$17 \cdot 6$ In triangle $ABC$, $\angle C=\theta, \angle B=2 \theta$, where $0^{\circ}<\theta<60^{\circ}$. A circle with center $A$ and radius $AB$ intersects $AC$ at $D$, and intersects $BC$ (if necessary, extend $BC$) at $B, E$ (E may coincide with B). Then $EC=AD$ is true if
(A) no value of $\theta$ is suitable.
(B) only when $\theta=45^{\circ}$.
(C) only when $0^{\circ}<\theta \leqslant 45^{\circ}$.
(D) only when $45^{\circ} \leqslant \theta<60^{\circ}$.
(E) for all $\theta$ satisfying $0^{\circ}<\theta<60^{\circ}$.
(26th American High School Mathematics Examination, 1975)
[Solution] Discuss the value of $\theta$ in three cases:
(1) If $0^{\circ}<\theta<45^{\circ}$ (see Figure (1)), then applying the exterior angle theorem to $\triangle E A C$ yields $2 \theta=\angle E A C+\theta$, so $\angle E A C=\theta$. Therefore, $\triangle E A C$ is an isosceles triangle. Hence, $E C=A E=A D$.
(2) If $\theta=45^{\circ}$, then $\triangle A B C$ is an isosceles right triangle. In this case, $E$ coincides with $B$, so
$$
E C=B C=A B=A D .
$$
(3) If $45^{\circ}<\theta<60^{\circ}$ (see Figure (2)), then
$$
\begin{aligned}
\angle E A C & =\angle E A B+\angle B A C \\
& =\left[180^{\circ}-2\left(180^{\circ}-2 \theta\right)\right]+\left[180^{\circ}-3 \theta\right] \\
& =\theta .
\end{aligned}
$$

Thus, $\triangle E A C$ is an isosceles triangle, and $E C=E A=A D$. Therefore, the answer is $(E)$.