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, construct $E M \perp B C$ 
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.
Also, $E F = B C$, then
$P F = M N$
$= \frac{1}{2} B C = \frac{1}{2} E F$
$\Rightarrow \angle P E F = 30^{\circ}$.
Combining $E F \perp A D$ and $E M \perp B C$, we get
$\angle A D C = \angle P E F = 30^{\circ}$.
From $\angle A D C = \angle B + \angle B A D$
$= \angle B + \frac{1}{2}(180^{\circ} - 2 \angle C)$
$\Rightarrow \angle C - \angle B = 90^{\circ} - \angle A D C = 60^{\circ}$.