7.(2025·龙华区期末)如图，AB=AD，AC=AE，∠BAE=∠DAC，下列结论不一定成立的是 （
A.AF=EF
B.∠C=∠E
C.BC=DE
D.∠B=∠D
A
）A.AF=EF
B.∠C=∠E
C.BC=DE
D.∠B=∠D
答案：7.A
8.如图,在△ACB中,∠ACB=90°,AC=BC,点C的坐标为(-2,0),点A的坐标为(-6,3),则点B的坐标为
(1,4)
.答案：8.$(1,4)$
9.如图，AD是△ABC的中线，过点C，B分别作AD的垂线，垂足分别为F，E.(1)求证：CF=BE；(2)若△ACF的面积为28，△CFD的面积为12，求△ABE的面积.
第9题图
第9题图答案：9.(1)证明:$\because CF⊥ AD,BE⊥ AD,\therefore ∠CFD=∠BED=90°.$
$\because AD$是$△ ABC$的中线,$\therefore BD=CD.$
在$△ CFD$和$△ BED$中,$\begin{cases} ∠CFD=∠BED, \\ ∠CDF=∠BDE, \\ CD=BD, \end{cases}$
$\therefore △ CFD≌ △ BED(\mathrm{AAS}),\therefore CF=BE.$
(2)解:$\because S_{△ ACF}=28,S_{△ CFD}=12,$
$\therefore S_{△ ACD}=S_{△ ACF}+S_{△ CFD}=40.$
$\because BD=CD,\therefore S_{△ ABD}=S_{△ ACD}=40.$
由(1)得$△ CFD≌ △ BED,\therefore S_{△ CFD}=S_{△ BED}=12,$
$\therefore S_{△ ABE}=S_{△ ABD}+S_{△ BED}=40+12=52.$
$\because AD$是$△ ABC$的中线,$\therefore BD=CD.$
在$△ CFD$和$△ BED$中,$\begin{cases} ∠CFD=∠BED, \\ ∠CDF=∠BDE, \\ CD=BD, \end{cases}$
$\therefore △ CFD≌ △ BED(\mathrm{AAS}),\therefore CF=BE.$
(2)解:$\because S_{△ ACF}=28,S_{△ CFD}=12,$
$\therefore S_{△ ACD}=S_{△ ACF}+S_{△ CFD}=40.$
$\because BD=CD,\therefore S_{△ ABD}=S_{△ ACD}=40.$
由(1)得$△ CFD≌ △ BED,\therefore S_{△ CFD}=S_{△ BED}=12,$
$\therefore S_{△ ABE}=S_{△ ABD}+S_{△ BED}=40+12=52.$
10. 如图①,点B,F,C,E在同一条直线上,AB//ED,AC//FD,AD交BE于点O.
(1)已知,求证:AD平分CF.
请在下列三个条件:①AB=DE;②AC=DF;③BF=EC中,选择一个补充到上面的横线上,并完成证明.
(2)若将△DEF的边EF沿BE方向移动,使BF=EC,如图②,则(1)中的结论是否仍然成立?若成立,请证明;若不成立,请说明理由.
(1)已知,求证:AD平分CF.
请在下列三个条件:①AB=DE;②AC=DF;③BF=EC中,选择一个补充到上面的横线上,并完成证明.
(2)若将△DEF的边EF沿BE方向移动,使BF=EC,如图②,则(1)中的结论是否仍然成立?若成立,请证明;若不成立,请说明理由.
答案：10.解:(1)选择①:$\because AB// ED,AC// FD,$
$\therefore ∠B=∠E,∠ACB=∠DFE.$
$\because AB=DE,\therefore △ ACB≌ △ DFE(\mathrm{AAS}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OF=OC$,即$AD$平分$CF$.
选择②:$\because AC// FD,$
$\therefore ∠OAC=∠ODF,∠OCA=∠OFD.$
$\because AC=DF,\therefore △ OAC≌ △ ODF(\mathrm{ASA}),$
$\therefore OF=OC$,即$AD$平分$CF$.
选择③:$\because AB// ED,AC// FD,$
$\therefore ∠B=∠E,∠ACB=∠DFE.$
$\because BF=EC,\therefore BF+CF=EC+CF$,即$BC=EF$,
$\therefore △ ACB≌ △ DFE(\mathrm{ASA}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OF=OC$,即$AD$平分$CF$.
(2)(1)中的结论仍然成立.证明如下:
$\because AB// ED,AC// FD,\therefore ∠B=∠E,∠ACF=∠DFC,$
$\therefore 180°-∠ACF=180°-∠DFC$,即$∠ACB=∠DFE.$
$\because BF=EC,\therefore BF-CF=EC-CF$,即$BC=EF,$
$\therefore △ ACB≌ △ DFE(\mathrm{ASA}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OC=OF$,即$AD$平分$CF$.
$\therefore ∠B=∠E,∠ACB=∠DFE.$
$\because AB=DE,\therefore △ ACB≌ △ DFE(\mathrm{AAS}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OF=OC$,即$AD$平分$CF$.
选择②:$\because AC// FD,$
$\therefore ∠OAC=∠ODF,∠OCA=∠OFD.$
$\because AC=DF,\therefore △ OAC≌ △ ODF(\mathrm{ASA}),$
$\therefore OF=OC$,即$AD$平分$CF$.
选择③:$\because AB// ED,AC// FD,$
$\therefore ∠B=∠E,∠ACB=∠DFE.$
$\because BF=EC,\therefore BF+CF=EC+CF$,即$BC=EF$,
$\therefore △ ACB≌ △ DFE(\mathrm{ASA}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OF=OC$,即$AD$平分$CF$.
(2)(1)中的结论仍然成立.证明如下:
$\because AB// ED,AC// FD,\therefore ∠B=∠E,∠ACF=∠DFC,$
$\therefore 180°-∠ACF=180°-∠DFC$,即$∠ACB=∠DFE.$
$\because BF=EC,\therefore BF-CF=EC-CF$,即$BC=EF,$
$\therefore △ ACB≌ △ DFE(\mathrm{ASA}),\therefore AC=DF.$
$\because AC// DF,\therefore ∠OAC=∠ODF,∠OCA=∠OFD,$
$\therefore △ OAC≌ △ ODF(\mathrm{ASA}),\therefore OC=OF$,即$AD$平分$CF$.