Wenzhou recruits dm

(1) proof: take the midpoint h of AD and connect FH and GH. Because GH∨DC∨EF, GH=EF=3, the quadrilateral EFGH is a parallelogram.

Therefore, there is EG∑FH,

EG again? Plane ADE, FH? Plane elevation

So EG∑ plane AFD

(2) solution: f is FM∨EC, m is MN⊥BD, and the vertical foot is n and FN connected.

Because FD⊥ plane ABCD, FD⊥MN, BD ∩ FD = D.

So MN⊥ aircraft BDF,

Therefore, MFN treatment is the angle between the European Community and BDF.

B is BO⊥DC, and the vertical foot is o,

Because the quadrilateral ABCD is an isosceles trapezoid

So OC= 1, BO=3, DO=3, DM= 1, BD=23.

Because △DBO∽△DMN

So MN=DM×BOBD= 12.

In Rt△FDM, FM=2.

So in Rt△FNM, FN=72.

So tan∠MFN=MNFN=77.

Therefore, the tangent of the angle formed by the straight line EC and the plane BDF is 77.