8.「2026 江苏连云港东海期末，★☆☆」如图，在$△ ABC$中，$∠ ACB = 90°$，$△ ABC$的角平分线$AD$，$BE$相交于点$O$，过点$O$作$OF ⊥ AD$交$BC$的延长线于点$F$，交$AC$于点$G$，下列结论：①$∠ BOD = 45°$；②$△ BOA ≌ △ BOF$；③$BD + AG = AB$.其中正确的结论是.(填序号)

答案 ①②③
解析 $\because △ ABC$的角平分线AD,BE相交于点O,
$\therefore ∠ ABO=∠ CBO=\dfrac{1}{2}∠ ABC$,$∠ BAO=∠ OAC=\dfrac{1}{2}∠ BAC$,$\because$在$△ ABC$中,$∠ ACB=90°$,$\therefore ∠ ABC+∠ BAC=90°$,$\therefore ∠ ABO+∠ BAO=\dfrac{1}{2}(∠ ABC+∠ BAC)=45°$,$\because ∠ BOD$是$△ AOB$的外角,$\therefore ∠ BOD=∠ ABO+∠ BAO=45°$,故①正确.
$\because ∠ BOD+∠ BOA=180°$,$\therefore ∠ BOA=135°$,$\because OF⊥ AD$,$\therefore ∠ DOF=90°$,$\therefore ∠ BOF=45°+90°=135°=∠ BOA$,
在$△ BOA$和$△ BOF$中,$\begin{cases} ∠ ABO=∠ FBO, \\ OB=OB, \\ ∠ BOA=∠ BOF, \end{cases}$
$\therefore △ BOA≌ △ BOF(\mathrm{ASA})$,故②正确.
如图,延长FO交AB于点H,
$\because OF⊥ AD$,$\therefore ∠ AOG=∠ AOH=90°$,
在$△ AOH$和$△ AOG$中,
$\begin{cases} ∠ HAO=∠ GAO, \\ AO=AO, \\ ∠ AOH=∠ AOG, \end{cases}$
$\therefore △ AOH≌ △ AOG(\mathrm{ASA})$,$\therefore AG=AH$,
$\because ∠ BOH=180°-∠ BOD-∠ DOF=45°$,$\therefore ∠ BOH=∠ BOD$,
在$△ BOD$和$△ BOH$中,$\begin{cases} ∠ OBD=∠ OBH, \\ BO=BO, \\ ∠ BOD=∠ BOH, \end{cases}$
$\therefore △ BOD≌ △ BOH(\mathrm{ASA})$,$\therefore BD=BH$,
$\because BH+AH=AB$,$\therefore BD+AG=AB$,故③正确.
综上所述,正确的结论是①②③.

9.「2026江苏无锡宜兴月考，★☆」已知$△ ABC$中，$∠ ACB=90°$，$AC=CB$，$D$为直线$BC$上一动点，连接$AD$，在直线$AC$右侧作$AE ⊥ AD$，且$AE=AD$.
(1)如图1，当点$D$在线段$BC$上时，过点$E$作$EH ⊥ AC$于点$H$，求证：$EH=AC$.
(2)如图2，当点$D$在线段$BC$的延长线上时，连接$DE$，连接$BE$交$CA$的延长线于点$M$.求证：$S_{△ ABM}=S_{△ AEM}$.
(3)当点$D$在直线$BC$上运动时，连接$BE$交直线$AC$于点$M$，若$AC=4CM$，则$\frac{S_{△ ADB}}{S_{△ AEM}}=$.

(1)证明:$\because AE⊥ AD,EH⊥ AC$,
$\therefore ∠ AHE=∠ EAD=∠ ACB=90°$,
$\therefore ∠ DAC+∠ ADC=90°$,$∠ DAC+∠ EAH=90°$,
$\therefore ∠ EAH=∠ ADC$,又$\because AE=DA$,$∠ AHE=∠ DCA=90°$,
$\therefore △ EAH≌ △ ADC(\mathrm{AAS})$,$\therefore EH=AC$.
(2)证明:如图,过点E作$EN⊥ AM$交AM的延长线于点N,$\because AE⊥ AD,EN⊥ AM$,$∠ ACB=90°$,
$\therefore ∠ ANE=∠ EAD=∠ DCA=90°$,
$\therefore ∠ DAC+∠ ADC=90°$,$∠ DAC+∠ EAN=90°$,
$\therefore ∠ EAN=∠ ADC$,
又$\because AE=DA$,$\therefore △ ANE≌ △ DCA(\mathrm{AAS})$,
$\therefore EN=AC$,$\because BC=AC$,$\therefore BC=EN$,
$\because S_{△ ABM}=\dfrac{1}{2}AM· BC$,$S_{△ AEM}=\dfrac{1}{2}AM· EN$,$\therefore S_{△ ABM}=S_{△ AEM}$.
(3)$\dfrac{2}{5}$或$\dfrac{2}{3}$.
详解:如图1,当点D在线段CB的延长线上时,过点E作$EN⊥ AM$交AM的延长线于点N,
$\because AC=4CM$,$\therefore$设$AC=4a$,则$CM=a$,$BC=4a$,
$\therefore AM=AC+CM=5a$,
同(1)(2)得$△ ANE≌ △ DCA(\mathrm{AAS})$,
$\therefore EN=AC=BC=4a$,$AN=CD$,
又$\because ∠ BMC=∠ EMN$,$∠ BCM=∠ ENM=90°$,
$\therefore △ BCM≌ △ ENM(\mathrm{AAS})$,$\therefore CM=NM=a$,
$\therefore CD=AN=AC+CM+MN=6a$,$\therefore BD=CD-BC=2a$,
$\because S_{△ ADB}=\dfrac{1}{2}BD· AC$,$S_{△ AEM}=\dfrac{1}{2}AM· EN$,$EN=AC$,
$\therefore \dfrac{S_{△ ADB}}{S_{△ AEM}}=\dfrac{BD}{AM}=\dfrac{2}{5}$.

如图2,当点D在线段BC上时,过点E作$EG⊥ AC$于点G,
同理可证$△ AEG≌ △ DAC(\mathrm{AAS})$,$△ BCM≌ △ EGM(\mathrm{AAS})$,
$\therefore CM=GM$,$CD=AG$,$AC=EG$,$\therefore GC=2CM$,
$\because AC=BC$,$\therefore GC=BD$,
设$CM=GM=n$,则$BD=CG=2n$,
$\because AC=4CM$,$\therefore GE=AC=4n$,$\therefore AM=3n$,
$\because S_{△ ADB}=\dfrac{1}{2}BD· AC$,$S_{△ AEM}=\dfrac{1}{2}AM· EG$,$AC=EG$,
$\therefore \dfrac{S_{△ ADB}}{S_{△ AEM}}=\dfrac{BD}{AM}=\dfrac{2}{3}$.

当点D在线段BC的延长线上时,如题图2,不满足$AC=4CM$,故此情况不存在.综上所述,$\dfrac{S_{△ ADB}}{S_{△ AEM}}=\dfrac{2}{5}$或$\dfrac{2}{3}$.

10. 核心 素养 推理能力 聚焦 中考 过程性学习 「2026 广东广州天河期中」如图①,OP是∠MON的平分线,点A是OP上任意一点,用圆规分别在OM,ON上截取OB=OC,连接AB,AC,则△AOB≌△AOC(SAS). 请你参考这个作全等三角形的方法,解答下列问题.
如图②,在△ABC中,∠ACB=90°,∠B=60°,AD,CE分别是∠BAC,∠ACB的平分线,AD,CE相交于点F.
(1)∠EFA的度数为.
(2)写出FD与FE之间的数量关系,并说明理由.
(3)如图③,若∠ACB不是直角,其他条件不变,则(2)中所得结论是否仍然成立?请说明理由.

(1)$60°$.
(2)$FE=FD$,理由如下:
如图1,在AC上截取$CH=CD$,连接FH,
$\because ∠ ACB=90°$,$∠ B=60°$,$\therefore ∠ BAC=30°$,
$\because AD$是$∠ BAC$的平分线,$\therefore ∠ BAD=∠ CAD=15°$,
$\therefore ∠ ADC=∠ BAD+∠ B=75°$,
$\because CE$是$∠ ACB$的平分线,$∠ ACB=90°$,
$\therefore ∠ FCD=∠ FCH=45°$,
又$\because CF=CF$,$CH=CD$,$\therefore △ FCD≌ △ FCH(\mathrm{SAS})$,
$\therefore FD=FH$,$∠ FHC=∠ FDC=75°$,
$\therefore ∠ AHF=180°-∠ FHC=105°$,
$\because ∠ AEF$是$△ BCE$的外角,
$\therefore ∠ AEF=∠ B+∠ BCE=105°$,$\therefore ∠ AEF=∠ AHF$,
又$\because AF=AF$,$\therefore △ AEF≌ △ AHF(\mathrm{AAS})$,
$\therefore FE=FH$,$\therefore FE=FD$.

(3)结论仍然成立.理由如下:
如图2,在AC上截取$CG=CD$,连接FG,
$\because ∠ B=60°$,$\therefore ∠ BAC+∠ BCA=120°$,
$\because AD,CE$分别是$∠ BAC,∠ BCA$的平分线,
$\therefore ∠ FCD=∠ FCG$,$∠ EAF=∠ GAF$,
$\therefore ∠ FAC+∠ FCA=\dfrac{1}{2}(∠ BAC+∠ BCA)=60°$,
$\therefore ∠ CFD=60°$,
$\because CF=CF$,$∠ FCD=∠ FCG$,$CD=CG$,
$\therefore △ FCD≌ △ FCG(\mathrm{SAS})$,
$\therefore FD=FG$,$∠ CFG=∠ CFD=60°$,
$\therefore ∠ AFG=180°-∠ CFG-∠ CFD=60°$,
$\because ∠ AFE=∠ CFD=60°$,$\therefore ∠ AFE=∠ AFG$,
又$\because AF=AF$,$\therefore △ AFE≌ △ AFG(\mathrm{ASA})$,
$\therefore FG=FE$,$\therefore FE=FD$.