3. In a right trapezoid $A B C D$, $A D / / B C$, $B C \perp C D$, $E$ is the midpoint of side $A B$, $B C=C D=C E$. Then the degree measure of $\angle B$ is ( ).
(A) $52.5^{\circ}$
(B) $62.5^{\circ}$
(C) $60^{\circ}$
(D) $75^{\circ}$
3.D.

As shown in Figure 6, draw perpendiculars from points $A$ and $E$ to $BC$, with the feet of the perpendiculars being $G$ and $F$, respectively. Clearly, quadrilateral $AGCD$ is a rectangle.
Thus, $AG = CD$.
Also, $EF \parallel AG$, and $E$ is the midpoint of $AB$, so
$$
EF = \frac{1}{2} AG = \frac{1}{2} CD = \frac{1}{2} CE.
$$

Therefore, $\angle ECF = 30^{\circ}$.
Given $BC = CE$, we know $\angle B = \angle BEC$.
Thus, $\angle B = \frac{1}{2}(180^{\circ} - \angle ECF) = 75^{\circ}$.