∵DE是△ABC的中位线，∴DE∥BC，DE=
| 1 |
| 2 |
若设△ABC的面积是1，根据DE∥BC，得△ADE∽△ABC，
∴S△ADE=
| 1 |
| 4 |
连接AM，根据题意，得S△ADM=
| 1 |
| 2 |
| 1 |
| 8 |
| 1 |
| 8 |
∵DE∥BC，DM=
| 1 |
| 4 |
∴DN=
| 1 |
| 4 |
∴DN=
| 1 |
| 3 |
| 1 |
| 3 |
∴S△DNM=
| 1 |
| 3 |
| 1 |
| 24 |
∴S四边形ANME=
| 1 |
| 4 |
| 1 |
| 24 |
| 5 |
| 24 |
∴S△DMN：S四边形ANME=
| 1 |
| 24 |
| 5 |
| 24 |
故选A．
A. 1：5∵DE是△ABC的中位线，| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 2 |
| 1 |
| 8 |
| 1 |
| 8 |
| 1 |
| 4 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 24 |
| 1 |
| 4 |
| 1 |
| 24 |
| 5 |
| 24 |
| 1 |
| 24 |
| 5 |
| 24 |