$17 \cdot 25$ In $\triangle A B C$, $\angle B A C=50^{\circ}$, and $B E=B D, C F=C D$. Then $\angle E D F$ equals
(A) $30^{\circ}$.
(B) $135^{\circ}$.
(C) $65^{\circ}$.
(D) $60^{\circ}$.
(China Anhui Province Hefei Junior High School Mathematics Competition, 1985)
[Solution] Given $\angle B A C=50^{\circ}$, we know $\angle B+\angle C=$ $130^{\circ}$.
Also, $B E=B D, C F=C D$, so
$$
\angle B D E=\angle B E D, \quad \angle C F D=\angle C D F \text {. }
$$

Therefore, $\angle B D E=\frac{1}{2}\left(180^{\circ}-\angle B\right), \angle C D F=\frac{1}{2}\left(180^{\circ}-\angle C\right)$.
Thus, $\angle B D E+\angle C D F=180^{\circ}-\frac{1}{2}(\angle B+\angle C)$,
Then
$$
\begin{aligned}
\angle E D F & =180^{\circ}-(\angle B D E+\angle C D F)=\frac{1}{2}(\angle B+\angle C) \\
& =65^{\circ} .
\end{aligned}
$$

Therefore, the answer is $(C)$.