27. $D, E$ are on the sides $A B, A C$ of $\angle C A B$, respectively. The angle bisector of $\angle B D E$ intersects the angle bisector of $\angle C E D$ at $F$. If $\angle B A C=38^{\circ}$, then $\angle D F E=$ ( ).
(A). $52^{\circ}$
(B). $71^{\circ}$
(C). $76^{\circ}$
(D). $57^{\circ}$
Answer: B.
Solution: As shown in the figure,
$$
\angle 1+\angle 2+\angle 3=180^{\circ}, \quad \angle 1=\angle 2, \angle 4+\angle 5+\angle 6=180^{\circ}, \angle 4=\angle 5,
$$

Therefore,
$$
2 \angle 2+\angle 3=180^{\circ},
$$

which means
$$
\begin{array}{l}
\angle 3=180^{\circ}-2 \angle 2, \\
2 \angle 5+\angle 6=180^{\circ},
\end{array}
$$

which means
$$
\angle 6=180^{\circ}-2 \angle 5 \text {. }
$$

Since
$$
\angle A+\angle 3+\angle 6=180^{\circ} \text {, }
$$

we have
$$
\angle A+\left(180^{\circ}-2 \angle 2\right)+\left(180^{\circ}-2 \angle 5\right)=180^{\circ} \text {, }
$$

thus
$$
2 \angle 2+2 \angle 5=\angle A+180^{\circ},
$$

which means
$$
\begin{aligned}
\angle 2+\angle 5 & =\frac{1}{2} \angle A+90^{\circ} \\
& =\frac{1}{2} \times 38^{\circ}+90^{\circ}=109^{\circ} .
\end{aligned}
$$

Since
$$
\angle F+\angle 2+\angle 5=180^{\circ},
$$

we have
$$
\begin{aligned}
\angle F & =180^{\circ}-(\angle 2+\angle 5) \\
& =180^{\circ}-109^{\circ} \\
& =71^{\circ} .
\end{aligned}
$$

Therefore, the answer is B.