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.