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Solution: (1) proves that ∵ quadrilateral ABCD is prismatic, ∠ ABC = 60,

△ ABC is an equilateral triangle,

E is the midpoint of ∴ae⊥bc BC province,

Also ∵ BC ∨ AD, ∴AE⊥AD,

∵PA⊥ plane ABCD, AE? Plane abcd plane,

∵PA？ Flat pAD, ad? Plane PAD, and PA∩AD=A,

∴AE⊥ flat pad,

* PD plane pad again, ∴ AE ⊥ PD

(2) Let AB=2 and H be any point on PD.

Connect, um, from (1) I know AE⊥ flat pad.

∴∠EHA is the angle formed by EH and the plane pad,

In Rt△EAH, AE=√3, so when AH is the shortest, that is, AH⊥PD, the angle between EH and planar PAD ∠EHA is the largest.

At this time tan∠EHA=l

Therefore, ah = AC1Σ surface CDB 1. And AD=2, so ∠ ADH = 45, so PA = 2.

At this time, the angle between AE and CH is 30.