(2010·徐汇区一模)如图:在△ABC中,∠C=90°,AC=12,BC=9.则它的重心G到C点的距离是
(2009·泉州质检)如图,在△ABC中,AM是中线,点G为重心,若AM=3,则MG=
(2009·普陀区一模)如图,已知点G是△ABC的重心,过点G作DE∥BC,分别交边AB、AC于点D、E,那么用向量| BC |
| ED |
| 2 |
| 3 |
| BC |
| 2 |
| 3 |
| BC |
| AB |
| a |
| AC |
| b |
| a |
| b |
| AG |
| AG |
| 1 |
| 3 |
| a |
| 1 |
| 3 |
| b |
| 1 |
| 3 |
| a |
| 1 |
| 3 |
| b |
| 5 |
| 3 |
| 5 |
| 3 |
如图,点O是△ABC的重心,若OD=1,则AD=