$17 \cdot 21$ In $\triangle A B C$, $A B=A C, \angle A=80^{\circ}$, $D$ is a point on $B C$, $B D=B E, C D=C F, \angle E F D$ to the degree: $\angle F E D$ to the degree $=3: 2$, then the degree of $\angle A E F$ is
(A) $78^{\circ}$.
(B) $63^{\circ}$.
(C) $52^{\circ}$.
(D) $37^{\circ}$.
[Solution] From the given, we have $\angle B=\angle C=\frac{1}{2}\left(180^{\circ}-\angle A\right)=50^{\circ}$,
$$
\angle B E D=\angle C F D=\frac{1}{2}\left(180^{\circ}-50^{\circ}\right)=65^{\circ} \text {. }
$$

If $\angle A E F=63^{\circ}$, then we have
$$
\begin{array}{l}
\angle A F E=180^{\circ}-80^{\circ}-63^{\circ}=37^{\circ}, \\
\angle E F D=180^{\circ}-37^{\circ}-65^{\circ}=78^{\circ}, \\
\angle F E D=180^{\circ}-63^{\circ}-65^{\circ}=52^{\circ},
\end{array}
$$

At this time, the degree of $\angle E F D$: the degree of $\angle F E D$ is $3: 2$. Therefore, the answer is $(B)$.