In the pyramid p-abcd as shown in the figure, the bottom surface ABCD is rhombic, the PA ⊥ plane ABCD, and E is the midpoint of PC. It is proved that: (1) Pa ⊥ plane BDE; & nbsp; & nbsp; (2) plane PAC ⊥ plane PBD

In the pyramid p-abcd as shown in the figure, the bottom surface ABCD is rhombic, the PA ⊥ plane ABCD, and E is the midpoint of PC. It is proved that: (1) Pa ⊥ plane BDE; & nbsp; & nbsp; (2) plane PAC ⊥ plane PBD

It is proved that: (1) connecting AC with BD at point O, connecting OE. ⊂ quadrilateral ABCD is a diamond, Ao = Co. ⊙ e is the midpoint of PC, EO ⊄ pa. ⊄ plane BDE, EO ⊂ plane BDE, ⊂ PA ⊉ plane BDE. (2) Pa ⊥ plane ABCD, BD ⊂ plane ABCD, ⊂ PA ⊥ BD, ⊥ quadrilateral ABCD is a rhomboid, BD ⊥ AC. ∩ PA = a, ⊥ BD ⊂ plane PAC, ⊂ BD ⊂ plane PBD, ⊂ plane PAC Plane PBD

In the cube abcd-a1b1c1d1, where e is the midpoint of dd1, the positional relationship between BD1 and the plane passing ace is ()
A. Intersection B. parallel C. vertical D. line in plane

Connect AC and BD, the intersection point is f, and connect EF ∵ in △ bdd1, e and F are the midpoint of dd1 and BD, so EF ∥ BD1, ∥ EF ⊂ plane ace, BD1 ⊄ plane ace, ∥ BD1 ⊂ plane ace, so select B