Question

如图,矩形ABCD中,E是AD的中点,将△ABE沿BE折叠后得到△GBE,延长BG交DC于点F,若

CF=1,FD=2,则BC长为

Answer 1

连接DG，==>△EGD等腰==>∠EGD=∠EDG
△ABE≌△BGE==>BG=AB=CF+DF=3
==>∠GEB=1/2*∠AEG=1/2(∠EGD+∠EDG）=∠EGD==>BE//DG
==>∠DGF=∠GBE=∠ABE=∠GDF
==>△DFG等腰
==>DF=GF=2
==>BF=BG+GF=3+2=5
==>BC^2=BF^2-CF^2=25-1=24==> BC=2√6

满意请采纳，谢谢了！