4. As shown in Figure $1, A B$ is the diameter of $\odot O$. The radius is $\frac{5}{2}, B C=3$, points $E$ and $F$ are taken on chords $A B$ and $A C$ respectively, such that line segment $E F$ divides $\triangle A B C$ into two parts of equal area. Then the minimum value of line segment $E F$ is ( ).
(A)2
(B) $2 \sqrt{2}$
(C) 3
(D) $3 \sqrt{2}$
4. (A).
$\because AB$ is the diameter of $\odot O$, $\therefore \angle C=90^{\circ}$.
Also, $\because$ the radius is $\frac{5}{2}$, $\therefore AB=5, BC=3$. Thus, $AC=4$. Therefore, $S_{\triangle ABC}=\frac{1}{2} \times 3 \times 4=6$. Let $AE=x, AF=y$.

Since $S_{\triangle AEF}=\frac{1}{2} x y \sin A=\frac{1}{2} S_{\triangle ABC}=3$, and $\sin A=\frac{3}{5}$, then $x y=10$.

By the cosine rule, we get
$$
\begin{aligned}
EF^{2} & =x^{2}+y^{2}-2 x y \cos A \\
& =(x-y)^{2}+2 x y(1-\cos A) \\
& =(x-y)^{2}+4 .
\end{aligned}
$$

Thus, when $x=y$, $EF_{\text {min }}=2$.