已知矩形ABCD中,E是CD上一点,AD:AE=1:2,CE:ED=1:3求证AE⊥BE；F是AB中点,DF交AE于G,若CE＝√3,求S△EFG

Question

已知矩形ABCD中,E是CD上一点,AD:AE=1:2,CE:ED=1:3求证AE⊥BE；
F是AB中点,DF交AE于G,若CE＝√3,求S△EFG

Answer 1

由CE=√3和条件可得
ED=3√3,AD=3=BC AE=6 BE=2√3 CD=AB=4√3
∠CBE=30° ∠EBA=60° 连接EF得
EF=AF=BF=2√3
可求S△AFE=3√3
DE:AF=3:2,H△DGE：H△AGF=3:2,
所以H△AGF=6/5
S△AGF=1/2×2√3×6/5=6√3/5
S△EFG =S△AFE-S△AGF=3√3-6√3/5=9√3/5