4. As shown in Figure 1, in trapezoid $A B C D$, $A B / / D C, A B$ $=A C, D A=DB, \angle A D B=90^{\circ}$. Then the degree measure of $\angle A C D$ is $\qquad$.
$4.30^{\circ}$.
As shown in Figure 4, draw $A E \perp C D$ intersecting the extension of $C D$ at $E$, and draw $D F \perp A B$ at $F$.

Since $D A = D B$ and $\angle A D B = 90^{\circ}$, we know that $\triangle A D B$ is an isosceles right triangle. Therefore,
$\angle D B A = \angle D A F = 45^{\circ}$.
Because $A B \parallel D C$, we have $\angle A D E = 45^{\circ}$.
Also, since $D F \perp A B$, then
$$
D F = \frac{1}{2} A B, \angle A D F = 45^{\circ} \text {. }
$$

Thus, $\angle F D E = 90^{\circ}$, which means $F D \perp C D$.
Since $A E \perp C D$, we get $A E \parallel F D$.
Also, $A F \parallel E D$, so quadrilateral $A F D E$ is a parallelogram.
Therefore, $A E = D F = \frac{1}{2} A B = \frac{1}{2} A C$.
Hence, $\angle A C D = 30^{\circ}$.