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Urgent, urgent, 20 1 1 a good supplement to the second model of mathematics for senior high school entrance examination in Dalian.
(1) according to the picture and the meaning of the question, EA⊥AC, and then ∠CAD= 1/2∠DAE, that is to say ∠CAD=30? ,∠DAE=60?

Because △ABC is an equilateral triangle, AB⊥AD. Then cross E to make eh ⊥ AD on AD, and cross AD to H, assuming AB=2.

In this way, through calculation, it can be concluded that EH=2, through BAF = EHF = 90? , ∠ AFB = ∠ EFH, AB = Eh, the principle that all two angles and a line are equal is an equal triangle. You can get BF=EF.

(2) Connect BH and AC to G because ∠CAD=2∠DAE, ∠CAD=60? ,∠DAE=30? Through calculation, it is found that AH=2, AB=AH, ∠BAC=∠CAD, ***AG are on the same side, so ∠ABC=∠ACB=30?

∠ADC=30? , so ∠ADC=∠ACB, that is, BH∪CD∪AE, and then ∠AFE=∠BFH, because it is diagonal.

So △AFE is similar to △BFH, so BF/EF=BH/AE=3/2.