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8. (2025·南京期末)如图,点A,B,C,D在一条直线上, $EA// FB,EC// FD,EA=FB$.
求证:$AB=CD$.
求证:$AB=CD$.
答案
$\because EA// FB,\therefore ∠A=∠FBD.$
$\because EC// FD,\therefore ∠D=∠ECA.$
在$△ EAC$和$△ FBD$中,$\begin{cases} ∠ECA=∠D, \\ ∠A=∠FBD, \\ AE=BF, \end{cases}$
$\therefore △ EAC≌ △ FBD(\mathrm{AAS}),\therefore AC=BD,$
$\therefore AB+BC=BC+CD,\therefore AB=CD.$
$\because EC// FD,\therefore ∠D=∠ECA.$
在$△ EAC$和$△ FBD$中,$\begin{cases} ∠ECA=∠D, \\ ∠A=∠FBD, \\ AE=BF, \end{cases}$
$\therefore △ EAC≌ △ FBD(\mathrm{AAS}),\therefore AC=BD,$
$\therefore AB+BC=BC+CD,\therefore AB=CD.$
9. 如图,已知 $AB// DE$,且 $AB=DE$. 请你只添加一个条件,使$△ ABC≌ △ DEF$,则添加一个条件是
答案不唯一,如$∠ACB=∠F$
,并给予证明.答案
答案不唯一,如$∠ACB=∠F$. 证明如下:
$\because AB// DE,\therefore ∠B=∠DEF.$
在$△ ABC$和$△ DEF$中,$\begin{cases} ∠ACB=∠F, \\ ∠B=∠DEF, \\ AB=DE, \end{cases}$
$\therefore △ ABC≌ △ DEF(\mathrm{AAS}).$
$\because AB// DE,\therefore ∠B=∠DEF.$
在$△ ABC$和$△ DEF$中,$\begin{cases} ∠ACB=∠F, \\ ∠B=∠DEF, \\ AB=DE, \end{cases}$
$\therefore △ ABC≌ △ DEF(\mathrm{AAS}).$
10. 一线三等角模型 教材 P26 例8·变式 如图,点 C 在
$BD$ 上,$AB ⊥ BD,ED ⊥ BD,AC ⊥ CE,AB=CD$.
求证:$△ ABC ≌ △ CDE$.
$BD$ 上,$AB ⊥ BD,ED ⊥ BD,AC ⊥ CE,AB=CD$.
求证:$△ ABC ≌ △ CDE$.
答案
$\because AB⊥ BD,ED⊥ BD,AC⊥ CE,$
$\therefore ∠B=∠D=∠ACE=90^{\circ },$
$\therefore ∠DCE+∠DEC=90^{\circ },∠BCA+∠DCE=90^{\circ },$
$\therefore ∠BCA=∠DEC.$
在$△ ABC$和$△ CDE$中,$\begin{cases} ∠BCA=∠DEC, \\ ∠B=∠D, \\ AB=CD, \end{cases}$
$\therefore △ ABC≌ △ CDE(\mathrm{AAS}).$
$\therefore ∠B=∠D=∠ACE=90^{\circ },$
$\therefore ∠DCE+∠DEC=90^{\circ },∠BCA+∠DCE=90^{\circ },$
$\therefore ∠BCA=∠DEC.$
在$△ ABC$和$△ CDE$中,$\begin{cases} ∠BCA=∠DEC, \\ ∠B=∠D, \\ AB=CD, \end{cases}$
$\therefore △ ABC≌ △ CDE(\mathrm{AAS}).$
11. (2024·扬州仪征期中)如图(1),在$△ ABC$中,$∠ A=$$∠ ABC$,延长$AC$到$E$,过点$E$作$EF ⊥ AB$交$AB$的延长线于点$F$,延长$CB$到$G$,过点$G$作$GH ⊥ AB$交$AB$的延长线于点$H$,且$EF=GH$.
(1)求证:$△ AEF ≌ △ BGH$;
(2)如图(2),连接$EG$与$FH$相交于点$D$,若$AB=4$,求$DH$的长.
精题详解
(1)求证:$△ AEF ≌ △ BGH$;
(2)如图(2),连接$EG$与$FH$相交于点$D$,若$AB=4$,求$DH$的长.
精题详解
答案
(1)$\because ∠A=∠ABC,∠ABC=∠GBH,$
$\therefore ∠A=∠GBH.$
$\because EF⊥ AB,GH⊥ AB,\therefore ∠AFE=∠BHG.$
在$△ AEF$和$△ BGH$中,$\begin{cases} ∠A=∠GBH, \\ ∠AFE=∠H, \\ EF=GH, \end{cases}$
$\therefore △ AEF≌ △ BGH(\mathrm{AAS}).$
(2)$\because △ AEF≌ △ BGH,$
$\therefore AF=BH,\therefore AB=FH=4.$
$\because EF⊥ AB,GH⊥ AB,\therefore ∠EFD=∠GHD.$
在$△ EFD$和$△ GHD$中,$\begin{cases} ∠EFD=∠GHD, \\ ∠EDF=∠GDH, \\ EF=GH, \end{cases}$
$\therefore △ EFD≌ △ GHD(\mathrm{AAS}),$
$\therefore DH=DF=\frac{1}{2}FH=\frac{1}{2}AB=2.$
$\therefore ∠A=∠GBH.$
$\because EF⊥ AB,GH⊥ AB,\therefore ∠AFE=∠BHG.$
在$△ AEF$和$△ BGH$中,$\begin{cases} ∠A=∠GBH, \\ ∠AFE=∠H, \\ EF=GH, \end{cases}$
$\therefore △ AEF≌ △ BGH(\mathrm{AAS}).$
(2)$\because △ AEF≌ △ BGH,$
$\therefore AF=BH,\therefore AB=FH=4.$
$\because EF⊥ AB,GH⊥ AB,\therefore ∠EFD=∠GHD.$
在$△ EFD$和$△ GHD$中,$\begin{cases} ∠EFD=∠GHD, \\ ∠EDF=∠GDH, \\ EF=GH, \end{cases}$
$\therefore △ EFD≌ △ GHD(\mathrm{AAS}),$
$\therefore DH=DF=\frac{1}{2}FH=\frac{1}{2}AB=2.$
12. (2024·镇江中考)如图，$∠ C=∠ D=90^{\circ },∠ CBA=$$∠ DAB$.
(1)求证:$△ ABC≌ △ BAD$;
(2)若$∠ DAB=70^{\circ }$,则$∠ CAB=\_\_\_\_\_\_^{\circ }$.
(1)求证:$△ ABC≌ △ BAD$;
(2)若$∠ DAB=70^{\circ }$,则$∠ CAB=\_\_\_\_\_\_^{\circ }$.
答案
(1)在$△ ABC$和$△ BAD$中,$\begin{cases} ∠C=∠D=90^{\circ }, \\ ∠CBA=∠DAB, \\ AB=BA, \end{cases}$
$\therefore △ ABC≌ △ BAD(\mathrm{AAS}).$
(2)20
解析 $\because ∠DAB=70^{\circ },∠D=90^{\circ },$
$\therefore ∠DBA=90^{\circ }-70^{\circ }=20^{\circ }.$
由(1)知$△ ABC≌ △ BAD,$
$\therefore ∠CAB=∠DBA=20^{\circ }.$
$\therefore △ ABC≌ △ BAD(\mathrm{AAS}).$
(2)20
解析 $\because ∠DAB=70^{\circ },∠D=90^{\circ },$
$\therefore ∠DBA=90^{\circ }-70^{\circ }=20^{\circ }.$
由(1)知$△ ABC≌ △ BAD,$
$\therefore ∠CAB=∠DBA=20^{\circ }.$
