优质解答
8、∵AC⊥CB,BD⊥CB,∴∠ACB=∠DBC=90°,
∵AB=CD,CB=BC,∴ΔACB≌ΔDBC(HL),
∴∠ABC=∠DCB,
∴∠ACB-∠DCB=∠DBC-∠ABC
即∠ABD=∠ACD.
9、∵BE=CF,∴BE+EF=CF+EF,即BC=EF,
∵AB=DE,AC=DF,
∴ΔABC≌ΔDEF(SSS),
∴∠A=∠D.
10、∵OA=OC,OB=OD,∠AOB=∠COD,
∴ΔAOB≌ΔCOD,
∴∠A=∠C,
∴DC∥AB(内错角相等,两直线平行)
11、∵BF=CE,∴BF+CF=CE+CF,即BC=EF,
∵AB∥CD,AC∥FD,∴∠B=∠E,∠ACB=∠DFE,
∴ΔABC≌ΔDEF(ASA),
∴AB=DE,AC=DF.
12、AE=CE.
理由:
∵FC∥AB,∴∠A=∠ECF,∠ADE=∠F,
∵DE=FE,
∴ΔADE≌ΔCFE(AAS),
∴AE=CE.
8、∵AC⊥CB,BD⊥CB,∴∠ACB=∠DBC=90°,
∵AB=CD,CB=BC,∴ΔACB≌ΔDBC(HL),
∴∠ABC=∠DCB,
∴∠ACB-∠DCB=∠DBC-∠ABC
即∠ABD=∠ACD.
9、∵BE=CF,∴BE+EF=CF+EF,即BC=EF,
∵AB=DE,AC=DF,
∴ΔABC≌ΔDEF(SSS),
∴∠A=∠D.
10、∵OA=OC,OB=OD,∠AOB=∠COD,
∴ΔAOB≌ΔCOD,
∴∠A=∠C,
∴DC∥AB(内错角相等,两直线平行)
11、∵BF=CE,∴BF+CF=CE+CF,即BC=EF,
∵AB∥CD,AC∥FD,∴∠B=∠E,∠ACB=∠DFE,
∴ΔABC≌ΔDEF(ASA),
∴AB=DE,AC=DF.
12、AE=CE.
理由:
∵FC∥AB,∴∠A=∠ECF,∠ADE=∠F,
∵DE=FE,
∴ΔADE≌ΔCFE(AAS),
∴AE=CE.