由题意得:BF=B′F,∠C=∠C,
若
CB′
CA
=
CF
CB
=
B′F
AB
,
则△CB′F∽△CAB,
设BF=x,则FC=BC-BF=8-x,
∴
8−x
8
=
x
5
,
解得:x=
40
13
;
若
CB′
CB
=
CF
CA
=
B′F
AB
,
则△CB′F∽△CBA,
设BF=y,则FC=BC-BF=8-y,
∴
8−y
5
=
y
5
,
解得:y=4.
∵此时CB′=6.5>5,
即B′不在AC上,舍去;
∴BF的长为
40
13
.
故答案为:
40
13
.