Question

21、如图,矩形ABCD中,AE⊥BD于E,CF⊥BD于F,连接CE,AF.

Answer 1

1)△ABE≌△CDF(角角边)=>AE=CF=>△AEF≌△CFE（边角边）

=>∠AFE=∠CEF=>AF‖CE

∠AEF=∠CFE=90°=>CF‖AE

=>四边形AECF是平行四边形

2)BD^2=3^2+(√3)^2=12=>BD=2√3

CF=CD*BC/BD=3*√3/2√3=1.5

DF=BE=√(CD^2-CF^2)=√3-2.25=√3/2

EF=BD-BE-DF=2√3-√3/2-√3/2=√3

=>CE=√(CF^2+EF^2)=√[(1.5)^2+3]=√21/2