3 |
4 |
13 |
4 |
13 |
5 |
(2)∵BF=t,
∴由△HBF∽△ABC,得到FH=
3 |
4 |
∴DH=4-
3 |
4 |
由△HDG∽△HBF,得DG=
16 |
5 |
3 |
5 |
∵点P到达G点,
∴
16 |
5 |
3 |
5 |
∴t=
9 |
2 |
(3)当0<t≤4时,
若∠PCF=∠B,则△PCF∽△ABC
∵PF=t,CF=8-t,
∴
t |
8−t |
3 |
4 |
∴t=
24 |
7 |
当4<t≤5时,作PK⊥BC于K,
若∠PCF=∠B,则△PCK∽△ABC,
∵PK=
4 |
5 |
3 |
5 |
∴
| ||
5−t−
|
3 |
4 |
解得t=
3 |
2 |
∴t=
24 |
7 |