The second is to prove with the method of space vector:

I use the first method to prove (1) as follows:? In a straight triangular prism; The upper and lower bottom surfaces are congruent triangles.

So AB=AC, and CC 1⊥ bottom ABC, so cc1⊥ ad;

And AD⊥DE, DE∩CC 1=E, so the face of AD⊥ BCC 1B 1.

And AD is in the plane ADE, so the plane ADE⊥ the plane bcc1b1;

(2) From (1), it can be known that the AD⊥ plane BCC 1B 1,? So BC ⊥;

Triangle ABC is an isosceles triangle; AB=AC, so D is the midpoint of BC. If DF is connected, AD=DF.

AA？ 1||DF, while AA 1=DF, so a 1F∑AD.

So a 1F∑ plane ADE