2. In $\triangle A B C$, it is known that $D$ is the midpoint of $A B$, point $E$ is on $A C$, and $\angle A E D=90^{\circ}+\frac{1}{2} \angle C$. Then $B C+2 A E-A C$ is ( ).
(A) Positive
(B) 0
(C) Negative
(D) Uncertain
2. B.

As shown in Figure 1, draw
$D F / / B C$ intersecting $A C$ at
point $F$. Then $\angle D F E=$
$$
\begin{array}{l}
\angle C, A C=2 A F, \\
B C=2 D F .
\end{array}
$$

In $\triangle D E F$, draw $F G \perp D E$ intersecting $D E$ at point $G$.
$$
\text { Given } \angle A E D=90^{\circ}+\frac{1}{2} \angle C=90^{\circ}+\angle E F G \text {, }
$$

It is easy to see that $E F=D F=\frac{1}{2} B C$.
Therefore, $A C=2 A F=2(A E+D F)=B C+2 A E$.