Question

如图,AE⊥BD,CF⊥BD,垂足分别是E、F,AB=CD,BF=DE.求证AB//CD

Answer 1

楼上的做错了 ∵BF=DE
∴BF+EF=DE+EF
∴BE=DF
∵AE⊥BD,CF⊥BD
∴角AED=∠CFD
在RT△ABE与RT△CDF中
∠AED=∠CFD
BE=DF
∴RT△ABE全等RT△CDF(HL)
∴∠CDF=∠ABE
∴AB∥CD

Answer 2

BF=DE
BF+EF=DE+EF
BE=DF
AE⊥BD,CF⊥BD
,AB=CD
三角形ABE与CDF全等
∠CDF=∠ABE
AB//CD