1.sΔABK=1/4sABCD
又E是中点.sΔAEG=1/4sΔABK
sΔAEG:S矩形ABCD=1:16
2.GF‖BK,不难证明ΔBKH∽ΔFGH
EG=1/2BK=1/4BC
BC=EF FG=EF-EG=3/4EF BK=1/2EF
EG是中位线,AG=GK
GH:HK=3:2 GH=3/5GK=3/10AK
GH:AK=3:10
3.AK⊥BF,不难证明ΔABK∽ΔAHB
AB²=AH*AK
设AK是10x,AH是8x
带入有AB²=80x²
AB=4√5
AK²-AB²=20=BK²
BK=2√5=1/2AB
正方形