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.