4. In Rt $\triangle A B C$, $\angle C=90^{\circ}, B C<A C$, $B C \cdot A C=\frac{1}{4} A B^{2}$. Then $\angle A$ equals $($ ).
(A) $15^{\circ}$
(B) $18^{\circ}$
(C) $20^{\circ}$
(D) $25^{\circ}$
4.A.

Draw $C D \perp A B$ at point $D$.
From $A B \cdot C D=A C \cdot B C=\frac{1}{4} A B^{2}$, we get
$$
C D=\frac{1}{4} A B \text {. }
$$

Take the midpoint $M$ of $A B$, and connect $C M$. Then $C M=\frac{1}{2} A B$. Therefore, $C D=\frac{1}{2} C M$.
Thus, $\angle C M D=30^{\circ}, \angle A=\frac{1}{2} \angle C M D=15^{\circ}$.