Question

相似三角形判定的题，求完整过程！！！

麻烦求下ae长

Answer 1

过A作AG∥DC分别交EF、BC于G、H点，则四边形AGED、四边形GHCF是平行四边形 ∴AG=DF=16／3，GH=FC=AE
∵ EF∥BC ∴⊿AEG∽⊿ABH ∴AE／AB=AG／AH即AE／﹙AE＋BE﹚=AG／﹙AG＋GH﹚

∴AE／﹙AE＋27／4﹚=﹙16／3﹚／﹙16／3＋AE﹚ 解得AE=±6 ﹙负数舍去﹚ ∴AE=6

Answer 2

作AG平行DC交BC于G EF于H
AH=DF
GH=FC
AE=FC
AB=AE+BE
AG=AH+FC=DF+AE
AH=
AE/AB=AH/AG
AE/(AE+BE)=DF/(DF+AE)

追问

亲，说下，什么条件

追答

不少 是漏看
作AG平行DC交BC于G EF于H
AH=DF
GH=FC
AE=FC
AB=AE+BE
AG=AH+FC=DF+AE

AE/AB=AH/AG
AE/(AE+BE)=DF/(DF+AE)

AE(DF+AE)=DF(AE+BE)
AEDF+AEAE=DFAE+DFBE

AEAE=DFBE=(6+3/4)(5+1/3)=36
AE=6

追问

后面的aeae是什么意思啊？

追答

AE*(DF+AE)=DF*(AE+BE)
AE*DF+AE*AE=DF*AE+DF*BE
AE*AE=DF*BE=(6+3/4)*(5+1/3)=36
AE=6

Answer 3

作AG平行DC交BC于G EF于H
AH=DF
GH=FC
AE=FC
AB=AE+BE
AG=AH+FC=DF+AE
AE/AB=AH/AG
AE/(AE+BE)=DF/(DF+AE)
AE(DF+AE)=DF(AE+BE)
AEDF+AEAE=DFAE+DFBE
AEAE=DFBE=(6+3/4)(5+1/3)=36
AE=6