6. A square piece of paper $A B C D$, folded along the diagonal $A C$, so that point $D$ is outside the plane $A B C$. At this time, the angle formed by $D B$ and the plane $A B C$ is definitely not equal to ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $90^{\circ}$
6.D.

As shown in the figure, if $D B \perp$ plane $B A C$, then $\angle D B A=90^{\circ}$. In $\triangle A D B$, $A D=A B$, $\angle A D B=\angle A B D=90^{\circ}$. The sum of the interior angles of $\triangle A B D$ will be greater than $180^{\circ}$, which contradicts the triangle angle sum theorem. Therefore, the angle formed by $D B$ and plane $A B C$ must not be $90^{\circ}$.