B. As shown in Figure 2, in quadrilateral $A B C D$, $A C$ and $B D$ are diagonals, $\triangle A B C$ is an equilateral triangle, $\angle A D C=$ $30^{\circ}, A D=3, B D=5$. Then the length of $C D$ is ( ).
(A) $3 \sqrt{2}$
(B) 4
(C) $2 \sqrt{5}$
(D) 4.5
B. B.

As shown in Figure 10, construct an equilateral $\triangle CDE$ with $CD$ as a side, and connect $AE$.
$$
\begin{aligned}
& \text{Given } AC=BC, \\
& CD=CE, \\
& \angle BCD \\
= & \angle BCA + \angle ACD \\
= & \angle DCE + \angle ACD = \angle ACE,
\end{aligned}
$$

we know that $\triangle BCD \cong \triangle ACE \Rightarrow BD=AE$.
Also, $\angle ADC = 30^\circ$, so $\angle ADE = 90^\circ$.
In the right triangle $\triangle ADE$, $AE=5$, $AD=3$, thus,
$$
DE = \sqrt{AE^2 - AD^2} = 4.
$$

Therefore, $CD = DE = 4$.