3. $A D 、 B E 、 C F$ are the angle bisectors of $\triangle A B C$. If $B D+B F=C D+C E=A E+A F$, then the degree measure of $\angle B A C$ is $\qquad$ .
3. $60^{\circ}$.

Let $B C=a, C A=b, A B=c$.
By the Angle Bisector Theorem, we have
$$
\begin{aligned}
B D & =\frac{a c}{b+c}, C D=\frac{a b}{b+c}, \\
B F & =\frac{a c}{a+b}, C E=\frac{a b}{a+c} .
\end{aligned}
$$

From $B D+B F=C D+C E$, we know
$$
\frac{c}{b+c}+\frac{c}{a+b}=\frac{b}{b+c}+\frac{b}{a+c} \text {. }
$$

Clearing the denominators and simplifying, we get
$$
\begin{array}{l}
a^{2} c+2 a c^{2}+2 b c^{2}+c^{3} \\
=a^{2} b+2 a b^{2}+2 b^{2} c+b^{3},
\end{array}
$$

which is $(c-b)\left(a^{2}+2 a c+2 a b+b^{2}+c^{2}+3 b c\right)=0$.
Clearly, $a^{2}+2 a c+2 a b+2 b c+b^{2}+c^{2}+b c$ $=(a+b+c)^{2}+b c>0$.
Thus, $c-b=0$, i.e., $b=c$.
Similarly, when $C D+C E=A E+A F$, we have $c=a$.
Therefore, $a=b=c$, and $\triangle A B C$ is an equilateral triangle.
Hence, $\angle B A C=60^{\circ}$.