∴∠AOB=45°,
∴CD=OD=DE=EF=t,
∴tan∠FOB=
| t |
| 2t |
| 1 |
| 2 |
(2)∵CF∥OB,
∴△ACF∽△AOB,
∴
2
| ||||
2
|
| t |
| OB |
∴OB=
| 2t |
| 2-t |
∴S△OAB=
| 2t |
| 2-t |
(3)要使△BEF与△OFE相似,
∵∠FEO=∠FEB=90°,
∴只要
| OE |
| EB |
| EF |
| EF |
| OE |
| EF |
| EF |
| EB |
即:BE=2t或EB=
| 1 |
| 2 |
①当BE=2t时,BO=4t,
∴
| 2t |
| 2-t |
∴t1=0(舍去)或t2=
| 3 |
| 2 |
∴B(6,0).(2分)
②当EB=
| 1 |
| 2 |
(ⅰ)当B在E的左侧时,

OB=OE-EB=
| 3 |
| 2 |
∴
| 2t |
| 2-t |
| 3 |
| 2 |
∴t1=0(舍去)或t2=
| 2 |
| 3 |
∴B(1,0).(2分)
(ⅱ)当B在E的右侧时,OB=OE+EB=
| 5 |
| 2 |
∴
| 2t |
| 2-t |
| 5 |
| 2 |
∴t1=0(舍去)或t2=
| 6 |
| 5 |
∴B(3,0).(2分)
综上,B(1,0)(3,0)(6,0).

