4. In a convex pentagon $A B C D E$,
$$
A B=B C=C D=D E=E A,
$$

and $\angle C A D=\angle B A C+\angle E A D$. Then the degree measure of $\angle B A E$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
4. C.

As shown in Figure 2, $\triangle A B C$ is rotated counterclockwise around point $A$ by the degree of $\angle B A E$ to the position of $\triangle A E F$, and $D F$ is connected. Then
$$
\begin{array}{l}
B C=E F, \\
A C=A F, \\
\angle B A C=\angle E A F .
\end{array}
$$

Since $\angle C A D=\angle B A C+\angle E A D$, we have $\angle C A D=\angle E A F+\angle E A D=\angle F A D$.
It is easy to see that $\triangle C A D \cong \triangle F A D$
$$
\Rightarrow C D=D F \Rightarrow D F=D E=E F
$$
$\Rightarrow \triangle D E F$ is an equilateral triangle
$$
\Rightarrow \angle D E F=60^{\circ} \text {. }
$$

Given $A E=D E=E F$, we can conclude that $E$ is the circumcenter of $\triangle A D E$.
According to the relationship between inscribed angles and central angles, we have
$$
\angle D A F=\frac{1}{2} \angle D E F=30^{\circ} \text {. }
$$

Therefore, $\angle B A E=2 \angle C A D=2 \angle D A F=60^{\circ}$.