∠ACB=π/2-∠BCE=∠CED,∠ABC=∠CDE=π/2,对应的斜边AC=CE,
∴Rt△ABC≌Rt△CDE,
∴CD=AB=3,DE=8.
∴AC=CE=√73,AE=√146,
sin∠DEB
=sin∠FAB
=sin(π/4-∠DEC)
=(5√146)/146,
∴DF
=DB+BF
=BE*sin∠BED+AB*sin∠FAB
=AE*sin∠BED
=5,
不懂的请再问!