(2014? Linyi Three-Model) As shown in the figure, it is known that the plane of the whip ABEF and the plane of the right-angled trapezoid ABCD are perpendicular to each other, AB=2AD=2CD=4, ∠BAD=∠C

Proof: (Ⅰ) Take the midpoint N of AD and connect FN and NG.

∵G is the midpoint of BC, ABCD is a right-angled trapezoid, AB=2CD=4,

∴NG∥AB, and NG=AB+CD2=3,

ABEF is a rhombus,

∴EF∥AB, and EF=AB= 4.

And ∵H is the midpoint of EF, M is the midpoint of HE,

∴FM=3, and FM∥NG,

∴ The quadrilateral FMGN is a parallelogram.

∴MG∥FN,

And ∵FN? Plane ADF, MG? Plane ADF,

∴MG∥Plane ADF.

(Ⅱ) Connect AE, because ABFE is a rhombus, ∠EFA=60°, H is the midpoint of EF,

∴AH⊥EF, that is, AH⊥AB, < /p>

∵Plane ABEF⊥Plane ABCD, AB is the intersection line of surface ABEF and surface ABCD,

∴AH⊥Plane ABCD,

∵BC?Plane ABCD,

∴AH⊥BC,

∵In the right-angled trapezoid ABCD, ∠BAD=∠CDA=90°, AB=4, CD=AD=2,

∵AB∥CD,

∴∠BCD=135°

And in △ADC, ∠ADC= 90°, AD=CD,

∴∠ACD=45°,

∴∠ACB=∠BCD-∠ACD=90°, that is, AC⊥BC,

Also AH? plane AHC, AC? plane AHC,

∴BC⊥ plane AHC,

∵BC? plane BCE,

∴ plane AHC ⊥Plane BCE.