Question

如图,已知△ABC,点D,E分别在BC,AC上,BD=2CD,AE=CE,AD,BE相交于F点,S△ABF=10,求S△

Answer 1

连接FC

得到△AFE

∵AE=EC

∴S△AFE=S△FEC

∵AE=EC

∴S△AEB=S△EBC

∵S△AFE=S△FEC

∴S△AFB=S△BFC=10

∵BD=2DC

∴S△DCF=1/3*S△BFC=10/3

设S△AFE为X

则(2X+10/3)*2=20-10/3

4X+20/3=50/3

4X=10

X=2.5