△ABC中,DE‖FG‖BC,S△ADE:S四边形DEGF:S四边形FGCB=4:5:16,且FD=2,则FB的长为.
3个回答

∵DE//FG//BC

∴△ADE∽△AFG∽△ABC

∵S△ADE:S四边形DEGF:S四边形FGCB = 4:5:16

∴S△ADE:S△AFG:S△ABC = 4:(4+5):(4+5+16)

= 4:9:16

∵相似三角形面积的比等于相似比的平方

∴(AD:AF)^2 = 4:9

(AF:AB)^2 = 9:16

∴ AD:AF = 2:3

AF:AB = 3:5

∴AD:(AF-AD)=2:(3-2)

AF:(AB-AF)=3:(5-3)

即AD=2DF=2*2=4

BF=2AF/3=2*(4+2)/3=4

2)延长BA,CD交于M点

则MA:MB=AD:BC=3:9=1:3

∴MA:(MB-MA)=1:(3-1)

即MA:4=1:2

∴MA=2

∴S△MAD:S△MBC=(1:3)^2=1:9