Question

如下图,三角形ABC的面积是1平方厘米,E是AC的中点,点D在BC上，且BD:DC=1:2，AD与BE交于点F，求四边形的面积

Answer 1

做EM∥DC交AD于M
∵E是AC的中点
∴DE=1/2DC，AM=MD=1/2AD
S△AEM/S△ADC=(AE/AC)²=1/4
S△AEM=1/4S△ACD
∵BD=1/2DC
∴DE=BD
∵EM∥BC，∠MEF=∠DBF，∠EMF=∠BDF
∴△BDF≌△EMF
∴DF=FM=1/2AM
S△BDF=S△EMF
∴DF=1/4AD
∴S△ABD/S△ABC=BD/BC=BD/(DB+CD)=1/3
S△ABD=1/3S△ABC=1/3
S△ACD=1+1/3=2/3
S△BDF/S△ABD=DF/AD=1/4
S△BDF=S△EMF=1/4S△ABD=1/4×1/3=1/12
S四边形CDFE
=S四边形CDME-S△EMF
=S△ACD-S△AEM-S△EMF
=2/3-1/4×2/3-1/12
=2/3-1/6-1/12
=5/12

Answer 2

dc什么