3. As shown in Figure 1, given that $E$ is the midpoint of side $AD$ of rectangle $ABCD$, $F$ is a point on $AB$, and $EF \perp CE$. If $AB$ : $BC = \sqrt{3} : 2$, then the number of pairs of similar triangles in the figure (there are 4 triangles) is ( ).
(A) 2
(B) 3
(C) 4
(D) 6
3.D.

Let $BC = 2a$. Then $AE = ED = a$.
From the problem, we know $AB = CD = \sqrt{3} a$,
$$
CE = 2a, \angle ECD = 30^{\circ} \text{.}
$$

Since $CE = CB$, then $\angle BCF = \angle ECF = 30^{\circ}$.
Thus, $\angle AEF = 30^{\circ}$.
Therefore, the four triangles in the figure are pairwise similar.
Hence, there are 6 pairs of similar triangles.