如图,在△ABC中,D、E两点分别在边BC、AC上,AE:EC=CD:BD=1:2,AD与BE相交于点F,若△ABC的面积为21

2024-12-04 05:09:01
推荐回答(1个)
回答1:

如图,过D作DG∥BE,角AC与G,
∵AE:EC=CD:BD=1:2,△ABC的面积为21,
∴S△ABE:S△BCE=S△ADC:S△ABD=1:2,
S△ABD

2
3
S△ABC=
2
3
×21
=14,
∵DG∥BE,
∴△CDG∽△CBE,△AEF∽△AGD,
CG
GE
DC
BD
1
2

GE=
2
3
CE,AE=
1
2
CE,
AE:EG=AF:FD=3:4,
AF:AD=3:7.
S△ABF:S△ABD=3:7,
S△ABF=
3S△ABD
7
=
3
7
×14
=6,
故答案为:6.