Perpendicular line to a plane ( cube )

Question

The Question

I'm trying to show that $A'C$ perpendicular on plane ($C'BD$). Found that $AB = 6$ and that $A'C'BD$ and $CC'BD$ are both a tetrahedron with the same base ( triangle $C'BD$). Still I can't explain why $A'C$ is perpendicular on plane ($C'BD$).

My Understanding

$AE=3\sqrt{6}$

$A'C\perp$(triangle $C' B)$?

I found that $AB=6$

Answer 1

HINT: $$ A'C'=A'B=A'D, \quad CC'=CB=CD \quad\text{and}\quad C'B=BD=DC'. $$