6. As shown in Figure 2, in $\triangle A B C$, $A B=A C, A D=A E$. $\angle B A D=60^{\circ}$. Then $\angle E D C=$ degrees.
II. 6. $30^{\circ}$.
Let $\angle C A D=2 \alpha$.
From $A B=A C$, we know
$$
\begin{array}{l}
\angle B=\frac{1}{2}\left(180^{\circ}-60^{\circ}-2 \alpha\right)=60^{\circ}-\alpha, \\
\angle A D B=180^{\circ}-\angle B-60^{\circ}=60^{\circ}+\alpha .
\end{array}
$$

From $A D=A E$, we know
$$
\angle A D E=90^{\circ}-\alpha \text {. }
$$

Therefore, $\angle E D C=180^{\circ}-\angle A D E-\angle A D B=30^{\circ}$.