$1$,$O是矩形ABCD$内一点,过点$O的直线EF \perp MN$,分别交矩形的边于点$E$,$F$,$M$,$N$。若$AD = 10$,$CD = 7$,$EF = 8$,则$MN = $____。
【类比探究】
(2) 如图$2$,在平行四边形$ABCD$中,点$E$,$M分别在边AB$,$BC$上,连接$DM与CE交于点O$。若$\angle DOE = \angle B$,求证：$CE \cdot AB = DM \cdot BC$。
【拓展延伸】
(3) 如图$3$,在四边形$ABCD$中,$BC = 17\frac{1}{5}$,$AB = 4$,$\angle B = \angle ADC = 120^{\circ}$,

$\frac{CD}{AD} = \frac{4}{5}$,点$M在边BC$上,连接$AC与DM交于点O$,当$\angle AOD = \angle B$时,求$\frac{AC}{DM}$的值。

答案：

(1)【解】如图，过点$M$作$MQ \perp CD$于点$Q$，过点$E$作$EP \perp BC$于点$P$。

则四边形$ABPE$，四边形$BCQM$是矩形，
$\therefore PE = AB = CD = 7$，$MQ = BC = AD = 10$。
易知，$\angle PEF = \angle QMN$。
$\because \angle EPF = \angle MQN = 90^{\circ}$，$\therefore \triangle EPF \sim \triangle MQN$，
$\therefore \frac{EF}{MN} = \frac{PE}{MQ}$，
$\therefore \frac{8}{MN} = \frac{7}{10}$，$\therefore MN = \frac{80}{7}$。
故答案为$\frac{80}{7}$。
(2)【证明】$\because \angle DOE = \angle COM$，$\angle DOE = \angle B$，
$\therefore \angle B = \angle COM$。
又$\because \angle MCO = \angle ECB$，$\therefore \triangle COM \sim \triangle CBE$，
$\therefore \frac{CO}{CM} = \frac{CB}{CE}$，$\angle DMC = \angle BEC$。
$\because$四边形$ABCD$是平行四边形，
$\therefore AB // CD$，$AB = CD$，$\therefore \angle DCO = \angle BEC$，
$\therefore \angle DMC = \angle DCO$。
又$\because \angle ODC = \angle CDM$，
$\therefore \triangle DOC \sim \triangle DCM$，$\therefore \frac{CO}{CM} = \frac{DC}{DM}$，$\therefore \frac{CB}{CE} = \frac{DC}{DM}$，
$\therefore CB \cdot DM = CE \cdot DC$，
$\therefore CB \cdot DM = CE \cdot AB$。
(3)【解】如图，过点$D$作$DE // BC$交$BA$的延长线于点$E$，过点$C$作$CF // AB$交$ED$的延长线于点$F$
$\therefore$四边形$BEFC$是平行四边形，
$\therefore BE = CF$，$BC = EF$，$\angle B = \angle ADC = \angle CFE = 120^{\circ}$，
$\therefore \angle E = 60^{\circ}$。
在$EF$上截取$EG = EA$，连接$AG$，则$\triangle EAG$为等边三角形，

$\therefore \angle AGD = 120^{\circ} = \angle CFE = \angle ADC$，
$\therefore \angle GAD + \angle GDA = \angle GDA + \angle FDC = 60^{\circ}$，
$\therefore \angle GAD = \angle FDC$，$\therefore \triangle DFC \sim \triangle AGD$，
$\therefore \frac{FC}{DG} = \frac{CD}{AD} = \frac{DF}{AG} = \frac{4}{5}$。
设$FC = 4x$，则$GD = 5x$，
$\therefore EB = FC = 4x$，
$\therefore AE = EG = AG = 4x - 4$，
$\therefore DF = \frac{4}{5}AG = \frac{16}{5}x - \frac{16}{5}$，
$\therefore EF = EG + GD + DF = 4x - 4 + 5x + \frac{16}{5}x - \frac{16}{5} = 17\frac{1}{5}$，
解得$x = 2$，
$\therefore EB = FC = 8$。
过点$F$作$FN // DM$交$BC$于点$N$。
$\therefore$四边形$DFNM$为平行四边形，$\therefore FN = DM$，
由
(2)可得，$BE \cdot CA = BC \cdot FN$，
$\therefore BE \cdot CA = BC \cdot DM$，$\therefore \frac{AC}{DM} = \frac{BC}{BE} = \frac{17\frac{1}{5}}{8} = \frac{43}{20}$。