EFHG is a parallelogram, such as //FH.

Because AB//EF,AB⊥BC of EF⊥BC.

And EF⊥FB,

So EF⊥ faces BCF.

So EF⊥FH

So AB⊥FH

And FH⊥BC

So FH⊥ surface ABCD

So EG⊥ surface ABCD

So EG⊥AC

Because AC⊥BD

So AC ⊥ Eboud

(2) by EF⊥FB, BF⊥FC,

So BF⊥ aircraft CDEF,

Let f be the extension line of FK⊥DE passing through DE at k and connect BK. From the theorem of three perpendicular lines, we can know that BK⊥DE.

Then ∠FKB is the plane angle of dihedral angle b-de-c.

Let EF= 1, then AB=2,

FC=BF=√2，

De = AE = √ 3 (one waist of right-angled trapezoid EFBA)

And EF//DC, ∴∠KEF=∠EDC,

∴ SIN ∠ EDC = SIN ∠ KEF = FK/EF = √ 2/√ 3 (EFCD is a right-angled trapezoid, which makes drawing a plane figure clearer).

∴FK=EFsin∠KEF=√6/3，tan∠FKB=BF/FK=√3

∴∠FKB=60

The dihedral angle is 60 degrees.