从而,在△CEG中,CF:FE=CD:DG=3:1,
∴S△DFC:S△DFE=3:1.
设S△DEF=x,则S△DFC=3x,S△DEC=4x.
由于AD:DC=2:3,
∴S△EAD:S△ECD=2:3,
∴S△EAD=
2 |
3 |
8 |
3 |
S△ACE=
8 |
3 |
20 |
3 |
又因为E是AB中点,
所以S△ACE=
1 |
2 |
∴
20 |
3 |
解得x=3,即S△DEF=3,
∴S△ADE=
8 |
3 |
∴S▱AEFD=S△ADE+S△DEF=8+3=11.
2 |
3 |
8 |
3 |
8 |
3 |
20 |
3 |
1 |
2 |
20 |
3 |
8 |
3 |