It is known that in the cube abcd-a1b1c1d1, e and F are the midpoint of CC1 and Aa1 respectively

It is known that in the cube abcd-a1b1c1d1, e and F are the midpoint of CC1 and Aa1 respectively

In the cube abcd-a1b1c1d1, BD ‖ b1d1, ∩ E and F are the midpoint of CC1 and Aa1 respectively, connect Ag (G is the midpoint of B1B), De, then the quadrilateral adeg is a parallelogram, ∩ d1b1 = D1, ∩ according to the inference of plane parallel, plane BDE ‖ plane b1d1f

As shown in the figure, in the cube abcd-a1b1c1d1, e is the midpoint of Aa1. Prove that A1C ‖ plane BDE, aa1c ⊥ plane BDE

Certification:
Connecting AC intersection BD with o point
In the triangle aa1c, EO is the median line
So EO parallels AC
Because EO is in the surface bed, AC is not in the surface bed
So AC is parallel to bed
BD⊥AC
BD⊥AA1
AC and Aa1 intersect with a
So the BD vertical plane aa1c
And because BD is in the surface bed
So surface bed vertical surface aa1c