4. In space, there are 4 points $A, B, C, D$, satisfying $A B=B C=C D$. If $\angle A B C=\angle B C D=\angle C D A=36^{\circ}$, then the angle formed by line $A C$ and line $B D$ is $\qquad$
When $A, B, C, D$ are coplanar, it is easy to see that $A D \parallel B C$. Extend $A C$ and $B D$ to intersect at point $E$. By $\angle A C B = \angle D B C = 72^{\circ} \Rightarrow \angle B E C = 36^{\circ}$, which means the angle formed by line $A C$ and line $B D$ is $36^{\circ}$.
When $A, B, C, D$ are not coplanar, let the midpoint of $B D$ be $F \Rightarrow C F \perp B D$. In $\triangle A B C$ and $\triangle A D C$, $\angle A B C = \angle A D C = 36^{\circ}, B C = D C$, thus $\triangle A B C \cong \triangle A D C$.

Therefore, $A B = A D \Rightarrow A F \perp B D \Rightarrow A C \perp B D$.
In summary, the angle formed by line $A C$ and line $B D$ is $36^{\circ}$ or $90^{\circ}$.