9. As shown in Figure 4, in quadrilateral $ABCD$, $AC$ bisects $\angle BAD$, $CE \perp AB$ at point $E$, and $AE = \frac{1}{2}(AB + AD)$. If $\angle D = 120^{\circ}$, then $\angle B$ equals
$9.60^{\circ}$. 
As shown in Figure 7, on $AB$, take $AF = AD$, and connect $CF$.

Since $AC$ bisects $\angle BAD$, then $\angle DAC = \angle FAC$.

Also, $AC$ is a common side, so,
$\triangle DAC \cong \triangle FAC$.
Thus, $\angle AFC = \angle D = 120^{\circ}$.
And $\angle AFC + \angle CFE = 180^{\circ}$, hence
$\angle CFE = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
Given that $AE = \frac{1}{2}(AB + AD)$, and
$$
AE = AF + FE, AB = AF + FE + EB, AD = AF,
$$

then $FE = BE$.
Since $CE \perp AB$ at point $E$, $\triangle BCF$ is an equilateral triangle. Therefore, $\angle B = \angle CFE = 60^{\circ}$.