7. As shown in Figure 3, in $\triangle A B C$, it is known that $\angle B=$ $40^{\circ}, \angle B A D=30^{\circ}$. If $A B$ $=C D$, then the size of $\angle A C D$ is $\qquad$ (degrees).
$7.40^{\circ}$. 
As shown in Figure 6, $\triangle A B D$ is folded along the line $A D$, making point $B$ land at point $E$, resulting in $\triangle A E D$, and $A E$ intersects $C D$ at point $O$.

Since $\triangle A E D \cong \triangle A B D$, we have
$\angle 1=\angle B A D=30^{\circ}, \angle 2=\angle B=40^{\circ}$.
By $\angle 4$ being an exterior angle of $\triangle A B D$, we get
$\angle 4=\angle B A D+\angle B=70^{\circ}$.
Therefore, in $\triangle A D E$,
$\angle 3=180^{\circ}-(\angle 1+\angle 2+\angle 4)=40^{\circ}$.
Thus, $\angle 2=\angle 3 \Rightarrow O D=O E$.
Since $A E$ is obtained by folding $A B$ along $A D$, we have $A E = A B$, and $A B = C D$, so $C D = A E$.
Hence, $C D - O D = A E - O E \Rightarrow O C = O A$.
Therefore, $\angle A C D = \angle 5$.
In $\triangle A B C$,
$$
\angle B + \angle B A D + \angle 1 + \angle 5 + \angle A C D = 180^{\circ},
$$

Thus, $\angle A C D = \frac{1}{2}\left(180^{\circ} - 40^{\circ} - 30^{\circ} - 30^{\circ}\right) = 40^{\circ}$.