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The last problem of mathematics
Solution: problem background: ef = be+df;

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.