4. As shown in Figure 1, in trapezoid $A B C D$, $A B$ // 
$D C, A B=B C=\frac{3}{2}$, 
$A D=A C=\sqrt{3}$, $E$ and $F$ 
are the midpoints of $C D$ and $A B$ respectively. Then the length of $E F$ is ( ).
(A) $\sqrt{2}$
(B) $2 \sqrt{2}$
(C) $\frac{\sqrt{41}}{4}$
(D) $\frac{\sqrt{41}}{5}$
4. C.

As shown in Figure 6, connect $A E$. Since $A B / / C D, A B=B C$, $A C=A D$, we have,
$$
\begin{array}{l}
\angle B C A=\angle B A C \\
=\angle A C D=\angle A D C .
\end{array}
$$

Therefore, $\triangle A D C \backsim \triangle B A C$. Hence, $\frac{D C}{A C}=\frac{A C}{B C}$.
Thus, $D C=2, E C=1$.
Since $D E=E C$, then $A E \perp C D, A E \perp A B$.
Therefore, $A E=\sqrt{A C^{2}-E C^{2}}=\sqrt{2}$,
$$
E F=\sqrt{A F^{2}+A E^{2}}=\frac{\sqrt{41}}{4} .
$$