解:取AD的中点G,并连接EG在△ABD中,E是AB的中点,由题知EG∥BD.又CD:DG=3:1,
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=
S
△DEC=
x,
S
△ACE=
x+4x=
x,
又因为E是AB中点,
所以S
△ACE=
S
△ABC=20,
∴
x=20,
解得x=3,即S
△DEF=3,
∴S
△ADE=
x=8,
∴S
?AEFD=S
△ADE+S
△DEF=8+3=11.