4. Given an acute triangle $\triangle A B C$ with the three interior angles $\angle A$, $\angle B$, $\angle C$ and the opposite sides of lengths $a$, $b$, $c$ respectively, and a point $O$ on the plane of $\triangle A B C$, and the distances from $O$ to the sides $B C$, $A C$, $A B$ are $x$, $y$, $z$ respectively. The following conclusions are given:
(1) When $O$ is the centroid of $\triangle A B C$, $x: y: z=b c: c a: a b$;
(2) When $O$ is the circumcenter of $\triangle A B C$, $x: y: z=\cos A: \cos B: \cos C$;
(3) When $O$ is the orthocenter of $\triangle A B C$, $x: y: z=\cos B \cdot \cos C: \cos C \cdot \cos A: \cos A \cdot \cos B$. Which of the following are correct?
$(\mathrm{A})(1)(\mathrm{B})(1)(2)(\mathrm{C})(2)(3)(\mathrm{D})(1)(2)(3)$
4. D.
(1) As shown in Figure 7 (a), when $O$ is the centroid of $\triangle ABC$, from $ax = by = cz$, we get
$$
x: y: z = \frac{1}{a}: \frac{1}{b}: \frac{1}{c} = bc: ca: ab.
$$
(2) As shown in Figure 7 (b), when $O$ is the circumcenter of $\triangle ABC$, let the circumradius of $\triangle ABC$ be $R$. Then
$$
\angle BOC = 2 \angle A, \angle BOD = \angle A.
$$

Thus, $\frac{x}{R} = \cos A$.
Similarly, $\frac{y}{R} = \cos B, \frac{z}{R} = \cos C$.
Therefore, $x: y: z = \cos A: \cos B: \cos C$.
(3) As shown in Figure 7 (c), when $O$ is the orthocenter of $\triangle ABC$, it is easy to see that points $O, D, C, E$ are concyclic, so
$$
\angle BOD = \angle C.
$$

Then $x = \frac{BD}{\tan \angle BOD} = \frac{BD}{\tan C}$
$$
= \frac{2R \sin C \cdot \cos B}{\tan C} = 2R \cos B \cdot \cos C.
$$

Similarly, $y = 2R \cos C \cdot \cos A$, $z = 2R \cos A \cdot \cos B$.
Therefore, $x: y: z = \cos B \cdot \cos C: \cos C \cdot \cos A: \cos A \cdot \cos B$.