$17 \cdot 20$ As shown in the figure, isosceles triangle $A B C, \quad A B=A C$, $\triangle D E F$ is an inscribed equilateral triangle, $\angle B F D=\alpha, \angle A D E=\beta$, $\angle F E C=\gamma$, then
(A) $\beta=\frac{\alpha+\gamma}{2}$.
(B) $\beta=\frac{\alpha-\gamma}{2}$.
(C) $\alpha=\frac{\beta-\gamma}{2}$.
(D) $\alpha=\frac{\beta+\gamma}{2}$.
(E) None of the above.
［Solution］By the exterior angle theorem of a triangle, we have

Therefore,
$$
\begin{array}{c}
\beta+60^{\circ}=\alpha+\angle B, \quad \alpha+60^{\circ}=\gamma+\angle C, \\
\beta-\alpha=\alpha-\gamma+\angle B-\angle C .
\end{array}
$$

Given $\angle B=\angle C$, then $\beta-\alpha=\alpha-\gamma$, which means $\alpha=\frac{\beta+\gamma}{2}$. Therefore, the answer is $(D)$.