Solution: (1)△ECF is an isosceles right triangle for the following reasons:

If A is AG⊥CD in G, then the quadrilateral ABCG is a rectangle. AB=CG

Because tan∠ADC=2 and AG=2DG.

DG = CG, CD= CB = CB because AB/CD = 0.5.

In △CDE and △CBF, CD = CB, ∠ CDE = ∠ CBF, DE = BF.

So△CDE?△CBF(SAS)So，△ ∠DCE = ∠BCF，CE = CF

And because ∠ DCE+∠ BCE = 90, ∠ BCF+∠ BCE = 90, that is ∠ FCE = 90.

So) △ECF is an isosceles right triangle.

(2)∫△ECF is an isosceles right triangle ∴ ∠ CEF = 45. EF = (root number 2)*CE.

∫≈bec = 135∴∠bef = 90。 And ∵BE/CE = 1/2 ∴EF = 2* (radical number 2)*BE.

In Rt△BEF, tan ∠ bfe = be/ef =1[2 * (root number 2)] = (root number 2)/4.