Question

正方形ABCD面积是264,E是AB中点,F在BC上 BC=4BF 连接BD ,DF.连接CE交BD与G交DF于H

Answer 1

S(BDC)=S(ABCD)/2=132

三角形EBG∽三角形CDG

所以DG/GB=DC/EB=2

所以S(GDC)=(2/3)S(BDC)=88

S(DFB)=(1/4)S(BDC)=33

过B作BI//FD交CE于I

则IG/GH=BI/ID=1/2

IH/HC=BF/FC=1/3

所以GH/HC=2/9

所以S(GDH)=(2/11)S(GDC)=16

S(BFHG)=S(DFB)-S(GDH)=33-16=17