1, ∫ plane D'AE is perpendicular to ⊥ plane ABCE, and these two planes intersect at AE,AE⊥BE.

It can be obtained from the vertical nature of the plane: BE ⊥ plane D'AE.

∴ Advertising' ⊥ EB Point D'

2. As the vertical line of the bottom ABCE, the vertical foot is O, then:

Because, D'A=D'E= 1.

Therefore, o is the midpoint of AE.

The intersection o is perpendicular to AB, the vertical foot is G, and OG passes through AC to F; Connect D'G, make f the vertical line of D'G, make h the vertical foot, and connect AH.

Because D'O⊥ surface ABCE, so: D'O⊥AB.

Once again, AB⊥OG

So, AB⊥ surfaced.

So, AB⊥FH

FH⊥D'G, D'G and AB intersect at point g.

So, FH⊥ facing Abd

However, the point f is on the straight line AC.

Therefore ∠FAH is the angle formed by AC and surface ABD'

In rectangular ABCD, AB=2, BC= 1.

Therefore, AC=√5.

Since O is the midpoint of AE, AG/AB= 1/4.

And FG//BC

Therefore, AF/AC=AG/AB= 1/4, FG/BC=AG/AB= 1/4.

So AF=AC/4=√5/4, FG=BC/4= 1/4.

Also, DO'=AO=EO=√2/2, OG= 1/2.

So in Rt△D'OG, there is Pythagorean theorem: D g =√3/2.

And, Rt△D'OG∽Rt△FHG.

Therefore, fg/d 'g = FH/d 'o.

Namely: (1/4)/(√3/2)=FH/(√2/2)

Therefore, FH = 6/12.

Therefore, in Rt△AHF, AF=√5/4, FH=√6/ 12.

That is, sin ∠ fah = FH/AF = (√ 6/12)/(√ 5/4) = √ 30/15.