4. As shown in Figure 1, in $\triangle A B C$, $D$ is a point on side $A C$. Among the following four situations, the one where $\triangle A B D \backsim \triangle A C B$ does not necessarily hold is ( ).
(A) $A D \cdot B C=A B \cdot B D$
(B) $A B^{2}=A D \cdot A C$
(C) $\angle A B D=\angle A C B$
(D) $A B \cdot B C=A C$
$\cdot B D$
4. (D).

It is obvious that (B) and (C) must hold. For (A), draw $B E \perp A C$ at $E$, and draw $D F \perp A B$ at $F$, then,
$$
D F=A D \sin A, B E=A B \sin A .
$$

From $A D \cdot B C=A B \cdot B D$, we get $D F \cdot B C=B E \cdot B D$.
Therefore, Rt $\triangle B D F \sim R t \triangle C B E$.
Thus, $\angle A B D=\angle A C B$, which implies $\triangle A B D \backsim \triangle A C B$.
Note: The Law of Sines can also be used.