全等三角形证明经典50题含答案
1：AB4，AC2，D是BC中点，AD是整数，求ADA
B
C
D
解：延长AD到E使ADDE∵D是BC中点∴BDDC在△ACD和△BDE中ADDE∠BDE∠ADCBDDC∴△ACD≌△BDE∴ACBE2∵在△ABE中ABBE＜AE＜ABBE∵AB4即42＜2AD＜421＜AD＜3∴AD2
2：D是AB中点，∠ACB90°，求证：CD1AB2
A
D
C
B
延长CD与P，使D为CP中点。连接APBP∵DPDCDADB∴ACBP为平行四边形又∠ACB90∴平行四边形ACBP为矩形∴ABCP12AB
3：BCDE，∠B∠E，∠C∠D，F是CD中点，求证：∠1∠2

v

f

A12
B
E
C
F
D
证明：连接BF和EF∵BCEDCFDF∠BCF∠EDF∴三角形BCF全等于三角形EDF边角边∴BFEF∠CBF∠DEF连接BE在三角形BEF中BFEF∴∠EBF∠BEF。∵∠ABC∠AED。∴∠ABE∠AEB。∴ABAE。在三角形ABF和三角形AEF中ABAEBFEF∠ABF∠ABE∠EBF∠AEB∠BEF∠AEF∴三角形ABF和三角形AEF全等。∴∠BAF∠EAF∠1∠2。4：∠1∠2，CDDE，EFAB，求证：EFAC
A12
F
CDEB
过C作CG∥EF交AD的延长线于点G
CG∥EF，可得，∠EFD＝CGDDE＝DC∠FDE＝∠GDC〔对顶角〕∴△EFD≌△CGDEF＝CG∠CGD＝∠EFD又，EF∥AB∴，∠EFD＝∠1

v

f

∠1∠2∴∠CGD＝∠2∴△AGC为等腰三角形，AC＝CG又EF＝CG∴EF＝AC5：AD平分∠BAC，ACABBD，求证：∠B2∠C
A
证明：延长AB取点E，使AE＝AC，连接DE∵AD平分∠BAC∴∠EAD＝∠CAD∵AE＝AC，AD＝AD∴△AED≌△ACD〔SAS〕∴∠E＝∠C∵AC＝ABBD∴AE＝ABBD∵AE＝ABBE∴BD＝BE∴∠BDE＝∠E∵∠ABC＝∠E∠BDE∴∠ABC＝2∠E∴∠ABC＝2∠C6：AC平分∠BAD，CE⊥AB，∠B∠D180°，求证：AEADBE

v

f

证明：在AE上取F，使EF＝EB，连接CF∵CE⊥AB∴∠CEB＝∠CEF＝90°∵EB＝EF，CE＝CE，∴△CEB≌△CEF∴∠B＝∠CFE∵∠B＋∠D＝180°，∠CFE＋∠CFA＝180°∴∠D＝∠CFA∵AC平分∠BAD∴∠DAC＝∠FAC∵AC＝AC∴△ADC≌△AFC〔SAS〕∴AD＝AF∴AE＝AF＋FE＝AD＋BE7：AB4，AC2，D是BC中点，AD是整数，求AD
A
B
C
D
解：延长AD到E使ADDE∵D是BC中点∴BDDC在△ACD和△BDE中ADDE∠BDE∠ADCBDDC∴△ACD≌△BDE∴ACBE2

v

f

∵在△ABE中ABBE＜AE＜ABBE∵AB4即42＜2AD＜421＜AD＜3∴AD2
8：D是AB中点，∠ACB90°，求证：CD1AB2
A
D
C
B解：延长AD到E使ADDE
∵D是BC中点∴BDDC在△ACD和△BDE中ADDE∠BDE∠ADCBDDC∴△ACD≌△BDE∴ACBE2∵在△ABE中ABBE＜AE＜ABBE∵AB4即42＜2AD＜421＜AD＜3∴AD2
9：BCDE，∠B∠E，∠C∠D，F是CD中点，求证：∠1∠2

v

f

A12
B
E
C
F
D
证明：连接BF和EF。∵BCEDCFDF∠BCr