∴∠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).