6. As shown in Figure 2, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and point $E$ is on $A C$ such that $\angle A E D=90^{\circ}+\frac{1}{2} \angle C$. Then $B C+2 A E$ equals ( ).
(A) $A B$
(B) $A C$
(C) $\frac{3}{2} A B$
(D) $\frac{3}{2} A C$
6. B.

Draw $B F / / D E$ intersecting $A C$ at point $F$. Then $\angle B F C=\angle D E F$.
From $D$ being the midpoint of $A B$, we know $E F=A E$.
Notice,
$$
\begin{array}{l}
\angle B F C=\angle D E F=180^{\circ}-\angle A E D \\
=180^{\circ}-\left(90^{\circ}+\frac{1}{2} \angle C\right)=90^{\circ}-\frac{1}{2} \angle C, \\
\angle F B C=180^{\circ}-\angle B F C-\angle C=90^{\circ}-\frac{1}{2} \angle C .
\end{array}
$$

Thus, $\angle F B C=\angle B F C \Rightarrow B C=F C$.
Therefore, $B C+2 A E=A C$.