3. Given that the circumcenter of acute $\triangle A B C$ is $O, A O$ intersects $B C$ at point $D, E$ and $F$ are the circumcenters of $\triangle A B D$ and $\triangle A C D$ respectively. If $A B>A C, E F=B C$, then $\angle C-\angle B=$ $\qquad$ .
3. $60^{\circ}$.

As shown in Figure 4, draw $E M \perp B C$ 
at point $M$, $F N \perp B C$ at point $N$, and $F P \perp E M$ at point $P$.

It is easy to see that $M$ and $N$ are the midpoints of sides $B D$ and $C D$, respectively.
$$
\begin{array}{l} 
\text { Also, } E F=B C, \\
P F=M N \\
= \frac{1}{2} B C=\frac{1}{2} E F \\
\Rightarrow \angle P E F=30^{\circ} .
\end{array}
$$

Combining $E F \perp A D$ and $E M \perp B C$, we get
$$
\begin{array}{l}
\angle A D C=\angle P E F=30^{\circ} . \\
\text { Since } \angle A D C=\angle B+\angle B A D \\
=\angle B+\frac{1}{2}\left(180^{\circ}-2 \angle C\right) \\
\Rightarrow \angle C-\angle B=90^{\circ}-\angle A D C=60^{\circ} .
\end{array}
$$