(1) proves that if A is AF⊥DC in F, then CF=DF=AF, so ∠ DAC = 90, that is, AC⊥DA …2 points.

PA⊥ ABCD, AC? Facing ABCD, so AC⊥PA …4 points.

Because of PA, AD Surface mat, and PA∩AD=A, so AC⊥ bottom mat …6 points.

And AC? Surface ABCD, so plane AEC⊥ plane pad ... 8 points.

(2) Solution: Connect BD to AC and EO at point O, because PD∑ plane AEC, PD? Facing PBD, facing PBD∩ facing AEC=EO, so PD∨EO... 1 1 min.

Then PE: EB = DO: OB, and DO: OB = DC: AB = 2, then PE: EB = 2 … 14 point.