∵正方形ABCD和AEFG,∴AG=AE,AD=AB,∠DAB=∠GAE=90°,
∴∠DAG=∠EAB,
∴△ADG≌△ABE,
∴DG=BE,
∵正方形ABCD和AEFG,
∴∠DAC=∠GAF=
| 1 |
| 2 |
∴∠DAG=∠FAC=∠EAB,
由勾股定理得:
| AF |
| AG |
| AC |
| AD |
| 2 |
∴△ABE∽△ACF,
∴
| BE |
| CF |
| AB |
| AC |
| 1 | ||
|
| ||
| 2 |
∴BE:CF:DG=1:
| 2 |
故选B.
A. 1:1:1| 2 |
| 3 |
∵正方形ABCD和AEFG,| 1 |
| 2 |
| AF |
| AG |
| AC |
| AD |
| 2 |
| BE |
| CF |
| AB |
| AC |
| 1 | ||
|
| ||
| 2 |
| 2 |