Exploration extension: EF=BE+DF still holds.

The proof is as follows: as shown in the figure, extend FD to G, make DG=BE, connect AG,

∠∠b+∠ADC = 180，∠ADC+∠ADG= 180，

∴∠B=∠ADG，

In Δ Abe and Δ △ADG,

DG=BE

∠B=∠ADG

AB=AD

∴△ABE≌△ADG(SAS)，

∴AE=AG,∠BAE=∠DAG，

∫∠EAF =

1

2

∠ No,

∴∠gaf=∠dag+∠daf=∠bae+∠daf=∠bad-∠eaf=∠eaf，

∴∠EAF=∠GAF，

At △AEF and △GAF,

AE=AG

∠EAF=∠GAF

AF=AF

∴△AEF≌△GAF(SAS)，

∴EF=FG，

∫FG = DG+DF = BE+DF，

∴ef=be+df；

Practical application: as shown in the figure, connecting EF and extending the intersection of AE and BF at point C,

∫∠AOB = 30+90+(90-70)= 140，

∠EOF=70，

∴∠EOF= ∠AOB，

OA = OB，

∠OAC+∠OBC =(90-30)+(70+50)= 180，

∴ Meet the requirements in exploration and extension,

∴ Conclusion EF=AE+BF holds,

That is, ef =1.5× (60+80) = 210 nautical mile.

A: At this time, the distance between the two ships is 2 10 nautical mile.