1( 1)PD⊥ planar ABCD, PD∈ planar ABCD, CD⊥AD, CD⊥ planar PDA, AP∈ planar APD, CD⊥AP, e,.

(2) Take the midpoint of AD as the GF⊥ plane PCB in the pad plane, connect PG and BG, PG=√5a/2,BG=√5a/2,△FG⊥BC is an isosceles triangle, F is the midpoint of PB, GF⊥PB is also called FH⊥ plane ABCD, H is a vertical foot, and is.

2.( 1), the bottom ABCD is rhombic CD=BC,

∠C 1CB=∠C 1CD =60 degrees, CD=BD⊥CO, CC 1=CC 1, △ c1CD △ c/kloc.

O is the midpoint of BD, C 1O⊥BD, CO∩C 1O=O, BD⊥ plane OCC 1, CC 1∈ plane OCC 1,

∴BD⊥CC 1。

(2) In the previous question, it is known that △BDC is positive △, c1d = c1b ≠ cc1,take the epicenter m of △BDC, and connect A 1M and CM, so that the CM⊥ plane BDC/kl. Ac1= a1d = a1b, where the edge is also a diamond, the triangle DBC 1 is a regular triangle, and CC 1=CD, that is, CD/cc1=/kloc-0.