∵CE=x,BE:CE=2:1,
∴BE=2x,AD=BC=CD=AD=3x;
∵BC∥AD∴∠EBF=∠ADF,
又∵∠BFE=∠DFA;
∴△EBF∽△ADF
∴S△BEF:S△ADF=(
BE |
AD |
2x |
3x |
4 |
9 |
9 |
4 |
∵S△BCD-S△BEF=S四边形EFDC=S正方形ABCD-S△ABE-S△ADF,
∴
9 |
2 |
1 |
2 |
9 |
4 |
化简可求出x2=
5 |
6 |
∴S△AFD:S四边形DEFC=
9 |
4 |
9 |
2 |
9 |
4 |
11 |
4 |
BE |
AD |
2x |
3x |
4 |
9 |
9 |
4 |
9 |
2 |
1 |
2 |
9 |
4 |
5 |
6 |
9 |
4 |
9 |
2 |
9 |
4 |
11 |
4 |