Question

如圖所示,已知點A、E、F、D在同一條直線上,AE=DF,BF⊥AD,CE⊥AD, 垂足分別為F、E,BF=CE,求證:AB∥CD

Answer 1

p;【答案】見解析解析:
∵AE=DF
∴AE+EF=DF+EF
即AF=DE-
∵BF⊥AD，CE⊥AD
∴∠AFB=∠DEC=90°
∵BF=CE
∴△AFB≌△DEC
∴∠A=∠D
∴AB∥CD