Question

7,如图,在等腰三角形ABC中,AB=AC,点D在边AB上,点F在边BC上,AF交CD于点E,∠CEF=∠BAC,BH⊥

Answer 1

（1）若∠ABC=60°,则ΔABC为等边三角形过A点做AI⊥CD交CD延长线于点I∵∠CEF=∠BAC=60°∴∠ACD=∠BAF,∠HAC=∠BCI∴ΔADC≌ΔBFA∴ΔAIC≌ΔBHA∴AI=BH又∵∠CEF=60°且AI⊥CD∴AI/AE=sin60°=√3/2∴BH=√3...