17-24 In $\triangle A B C$, $\angle A=80^{\circ}$, points $D, E, F$ are on sides $B C, A B$, $A C$ respectively, and $B D=B E, C D=C F$ (as shown in the figure).
Then $\angle E D F$ equals
(A) $30^{\circ}$.
(B) $40^{\circ}$.
(C) $50^{\circ}$.
(D) $65^{\circ}$.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1990)
[Solution] Let $\angle B D E=\alpha, \angle F D C=\beta$. Given $B E=B D, C D=C F$, then in $\triangle B E D$ and $\triangle C D F$ we have
$$
2 \alpha=180^{\circ}-\angle B, \quad 2 \beta=180^{\circ}-\angle C .
$$

Thus,
$$
\begin{aligned}
2(\alpha+\beta) & =360^{\circ}-(\angle B+\angle C) \\
& =180^{\circ}+\left[180^{\circ}-(\angle B+\angle C)\right] \\
& =180^{\circ}+80^{\circ}=260^{\circ},
\end{aligned}
$$

Therefore, $\alpha+\beta=130^{\circ}$, and $\angle E D F=180^{\circ}-(\alpha+\beta)=50^{\circ}$. Hence, the answer is (C).