$17 \cdot 23$ In $\triangle A B C$, $A B=A C, \angle A=80^{\circ}$. If points $D, E$, and $F$ are on sides $B C, A C$, and $A B$ respectively, and $C E=C D, B F=B D$, then $\angle E D F$ equals
(A) $30^{\circ}$.
(B) $40^{\circ}$.
(C) $50^{\circ}$.
(D) $65^{\circ}$.
(E) None of these.
［Solution］Given $\angle A=80^{\circ}$, and $A B=A C$, then

Similarly,
$$
\angle B=\angle C=50^{\circ} \text {, }
$$

Therefore,
$$
\angle C D E=\angle F D B=65^{\circ} \text {. }
$$

Thus, the answer is (C).
Note: Even when $A B \neq A C$, the above conclusion still holds:
$$
\begin{aligned}
\angle E D F & =180^{\circ}-\angle C D E-\angle F D B \\
& =180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle C\right)-\frac{1}{2}\left(180^{\circ}-\angle B\right) \\
& =\frac{1}{2}(\angle B+\angle C)=\frac{1}{2}\left(180^{\circ}-\angle A\right)=50^{\circ} .
\end{aligned}
$$