4. Given that in the acute triangle $\triangle A B C$, the distance from vertex $A$ to the orthocenter $H$ is equal to the radius of its circumcircle. Then the degree of $\angle A$ is ( ).
(A) $30^{\circ}$
(B) $45^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
4.C.

In the acute triangle $\triangle ABC$, the orthocenter is inside the triangle. As shown in Figure 2, let the circumcenter of $\triangle ABC$ be $O$, and $D$ be the midpoint of $BC$. The extension of $BO$ intersects $\odot O$ at point $E$. Connecting $CE$, $AE$, and $CH$, we know that $CE \parallel AH$, $AE \parallel CH$, thus $OB = AH = CE = 2OD$. Therefore,
$$
\begin{array}{l}
\angle O B D=30^{\circ}, \\
\angle B O D=60^{\circ} . \\
\text { Hence } \angle A=\angle B O D=60^{\circ} .
\end{array}
$$